Which polynomials fix the unit circle? Find all polynomials $P(x)$ with real coefficients such that for every $x,y\in \mathbb{R}$ satisfying $x^2+y^2=1$ we have

$$P(x)^2+P(y)^2=1$$

Edit: now a complete solution, though the D case is rather "hairy". It's conceptually very simple, as we basically only develop the squares and solve the equations they yield, but one has to be careful when doing this. This part is much longer than the preceding three.

We will use some basic facts about polynomials:

Two polynomials are equal iff they have the same coefficients.
The monomial of highest degree of $(\alpha_0+\alpha_1x+\cdots+\alpha_nx^n)^2$ is $\alpha_n^2x^{2n}$.
If a polynomial of degree $n$ has more than $n$ roots, it's the null polynomial. Especially true if there are infinitely many roots.

Now, let's hunt the Snark...

A. A solution $P$ is either an odd polynomial or an even polynomial

We suppose we have a polynomial $P$ satisfaying $P(x)^2+P(y)^2=1$ whenever $x^2+y^2=1$. Notice that $-P$ is also a solution.

The first thing to notice, is that for any $x\in[-1,1]$, you can find an $y$ such that $x^2+y^2=1$. There are two solutions: $y=\pm\sqrt{1-x^2}$. For such an $(x,y)$, you have an equally valid $(-x,y)$.

This means that for any $x\in[-1,1]$, there is an $y$ such that

$$P(x)^2=P(-x)^2=1-P(y)^2$$

Hence, for any $x\in [-1,1]$,

$$P(x)^2-P(-x)^2=(P(x)+P(-x))(P(x)-P(-x))=0$$

You have a product that vanishes for infinitely many $x$, so (pigeonhole principle), there is at least one factor that vanishes for infinitely many $x$.

But both factors are polynomials, so one of them is null, and

$$P(x)=P(-x)$$

or

$$P(x)=-P(-x)$$

So $P$ is either even or odd. It can't be both, since it would be null, and the null polynomial is not a solution. That is, either $P$ has only coefficients of even degree, either it has only coefficients of odd degree.

B In the even case, $P$ must be a constant.

Let's suppose the $\deg P=2n\gt 0$, and write, with $y=\sqrt{1-x^2}$,

$$P(x)^2+P(y)^2=\left(\sum_{k=0}^na_{2k} x^{2k}\right)^2+\left(\sum_{k=0}^na_{2k} (1-x^2)^k\right)^2=1$$

And with our hypothesis on degree, we must have $a_{2n}\neq0$.

Now, let's find the coefficient of highest degree of the left polynomial.

Inside the first square we have highest monomial $a_ {2n}x^{2n}$, and inside the second one, we have after expanding, $(-1)^na_{2n}x^{2n}$.

After squaring and summing, the monomial of highest degree on the left is thus $2a_{2n}^2x^{4n}$. But, it must be null, since the polynomial on the right of the equality is a constant, and we have supposed $n>0$.

Contradiction! So if $P$ is even, its degree must be $0$.

Incidentally, a solution of degree $0$ satisfies $P(x)=a$, for all $x$, and

$$P(x)^2+P(y)^2=1=2a^2$$

Hence $P(x)=\frac{\sqrt2}{2}$ or $P(x)=-\frac{\sqrt2}{2}$

These are the only solutions if $P$ is even.

Notice that the same argument does not work for the odd case, because, when you write $P(x)^2+P(y)^2=1$ with $y=\sqrt{1-x^2}$, you get

$$P(x)^2+P(y)^2=\left(\sum_{k=0}^na_{2k+1} x^{2k+1}\right)^2+\left(\sum_{k=0}^na_{2k+1} \left(\sqrt{1-x^2}\right)^{2k+1}\right)^2$$
$$=\left(\sum_{k=0}^na_{2k+1} x^{2k+1}\right)^2+(1-x^2)\left(\sum_{k=0}^na_{2k+1} (1-x^2)^k\right)^2=1$$

And the coefficient of degree $4n+2$ vanishes.

C. There is an infinite sequence of solutions with odd $P$.

You have a general solution with $P(x)=T_{2n+1}(x)$, the Chebyshev polynomial of the first kind, for any $n$.

You have

$$T_{2n+1}(\cos \theta)=\cos (2n+1)\theta$$
$$T_{2n+1}(\sin \theta)=(-1)^n\sin (2n+1)\theta$$

For the cosine, it's a well known property (and sometimes the definition). For the sine it's not very difficult:

$$T_{2n+1}(\sin(\theta))=T_{2n+1}\left(\cos\left(\frac{\pi}2-\theta\right)\right)=\cos\left(\frac{(2n+1)\pi}2-(2n+1)\theta\right)$$
$$=\cos\left(\frac{(2n+1)\pi}2\right)\cos (2n+1)\theta+\sin\left(\frac{(2n+1)\pi}2\right)\sin(2n+1)\theta$$
$$=\sin\left(\frac{(2n+1)\pi}2\right)\sin(2n+1)\theta$$

And $\sin\left(\frac{(2n+1)\pi}2\right)=\sin\left(n\pi+\frac{\pi}2\right)=(-1)^n$, so

$$T_{2n+1}(\sin(\theta))=(-1)^n\sin(2n+1)\theta$$

With this $P$, you have obviously that for $x=\cos\theta$ and $y=\sin\theta$, which is always possible if $x^2+y^2=1$,

$$P(x)^2+P(y)^2=\cos^2 (2n+1)\theta+\sin^2 (2n+1)\theta=1$$

Again, the polynomials $-T_{2k+1}$ are also solutions.

D. There is no other solution for odd $P$

This one is a bit tricky.

We will need another fact about polynomials: any sequence $Q_n$ of polynomials of degree $\deg Q_n=n$, form a basis of the linear space of polynomials, $\Bbb R[X]$.

To see why, we just have to notice that then you can write $x^n$ as a linear combination of $(Q_0,Q_1,\cdots,Q_n)$, uniquely. To this end, annihilate first the monomial in $x^n$ with some $\alpha Q_n$, and continues with $x^n-\alpha Q_n$, and with subsequent differences.

It's true for Chebyshev polynomials of the first kind, and we have even a bit more: since $T_n$ is even or odd according to parity of $n$, an odd polynomial can be written uniquely as a linear combination of $\{T_{2n+1};n\in\Bbb N\}$, while an even polynomial can be written uniquely as a linear combination of $\{T_{2n};n\in\Bbb N\}$.

We will also need a similar statement about trigonometric polynomials.
If two polynomials of the form $\alpha_0+\sum_{k=1}^n \alpha_k\cos k\theta + \beta_k\sin k\theta$ are equal, then all corresponding coefficients are equal. This stems from the fact that the $\cos kx, \sin kx$ are orthogonal w.r.t the inner product $(u|v)=\frac1\pi\int_{-\pi}^\pi u(\theta)v(\theta) \,\mathrm{d}\theta$.

Thus, if you have a solution $P$ of your problem $P(x)^2+P(y)^2=1$, which is an odd polynomial, then we can write

$$P(x)=\sum_{k=0}^n a_{2k+1}x^{2k+1}$$

And, in the Chebyshev basis, there are unique $\{b_{2k+1}\}$ such that

$$P(x)=\sum_{k=0}^n b_{2k+1}T_{2k+1}(x)$$

And we can assume $a_{2n+1}\neq0$, or equivalently $b_{2n+1}\neq0$.

With $x=\cos\theta$ and $y=\sin\theta$, we have then (see the preceding part C for $T_{2k+1}(\sin\theta)$):

$$P(x)=P(\cos\theta)=\sum_{k=0}^n b_{2k+1}T_{2k+1}(\cos \theta)=\sum_{k=0}^n b_{2k+1}\cos (2k+1)\theta$$

$$P(y)=P(\sin\theta)=\sum_{k=0}^n b_{2k+1}T_{2k+1}(\sin \theta)=\sum_{k=0}^n (-1)^k b_{2k+1}\sin (2k+1)\theta$$

Write that $P$ is a solution:

$$\left(\sum_{k=0}^n b_{2k+1}\cos (2k+1)\theta\right)^2+\left(\sum_{k=0}^n (-1)^k b_{2k+1}\sin (2k+1)\theta\right)^2=1$$

Now, we are going to develop this beast carefully.

We develop the square like this:

$$(\alpha_0+\alpha_2+\cdots+\alpha_n)^2=\sum_{k=0}^n \alpha_k^2+2\sum_{i\lt j} \alpha_i\alpha_j$$

Where the second sums runs on all $i,j$ such that $0\leq i\lt j\leq n$.

For our sum of two squares, this yields

$$\sum_{k=0}^n b_{2k+1}^2\left(\cos^2 (2k+1)\theta + \sin^2 (2k+1)\theta\right)+\\
2\sum_{i\lt j} b_{2i+1}b_{2j+1} \cos (2i+1)\theta \cos (2j+1)\theta+\\
2\sum_{i\lt j} (-1)^{i+j} b_{2i+1}b_{2j+1} \sin (2i+1)\theta \sin (2j+1)\theta=1$$

We want to "linearize" this, that is use trigonometric identities to transform the expression so that there is no product of trigonometric functions. Then we will apply what we know on trigonometric polynomials, to annihilate all coefficients, in some interesting order.

The first sum, let's call it $U$, is simply a constant,

$$U=\sum_{k=0}^n b_{2k+1}^2\left(\cos^2 (2k+1)\theta + \sin^2 (2k+1)\theta\right)=\sum_{k=0}^n b_{2k+1}^2$$

The second and the third sum may be simplified by remembering that

$$\cos(a+b)=\cos a\cos b- \sin a\sin b$$
$$\cos(a-b)=\cos a\cos b+ \sin a\sin b$$

So, taking corresponding terms, according to whether $(-1)^{i+j}$ is $+1$ or $-1$, you will simplify to respectively $b_{2i+1}b_{2j+1}\cos (2j-2i)\theta$ or $b_{2i+1}b_{2j+1}\cos (2i+2j+2)\theta$.

Hence, the second and third sum in our preceding expansion, simplify to $V+W$ with

$$V=\sum_{i\lt j, \ i+j\ \mathrm{even}} b_{2i+1}b_{2j+1}\cos (2j-2i)\theta$$
$$W=\sum_{i\lt j, \ i+j\ \mathrm{odd}} b_{2i+1}b_{2j+1}\cos (2i+2j+2)\theta$$

We must be careful, because among the factors $2j-2i$ and $2i+2j+2$ in the cosines, many overlap. But we have, with $x=\cos\theta, y=\sin\theta$:

$$P(x)^2+P(y^2)=U+2V+2W=1$$

Let's have a closer look at those cosines. From now on, we will suppose $n$ is even. The proof will be similar for odd $n$. We have always $0\leq i\lt j\leq n$, and:

$U$ is a constant, so there is no cosine at all
In $V$, they appear with factor $2j-2i$, and $i+j$ is even. So it can run from $4=2\cdot2-2\cdot0$ to $2n=2\cdot n-2\cdot 0$.
In $W$, they appear with factor $2i+2j+2$, and $i+j$ is odd. So it can range from $4=2\cdot1+2\cdot0+2$ to $4n=2\cdot n+2\cdot(n-1)+2$.

Since there is no constant in $V$ or $W$, it means $U=1$ and $V$ and $W$ must be zero. Moreover, we can see that the "highest cosine", that is $\cos 4n\theta$, appears only when $i=n-1$ and $j=n$, thus with only coefficient $b_{2n+1}b_{2n-1}$.

It's here we use our hypothesis that $b_{2n+1}\neq0$, so $b_{2n-1}=0$.

The latter developments are devoted to proving that all coefficients must vanish except $b_{2n+1}$. And we will have at each step a "highest remaining cosine" (that is, cosine with the highest integer in factor of $\theta$, we will call this factor the $\theta-factor$ for short), with in its coefficients a single remaining term, even though they originally overlap. That is, by annihilating the $b_{2k+1}$ in the right order, the problem is much simplified.

To give an idea, the order for $k$ will be (for even $n$):

$$n-1,n-3,n-5,\cdots,3,1,0,2,4,6,\cdots,n-2$$

We have already proved $b_{2n-1}=0$, that's for $k=n-1$.
Notice also that we can consider only $W$, as long as our $\theta$-factor is greater than $2n$, since there is no factor beyond this one in $V$.

Let's represent our $\theta$-factors in a table, as function of indices $i$ and $j$, for $n=8$, in order to show how (and why) the following proof works.

$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|}
\hline
j\backslash i  & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\
\hline
0 &  * \\
\hline
1 &  4 &  * \\
\hline
2 &  * &  8 &  * \\
\hline
3 &  8 &  * & 12 &  * \\
\hline
4 &  * & 12 &  * & 16 &  * \\
\hline
5 & 12 &  * & 16 &  * & 20 &  * \\
\hline
6 &  * & 16 &  * & 20 &  * & 24 &  * \\
\hline
7 & 16 &  * & 20 &  * & 24 &  * & 28 &  * \\
\hline
8 &  * & 20 &  * & 24 &  * & 28 &  * &  32 &  * \\
\hline
\end{array}$$

The table reads like this: for $i=2$, $j=3$, there is a $b_{2i+1}b_{2j+1}\cos k\theta$ term in $W$, with $k=12$ (the $\theta$-factor, from table). Also, elements in each growing diagonal belong to the same $k$. Hence, given that the overall coefficient before each cosine must vanish, a growing diagonal gives rise to an equation between the $b_{2i+1}$.
For example, for $i+j=9$, you get $(b_{17}b_{3}+b_{15}b_{5}+b_{13}b_{7}+b_{11}b_{9})\cos 20\theta =0$.
Since $i+j$ must be odd, all stars in the table correspond to zero coefficients, and they don't appear in $W$.

For a given $n$ like here, we can continue by hand: we have supposed $b_{2n+1}=b_{17} \neq 0$. Also we already saw that $b_{2n-1}=b_{15}=0$. It's really because $32$ is alone on its diagonal. On the next diagonal, there are two terms, $b_{17}b_{11}$ and $b_{15}b_{13}$, and the sum must be $0$. But $b_{15}=0$, so $b_{11}=0$.
And we go on, proving successively that $b_{15}$, $b_{11}$, $b_{7}$, $b_{3}$, that is the $b_{2k+1}$ for $k=7,5,3,1$, are null. Then, we are left with only even $k$, and we will attack them with $V$, a bit later.

For now, we prove this approach actually always works, for any even $n$, by induction.
We suppose $b_{2n+1}\neq 0$, as before, and also $b_{2(n-1)+1}=b_{2(n-3)+1}=\cdots=b_{2(n-2p+1)+1}=0$,
and we want to prove $b_{2(n-2p-1)+1}=0$, working on the diagonal
that contains $b_{2(n-2p-1)+1}b_{2n+1}=b_{2i_0+1}b_{2j_0+1}$, with $i_0=n-2p-1$ and $j_0=n$ (on the bottom line).

Consider all coefficients before $\cos (2i+2j+2)\theta=\cos 4(n-p)\theta$ (that is, the whole diagonal), they yield the equation:

$$\sum_{i+j=2n-2p-1} b_{2i+1}b_{2j+1}=\sum_{i=n-2p-1}^n b_{2i+1}b_{2(2n-2p-1-i)+1}=0$$

Where the sum is done on valid $i,j$ that satisfy also $0\leq i\lt j\leq n$.

We notice that $i$ and $2n-2p-1-i$ don't have the same parity (by construction, since $i+j$ must be odd), so one of them is odd, for any $i$.
If it's $i$, appart from the first that corresponds to $b_{2(n-2p-1)}$, all other are greater, so $b_{2i+1}=0$ by hypothesis.
And if it's $2n-2p-1-i$, it's also greater than $n-2p-1$, since $i\lt j\leq n$ implies $i\lt n$, so again $b_{2(2n-2p-1-i)+1}=0$, by hypothesis.
Hence, appart for the term $b_{2(n-2p-1)+1}b_{2n+1}$, all other are zero. But since the sum must vanish, and $b_{2n+1}\neq0$, we must have $b_{2(n-2p-1)+1}=0$, and the induction step is proved.

We can continue the induction as long as $4(n-p)\gt2n$, otherwise there would be terms from $V$ in the equation, and we would not be able to conclude.
That is, $p<\frac{n}2$, or until $p=\frac{n}2-1$, included.

Hence it is proved that $b_{2(n-2p-1)+1}=0$ for all $p$ such that $0\leq p\leq \frac{n}2-1$.
And $n-2(\frac{n}2-1)-1=1$. Thus, simplifying a bit, we have

$$b_{2i+1}=0$$

For all odd $i$ such that $1 \leq i \leq n-1$, that is, all odd $i$.

Now, we will do the same on $V$.

Again, let's have a look at $\theta$-factors for $i,j$: the corresponding term in $V$ is
$b_{2i+1}b_{2j+1} \cos 2(j-i)\theta$, and now $i+j$ must be even, and we have still $0\leq i \lt j \leq n$.

$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|}
\hline
j\backslash i  & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\
\hline
0 &  \\
\hline
1 &  * &  \\
\hline
2 &  4 &  * &  \\
\hline
3 &  * &  4 &  * &  \\
\hline
4 &  8 &  * &  4 &  * &  \\
\hline
5 &  * &  8 &  * &  4 &  * &  \\
\hline
6 & 12 &  * &  8 &  * &  4 &  * &  \\
\hline
7 &  * & 12 &  * &  8 &  * &  4 &  * &  \\
\hline
8 & 16 &  * & 12 &  * &  8 &  * &  4 &  * &  \\
\hline
\end{array}$$

This time, $\theta$-factors are constant along decreasing diagonals, and we will start with the bottom-left corner, where we have $b_{2n+1}b_{2\cdot0+1}=0$ thus $b_1=0$.
This works for any even $n$, because $2(n-0)$ is the maximum value that can be taken by $2(j-i)$, and it appears only for $j=n$ and $i=0$, thus there is always only one term $b_{2n+1}b_{2\cdot0+1}$ before $\cos 2n\theta$.

We continue with induction, supposing that $b_{2\cdot0+1}=\cdots=b_{2(2p-2)+1}=0$, and we want to prove that $b_{2(2p)+1}=0$.
We work on the diagonal starting on the bottom line at $j_0=n$ and $i_0=2p$. The whole diagonal yields the equation

$$\sum_{i=0}^{2p} b_{2i+1}b_{2(n-2p-i)+1}=0$$

Considering $b_{2i+1}b_{2(n-2p-i)+1}$, and appart from $b_{2i_0+1}b_{2j_0+1}$, either both $i$ and $n-2p-i$ are odd,
and the term is null for we have proved earlier that $b_{2i+1}=0$ in this case (when working with $W$).
Either they are both even, and one of them is smaller than $i_0$, thus $b_{2i+1}=0$ or $b_{2(n-2p-i)+1}=0$.
Again, there is only one term remaining in the sum, and it's $b_{2i_0+1}b_{2j_0+1}=b_{2(2p)+1)}b_{2n+1}=0$.
Since we suppose $b_{2n+1}\neq0$, we must have $b_{2(2p+1)}=0$, and the induction step is proved.

We can follow the induction as long as $i_0 \lt j_0 \leq n$, that is $2p \lt n$. Hence we have proved that $b_{2i+1}=0$ for all even $i$ below $n$.

To sum up, we have proved that all $b_{2i+1}$ are null, except $b_{2n+1}$. We can now rewrite the equation $P(x)^2+P(y)^2=1$ with $x=\cos \theta$, $y=\sin\theta$ and

$$P(x)=\sum_{k=0}^n b_{2k+1}T_{2k+1}(x)=b_{2n+1}T_{2n+1}(x)$$

Or

$$P(x)=b_{2n+1}\cos (2n+1)\theta$$
$$P(y)=(-1)^nb_{2n+1}\sin(2n+1)\theta$$

Then

$$P(x)^2+P(y)^2=b_{2n+1}^2=1$$

And finally $b_{2n+1}=\pm1$.

So, we have proved that if $n$ is even and $\deg P=2n+1$, then $P=\pm T_{2n+1}$.

I'll leave the case when $n$ is odd as an "exercise", as the proof would be exactly similar, with a little modification in the order of elimination of coefficients $b_{2i+1}$.
Above it was $n-1,n-3,\cdots,3,1,0,2,\cdots,n-2$, and if $n$ is odd, it will be in the order $n-1,n-3,\cdots,2,0,1,3,\cdots,n-2$.

So, the conclusion of the whole answer is:

If $P(x)^2+P(y)^2=1$ whenever $x^2+y^2=1$, then either $P=\pm \frac{\sqrt{2}}2$, either $P=\pm T_{2n+1}$ for some $n$.


Find the square root of a matrix Let $A$ be the matrix
$$A = \left(\begin{array}{cc}
41 & 12\\
12 & 34
\end{array}\right).$$

I want to decompose it into the form of $B^2$.

I tried diagonalization , but can not move one step further.

Any thought on this? Thanks a lot!

ONE STEP FURTHER:

How to find a upper triangular $U$ such that $A = U^T U$?

This is an expansion of Arturo's comment.

The matrix has eigenvalues $50,25$, and eigenvectors $(4,3),(-3,4)$, so it eigendecomposes to $$A=\begin{pmatrix}4 & -3 \\ 3 & 4\end{pmatrix} \begin{pmatrix}50 & 0 \\ 0 & 25\end{pmatrix} \begin{pmatrix}4 & -3 \\ 3 & 4\end{pmatrix}^{-1}.$$

This is of the form $A=Q\Lambda Q^{-1}$. If this is $B^2$, then there will be a $B$ of the form $Q\Lambda^{1/2} Q^{-1}$ (square this to check this is formally true). A square root of a diagonal matrix is just the square roots of the diagonal entries, so we have

$$B=\begin{pmatrix}4 & -3 \\ 3 & 4\end{pmatrix} \begin{pmatrix}\sqrt{50} & 0 \\ 0 & \sqrt{25}\end{pmatrix} \begin{pmatrix}4 & -3 \\ 3 & 4\end{pmatrix}^{-1}$$

$$=\frac{1}{5}\begin{pmatrix}9+16\sqrt{2} & -12+12\sqrt{2} \\ -12+12\sqrt{2} & 16+9\sqrt{2}\end{pmatrix}.$$

Here we used $\sqrt{50}=5\sqrt{2},\sqrt{25}=5$, and a quick formula for the inverse of a $2\times 2$ matrix:

$$\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}.$$

Keep in mind that matrix square roots are not unique (even up to sign), but this particular method is guaranteed to produce one example of a real matrix square root whenever $A$ has all positive eigenvalues.

Finding an upper triangular $U$ such that $A=U^TU$ is even more straightforward:

$$A=\begin{pmatrix} a&0 \\ b&c \end{pmatrix} \cdot  \begin{pmatrix} a&b \\ 0&c \end{pmatrix} $$

This is $a^2=41$ hence $a=\sqrt{41}$, $ab=12$ hence $b=\frac{12}{41}\sqrt{41}$, and $b^2+c^2=34$ hence $c=25\sqrt{\frac{2}{41}}$.

In other words,

$$U=\sqrt{41}\begin{pmatrix}1&\frac{12}{41}\\0&\frac{25}{41}\sqrt{2}\end{pmatrix}. $$

As traditional known the square root of any number would have two result, its cube root would have three results and so on
The matrix $$A^2 = $$
$$
\begin{pmatrix}
41 & 12\\
12 & 34
\end{pmatrix}
$$
Has solutions as
Matrix $$A =$$
$$
\begin{pmatrix}
6.3254834 & 0.9941125\\
0.9941125 & 5.7455844
\end{pmatrix}
$$
And
Also
$$
\begin{pmatrix}
2.7254834 & 5.7941125\\
5.7941125 & -0.6544156
\end{pmatrix}
$$
This is a clear solution to the question, if you really checked well you'll see it.
I don't know why anyone here can block someone's comment, claiming it to be wrong


A question about percents This is the body in the question about percents.

$8,000$ is $40\%$ of the whole

Divide by $4$

$2,000$ is $10\%$ of the whole

Muliptly by $10$

$20,000$ is $100\%$ of the whole.

This answer doesn't really depend on algebra (but essentially the same) ,  using just your head.  But, like other answers suggest, mastering algebra would help you better to solve another problem.


Probability Urn Question(Recurrence Model) From an urn containing a white and b black balls, balls are drawn one
by one at random according to the following rules:

(i) at any drawing, if the ball drawn is white, then it is returned to the urn,

(ii) if it is black, it is replaced by a white ball (from another collection of
balls).

After n such operations, a ball is drawn from the urn. Find the probability
that it will be white.

my Approach:
Let w(k) denote that ball from kth selection is white,k = 1,2,...,n

suppose p_k = P(w(k)),we have to find p_n = P(w(n))..
i know surely that,we have to obtain Recurrence relation,but don't know how to manipulate p_n.

Correct Answer:$1-($($b/(a+b)$)$(1-1/(a+b))^{n}$)

There are $w+b$ balls.  Number them from $1$ to $w+b$.  When a black ball is replaced, give the new white ball the same number.
For the ball to be black on the $k^{th}$ draw, it must be the first time that number was drawn.  The number of ways to do that is $(w+b-1)^{k-1}$.
Can you finish the question?

At each draw when a black ball is drawn the probability of selection of white increases in the consecutive  draw.
We can have the following $n+1$ different combinations of drawings of $n$-balls,(where $w_k$ represents a white ball is drawn on $k$th draw and $B_k$ represents a black ball is drawn at $k$th draw) after which we have  to obtain the probability of getting a white ball.

$w_1w_2w_3w_4\dots  w_n$
$w_1w_2w_3w_4 \dots B_n$
$w_1w_2w_3w_4\dots B_{n-1}B_n$

and so on...

n)     $w_1B_2B_3B_4\dots B_{n-1}B_n$

n+1)   $B_1B_2B_3B_4\dots B_{n-1}B_n$

The probability of getting the white ball in $n+1$th drawing by 1) will be $a/(a+b)$
by 2) will $(a+1)/(a+b)$ and so on.............. by n+1) will be $(a+n)/(a+b)$

Thus the required probability should be $(a+a+1+a+2+........+a+n)/(a+b)$ (since each of the ways are independent of each other.


Is it possible to construct probability theory which is not based on measure theory but on logic? A probability space is a triplet $(\Omega, F, P)$, where $\Omega$ is a set, $F$ is a $\sigma$-algebra on $\Omega$ and $P$ is a probability measure on $(\Omega, F)$.

However, do we need $\Omega$?

What if we define as follows:

A probability set $G$ is a set of propositions which satisfies

$\bot \in G$
If $p \in G$, then $\lnot p \in G$
If $p_1, p_2, \dots \in G$, then $\lor_{i=1}^\infty p_i \in G$

A probalitiy function $P$ is a function from $G$ to $[0, 1]$ which satisfies

$P(\top)=1$
$P(\lor_{i=1}^\infty p_i) = \sum_{i=1}^\infty P(p_i)$ if $p_i \land p_j$ is false when $i \neq j$

Yes.  If you think about it, you need $\Omega$ and $P$ first. Or else for a given $p\in\Omega$, $\neg p$ does not make sense.  Since $$\neg p = \{q\in\Omega | q \neq p\}$$
The $\sigma$-algebra $F$ is in some sense "What events in $\Omega$ satisfy conditions of $P$"


Can the third derivative tell me anything about the original function? The first derivative can tell me about the intervals of increase/decrease for $f(x)$. The second derivative can tell me about the concavity of $f(x)$.

So can the third derivatives, and any derivatives beyond, yield any useful piece of information for graphing the original function?

Through supremely literal, I guess just as we can think of the first derivative as 'how quickly the function changes' and the second as 'how quickly the function of how quickly the function changes changes', we can say that the third is just 'how quickly the function of how quickly the function of how quickly the function changes changes changes'.

I would say that concavity and slope only seem significant because we gave them very visual names. We gave them names because they were used in visualising functions, but they are nothing more than a geometrical interpretation of the above quoted statements.

If we insert the names of the previous iterations into those statements, we get that the second derivative is 'how quickly the slope changes' and the third derivative is 'how quickly the function of how quickly the slope changes changes', which is just 'how quickly the concavity changes'.

Maybe we can give this one a nice geometrical name too, something like 'flexion', because it's describing whether the concavity is becoming tighter or looser and how quickly it is happening.

But there's nothing particularly remarkable about this property, just as there is nothing inherently remarkable about slope or curvature (just that we have given them names).

When in college, I was told that the order of the derivative is related to the "dimension."

I think this is easier to see once you take integration. Double integrals give us regions in the two dimensions while triple integrals give us regions in three dimensions.


A problem on order of a Group. Let $G$ be a group of order $8$ and $x$ be an element of $G$ of order $4$. Show that $x^2 \in Z(G)$, the center of $G$. How this result can be proved?

Put $N=\langle x \rangle$. Then $|N|=4$ and index$[G:N]=2$ so, $G=N \cup gN$, with $g \notin N$. $N$ is normal, so $g^{-1}x^2g \in N$. But this element, being conjugate to $x^2$, has order equal to that of $x^2$, which is $2$. Since $x^2$ is the unique element of order $2$ in $N$, it follows that $g^{-1}x^2g=x^2$. So $x^2$ commutes with $g$ and of course with any power of $x$. Hence $x^2 \in Z(G)$.

If x^2 is not in the center then the subgroup  does not intersect the center except in e. It follws that .Z=G and G is Abelian but then x^2 is in the center.


Let $g:\mathbb{R}\to\mathbb{R}$ be a measurable function such that $g(x+y) =g(x)+g(y).$ Then $g(x) = g(1)x$ . Let $g:\mathbb{R}\to\mathbb{R}$ be a measurable function such that
  $$g(x+y) =g(x)+g(y).$$
  How to prove that $g(x) = cx$ for some $c\in \mathbb{R}?$

The main thing to do here relies upon the fact that such function should be continuous and therefore by natural argument the answer will follow.

Using this
Additivity + Measurability $\implies$ Continuity

Therefore I found out that there is nothing missing in this question.

For $x_1,....,x_n \in \mathbb R$ we have, by induction, $g(x_1+...+x_n)=g(x_1)+...+g(x_n)$

Let $c=g(1)$.

Then prove in the following order:

$g(n)=cn$   $\quad$for all $ n \in \mathbb N$,
$g(1/n)=c\frac{1}{n}$ $\quad$ for all $ n \in \mathbb N$,
$g(\frac{m}{n})=c\frac{m}{n}$ $\quad$ for all  $n,m \in \mathbb N$,
$g(r)=cr$  $\quad$ for all $r \in \mathbb Q$.

Now use that $f$  is measurable to show $g(x)=cx$ $\quad$ for all $x \in \mathbb Q$.


What is a function to represent a diagonal sine wave? I need to be able to plot pixels in this pattern.
To me, it looks like a sine wave pattern that is both diagonal and convergent.

What would a function for that look like?  Thanks.

How about $x+\exp(-a x)\sin(b x)$ for suitable values of $a$ and $b$?

Here is a sample plot for $a=0.2$ and $b=10$.

It looks like you want almost a staircase type of line, so use this.

f(n)=sin(x)+x

It won't let me post unless I have more characters, so I'll throw in

f(n)=sin(wx)a+dx where w is wavelength, a is amplitutde, and d is degree.


Find the value of k that bisects the area Find the value of $k$ for which the line $ y= kx$ bisects the area enclosed by the curve $4y=4x-x^2$ and the $x$ - axis.

I have tried to solve this and the solution seems odd... the solution that came out was $k=1 - \sqrt [3]{2}$

The step I took was to find the interval of the curve $4y=4x-x^2$ (which I got 0 to 4) then integrated after subtracting the line from the curve but the solution I got was $4/3$ or something similar to that.

Any help would be appriciated.

Simple. If the area is bisected by a line, there will be 2 sections. The area above the line but under the curve, and the rest of the area. And since its being bisected, one area will always be half of the whole thing. So to get the area of the first section, we must set up an integral from 0 to the intersection point of the 2 lines.

To get the intersection point, we will have to solve this equation for x $$x-\frac{x^2}4=kx$$
You should get this as your answer: $$x=4-4k$$

Then we set up the integral: $$\int_0^{4-4k} (x-\frac{x^2}4-kx) \,dx= \frac12\int_0^4(x-\frac{x^2}4) \,dx$$
Then solving the specific integral we are left with:
$$\frac{(4-4k)^2}2-\frac{(4-4k)^3}{12}-\frac{k(4-4k)^2}2=\frac43$$
From there, we can solve for k:$$k=1-\sqrt[3]{\frac12}$$

Hint:

The abscissa of intersections of $y=0, 4y=x^2-4x$ are the roots of $$x^2-4x=0$$

Total area $$\int_0^4\dfrac{4x-x^2}4dx=?$$

The abscissa of intersections of $y=kx, 4y=x^2-4x$ are the roots of $$x^2-4x=-4kx\iff x=0,x=4-4k$$

$$\int_0^{4-4k}\dfrac{4x-x^2}4dx=?$$


Hyberbolic and Circular (Trig) Functions: Why no parabolic? There are circular (trig) functions which determine all the points on a unit circle:

and which relate to the area swept out by an angle subtended on the circle. -- These functions can of course be extended to relations to ellipses as well.

There are also hyperbolic functions which determine all the points on a hyperbola:

My question is why there are no analogs of these functions for parabolas (the other type of conic section):

Here I have defined $\mathrm {sinp}(\theta)$ and $\mathrm {cosp}(\theta)$ to be the x- and y-coordinates of points on a "unit parabola".

Is there any good reason why we should have these extremely useful transcendental functions (sin, cos, sinh, cosh, etc), but we can't (or don't) define analogous functions for parabolas?

NOTE: I recommended that this post get deleted because Henning's response in Do "Parabolic Trigonometric Functions" exist? explained theoretically why a "parabolic trigonometric function" is different than the circular and hyperbolic trigonometric functions.  However johannesvalks' answer to this was very interesting, as well, and probably shouldn't be deleted.

Such functions do exist! They're called cubic polynomials:

$$\int_a^b x^2 dx=\frac13(b^3-a^3)$$

Note that

(1) For $\cos(\phi)$ and $\sin(\phi)$ the argument is a 'surface'.

(2) For $\cosh(\zeta)$ and $\sinh(\zeta)$ the argument is a 'surface'.

So you should look for functions where the argument is a 'surface'.

Consider $y = x^2 - c$, where $c$ is positive.

For a given $x = \xi$ we have an arc formed by:

(3a) the curve $(0,-c) \rightarrow (\xi,\xi^2-c)$

(3b) the line $(\xi,\xi^2-c) \rightarrow (0,0)$

(3c) the line $(0,0) \rightarrow (0,-c)$

The surface of this arc is given by

(4) $\displaystyle A =
\int_0^\xi \left( \frac{\xi^2 - c}{\xi} x + c - x^2 \right) dx =
\frac{1}{6} \xi^3 + \frac{1}{2} c \xi$

So for a point $(x,y)$ on the curve $y = x^2 - c$ we obtain

(5) $A = \frac{1}{6} x \Big( x^2 + 3c \Big)$

And what we want is

(6a) $x = x(A)$

(6b) $y = y(A)$

To find the inverse for $A = \frac{1}{6} x \Big( x^2 + 3c \Big)$, we write $\displaystyle x = \frac{c}{\zeta} - \zeta$, then we get the equation

(7) $\Big[ \zeta^3 \Big]^2 + 6 A \Big[ \zeta^3 \Big] + c^3 = 0$,

whence

(8) $ \zeta^3 = \sqrt{9A^2 + c^3} - 3A $,

therefore

(9a) $\displaystyle x(A) = \frac{c}{\sqrt[3]{\sqrt{9A^2 + c^3} - 3A}} - \sqrt[3]{\sqrt{9A^2 + c^3} - 3A}$

The 'parabolic' functions are given by

(10a) $\displaystyle x(A) = \frac{c}{\sqrt[3]{\sqrt{9A^2 + c^3} - 3A}} - \sqrt[3]{\sqrt{9A^2 + c^3} - 3A}$,

(10b) $\displaystyle y(A) = \frac{c^2}{\sqrt[3]{\sqrt{9A^2 + c^3} - 3A}^2} + \sqrt[3]{\sqrt{9A^2 + c^3} - 3A}^2 - 3 c$,

but they can be expressed using radicals, as is not the case for goniometric and hyperbolic functions.

The case $c=1$ gives

(11a) $\displaystyle x(A) = \frac{1}{\sqrt[3]{\sqrt{9A^2 + 1} - 3A}} - \sqrt[3]{\sqrt{9A^2 + 1} - 3A}$,

(11b) $\displaystyle y(A) = \frac{1}{\sqrt[3]{\sqrt{9A^2 + 1} - 3A}^2} + \sqrt[3]{\sqrt{9A^2 + 1} - 3A}^2 - 3$.


For what real values of $a$ does $\sum_{n=1}^{\infty}{(\frac{an}{n +1})^n}$ converge? For what real values of $a$ does the series
$$\sum_{n=1}^{\infty}\left(\frac{an}{n
+1}\right)^n$$
converge?

At first, I wanted to do the root test, but the problem is that $a$ could be negative.  If I take it out of the fraction to get $$\left(\frac{a}{n+1}\right)^n$$ it doesn't meet the criteria of either the Dirichlet or the Abel tests.

If $a\ne0$ then
$$\left(\frac{an}{n+1}\right)^n=a^n \cdot \left(\frac{n}{n+1}\right)^n = a^n \cdot \left(
\frac{n+1}{n} \right)^{-n}
=a^n\left(1+\frac{1}{n}\right)^{-n}\sim e^{-1}a^n$$
so the given series is convergent if and only if $|a|<1$. The result is clear for $a=0$.

The question asks about real values of $a$, but, in order to make this answer more general (so that other versions of this question may be linked to this question as duplicates), I will assume that $\alpha \in \mathbb{C}$.  An expedient approach to the problem is to consider various tests of convergence.  Personally, I like to start with the simple ones, and work my way up.  To my mind, the simplest test of convergence is the ratio test:

Theorem (Ratio Test):  Let $(a_n)_{n=1}^{\infty}$ be a sequence of complex numbers, define
$$L = \lim_{n\to \infty} \left| \frac{a_{n+1}}{a_n} \right|, $$
and let $S$ denote the infinite series $\sum_{n=1}^{\infty} a_n. $

If $L < 1$, then the series $S$ is absolutely convergent (and, therefore, convergent).
If $L > 1$, then the series $S$ is divergent.
If $L = 1$, then the ratio test is inconclusive—the series $S$ may be divergent, conditionally convergent, or absolutely convergent.

In the case of the series in this question, $a_n = \left( \frac{an}{n+1} \right)^n$.  Thus
$$ L
= \lim_{n\to \infty} \left| \frac{ a^{n+1}(n+1)^{n+1} / (n+2)^{n+1} }{ a^n n^n / (n+1)^n } \right|
= \lim_{n\to\infty} \left| \frac{a(n+1)^{2n+1} }{ n^n(n+2)^{n+1} } \right|. \tag{1}$$
It can actually be shown that this limit is $|a|$, but the argument is a little tedious.  If we allow ourselves the use of big-$O$ notation,
$$ \frac{(n+1)^{2n+1}}{n^n(n+2)^{n+1}}
= \frac{n^{2n+1} + O(n^{2n})}{n^n(n^{n+1} + O(n^n))}
= \frac{n^{2n+1} + O(n^{2n}) }{n^{2n+1} + O(n^{2n})}
= \frac{n^{2n+1}}{n^{2n+1} + O(n^{2n})} + \frac{O(n^{2n}) }{n^{2n+1} + O(n^{2n})}. $$
As $n$ goes to infinity, the first term goes to $1$, and the second term vanishes.  Thus
$$ L
= \lim_{n\to\infty} \left| \frac{a(n+1)^{2n+1} }{ n^n(n+2)^{n+1} } \right|
= |a|.$$
That being said, this looks like the kind of problem which is often given to students in a first year calculus course (or, perhaps, a third year complex analysis course), which means that big-$O$ notation may not feel very rigorous.  A more convincing argument might be to compute the limit via binomial expansion of the terms in the fraction, but this is going to be quite tedious, as well.  Doable, but tedious.
Personally, I probably would have looked at the limit in (1) and decided to move on.  The next test I usually choose to apply is the root test.

Theorem (Root Test): Let $(a_n)_{n=1}^{\infty}$ be a sequence of complex numbers, define
$$ L = \lim_{n\to \infty} \sqrt[n]{|a_n|^{n}}, $$
and let $S$ denote the series $\sum_{n=1}^{\infty} a_n$.

If $L < 1$, then the series $S$ is absolutely convergent (and, therefore, convergent).
If $L > 1$, then the series $S$ is divergent.
If $L = 1$, then the ratio test is inconclusive—the series $S$ may be divergent, conditionally convergent, or absolutely convergent.

With $a_n = \left( \frac{an}{n+1} \right)^n$, the limit of interest is
$$
L = \lim_{n\to\infty} \left| \left( \frac{a n}{n+1} \right)^n \right|^{1/n}
= |a| \lim_{n\to\infty} \left| \frac{n}{n+1} \right|
= |a|. $$
This is, I think, a much more straight-forward computation, but it gives us the same result as the ratio test; namely, the series of interest converges absolutely when $|a| < 1$, and diverges when $|a| > 1$.
On the other hand, when $|a| = 1$, both tests are inconclusive.  In general, there is not really a good way of checking for convergence on the boundary of the disc of convergence.  Indeed, there are examples of series which converge for some points on this boundary, and diverge for others (as a simple example, consider the harmonic series, as compared with the alternating harmonic series).  In this case, however, we can do a little better.  Observe that if $|a| = 1$, then
$$ \lim_{n\to \infty} \left|  \frac{an}{n+1} \right|^{n}
= \lim_{n\to\infty} \left[ |a|^n \left( 1 + \frac{1}{n} \right)^{-n} \right]
= \frac{1}{\mathrm{e}}.
$$
In other words, we know that when $|a| = 1$, the general term of the series fails to converge to $0$, which means that the series fails to converge (by the divergence test).  Therefore, we may conclude:

If $|a| < 1$, then the series converges absolutely.
If $|a| \ge 1$, then the series diverges.


Complex number $\ln(i)$ into rectangular form? I am going through the Mathematics of the DFT book by Julius O. Smith III.

One of the questions ask the following:

  How would you convert the complex number $\ln(i)$ into the form $x + yi$, where $x$ is the real part and $y$ is the imaginary part?

I am stumped.

Remember that any non-zero complex number $z = x+iy$ can be written in the form $z = r e^{i\theta}$, where $r = |z|$ and $\theta$ is the "argument" of $z$, i.e. the angle between $z$ and the positive real axis when we identify $\mathbb C$ with $\mathbb R^2$. Since $e^{i\theta} = \cos\theta + i\sin\theta$, this is just writing $z$ in polar coordinates.  $\theta$ is a priori defined only up to a multiple of $2\pi$, but we can get a unique value of $\theta$ by requiring $\theta\in [0,2\pi)$.

Now, we can also write $re^{i\theta} = e^{i\theta + \ln r}$. So, it is natural to define the logarithm of $z$ as $\ln z = i\theta + \ln r$, so that it cancels the exponential.

In the case $z= i$, we have $r = |i| = 1$ and $\theta = \pi/2$, since $i$ is on the positive imaginary axis, which makes a right angle with the positive real axis. Therefore $\ln(i) = i\pi/2 + \ln 1 = i\pi/2$.


Where to find the paper of G.H. Hardy called "Sur les zéros de la fonction Zeta de Riemann" please? Where to find the paper of G.H. Hardy called "Sur les zéros de la fonction Zeta de Riemann" please?

I've been looking for a while now, and I just don't find it, the basics of the hypothesis !

If anyone can help !

The paper is listed in the Journal Comptes Rendus de l'Académie des Sciences and the full title of the article is Sur les zéros de la fonction $\zeta(s)$ de Riemann. The full reference is

Hardy G H. Sur les zéros de la fonction $\zeta(s)$ de Riemann. Comptes Rendus de l'Académie des Sciences,
1914, 158: 1012-1014

and you can access this paper on

www.academie-sciences.fr with free access.
EDIT (added another source): thanks to @Raymond Manzoni, who found this resource (also free access) archive.org

this paper is included in book by peter borwein "the riemann hypothesis"


Arrangement of $10$ people given $3$ cannot be next to each other. Question: How many ways can you arrange a group of $10$ students given that $3$ of them cannot sit next to each other (they cannot sit as a group of $3$ nor can $2$ of them sit next to each other $-$ the three students need to be separated completely).
My attempt: Naturally, I went to counting possible cases and I decided to visualise different combinations if they were arranged in a line. I ended up with $15$ combinations that they could sit in a line such that they are unique and follow the condition if each line was a circle. From there it was pretty easy to see that $\dfrac{15 \times 3! \times 7!}{10}$ was the resulting answer. I have tried to justify the $15$ without counting situations, although I have been unsuccessful. Is this answer correct? If it is, how can I properly justify it? If it isn't, how would I approach a problem like this?
Any guidances or help would be greatly appreciated!
EDIT: I know for a fact $15$ is incorrect. I feel like it may be $12$ but still I'm not entirely sure.

(1) Assuming they in a line, not a circle, then imagine $7$ stars a row (the "other" students). Among the $8$ gaps in between or outside of those stars, choose $3$ for the locations of the special $3$ students. This can be done in $C(8,3)=56$ ways.
(2) Now order (say, left to right) both of the two subsets of students arbitrarily, this can be done $(7!)(3!)$  ways.
Each admissible arrangement of students corresponds to a unique sequence of choices in (1) and (2). So there are $56(7!)(3!)$ ways to do it.

The assumption in crush3dice's answer that the students sit in a circle is just as reasonable as the assumption that I am going to make, that the students sit in a row of chairs numbered, from left to right as 1, 2, ..., 10.
My approach is to favor brute force over elegance of thought in this one-off (small) setting.  If instead the # of chairs was (for example) 1000, then I would look for a pattern and then try for elegance.
Suppose that you select chairs 1, 3, 5, and decide that the 3 troublesome students (refer to them as a, b, and c) will sit in those 3 chairs.  There are 3! ways of permuting these 3 students among these 3 chairs and 7! ways of permuting the other 7 students among the other 7 chairs.
Therefore, the exact answer is $3! \times 7! \times T$, where $T$ represents the total # of ways of selecting 3 chairs out of 10 so that none of the three chairs is next to each other.
Now for the brute force:
Rather than try for elegant analysis, why not manually enumerate $T$.
Let $L$ = the # of the lowest of the three chairs.
Let $M$ = the # of the middle of the three chairs.
Let $N$ = the # of the largest of the three chairs.
Clearly, $1 \leq L \leq 6.$
That is if $L = 7$, there is no way of selecting 2 chairs among chairs 8, 9, and 10 so that the 2 chairs are apart both from each other and chair 7.
If $L$ and $M$ are set then $N$ must be in $\{(M+2), \cdots, 10\}.$
If $L$ is set, then $M$ must (similarly) be in $\{(L+2), \cdots, 8\}.$
Since you know that $L$ is in $\{1,2,3,4,5,6\},$ then all you have to do to enumerate $T$ is to count.
Addendum : elegance rears its ugly head
Since the following idea just came to me, I might as well share it.  There is a shortcut to enumerating $T.$
As analyzed, $M$ must range from 3 through 8, inclusive.
For any specific $M$ in this range, $L$ must range from 1 through $M-2$ and $N$ must range from $M+2$ through 10.
Therefore, $T = \sum_{M=3}^8 \,[(M-2) \times (9-M)].$
Addendum-2 : A symmetry argument.
By symmetry, the # of arrangements with $M=3$ equals the # of arrangements with $M=8.$  A similar argument relates $M=4$ to $M=7$ and $M=5$ to $M=6.$
Therefore, $T = 2 \times \left\{\sum_{M=3}^5 \,[(M-2) \times (9-M)]\right\}.$


Complex function not continuous at $z_0$ If I have a function $f(z)$ defined on a domain $D$ in the complex plane that is not continuous at a point $z_0 \in D$, can $f$ be analytic in the region $D$? or I guess another way to phrase this would be can $f$ have a derivative at $z_0$?

$\ f\ is\ analytic \ on \ region \ D \Rightarrow\ f \ is \ continuous  \ on \ D $. Therefore $\ f $ cannot be analytic on D if there's a point in D in which $\ f$ is not continuous.

Generally speaking, you can always rewrite f(z) as: u(x,y) + i*'v(x,y), and see (at any point z = x + iy you like) if:
du/dx +idv/dx = du/dy - i*dv/dy.
If yes then it is analytic at that point.


"Why do I always get 1 when I keep hitting the square root button on my calculator?" I asked myself this question when I was a young boy playing around with the calculator.

Today, I think I know the answer, but I'm not sure whether I'd be able to explain it to a child or layman playing around with a calculator.

Hence I'm interested in answers suitable for a person that, say, knows what the square root of a square number is, but doesn't know about sequences, functions, convergence and the like.

Here is how I had justified it to myself when I was a kid (I was convinced :-))

If $x > 1$ then $\sqrt x > 1$ and $x > \sqrt x$.

So we keep reducing the number while still being $> 1$.

Also, we cannot end up at a number $>1$ as then taking the square root would reduce it.

If we do end up at $1$, we stay there. Since the calculator has a limited precision, we end up at $1$, and pretty quickly.

Of course, the assumption here is that we do end up somewhere :-)
(Which seemed justified by the fact that the first non-zero digit after the decimal point always seemed to reduce)

if(x == 0): sqrt(x) == 0
if(0 < x < 1): x < sqrt(x) < 1
if(x == 1): sqrt(x) == 1
if(1 < x ): 1 < sqrt(x) < x

Except for the cases x == 0 and x == 1, x != sqrt(x) so it must move, and it must move towards 1, so it will approach 1 to any accuracy you choose.


Solving this Integral with Bessel Functions Any suggestions on solving this (J0,J1 Bessel function of first kind, 0th and 1st order, respectively) :
$$
T = \int_0^a \int_0^\infty J_0(\lambda r) J_1(\lambda a) e^{-\sqrt{\frac{s}{\alpha}+\lambda^2} z} \frac{1}{\sqrt{\frac{s}{\alpha}+\lambda^2} }  d\lambda dr\
$$

Maple gives the result as
$$
{\frac {a \alpha\,{{ J}_{1}\left(\lambda\,a\right)} \left( {
{ J}_{1}\left(\lambda\,a\right)}\pi \,{{  H}_{0}\left(\lambda
\,a\right)}-{{ J}_{0}\left(\lambda\,a\right)}\pi \,{{  H}_{1
}\left(\lambda\,a\right)}+2\,{{ J}_{0}\left(\lambda\,a\right)}
 \right) }{2({\lambda}^{2}\alpha+s)}}
$$
where $H_0$ and $H_1$ are Struve functions.

But did you really mean what you wrote?  Note that the $J_1(\lambda a)$ is just a constant, the integral over $z$ is easy, and the only nontrivial part is integrating
$J_0(\lambda r)$ from $r=0$ to $a$.

EDIT: With the integral over $\lambda$ rather than $z$, you can still do the integral over $r$, with a result involving Struve functions, but
the integral over $\lambda$ seems to be difficult.


Determine the Galois group of the splitting $x^4 - 4x +2$ over $\mathbb{Q}$ By Eisenstein's criterion this is irreducible over Q, and I know it has only 2 real roots. Is this actually solvable by radicals? I know the Galois group isomorphic to some subgroup of $S_4$, which is a solvable group, which means the Galois group is solvable, but I can't figure out how to do it.

Well, when you derive your polynomial $f$ you get only one real root $1$ for $f'$. Then you see that $f$ is decreasing from $-\infty$ to $1$ and increasing from $1$ to $+\infty$. Now since $f(1)=-1$. We see that $f$ has two real roots and two  non-real conjugate complexes.

Now it follows that the coomplex conjugation will fix two roots (the real one) and exchange the two non-real ones. That is the Galois group induces a transposition of the set of roots. It follows that your Galois group as a subgroup of the permutations of the roots $\mathfrak{S}_4$ will contain a transposition. Up to conjugation there are only two transitive subgroups which contain a transposition :

$$\mathfrak{S}_4\text{ and } D_4 $$

Where $D_4$ is one $2$-Sylow of $\mathfrak{S}_4$.

Reducing your polynomial mod $5$ you get $x^4+x+2$. Mod $5$ this factorizes :

$$(x-2)(x^3+2x^2-x-1) $$

Remarking that both polynomials in the factorization are irreducible mod $5$, you have a theorem in your course that say this implies that you have a $3$-cycle in your Galois group. Because of the alternative we already highlighted the Galois group cannot but be $\mathfrak{S}_4$.

It is of degree 4, so if all else fails (guessing roots, other tricks) you can always use the quartic equation. http://mathworld.wolfram.com/QuarticEquation.html


Definition of one-sample U-statistics and conditional expectation Could you help me to understand the definition of one-sample U statistic? I'm following the notation in van der Vaart "Asymptotic statistic" p. 161 here

In a particular I have a doubt regarding the following point (at the end of page 161):

Why can we write $U$ as a conditional expectation, i.e. $U=E(h(X_1,...,X_r)|X_{(1)},...,X_{(n)})$?

As an example suppose $r=1$. The conditional expectation now reads
$E[h(X_1)|X_{(1)},\ldots,X_{(n)}]$
meaning that you take the expectation condition on your sample of size $n$. Since the sample is stripped of the order, according to the definition, this expectation is not just equal to $h(X_1)$. Instead it will be equal to th e sample average of the $h(X_i)$s.

Edit to clarify: if the set on which the expectation is conditioned on weren't "stripped of the order" (expression written in the source linked in the question), the definition wouldn't make sense. Suppose for example we have two observations, $X_1=-1,X_2=1$ and if $h$ is the identity function then $E[X_1|X_1,X_2]=X_1=-1$ and it's questionable whether this is an interesting statistic. However, $E[X_1|X_{(1)},X_{(2)}]=(-1+1)/2=0$ (conditional on the empirical distribution but we don't know to which index, (1) or (2), the index 1 corresponds) which is the sample mean and which can be a useful statistic. The sample mean is further a U statistic. According to the linked source U statistics can be written as a conditional expectation and to make that definition meaningful the "strip of the order" is done.


For what integer $n$ is $1+n$ divisible by $1+n^2$? For what integer $n$ is $1+n$ divisible by $1+n^2$?

I tried it in modulo $3$ and $8$ but to no avail.

Trial and error gave me answers $0$,$1$,$-2$ and $-3$.


How can i show $\prod a_i<1$? Suppose we have $a_1>1$ and for $i>1,0<a_i<1$ how can i show that the product of infinite numbers less than 1 and $a_1$ is less than 1?
$$\prod_{i=1}^\infty a_i<1.$$

Well, $$\prod_{i=1}^\infty a_i= a_1 \prod_{i=2}^\infty a_i$$ so you want to prove that
$\prod_{i=1}^\infty b_i =0$ if $0<b_i<1$. Which is false.

Taking logarithms, that would be equivalent to
$ \sum_{i=1}^\infty \log b_i  = \sum_{i=1}^\infty c_i = - \infty$ if $c_i<0$.
Or $\sum_{i=1}^\infty d_i = + \infty$ if $d_i>0$

If you know about convergent series, you know this is false; for example take $d_i=2^{-i}$


Every Transitive Permutation Group Has a Fixed Point Free Element If $G$ acts transitively by permutations on a finite set $A$ with more than one element (i.e. $G$ is a transitive permutation subgroup of the symmetric group $S_A$). Why does $G$ necessarily contain an element which has no fixed points (i.e. $g$ such that $g \cdot a \neq a$ for any $a \in A$)?

The hint I have is to think about, given $a \in A$, what fraction of elements of $G$ fixes $a$. I'm not sure how to go about this hint...

By Burnside's lemma, you have

$$\frac{1}{|G|}\sum_{g \in G}|\text{fix }g| = |A/G|=1$$

Since $1 \in G$ has $|A|>1$ fixed points, at least one of the terms in the sum must be $0$.

It suffices to show that $\bigcup_{a \in A} \operatorname{Stab}_a \neq G$ where $G$ is the transitive permutation group on $A$, so there is some $\sigma \in G$ not in $\bigcup_{a \in A} \operatorname{Stab}_a$, meaning $\sigma(a) \neq a$ for all $a \in A$.
As $G$ is transitive, the Orbit-Stabilizer theorem says that $|\operatorname{Stab}_a| = \frac{|G|}{|A|}$ for all $a$, which gives $$\left| \bigcup_{a \in A} \operatorname{Stab}_a \right| \le 1 + |A|\left(\frac{|G|}{|A|} - 1\right) < |G|,$$
which is what we want.


Show a function is continuous based on its properties relative to a covering map Let $p:E\to B$  be a covering map, let $Y$ be locally path-connected, and let $g:Y\to E$ be a function such that

$p\circ g$ is continuous
$g \circ \gamma $  is continuous for every path $\gamma$ in $Y$

Prove that $g$ is continuous.

I've been stuck on this problem for a while; some people have told me that the hypotheses that $Y$ be locally path-connected and that $g \circ \gamma $  is continuous for every path $\gamma$ in $Y$ are unnecessary.  I've proven a Lemma that says that if $y_0 $ and $y$ are path-connected in $Y$, there is a unique path in $E$ from $g(y_0)$ to $g(y)$ by using the uniqueness of path liftings in $B$.  I imagine this can be used to prove that $g$ is continuous, but I haven't made any progress on that front.

Any help is appreciated.

You have to show that the inverse image $g^{-1}(U)$ of an open subset $U$ of $E$ is open.

Let $x\in U$,  there exists a neighborhood $V_x$ of $p(x)$ such that for every $y\in p^{-1}(p(x))$, there exists a neighborhood $U_y$ of $y$ such that $p_{\mid U_y}$ the restriction of $p$ to $U_y$ is an homeomorphism onto its image $V_x$, and $U_y\cap U_z$ is empty if $y\neq z$. $(p\circ g)^{-1}(V_x) =\cup_{y\in U_x}g^{-1}(U_y)$ is an open subset. Remark that $g^{-1}(U_y)\cap g^{-1}(U_z)$ is empty if $y\neq z$ since $U_y\cap U_z$ is empty.

Let see that $g^{-1}(U_y)$ is open for every $y\in p^{-1}(p(x))$. Let $z\in g^{-1}(U_y)$,  since  $z\in g^{-1}(V_x)$ there exists a neighborhood $W_z$ of $z$ contained in $g^{-1}(V_x)$. We can suppose that $W_z$ is connected by arcs, for every arc $c:I=[0,1]\rightarrow W_z$ such that $c(0)=z$, $g\circ c$ is an arc such that $g(c(0))\in U_y$, since $g\circ p$ is continuous, there exists $I'=[0,d]\subset I,d>0$ such that $(g\circ c)_{\mid I'}\subset U_y$ this implies that $g^{-1}(U_y)$ contains a neighborhood of $z$, thus is open. In particular, $g^{-1}(U_x)$ is open.

We can write $U=\cup_{x\in U} U_x$ such that $p_{\mid U_x}$ is an homoeomorphism onto its image, since $g^{-1}(U)=\cup_{x\in U}g^{-1}(U_x)$ is open since $g^{-1}(U_x)$ is open.


Computing the sum of an infinite series I am confused as to how to evaluate the infinite series $$\sum_{n=1}^\infty \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n^2+n}}.$$
I tried splitting the fraction into two parts, i.e. $\frac{\sqrt{n+1}}{\sqrt{n^2+n}}$ and $\frac{\sqrt{n}}{\sqrt{n^+n}}$, but we know the two individual infinite series diverge. Now how do I proceed?

Hint:

$$\frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n^2+n}} = \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n+1}\sqrt{n}} = \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}}$$

This kind of series is called Telescoping Series

Your problem may be converted to the following formula:
\begin{align}
& \lim_{N\to\infty}\sum_{n=1}^{N}\left({\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}}\right) =
\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)+\left(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}\right)+...+\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{N+1}}\right) \\
& \hspace{5mm} =
\lim_{N\to\infty}\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{N+1}}\right)=
\lim_{N\to\infty}\left(\frac{\sqrt{N+1}-\sqrt{2}}{\sqrt{2}\sqrt{N+1}}\right) = 1
\end{align}


What's the proper way to prove P$\rightarrow$Q is not tautologically equivalent to Q$\rightarrow$P I want to rigorously show that [$(p \rightarrow q)\equiv(q \rightarrow q)$] is not true.

$Let A =(p \rightarrow q), B =(q \rightarrow q)$

I tried to prove this by supposing $A$ and showing that $B$ does not follow.

[$(p \rightarrow q) \rightarrow (q \rightarrow p)$]

I expanded that but it didn't $= 0$

I was told to look at Reductio Ad Absurdum, but I don't see how that applies in this case.

This is a real-world example.  This is not homework.

The most straightforward way is to construct the truth tables and see that they are not equal. Or you could show values of $P$ and $Q$ that make them different (which would amount to mention the specific line of the truth tables where they differ).

If $P$ is true and $Q$ is false, then $P\to Q$ is false, while $Q\to P$ is true.

Edit: Changed $Q\to Q$ to $Q\to P$, to go with the title.

Assume $B$.  Show that $A$ could be false.

By a similar method you can show that if B holds true, then (p $\rightarrow$ r) could be false.


Proving the de Rham cohomology of $M$ and $M \times \mathbb R$ are isomorphic without homotopy Given a smooth manifold $M$, is there an elementary way of showing $H^k_{dR} (M) \cong H^k_{dR} (M \times \mathbb R)$, i.e. straight from the definitions of cohomology being the quotient of closed forms by exact forms, without homotopy or integration on $M$?

The idea I had was to write forms on $M \times \mathbb R$ in terms of those on $M$, however it is not entirely obvious to me how to go about this and convert to a statement about cohomology.

The existing answer is worthless! Forms on $\Bbb R \times M$ do not take the shape "$dt \wedge \alpha$", they are much more complicated. So nothing has been proved.

A $k$-form on $N = \Bbb R \times M$ takes the shape $dt \wedge \alpha_t + \beta_t$, where $\beta_t$ is a time-varying $k$-form on $M$ and $\alpha_t$ is a time-varying $(k-1)$-form.

Then $$d_N(dt \wedge \alpha_t + \beta_t) = dt \wedge (\beta'_t - d_M \alpha_t) + d_M \beta_t.$$

For our form to be closed, then, we would have $\beta'_t = d_M \alpha_t$ and $d_M \beta_t = 0$. Thus we may write $\beta_t = \beta_0 + \int_0^t d_M \alpha_s ds$, where $d_M \beta_0 = 0$.

Claim: suppose $\beta_0 = d_M \eta_0$. Then what we seek are time-varying forms $\gamma_t$ and $\eta_t$ with $\eta'_t - d_M \gamma_t = \alpha_t$ and $\beta_t = d_M \eta_t.$ This is simple enough: take $\eta_t = \eta_0 + \int_0^t \alpha_s ds.$ Now we have the equation $d_M \gamma_t = \eta'_t - \alpha_t$; now one needs to know that there is a left-inverse $A$ to $d_M$ on the space of exact forms. This would give $\gamma_t = A(\eta'_t - \alpha_t)$.

Usually this operator $A$ comes from Hodge theory, which says that $\text{ker}(d_M^*) \xrightarrow{d_M} \text{im}(d_M)$ is surjective with finite-dimensional kernel; here $d_M^*$ is the adjoint operator. Take the existence of this operator $A$ for granted.

What we have shown is that every closed form with $\beta_0$ exact is in fact an exact form. This means that the map $H^k(M \times \Bbb R) \to H^k(M)$ given by restricting to the time-0 slice is injective. But surjectivity is clear because if $\alpha$ is a closed form on $M$ then $dt \wedge \alpha$ is a closed form on $M \times \Bbb R$.

Hint: Let $\alpha$ be a form of $M$, suppose that $dt\wedge \alpha$ is closed, it implies that $\alpha$ is closed, $d(t.\alpha)=dt\wedge\alpha$


Prove or give an counter example of the following statement. Let $\emptyset \neq U \subset \mathbb{R}^N$ open, let $x_0 \in U$, and let $f\in C^1(U, \mathbb{R}^N)$ be such that $f(V)$ is open for each open neighborhood $V\subset U$ of x_0. Then $\det J_f(x_o)\neq 0$.

Can anyone tell me how to start with this?

I think that if $n=2$ and $p=x^2-45=\frac{4}{70}$, then you will achieve a pen15 ratio in the form of $e^{35b}$.


folland Exercise 2.24 folland Exercise 2.24. Let $(X, \mathcal{M}, \mu)$ be a measure space with $\mu(X)<\infty,$ and let $(X, \overline{\mathcal{M}}, \bar{\mu})$ be its completion. Suppose $f: X \rightarrow \mathbb{R}$ is bounded. Then $f$ is $\overline{\mathcal{M}}$ -measurable (and hence in $\left.L^{1}(\bar{\mu})\right)$ iff there exist sequences $\left\{\phi_{n}\right\}$ and $\left\{\psi_{n}\right\}$ of $\mathcal{M}$ -measurable simple functions such that $\phi_{n} \leq f \leq \psi_{n}$ and $\int\left(\psi_{n}-\phi_{n}\right) d \mu< \frac{1}{n} .$ In this case, $\lim \int \phi_{n} d \mu=\lim \int \psi_{n} d \mu=\int f d \bar{\mu}$ .

also show that for case $\mu(X)=\infty$ or $f$ is $un$bounded this is not true .

Suppose $f$ is $\bar{\mathcal{M}}$-measurable and non-negative.  Then there is an $\mathcal{M}$-measurable $g$ such that $f = g$ up to an $\mathcal{M}$-null set $N$.  If $s = \sum_1^k a_j\cdot \chi_{A_j}$ is an arbitrary simple function, then $s^- = (\sum_1^k a_j\cdot \chi_{A_j \setminus N}) + 0\cdot \chi_N$ and $s^+ = (\sum_1^k a_j\cdot \chi_{A_j \setminus N}) + M \cdot \chi_N$ are simple functions, where $M$ is a bounding constant for $f(x)$.

As in Theorem $\textbf{2.10}$, write simple functions
$$\phi_n = \sum_{k = 0}^{2^{2n}-1} k2^{-n}\cdot \chi_{E_n^k} + M \cdot \chi_{F_n}, \hspace{1cm} \psi_n =  \sum_{k = 0}^{2^{2n}-1} (k+1)2^{-n}\cdot \chi_{E_n^k} + M \cdot \chi_{F_n}$$
$$E_n^k = g^{-1}((k2^{-n}, (k+1)2^{-n}]), \hspace{1cm} F_n = g^{-1}((2^n, M))$$
Then these are simple functions converging pointwise to $g$ from above and below, respectively, in monotone fashion.  Note that for $n$ large enough, $2^n > M$ so that for $n$ sufficiently large we can simply write
$$\phi_n^- = \sum_{k = 0}^{2^{2n}-1} k2^{-n}\cdot \chi_{E_n^k \setminus N}, \hspace{1cm} \psi_n^+ =  \sum_{k = 0}^{2^{2n}-1} (k+1)2^{-n}\cdot \chi_{E_n^k} + M \cdot \chi_N$$
Since $f = g$ away from $N$, $\phi_n^- \leq f \leq \psi_n^+$ for all $n$, and on $N$ itself, $\phi_n^- \equiv 0 \leq f \leq M \equiv \psi_n^+$.

$f: X \rightarrow \mathbb{R}$ is bounded, say $|f| \leq M, \quad$  Suppose $f$ is $\bar{m}$ -measurable. Let $g_{+}(x)=f(x)+M$ and $g_{-}(x)=-f(x)+M .$ Then $g_{\pm} \in L^{+}(\bar{\mu})$ so there are increasing sequences $\left\{\phi_{j}^{+}\right\}$ and $\left\{\phi^- _{j}\right\}$ of simple functions in $L^{+}(\bar{\mu}) , \phi_{j}^{+} \rightarrow g_{+}$ and $\phi_{j}^{-} \rightarrow g_{-} b y$ Theorem 2.10. By the Monotone Convergence Theorem, $\int \mathrm{\phi}_{j}^{\pm} d \bar{\mu} \rightarrow \int g_{\pm} d \bar{\mu}$. Given $n \in \mathbb{N},$ since $g_{\pm} \in \mathrm{L}^{\prime}$
$\exists j \in \mathbb{N} , 0 \leq \int g_{+} d \bar{\mu}-\int \phi_{j}^{+} d \bar{\mu}<\frac{1}{2 n}, \quad 0 \leq \int g_{-} d \bar{\mu}-\int \phi_{j}^{-} d \bar{\mu} .<\frac{1}{2 n} . \quad$ Suppose the standard representation of
$\phi_{j}^{+}$ is $\sum_{k=1}^{m^{k}} a_{k}^{\pm} \chi_{A_{k}^{+}}$ where $A_{k}^{\pm} \in \bar{m} ;$ for each $k$, choose $E_{k}^{\pm} \in M$ 3 $E_{k}^{\pm} \subset A_{k}^{\pm}$ and $\bar{\mu}\left(A_{k}^{\pm}-E_{k}^{\pm}\right)=0$
(which we can do by Theorem( 1.9)). Then $0 \leq \sum a_{k}^{\pm} x_{E_{k}^{\pm}} \leq \phi_{j}^{\pm}+M . \quad$ Let $\varphi_{n}=\sum q_{k}^{+} x_{E_{k}^{+}}-N$
and $\psi_{n}=M-\sum a_{k}^{-} X_{E_{k}}^{-} .$ Then $\varphi_{n}$ and $\Psi_{n}$ are simple functions which are $m$ -measurable, $\varphi_{n} \leq f \leqslant \Psi_{n}$ and $\int\left(\psi_{n}-\varphi_{n}\right) d \mu=\int\left(\psi_{n}-f\right) d \bar{\mu}+\int\left(f-\varphi_{n}\right) d \bar{\mu}=\int\left(g_{-}-\rho_{j}^{-}\right) d \bar{\mu}+\int\left(g_{+}-\rho_{j}^{+}\right) d \bar{\mu}<\frac{1}{2 n}+\frac{1}{2 n}=\frac{1}{n}$
Conversely. Let $g=\sup \varphi_{n}$ and $h=\inf \psi_{n}$ gand $h$ are $m$ -measurable, $g \leq f \leq h$ and $(\forall n) \quad \int(h-g) d \mu \leq \int\left(\psi_{n}-\varphi_{n}\right) d \mu<\frac{1}{n} \quad$ So $\quad h-g \in L^{+}(\mu) \quad$ and $\quad \int(h-g) d \mu=0,$ so by
Prop. $2.16, \quad g=h \quad$ a.e. $(\mu) . \quad$ Let $E=\{x: g(x)<h(x)\} .$ Then $E \in m$ and $\mu E=0 .$ Let $F=\{x: g(x)<f(x)\}$
Then $F \subset E,$ so $F \in \bar{m}$ and $\bar{\mu} F=0 .$ since $g$ is also $\bar{m}$ -measurable and we now have $f=g$ a.e. $(\bar{\mu})$
$f$ is $\bar{m}$ -measurable by Prop. $2.11(a)$ Moreover, since $\int f d \bar{\mu} \leq \int \psi_{n} d \bar{\mu}=\int \psi_{n} d \mu<\int \varphi_{n} d \mu+\frac{1}{n}=\int \varphi_{n} d \bar{\mu}+\frac{1}{n} \leq \int f d \bar{\mu}+\frac{1}{n}$
$\lim \int \varphi_{n} d \mu=\lim \int \psi_{n} d \mu=\int f d \bar{\mu}$


What is the result of $\infty - 1$? I was wondering after reading "What is the result of infinity minus infinity", is there any logical result for $\infty - 1$ ?

Most usually, the answer will be "it doesn't make sense to write that down". The operation of subtraction is something defined only for certain classes of numbers. The class you are probably most familiar with are the Real numbers (denoted by $\mathbb{R}$), which you can think of intuitively as numbers that you can point to on a number line. These include things like $2.5$, $\pi$, and $\sqrt{2}$, but it does not include $\infty$. Since subtraction is defined for these numbers only, to then ask "What is $\infty - 1$ ?" is like asking "What is Cat - 1 ?".

In more advanced mathematics (for example, in Measure theory) we do allow $\infty$ to denote a certain object that can interact with the Real numbers. In those situations, it usually follows your intuition and $\infty - 1$ is defined to be $\infty$ again. Another example of when mathematicians consider the idea of "$\infty - 1$" is in introductory calculus, when one first learns about limits. In that case however, the usage of the symbol $\infty$ is purely short hand notational- it doesn't actually denote any type of number. If a function $f$ becomes arbitrarily large as $n$ becomes large, we may write $\displaystyle\lim_{n\to\infty} f(n) = \infty$ and then for shorthand a teacher may write on the board $$\displaystyle\lim_{n\to\infty} f(n)-1 = \infty - 1 = \infty $$ but strictly that equation is invalid. So indeed, the idea does come up in valid forms, but what you should take away from this post is mainly contained in the first paragraph.

In layman's terms, infinity is something which you cannot count. Take out $1$ from it and you still cannot count it, so the result is infinity.


$\mathbb{C}$ Forms a Vector Space Over $\mathbb{R}^2$ Proof Question In my Mathematical Techniques course we've been talking about vector spaces, bases, etc.

There is one problem however that I cannot get my head around and that is to prove that $\mathbb{C}$ can be represented as a 2 dimensional vector space over the reals, $\mathbb{R}^2$. I'm fine with proving the associativity, commutativity and stuff however I'm not sure whether the basis vectors should have one or two elements.

What are the basis vectors? $$\begin{bmatrix} 1 \end{bmatrix} \begin{bmatrix} i \end{bmatrix} or \begin{bmatrix} 1 \\ 0 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} or \begin{bmatrix} 1 \\ 0 \end{bmatrix} \begin{bmatrix} 0 \\ i \end{bmatrix}$$ Such that: $$ a + bi  \rightarrow a \begin{bmatrix} 1 \end{bmatrix} +b\begin{bmatrix} i \end{bmatrix} ~or~ \begin{bmatrix} a \\ 0 \end{bmatrix} +\begin{bmatrix} 0 \\ b \end{bmatrix} ~or~ \begin{bmatrix} a \\ 0 \end{bmatrix} +\begin{bmatrix} 0 \\ bi \end{bmatrix}$$

For some reason I can't get my head around this part of the problem. My lecturer was very vague when he covered this bit which has left me a little dubious.

Thanks,
Sean.

In order to identify $\mathbb{C}$ as a $2$-dimensional vector space over $\mathbb{R}$ (note that we do not say it is a vector space over $\mathbb{R}^2$---"over $\mathbb{R}$" simply reminds us what set we're using for scalar multiplication), we need to specify a basis (over $\mathbb{R}$), necessarily of two elements. We usually write elements of $\mathbb{C}$ as
$$a + bi,$$
where $a$ and $b$ are (unique) real numbers which implicitly defines the basis $(1, i)$ of $\mathbb{C}$ over $\mathbb{R}$.
On the other hand, we know that any $n$-dimensional vector space over $\mathbb{R}$ is isomorphic to $\mathbb{R}^n$, and we usually think of $\mathbb{R}^n$ as the set of vectors with $n$ real entries.

We can connect this with our usual complex notation by writing down an explicit isomorphism. As usual, it is enough to specify what happens to basis elements. One natural choice for the isomorphism is the map $\Phi$ defined by the assignments
$$1 \mapsto \begin{pmatrix}1 \\ 0\end{pmatrix}, \qquad i \mapsto \begin{pmatrix}0 \\ 1\end{pmatrix}.$$
By linearity (over $\mathbb{R}$), the general formula for $\Phi$ must then be
$$\Phi: a + bi \mapsto \begin{pmatrix}a \\ b\end{pmatrix}.$$

Remark Note that we can also represent complex numbers $a + bi$ as matrices, via the isomorphism
$$\Psi: a + bi \mapsto \begin{pmatrix} a & -b \\ b & a \end{pmatrix}.$$
The repeated entries may seem redundant but this map has the additional benefit that $\Psi$ turns multiplication of complex numbers into multiplication of matrices, that is, $\Psi$ respects the addition and multiplication operations of complex numbers and of complex matrices, and we call such maps ring isomorphisms.

You know that every field is an Integral domain, but in $\mathbb{R}^2$, what is $(1,0).(0,1)?$ Aren't they zero divisors, so how can $\mathbb{R}^2$ be a field if not an integral domain, so how can $\mathbb{C}$ be  a vector space?

It makes sense over $\mathbb{R}$ as a vector space but not over $\mathbb{R}^2$, rethink what your professor did?


computability, number theory, and continuum hypothesis Is there any similarity between proofs of unsolvability results in computability theory (and decision procedures for identities in number theory) and the independence of the continuum hypothesis in set theory? See e.g.,

  A. A. Muchnik, On the unsolvability of the problem of reducibility in the theory of algorithms. (in Russian) Doklady Akademii Nauk SSSR (N.S.), vol. 108 (1956), pp. 194–197.

Note 1. Relevant links were pointed out by Noah Schweber: on computability and Ethan Bolker: on number theory.

No, there isn't really. The independence of the continuum hypothesis means that (assuming ZFC is consistent) we can find models $A$ and $B$ of ZFC such that CH is true in $A$ but false in $B$, and is proved by developing a method of constructing models of ZFC. The unsolvability of the word problem, meanwhile, just asserts that a certain set is not computable, and is proved by showing that we can code the halting problem into the word problem - this does involve a construction (understanding how to make groups and words "to order"), but the construction is completely unrelated to forcing. These results and proofs don't really have anything to do with each other.

There is a vague connection between forcing and oracle computations in computability theory, where we think of "in $V$" as "computable" and "in $V[G]$" as "computable from $G$" (and I suspect the source you're thinking of is one where Cohen mentioned that this analogy, following his learning of the Friedberg-Muchnik theorem, helped conceptualize forcing - although as I recall, this statement was made long after the discovery of forcing, and contradicts earlier statements of Cohen; I'll try to track down the citation and commentary). However, this doesn't actually relate CH-independence to word problem-unsolvability, it just says that we should expect computability-theoretic intuition to sometimes be useful in understanding forcing arguments (and vice versa).

EDIT: Re: the possibility of a connection with Friedberg-Muchnik (and priority arguments in general) in place of group-theoretic unsolvability, this was addressed at this mathoverflow question. Incidentally, note that there I was unable to track down the reference I remember in which Cohen spoke of this connection, so even that might not have ever been made by Cohen. That said, an analogy was drawn by others (e.g. Kreisel, Nerode, Remmel), and I think it is a very fruitful one. So I would say yes, there is a strong analogy here; however, it may not have ever been noted by Cohen, contrary to my prior claim.

Both belong to a loose class of mathematical results saying that something is not achievable by such-and-such kit of tools. The irrationality of $\sqrt{2}$ is most likely the first result in this class. Then follow the angle-trisection and a few other similar ancient problems. And a zillion more of the same has come with the modern mathematics, including the CH problem.


Prove that $\displaystyle \sum_{k=1}^n \bigg(\dfrac{1}{k}+\dfrac{2}{k+n}\bigg ) \leq \ln(2n) + 2 -\ln(2)$ Prove that $$\displaystyle \sum_{k=1}^n \bigg(\dfrac{1}{k}+\dfrac{2}{k+n}\bigg ) \leq \ln(2n) + 2 -\ln(2).$$

I was thinking of using mathematical induction for this. That is,

We prove by induction on $n$. The case $n=1$ holds trivially since $2 \leq 2$. Now assume the result holds for some $m$. Then by assumption we know that $$\displaystyle \sum_{k=1}^{m+1} \bigg(\dfrac{1}{k}+\dfrac{2}{k+m}\bigg ) \leq \ln(2m) + 2 -\ln(2)+\dfrac{1}{m+1}+\dfrac{2}{2m+1}. $$

We must relate this somehow to $\ln(2(m+1)) + 2 -\ln(2)$.

First note, that we can rewrite the RHS of your inequality as $\ln(n) +2$.

We proceed by induction. We compute

$$ \sum_{k=1}^{n+1} \left( \frac{1}{k} + \frac{2}{k+n+1}\right)
= \frac{1}{n+1} + \frac{2}{2n+2} + \sum_{k=1}^n \frac{1}{k} + \sum_{k=1}^{n} \frac{2}{k+n+1}$$

shifting the index yields

$$ =\frac{2}{n+1} + \sum_{k=1}^n \frac{1}{k} + \sum_{k=2}^{n+1} \frac{2}{k+n}
= \sum_{k=1}^n \frac{1}{k} + \sum_{k=1}^{n+1} \frac{2}{k+n}
= \sum_{k=1}^n\left( \frac{1}{k} + \frac{2}{k+n}\right) + \frac{2}{2n+1}$$

applying the induction hypothesis gives

$$ \leq  \ln(n) + 2 + \frac{2}{2n+1} = \ln(n+1) +2 + \left(  \frac{2}{2n+1} + \ln(n) - \ln(n+1) \right).$$

Hence, we are left to show that

$$ \frac{2}{2n+1} - \ln\left(1 + \frac{1}{n}\right) = \frac{2}{2n+1} +  \ln(n) - \ln(n+1) \leq 0. $$

This is already done in a previous answer.

If you want to do a proof by induction.
you have covered the base case.

Inductive hypothesis:

$\sum_\limits{k=1}^n \bigg(\dfrac{1}{k}+\dfrac{2}{k+n}\bigg ) \leq \ln(2n) + 2 -\ln(2)$.

We need to show that:

$\sum_\limits{k=1}^{n+1} \bigg(\dfrac{1}{k}+\dfrac{2}{k+n+1}\bigg ) \leq \ln(2n) + 2 -\ln(2)$

thoughts:

$\ln(2n) + 2 -\ln(2) = \ln (n) + 2$

$\sum_\limits{k=1}^{n+1} \bigg(\dfrac{1}{k}+\dfrac{2}{k+n+1}\bigg ) = \sum_\limits{k=1}^{n} \bigg(\dfrac{1}{k}+\dfrac{2}{k+n}\bigg ) + \dfrac{2}{2n+1}$

$\sum_\limits{k=1}^{n} \bigg(\dfrac{1}{k}+\dfrac{2}{k+n}\bigg )\leq  \ln n + 2$ by the inductive hypothesis.

So, if $\dfrac{2}{2n+1}-\ln (n) + 2 \leq \ln(n+1) + 2$, we are done.

$\dfrac{2}{2n+1} < \ln(1+\frac{1}{n})$

$\sum_\limits{k=1}^{n+1} \bigg(\dfrac{1}{k}+\dfrac{2}{k+n+1}\bigg ) \leq \ln(2n) + 2 -\ln(2)$

QED


solve system equation: $ 2a^2 - 1 = b, 2b^2 - 1 = c, 2c^2 - 1 = a $ I have this system equation:

$$ 2a^2 - 1 = b $$
$$ 2b^2 - 1 = c $$
$$ 2c^2 - 1 = a $$

From system equation we see that $ a \neq 0 , b \neq 0, c \neq 0 $ , so :
$ 2a^2 - 1 \neq 0 => a \neq \sqrt{\frac 1 2} $ .
Also we can write: $ b \neq \sqrt{\frac 1 2} $ And $ c \neq \sqrt{\frac 1 2} $

But I don't know if this is the answer or I need to find exact numbers that satisfy this system (btw, I found an answer and that's : $a = 1, b = 1, c = 1 $).

The given system of equations is symmetric, which means that by substituting the variables in the equations by any permutation of the set $\left (a,b,c \right )$ we get the same system all over again. Using that fact, equality $a=b=c$ must hold.

Now simply by using $a=b$ from the first equation of the system we obtain:
$$2a^2-a-1=0$$ Using quadratic formula we get $a_{1}=1$ and $a_{2}=-\frac{1}{2}$.

Therefore, the solution of the system is:

$$\left ( a,b,c \right )=\left \{ \left ( 1,1,1 \right ), \left ( -\frac{1}{2}, -\frac{1}{2},-\frac{1}{2} \right)  \right \}$$


Determine if $ \int_3^{\infty}\frac{x+1}{\sqrt{x^4-x}}\,dx $ converges Determine whether the following integral is convergent or divergent without evaluating it.
￼￼￼(Whichever answer is correct, you must show why it is true.)
$$
\int_3^{\infty}\frac{x+1}{\sqrt{x^4-x}}\,dx
$$

I know that convergence depends on the existence of the limit. How do you find the limit of this equation? Is there a short cut (theorem, principle, memorizable equation) that simplifies this?

This is the first step...

$$\lim\limits_{t \to \infty} \int_3^t\frac{x+1}{\sqrt{x^4-x}}$$

\begin{equation*}
\int_3^{\infty}\frac{x+1}{\sqrt{x^4-x}}\,dx \;>\;
\int_3^{\infty}\frac{x+1}{\sqrt{x^4}}\,dx  \;=\;
\int_3^{\infty}\frac{x+1}{x^2}\,dx \;>\;
\int_3^{\infty}\frac{x}{x^2}\,dx \;=\;
\int_3^{\infty}\frac{1}{x}\,dx
\end{equation*}
The last integral diverges.

HINT:

$\displaystyle \frac{1}{2^{x}} < \frac{(x+1)}{\sqrt{x^4 - x}}$

Does $\displaystyle \int_{3}^{\infty} 2^{-x} dx$ converge?

Then find a function greater.


logistic differential equation problem setup The Pacific halibut fishery has been modeled by the differential equation $$\frac{dy}{dt} = ky(1-\frac{y}{m}) $$ where $y(t)$ is the biomass in kg at time $t$ years. The carrying capacity is estimated to be $M = 8x10^7$kg and $k = 0.71/yr$. If $y(0) = 2x10^7$kg, find the biomass a year later.

I am having a difficult time calculating this problem so I would just like some input on whether I setup the problem correctly.

$$y(t) = \frac{8x10^7}{1 + 3x10^7e^{-0.71}}, ~~A = \frac{8x10^7 - 2x10^7}{2x10^7} = 3x10^7$$

I think it is just a simple plugging into the equation given in the reading which is $$P(t) = \frac{M}{1+Ae^{-kt}}, ~~ A = \frac{M-P_0}{P_0} $$

Let me k now if this is correct or not, thanks


What is the term for when 2 numbers cannot be factors of one another? Lets say I have 2 numbers, A and B, In my scenario A cannot be multiplied by a integer constant (k) in order to make B. Is there a defined term or symbol for this to make it work?.

I assume this symbol does exist since vectors can be scalar products of one another, and it seems easier just to say something like B/A ≠ ℤ.

This situation is usually written as $A\nmid B$

You can write that:

$\frac BA\notin\Bbb Z$


Definitions of an additive functor In my Rings and Modules class, we defined additive functors this way:

A (covariant/contravariant) functor $F$ is called additive if:

(i). $F0 = 0$, where $0$ is the zero-module.

(ii). For any modules $M, N$ we have $F(M \oplus N) = FM \oplus FN$.

I looked around online and I saw this definitions, which looks the same but uses morphisms instead of modules:

A (covariant/contravariant) functor $F$ is called additive if:

(i). $F0 = 0$, where $0$ is the zero-map.

(ii). For any modules $M, N$ and morphisms $\phi, \psi: M \longrightarrow N$ we have $F(\phi + \psi) = F \phi + F \psi : FM \longrightarrow FN$.

Now it seems clear, intuitively, how these two concepts are equivalent. However, I always have a lot of trouble proving elementary stuff with functors, because I always think if what I'm doing is allowed.

I was trying to prove that the first two conditions imply the other two. So I tried to rephrase them in terms of compositions of homomorphisms.

(i). Let $\phi : M \longrightarrow N$ be the zero-map. Then $\phi$ is the same as $\iota \circ \phi_0$, where $\phi_0 : M \longrightarrow 0$ and $\iota : 0 \longrightarrow N$. Then $F \phi = F(\iota \circ \phi_0) = F \iota \circ F \phi_0$ is the zero-map, because we are assuming that $F0 = 0$.

(ii). Let $\phi, \psi: M \longrightarrow N$ be homomorphisms. Then $\phi + \psi$ is the same as the following sequence:
$$ M \overset{id_M \oplus id_M}{\longrightarrow} M \oplus M \overset{(\phi, \psi)}{\longrightarrow} N \oplus N \overset{id_N + id_N}{\longrightarrow}N.$$
At this point, however I am stuck. I don't know how to justify that applying $F$ to this sequence gives $F\phi + F\psi$, since natural stuff such as inclusions and projections are not necessarily preserved. How should I go on?

I haven't yet tried proving the other direction...

Your first definition is imprecise. What you should say is that the universal property of coproducts furnishes a natural map $F(M) \oplus F(N) \to F(M \oplus N)$ and you want these natural maps to be isomorphisms. Anyway, see this blog post for a thorough discussion. The upshot is that in an additive category direct sums determine the abelian group structure on morphisms.

To show the first definition implies the second, see The Stacks Project, Lemma 12.7.1.
To show the second fefinition implies the first, see The Stacks Project, Lemma 12.3.7. Note that now products equal to coproducts by The Stacks Project, Lemma 12.3.4. The key here is to use an equivalent definition of binary (co)products. Namely, as in The Stacks Project, Remark 12.3.6, if $f:x\to z$, $j:y\to z$, $p:z\to x$ and $q:z\to y$ satisfy $p\circ i=id_{x}$, $q\circ j=id_{y}$, $p\circ j=0$, $q\circ i=0$ and $i\circ p + j\circ q=id_{z}$, then $z$ with these maps is the (co)product of $x$ and $y$.
To see this, we verify for being the product. For any $f:w\to x$ and $g:w\to y$, if there is some $h:w\to z$ s.t. $p\circ h=f$ and $q\circ h=g$, then we have $$ i\circ f+j\circ g=(i\circ p)\circ h+(j\circ q)\circ h= (i\circ p + j\circ q)\circ h = h. $$ And $i\circ f+j\circ g$ can be such $h$ since $$p\circ (i\circ f+j\circ g)=(p\circ i)\circ f +(p\circ j)\circ g= f+0=f $$ and also $q\circ(i\circ f+j\circ g)=g $. So there is a unique morphism from $w$ to $z$ as desired. As for being the coproduct, the proof is similar. Then we see that being additive on the hom sets can preserve these equalities of morphisms. So $F$ will send the binary (co)product to a binary (co)product.
As for the zero object, see the third characterization in The Stacks Project, Lemma 12.3.2 and use the same reasoning.


there is any relation between $\pi$, $\sqrt{2}$ or a generic polygon? I'm a programmer, I'm always looking for new formulas and new way of computing things, to satisfy my curiosity I would like to know if there are any formulas, or I should say equalities, that make use of both $\pi$ and $\sqrt{2}$ .

I would also like to know if it's possible to generalize this relatively to any n-sided polygon ( even a 3D figure ), $\sqrt{2}$ that usually appears in quadrilaterals only.

Of course I would like to know about any possible domain, but since we should start from something, I would say that the domain of polygons and polyhedron triggers my interest in the first place.

$$
\pi = 2\sqrt 2 \cdot \frac{2}{\sqrt{2+\sqrt2}} \cdot \frac{2}{\sqrt{2+\sqrt{2+\sqrt2}}} \cdots.
$$

Find the square root of 2, and call that A. Then find the square root of 2 plus A, and call that B. Then find the square root of 2 plus B, and call that C.  Then find the square root of 2 plus C,  and call that D. Then find the square root of 2 plus D,  and call that E. Then find the square root of 2 plus E,  and call that F. Then find the square root of 2 minus F, and call that G. Then multiply G by 2 to the 7th power, and call that $\pi.$


How to prove the convergence of an integral? I need to prove the convergence of the integral $$\int_{1}^{\infty} \frac{x}{(5x^{2}-1)^3}dx $$
I tried all the tests but could not prove its convergence. I was trying the comparison test and was able to prove that $\frac{x}{(5x^{2}-1)^3} \leq \frac{x}{5x^{2}-1}$ but that wasn't much of help. Any suggestions on how to prove this integral to be convergent?

Do a standard substitution: $u = 5x^2-1 \implies xdx = \dfrac{du}{10}\implies I = \displaystyle \int_{4}^\infty \dfrac{du}{10u^3}=...$ I hope you can find the answer from this point.

If you substitute $v=5x^2-1$ you get an easier integrand (fro some constant $a$ after substitution):
$$
a\int_{1}^{\infty}\frac{dv}{v^3}
$$
which converges


Representable homology classes on smooth manifolds Let $X$ be a closed (compact without boundary) smooth manifold. We can consider its singular homology $H_*(X,\mathbb{Z})$. Let $H_{k}(X,\mathbb{Z})$ be the $k$-th singular homology group of $X$ and let us consider $[\alpha]\in H_{k}(X,\mathbb{Z})$. I have seen in various places that it is sometimes possible to "represent" $[\alpha]$ as a smooth $k$-dimensional submanifold of $X$.

How does this correspondence between elements in $H_{k}(X,\mathbb{Z})$ and $k$-dimensional submanifolds of $X$ exactly works?
When exactly elements in $H_{k}(X,\mathbb{Z})$ can be represented by $k$-dimensional submanifolds?
How is this related to intersection theory, the cup product in cohomology, Poincare duality and the intersection form of $X$ (when $X$ is even-dimensional)?
Does this correspondence work for other homology theories?

Thanks.

Fundamental Classes

Let $Y$ be a connected, closed, orientable $k$-dimensional manifold, then $H_k(Y; \mathbb{Z}) \cong \mathbb{Z}$. The choice of a generator of $H_k(Y; \mathbb{Z})$ is equivalent to a choice of orientation. One often denotes the generator by $[Y]$; we call this a fundamental class of $Y$. If we consider $Y$ with it's other orientation, the corresponding fundamental class is $-[Y]$.

If we drop the orientability hypothesis, $H_k(Y; \mathbb{Z}) = 0$ so the above construction fails. However, $H_k(Y; \mathbb{Z}_2) \cong \mathbb{Z}_2$, so we define the $\mathbb{Z}_2$-fundamental class to be the generator of $H_k(Y; \mathbb{Z}_2)$.

We can also define the fundamental class if $Y$ is only compact instead of closed (i.e. $Y$ has boundary). In that case, we use the fact that $H_k(Y, \partial Y; \mathbb{Z}) \cong \mathbb{Z}$ in the orientable case and $H_k(Y, \partial Y; \mathbb{Z}_2) \cong \mathbb{Z}_2$ in the non-orientable case to define fundamental classes as relative homology classes.

All of the above is discussed in detail on the Manifold Atlas Project's fundamental class page.

Submanifolds and Homology

Now suppose $Y$ is a connected, closed, orientable $k$-dimensional submanifold of $X$ and let $i : Y \to X$ denote the inclusion, then $i_* : H_k(Y; \mathbb{Z}) \to H_k(X; \mathbb{Z})$, in particular $i_*[Y] \in H_k(X; \mathbb{Z})$. Note, the element $i_*[Y]$ might be zero. For example, $S^1$ can be realised as a submanifold of $S^2$ via the equator, but $i_*[S^1] \in H_1(S^2; \mathbb{Z}) = 0$.

We can create homology classes on $X$ from different types of submanifolds as above using the different notions of fundamental class.

If $Y$ is a non-orientable submanifold of $X$, using the $\mathbb{Z}_2$-fundamental class, $i_*[Y] \in H_k(X; \mathbb{Z}_2)$.
If $X$ is a manifold with boundary and $Y$ is an orientable submanifold with boundary (so that $\partial Y \subseteq \partial X$), $i_*[Y, \partial Y] \in H_k(X, \partial X; \mathbb{Z})$.
If $X$ is a manifold with boundary and $Y$ is a non-orientable submanifold with boundary, using the $\mathbb{Z}_2$-fundamental class, $i_*[Y, \partial Y] \in H_k(X, \partial X; \mathbb{Z}_2)$.

Realisable Classes and The Steenrod Problem

An element $\alpha \in H_k(X; \mathbb{Z})$ is said to be realisable if there is a $k$-dimensional connected, closed, orientable $k$-dimensional submanifold $Y$ such that $\alpha = i_*[Y]$. The Steenrod Problem was whether every class is realisable. Thom showed in $1954$ that this is not true in general. However, it is true if $0 \leq k \leq 6$. What is true however is that for any class $\alpha$, there is an integer $N$ such that $N\alpha$ is realisable.

We could ask the analogue of Steenrod's Problem for $\mathbb{Z}_2$ coefficients: is every class in $H_k(X; \mathbb{Z}_2)$ realisable? The answer is yes!

I don't know about the analogue of Steenrod's Problem for manifolds with boundary.

These classes are not always representable, but classes of $H_l(M,Z/2)$ are representable. Positive multiple of elements of $H_l(M,Z)$ can be represented by smooth manifolds, this is shown in the thesis of Rene Thom.


proving that $a\mid b$ I'm trying to prove that $83\mid 3^{123}-1$, or in other words $3^{123}\equiv 1 \mod{83}$ .

Of course $\gcd(83,3)=1$ so from Euler's theorem $3^{\phi(83)}\equiv1 \mod{83}$,
 and $83$ is prime so $\phi(83)=82$, giving us that $3^{82}\equiv 1 \mod{83}$.

This is where I am stuck. I tried to show that $3^{41} \equiv 1 \mod{83}$ but I couldn't manage to do so. Also $83$ is prime so I can't try to prove this on each of his prime factors separately.

I tried out a few steps up the numbers $3+83k$ to see if there is a nearby square - because if there is such a number $a^2\equiv 3 \bmod 83$, then $3^{41} \equiv a^{82} \equiv  1 \bmod 83$ by Fermat's little theorem.

And luckily, $3+2\cdot 83 =169 = 13^2$. So $13^2\equiv 3$ and $3^{123} \equiv 13^{246}\equiv (13^{82})^3 \overset{\small{FLT}}\equiv  1 \bmod 83$

Another "finding convenient numbers" approach: note that $83+1=84=2^2\cdot3\cdot 7$:

$3^4 = 81\equiv -2$
$3^{32} \equiv 2^8 \equiv 256 \equiv 7$
$\to 3^{41} \equiv 7\cdot(-2)^2\cdot 3 \equiv 84\equiv 1 \bmod 83$

Without Euler's Theorem:

$\color\red{3^{4}}\equiv81\equiv\color\red{-2}\pmod{83}\implies$

$3^{123}\equiv3^{4\cdot30+3}\equiv(\color\red{3^{4}})^{30}\cdot3^{3}\equiv(\color\red{-2})^{30}\cdot3^{3}\equiv28991029248\equiv1\pmod{83}$


Show that $\log (xy) \leq \log\Big(\frac{1}{2}\Big)$ I'm trying to sovle the following question:

  Show that $\log (xy) \leq \log\Big(\frac{1}{2}\Big)$ if $(x, y)$ is a point of the unit circle $x^2 + y^2$ = 1 in the open first quadrant $x > 0, y > 0$.

I have used the Lagrange multiplier method to show that the point $\Big (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\Big)$ satisfied the constraint equation, whch yeilds the value $\log\Big(\frac{1}{2}\Big)$. I am not sure, however, how to to argue that this is the (absolute maximum) of the function $\log(xy)$ on the given curve. I don't think I can compactify the region by considering the unit circle in the first quadrant, because points on the coordinate axes have to be excluded.

Any suggestions? I'm looking for a solution which doesn't involve converting this problem to single variable problem.

\begin{align}
   &(x-y)^2\ge0\\
\implies&x^2-2xy+y^2\ge0\\
\implies&-2xy\ge-1
\end{align}

Do you know how to continue?

ln(xy) ≤ ln(1/2)
_ as ln is a bijective, increasing function, we know that:
xy ≤ 1/2. making the reasonable sub: (x,y) = (cos(t), sin(t))

for it satisfies the equation: $xˆ2 +yˆ2 = 1$

We find that: $2xy ≤ 1 \rightarrow 2 \sin(t)\cos(t) ≤ 1\rightarrow \sin(2t) ≤ 1$ which is true because we know that $\sin(t)$ is bounded between 1,-1.


Packing two different circles in a rectangle Given the dimension of a rectangle and radii of two circles, how can I decide if these two circles can fit in the rectangle?

I don't know if there is a formula to compute such a thing!

thank you :)

In the figure below, let circle $E$ have radius $r_1$, circle $G$ have radius $r_2$ and the rectangle be $H$ high and $W$ wide.  You get the best fit by putting the circles in opposite corners.  First you need $H,W \gt 2r_1, 2r_2$ or one of the circles won't fit all by itself.  Then the coordinates of $E$ are $(r_1,H-r_1)$ and the coordinates of $G$ are $(W-r_2,r_2)$.  You need them to be at least $r_1+r_2$ apart, so the requirement is $$\sqrt{(W-r_1-r_2)^2+(H-r_1-r_2)^2} \gt r_1+r_2$$

To pack two circles tightly together is to touch them tangently.  The rectangle must be as wide as the longer diameter and the longer side as long as the sum of the diameters.

So if circle one  has diameter = $d_1$ and circle two has diameter = $d_2$ and $d_1 \ge d_2$ and the rectangle has sides $l$ and $w$ and $l \ge w$, then the two circles will fit if $d_1 \le w$ and $d_1 + d_2 \le l$.


How to find ALL cluster points of a sequence? The task of finding a limit point of a sequence is 'simple'; if the sequence converges, we locate the limit point by inspection and proved that it the sequence converges to that point. By uniqueness, there is no other limit point.

For cluster points or accumulation points, since they are not unique, how do I know for sure that I have found them all? Must I prove that there are no other cluster points besides those I have found? Am I being too cautious? What is the importance of cluster points other than the limit point of some subsequence?

Yes, in order to be sure that you have found them all, you must show no other cluster points exist. This will most likely be difficult in practice as you need to find a neighborhood of the point which contains no points of the sequence beyond a certain index.

Intuitively, I think about cluster points as the set of points that sequence continues to get near as the index increases. They are also the idea that leads to limit points in topology.

A special case:Suppose a sequence {xn=x} is constant then its cluster point is same x


The calculation is correct.

You might note instead that in the interval you are integrating over, our function is well-behaved (integrable). It is an odd function, and we are integrating over an interval symmetric about $x=0$. So we can conclude that the integral is $0$ without doing any calculation at all.

Recall that a function $f(x)$ is odd if $f(-x)=-f(x)$ for all $x$.

You might want to sketch the curve, or have software do it for you. The negative "area" from $-1$ to $0$ exactly cancels the positive area from $0$ to $1$.

Not quite. Use u-substitution and set u=x^2-9.


What is a good reference for automatic (algorithmic) differentiation? Could someone kindly provide me with a good reference about automatic  differentiation? I have already had a look at the wikipedia article, but I would need a book or article, which gives a better theoretical understanding. It would also be helpful if I had a step by step guide to understand the concepts.

Try also these simpler papers:

The arithmetic of differentiation by L. B. Rall
A simple automatic derivative evaluation program by R. E. Wengert

You won't find a reference because for a programmer the problem is almost trivial. However, you can find many students projects implementing this problem in every language.

The steps are :

parsing, transforming the text $\sin(z)+4 \exp(2\log(z))$ to a binary tree. At first, you can avoid this step by constructing binary trees directly :

 new add(new sin(z),new mult(4,new exp(new mult(new constant(2),new log(new var('z')))))

differentiating the binary tree to return a new binary tree
simplifying the obtained mathematical expression (this step is of course the most difficult, and is still a research problem)

Once you have your binary tree of mathematical expressions, differentiating is really not complicated.

A node will be of the form $u+v, u-v, uv, u/v, u^v$ where $u,v$ are two mathematical expressions, or $f(u)$ where $f$ is an elementary function. Each of those cases is easy to differentiate :

$$(u+v)' = u'+v', \ (uv)' = u'v+v'u, \ (u/v)' = u'/v - uv'/v^2,\  u^v = u' u^{v-1} + v' \log(u) u^v , \  (f(u))' = u' f'(u)$$

Programatically in C++ you'll have a classadd and a class sin

 class add : expression {
     expression * u,* v;
     void print() {  .... }
     expression * differentiate() {
         return new add(u->differentiate(),v->differentiate());
     }
 };

  class sin : expression {
     expression * u;
     expression * differentiate() {
        return new mult(u->differentiate(),new cos(u));
     }
 };

And the other operators and elementary functions follow the same idea.


Intuition for partial averaging eqution I just learnt about the condition expectation and as is known, the definition is as follows:

My question is for the second property (partial averaging property), what kind of intuition does it express? How can I understand in a more perceptual way. Thanks so much!

Well, for $A$ with positive probability, we can define $\frac{1}{P(A)}\int \limits_{A} X\,dP$ as the partial average of $X$ given that the event $A$ occurs.  In other words, this quantity is our best guess for the value of $X$ if all we know is that the event $A$ has occurred.

The partial averaging property you listed is equivalent to:

$$\begin{cases} \frac{1}{P(A)}\int \limits_{A} E[X|\mathcal{G}]\,dP = \frac{1}{P(A)}\int \limits_{A} X\,dP & P(A) \neq 0 \\ \int \limits_{A} E[X|\mathcal{G}]\,dP = \int \limits_{A} X\,dP = 0 & P(A) = 0 \end{cases}.$$

Note that the second line (over probability $0$ events) is trivial.  For any random variables $X$ and $Y$, if $P(A) =0$, $\int_{A} X \,dP = \int_{A} Y \,dP$, since both the left hand side and right hand side equal $0$.  So, we are really interested in the first equation which is over non-zero probability events.  All I'm trying to say is that the following statements are equivalent:

$$(S1) \,\,\,\,\,\, \frac{1}{P(A)}\int \limits_{A} E[X|\mathcal{G}]\,dP = \frac{1}{P(A)}\int \limits_{A} X\,dP \text{ for all events } A \text{ with } P(A) \neq 0$$
and
$$(S2) \,\,\,\,\,\,\int \limits_{A} E[X|\mathcal{G}]\,dP = \int \limits_{A} X\,dP \text{ for all events } A.$$

The latter ($S2$) is typically taken as the axiom defining conditional expectation, but the former ($S1$) is the one where the intuition/interpretation comes into play.  With the idea that $\frac{1}{P(A)} \int_{A} Y \,dP$ is our best guess for the random variable $Y$ over the event $A$ (i.e., given the event $A$ occurs), we can interpret the "intuitively nice" form ($S1$) from above as:

The conditional expectation of $X$ over the collection of events $\mathcal{G}$ is our best guess of the value $X$ given the information in $\mathcal{G}$, and so should be the random variable for which our best guess of it, given any piece of information in $\mathcal{G}$ (i.e., given any event $A \in \mathcal{G}$), equals our best guess of $X$ given that same piece of $\mathcal{G}$-information.  Keep reading this because I did in fact talk about the best guess of a best guess, which can take some getting used to.  The important point is that the best guess of a random variable given information in a $\sigma$-algebra is itself a random variable, so it makes sense to talk about the best guess of the best guess.

It can easily be proved from this that there is a unique $\mathcal{G}$-measurable function such that the partial averaging property holds (and intuitively it's clear that there should only be one best guess of a random variable -- if there were more than one, then they can't both be the "best").  So, that tells us there is a unique random variable the value of which is known from the information in $\mathcal{G}$ and which has the same best guess over each piece of information in $\mathcal{G}$ as the random variable $X$.

Let $A$ have the following elements in it $(w_1,w_2,w_3)$

$E[X|G]$ is a random variable measurable in $G$.

In simplest terms, for a set $A$ in $G$, you know the value of
$$E[X|G] = \frac{ X(w_1) * P(w_1) + X(w_2) * P(w_2) + X(w_3) * P(w_3)}{ P(w_1) + P(w_2) + P(w_3)}$$

i.e., expected value of $X$ , given $A$ event has occurred is product of $X(w)*P(w) $, divided by $P~$ ($~A$ event happening)

Now integral of this $E[X|G]$ which has one value for set $A$, over $A$ is :

$$4~\frac{ X(w_1) * P(w_1) + X(w_2) * P(w_2) + X(w_3) * P(w_3) } { (P(w_1) + P(w_2) + P(w_3)) * (P(w_1) + P(w_2) + P(w_3))}$$

which is nothing but

$$X(w_1) * P(w_1) + X(w_2) * P(w_2) + X(w_3) * P(w_3)$$


Proving that $f(x) = \frac{1}{x}$ is not uniformly continuous over $(0,1)$ - Approach to choosing the correct $x$ and $y$ I've looked at the various other posts on this question and I've also worked on uniform continuity in the past. But right now as I work through Spivak's Calculus, I'm seeing that I haven't actually learned how to find choices for my $x$ and $y$ elements in a structured well thought out way. In the past what I thought was doing this was actually me just regurgitating what a professor did in their proof without any understanding of how $x$ and $y$ were chosen.
To illustrate where I'm stuck:
Let $\epsilon = 1$. Taking arbitrary $\delta > 0$. I encounter the following:
$$\bigg|\frac{1}{x} - \frac{1}{y} \bigg| > \epsilon = 1$$
I've seen prior choices such as letting $x = \delta$ and $y = \frac{\delta}{2}$ or letting $x < \delta$ and $y = \frac{x}{2}$. But I want to understand how these choices were made initially.
I know that $x(\delta)$ and $y(\delta)$ as well if I fiddle with the the expression above I have
$$\bigg|\frac{x-y}{xy}\bigg|$$
I also know that since $x,y \in (0,1)$ that $xy < 1$ (but this now won't help me when I want to extend the interval to $(0, \infty)$.
It appears I'm throwing a lot of stuff at the wall hoping for something to stick and this isn't very conducive. How should I be reasoning these steps out?

Some rough steps :

Visualize the graph of this function : since you are asked to prove that the function is not uniformly continuous, you must look for a place where the function is increasing really fast (or has steep slope) . (Note : a heuristic is that this can happen near a point where the function takes the value infinity : that point won't be there in the domain, but points close to it will be there)

Why? Because, at these points, the $x$ coordinates are really close, but the $y$ coordinates are far apart. Remember uniform continuity says that the $y$ - coordinate difference between two points on the function's graph can be uniformly controlled by controlling their $x$-coordinate difference : so the points lying on these high-slope regions are most likely to provide a contradiction.

After doing this, look for points $x,y$ for which the expression for $f(x)$ is very simple. For example, if $f(x) = \sin x$, then for $x$ being multiple of $\frac {pi }4$, for example, the expression is simple.

Sometimes, you will be able to do both : find a sequence of points in the slopy region, for which the function values are really easy to evaluate. Then you can compare these values and come to a contradiction.

Let us apply these principles for $f(x) = \frac 1x$. Drawing $\frac 1x$ in your head, you see that if $x = 0$ the expression is infinite. So points near $0$ is where you can try to look.
Note that if $x$ is the reciprocal of an integer, then $f(x) $ is that integer! So taking $x = \frac 1n$ for $n \in \mathbb N$ gives simple values of $f(x)$. Thus, we don't need to complicate matters by looking at the general expression $|f(x)-f(y)|$.
Finally, we notice that if $x_n = \frac{1}{n}$ and $y_n = \frac 1{n+1}$, then $|x_n-y_n|$ can be made small enough, but we always have $|f(x_n)-f(y_n)| = 1$. So, for a given $\delta$, we can easily take such large $N$ that $\frac 1N < \delta$ (and $|x_N - y_N| =x_N-y_N < x_N < \delta$) but $|f(x_n)-f(y_n)| = 1$.
Thus uniform continuity is contradicted with $\epsilon = 1$, because we can find for any $\delta$ a pair of points differing by less than $\delta$ but whose function values have difference $1$.
You can keep these points in mind while approaching "non-uniform continuity" questions.

Simply let $ x = 2y \implies \left|\dfrac{x-y}{xy} \right| = \dfrac{1}{x} \ge \epsilon = 1$ when $0 < x < \delta < 1$.