Solving $\lim_{x\to 0}\frac{\ln(1+x)}x$ without De L'Hospital Trying to solve this limit without derivatives I found this answer that is pretty straightforward and I can easily follow the flow. I can understand why ${u\to \infty}$ because:

$$\lim_{u\to\infty}(1 + \frac{1}u)^u = e $$

but how there is a relation to ${x\to 0}$ with ${u\to \infty}$ when we need to find the limit that approaches $0$.

We can squeeze $\ln(1+x)$ between integrals.

For upper bound we use $\frac 1{1+t}\le \frac 1{1+t_{min}}=1$
For lower bound $\frac 1t\ge \frac 1{t_{max}}=\frac 1{1+x}$.

$$\dfrac x{1+x}=\int_1^{1+x}\dfrac{\mathop{dt}}{1+x}\le\int_1^{1+x}\dfrac{\mathop{dt}}{t}=\ln(1+x)=\int_0^x\dfrac{\mathop{dt}}{1+t}\le\int_0^x\mathop{dt}=x$$

Now dividing by $x$ we get

$$\dfrac 1{1+x}\le\dfrac{\ln(1+x)}x\le 1$$

Since both sides converge to $1$, this is also the limit for our ratio.

When $u \rightarrow \infty$, $x = \frac 1u \rightarrow 0$, as needed.


Prime Numbers: 6k-1 mod rule (New Discovery?) I've noticed that although all primes follow the pattern of $6k - 1$ and $6k + 1$ which seems to be a somewhat known fact.

However, I also noticed that all the primes of the pattern of $6k - 1$ only need to be tested against those that are also $6k - 1$ (beyond the square root, to a sixth of its size). Has this ever been discovered before? Would this algorithm in theory be faster to find the largest prime faster than current algorithms?

In theory, all one would need to do is multiply six by billions of digits, and then mod once across the spectrum by all products of $6k - 1$, faster even if we know if those $6k - 1$ numbers are already prime.

For instance $6k - 1$ where $k = 100$, therefore we are testing if 599 is prime.
For this number we only need to test all $6k - 1 \times 6 \leq 599$.

Here is Java code that does a half-sieve and performs fairly fast since it doesn't mod, but merely crosses off multiples of each number in the sieve.

public class primesUnder {

public static void main(String[] args) {

    int maxNumber = 100;
    int[] primesUnder = new int[maxNumber];

    for (int i=5, count=0; count < maxNumber; i+=6) {
        primesUnder[count++]=i;
    }

    for (int index=0; index < maxNumber; index++) {

        if (primesUnder[index] != 0) {

            int x = primesUnder[index];
            int z = x+index;

            while (z < primesUnder.length) {

                primesUnder[z] = 0;
                z+=x;

            }
        }
    }

  }

}

If a number $n=6k-1$ is decomposed as a product of not necessarily distinct primes
$$
n=6k-1=p_1\cdot\cdots\cdot p_t
$$
at least one of the $p_i$ must be of the form $6k^\prime-1$. This is actually pretty obvious using modular arithmetic for the modulus $6$.

Indeed, no $p_i$ can be either $2$ or $3$ since the lhs is not divisibile by $2$ or $3$. Thus the displayed formula reads
$$
n\equiv-1\equiv(\pm1)\cdot\cdots\cdot (\pm1)\bmod 6
$$
and for it to hold at least one factor on the rhs (and in fact an odd number of factors) must be $-1$.

With those formulas you get not-prime numbers i.e 6*8+1=49=7*7. also it is a known fact that it does not exist a polynomial formula to get prime numbers. However,this formula produces good results. I recommend you to read the wikipedia article about it. It will give you a better approach on this subject


Is it true that $\langle x^2+y^2-1\rangle$ is prime in $\mathbb{C}[x,y]$? Is it true that $\langle x^2+y^2-1\rangle$ is prime in $\mathbb{C}[x,y]$ ?

One way to see that this ideal is prime is to show that the quotient ring is an integral domain. By change of variables $u=x+iy, v=x-iy$ you get that $$\mathbb{C}[x,y]/\left<x^2+y^2-1\right>\cong \mathbb{C}[u,v]/\left<uv-1\right>\cong \mathbb{C}[t^{\pm 1}].$$

Yes. Since the  conic $x^2+y^2=1$ is  an irreducible curve in $\mathbb{C}^2$.

More than irreducible, its projective closure is smooth. One can simply check the vanishing of the partial derivatives.


Show that $2x^2 + 4$ is reducible in $\Bbb C$ Show that $2x^2 + 4$ is reducible in $\Bbb C$.

By the definition of irreducible, $2x^2 + 4$ is reducible if I can find $f(x)$ and $g(x)$ such that $2x^2 + 4 = f(x)g(x)$ and $f(x)$ and $g(x)$ are both not units.  But I can't seem to find a non-unit product of $2x^2 + 4$ in $\Bbb C$.

A polynomial of degree $2$ over a field (like $\mathbb{C}$ or $\mathbb{R}$ or $\mathbb{Q}$) is reducible exactly when it has a root.

In your case the polynomial does have a root in $\mathbb{C}$ so it is reducible.

$$
2x^2 + 4 = 2(x- i\sqrt{2})(x + i\sqrt{2})
$$

In fact the Fundamental Theorem of Algebra states that all polynomials over $\mathbb{C}$ are reducible if their degree is greater than $1$.

If you consider the polynomial over $\mathbb{R}$ (or $\mathbb{Q}$), then the polynomial is irreducible because it does not have any roots.

polynomial over C[x] does not have a unit when x is put to be roots...
As 2(x-i√2)(x+i√2) for x to be +-i√2 factors are not unit over C[x]


Why diagonal matrix SVD sorted from largest to smallest value? Why diagonal matrix SVD sorted from largest to smallest value?

D is diagonal matrix, $D=(d_1 \ge ,d_2 \ge ,..., \ge d_L)$.

Whether there is a journal that could explain this?

There is no reason other than convention. You can order the eigenvalues anyway you want, as long as you order the rows and columns of $U$ and $V$ accordingly.

The reason this is the convention, is that SVD is mostly used in applications for finding the least squared solution, given by the eigenvector corresponding to the smallest eigenvalue. Since developers don't want to perform a costly sort to find the smallest eigenvalues every time, most implementations output sorted SVD matrices.

A simple proof this is so:

Note that if $A=U\Sigma V$ the element $a_{ij}$ is:
$$a_{ij} = \sum_k \sigma_k u_{ik}v_{kj}$$
Note that the order of the sum over $k$ doesn't matter, as long as you reorder the rows of $U$, the columns of $V$ and the eigenvalues of $\Sigma$ in the same order.

I think that the purpose behind sorting the singular value is to show the maximum and minimum energy sensed in direction of the left or right singular vector.
but the question is this singular values came from where, is it the eigen value of (A.A Transpose) or from (A Transpose. A)  and this value corresponds to which column vector from the original matrix A.


"Where" exactly are complex numbers used "in the real world"? I've always enjoyed solving problems in the complex numbers during my undergrad. However, I've always wondered where are they used and for what? In my domain (computer science) I've rarely seen it be used/applied and hence am curious.

So what practical applications of complex numbers exist and what are the ways in which complex  transformation helps address the problem that wasn't immediately addressable?

Way back in undergrad when I asked my professor this he mentioned that "the folks in mechanical and aerospace engineering use it a lot" but for what? (Don't other domains use it too?). I'm well aware of its use in Fourier analysis but that's the farthest I got to a 'real world application'. I'm sure that's not it.

PS: I'm not looking for the ability to make one problem easier to solve, but a bigger picture where the result of the complex analysis is used for something meaningful in the real world. A naive analogy is deciding the height of tower based on trigonometry. That's going from paper to the real world. Similarly, what is it that is analyzed in the complex world and the result is used in the real world without imaginaries clouding the problem?

The question: Interesting results easily achieved using complex numbers is nice but covers a more mathematical perspective on interim results that make solving a problem easier. It covers different ground IMHO.

Complex numbers are used in electrical engineering all the time, because Fourier transforms are used in understanding oscillations that occur both in alternating current and in signals modulated by electromagnetic waves.

Complex analysis (transformation or mapping) is also used when we launch a satellite and here on earth we have $z$-plane but in space we have $w$-plane as well. So to study various factors we use transformation.


Square root is operator monotone This is a fact I've used a lot, but how would one actually prove this statement?

Paraphrased: given two positive operators $X, Y \geq 0$, how can you show that $X^2 \leq Y^2 \Rightarrow X \leq Y$ (or that $X \leq Y \Rightarrow \sqrt X \leq \sqrt Y$, but I feel like the first version would be easier to work with)?

Note: $X$ and $Y$ don't have to commute, so $X^2 - Y^2$ is not necessarily $(X+Y)(X-Y)$.

If $X$ and $Y$ are $n\times n$ positive semidefinite matrices over $\mathbb{C}$ or $\mathbb{R}$, we can prove your statement as follows.

We first assume that $X$ is positive definite. Let $A=YX^{-1}$. Since $A$ is similar to $X^{-1/2}YX^{-1/2}$, all eigenvalues of $A$ are positive. Let $v$ be a unit eigenvector corresponding to $\lambda_\min(A)$. As $X^2\le Y^2$, we have $I \le X^{-1}Y YX^{-1} = A^TA$ and hence $1\le v^TA^T Av=\lambda_\min(A)^2$. Thus all eigenvalues of $X^{-1/2}YX^{-1/2}\sim A$ are bounded below by $1$, i.e. $I\le X^{-1/2}YX^{-1/2}$. Hence $X\le Y$.

Now the positive semidefinite case can be obtained as a limiting case of positive definite cases.

Afternote: In the above proof, the only irreversible step is $I\le A^TA\Rightarrow1\le\lambda_\min(A)$. This is because we have $\sigma_\min(A)\le|\lambda|_\min(A)$ in general, but strict inequality can hold. The irreversibility of this step indicates that, while the square root function is operator monotone in our case, the square function is not. (Thanks for julien for pointing out that.)

Define $Y = X + k$, where $k$ is $\ge 0$. Then $Y^2 = X^2 + 2kX + k^2$


Finding the largest divider of big number without fractions Let's say I have a list of random numbers, all very big;

$$A: 1304810398401344441324$$

$$B: 5641893748137481374813764$$

$$C: 33322299992418948194899999$$

etc..

I want to be able to divide each and every number with another number, let's call it a divider, which will be unique for each number, however, I would like to find the biggest "divider" which will divide a number without leaving fractions.

How can I go about doing this in an efficient manner without trying every possible combination of random numbers???

to make it less complicated I'm making the "divider" at least $\frac 13$ as long as the number that's being divided by it.

Is this possible at all?

Thanks.

A thorough answer to your question is impossible on this site. Whether one can factor large numbers efficiently is an important open question. If you could do it much of the security on the internet would be at risk.

Your examples aren't particularly large for fast computers using the most efficient algorithms known. Wolfram alpha does the first one pretty much instantaneously, yielding
$$
1304810398401344441324 = 2^2×47×6940480842560342773
$$
https://www.wolframalpha.com/input/?i=factor+1304810398401344441324

You can start reading at

https://en.wikipedia.org/wiki/Integer_factorization

Just divide each $n$ by $n$ itself (or $n/2)$.  That trivially satisfies all your stated requirements.


How to prove $\ln N\ge 1/2+1/3+1/4+\cdots+1/N$ Let $\ln(x)=\int_1^x\frac{1}{t}dt$. I want to show that $\ln N\ge 1/2+1/3+1/4+\cdots+1/N$. The hint is Riemann sum, but I can only get $\int_1^N\frac{1}{t}dt>1\cdot\ln 1+1\cdot\ln2+\cdots+1\cdot\ln N$ (use the fact that integral $\ge$ lower sum). How to prove that? Or is my attempt wrong?

And also for $2^k$, how to show that $1/2+1/3+\cdots+1/N\ge k/2$?

Hint: If $f(t)$ is decreasing,
$\int_n^{n+1} f(x)\; dx \ge f(n+1)$.  Add these for $n$ from $1$ to $N-1$.

for some function $f(x)$:
$$\int_a^bf(x)dx\ge\sum_{x=a}^bf(x)$$
so we know that:
$$\int_1^N\frac1xdx\ge\sum_{i=1}^N\frac1i$$
and we know that:
$$\int_1^N\frac1xdx=\ln(N)$$


Finding prime factors of large expression (without calculator) Show that: $5^3(5^3(253)+3)+1 = 19 \times 251 \times 829$.

I tried setting $n=5$, so that $253 = 23 \times 11 = (4n+3)(2n+1)$ and going from there, but the resulting polynomial in $n$ was $8n^6 + 10n^7 + 3n^6 + 3n^3 + 1$, which turns out to be irreducible over the integers, so this doesn't help at all.

My eye is drawn to the $253$ in the expression being close to the factor $251$ and the fact that $5^3$ is very close to half of $251$.  Given the known result I would write
$$5^3(5^3(253)+3)+1=5^3(5^3\cdot (251+2)+3)+1\\=5^6\cdot 251+5^3\cdot253+1\\=(5^6+5^3+1)251$$
then just compute $5^6+5^3+1=15625+125+1=15751$ and divide by $19$


If $|A|\leq|B|$, then $|\mathscr{P}(A)|\leq|\mathscr{P}(B)|$ I'm trying to prove that if $|A|\leq|B|$, then $|\mathscr{P}(A)|\leq|\mathscr{P}(B)|$.
I know that $|A|\leq|B|$ implies that there is an injection $f:A\to B$. I also know that for any set $S$, $|S|<|\mathscr{P}(S)|$.
Clearly I need to show that there is also an injection from $\mathscr{P}(A)$ to $\mathscr{P}(B)$ but I'm really struggling to think of a way to get there.
I'm not asking for a solution but if anybody has a small hint to point me in the right direction I would really appreciate it. Thank you!

It is enough to Know that $|\mathscr{P}(A)|=2^{|A|}$ so $$ |\mathscr{P}(A)|=2^{|A|}\leq 2^{|B|}=|\mathscr{P}(B)|$$ It remains to show  $$ |\mathscr{P}(A)|=2^{|A|}$$
define a bijection function $$f\colon \mathscr{P}(A)\to 2^{A}$$ by $f(C)=\chi_{C}$ where $\chi$ is the characteristic function of $C$, that is , $\chi_{C} (x)=1$ for $x\in C$ and $\chi_{C} (x)=0$   for $x\in A\setminus C.$ I hope it is clear now


Why do elements in $\mathbb R[x]/(x^2)$ look like at most degree one polynomials? What do elements in $\mathbb R[x]/(x^2)$ look like? Intuitively, I know they are polynomials of degree one but if we stick with the definition of quotient ring, the elements look like $f(x)+(x^2)g(x)$.

  How does this show that the elements are polynomials of degree at most $1$?

Because every polynomial $P(x)$ can be written in one and only one way as $x^2Q(x)+R(x)$, where $\deg R(x)\leqslant1$. So, it is natural to identify $P(x)+x^2\mathbb{R}[x]$ with $R(X)$, whose degree is $0$ or $1$.

A nicer way of looking at the quotient $\mathbf{R}[x]/(x^2)$ is the following: $\mathbf{R}[x]$ is just a free $\mathbf{R}-$algebra (with the usual multiplication) on elements $\{1,x,x^2,\ldots\}$. Now, by quotienting by $(x^2)$, we are declaring that any element of our quotient ring from which $x^2$ can be factored out is trivial. You can think about this as imposing an algebraic relation of $x^2=0$. Then, the polynomials of $\mathbf{R}[x]$ are of the form
$$ a_nx^n+\cdots+a_1x^{1}+a_0$$
After quotienting, we have that all of the monomial terms with degree $\ge2$ are killed, so that we have a resulting polynomial of $a_1x^1+a_0$. Constructing the isomorphism between these two rings makes this formal.


Example of a sequence needed that satisfies certain conditions Can you please provide examples of sequence $\{a_1,a_2,\cdots\}$ such that $\sum_{i=1}^{\infty}a_i\to \infty$ while $\sum_{i=1}^{\infty}(a_i)^2<\infty$ with each $a_i\in[0,1)$. Thank you.

A classical example is the sequence
$$ a = (1, 1/2, 1/3,1/4...) .$$
It is a well known result that
$$ \sum_{k=1}^{\infty} a_k = \sum_{k=1}^{\infty} 1/k = \infty. $$
On the other hand, it holds that
$$ \sum_{k=1}^{\infty} a_k^2 = \sum_{k=1}^{\infty} 1/k² = \pi²/6 \quad (\dagger). $$
Showing the equality in $(\dagger)$, hoewever, is nontrivial and is known as the Basler problem.

If you want the members of the sequence to be in $(0,1)$, then you can let it start at $a_2$ by defining
$$ b = (1/2,1/3,...). $$
Obviously, $b$ is now another sequence fulfilling your requirements.

An interesting one is an = $\frac{1} p_{n}$, where pn is the n'th prime. Then $\sum_{}^{} \frac{1} p_{n}$ diverges and $\sum_{}^{} \frac{1}p_{n}^2 $ converges by comparison to the reciprocals of squares.


Trigonometry equation $\tan2x+3 \sec x +3=0$ for $0\leq x\leq 360$ Solve the trigonometry equation for $0\leq x\leq 360$

$$\tan2x+3 \sec x +3=0$$

I've no idea how should I start. Is there any identity I've to use? Hope someone can show out the working and explain for it. Thanks in advance.

Clearly, $\cos2x\cos x\ne0$

Multiply both sides of the given equation $$\cos x\sin2x+3(1+\cos x)\cos2x=0$$

$$0=2\sin x\cos^2x+3(1+\cos x)\cos2x$$

$$=4\sin\dfrac x2\cos\dfrac x2\cos^2x+3\cdot2\cos^2\dfrac x2\cos2x$$

$$=2\cos\dfrac x2\left(2\sin\dfrac x2\cos^2x+3\cos\dfrac x2\cos2x\right)$$

If $\cos\dfrac x2=0,\dfrac x2=(2n+1)90^\circ\implies x=(2n+1)180^\circ\equiv180^\circ\pmod{360^\circ}$

Else  $2\sin\dfrac x2\cos^2x+3\cos\dfrac x2\cos2x=0$

$-2\tan\dfrac x2=\dfrac{3(2\cos^2x-1)}{\cos^2x}$

Use $\cos x=\dfrac{1-\tan^2\dfrac x2}{1+\tan^2\dfrac x2}$ which unfortunately leaves us with a bi-quadratic equation in $\tan\dfrac x2$


stuck in Foundation of Mathematics course I'm studying Foundation of Mathematics on my own due to personal situation and can't go to class

In the Reference book that I'm studying there is one portion of problem solution I can't understand.
here is the problem.
$$\{x^2 -3 | x \in [-5, 1)\} = [-3, 22]$$

I understand whole proof and it's not hard at all except the part that:
$$-5 \leq x \lt 1 \space \text{ then } -5 \leq x \leq 5$$
I don't know why ?

$1\lt 5$, so if $x<1$ then surely $x<5,$ which implies $x\le 5$

Let $A= \{ x| -5 \le x < 1 \}$ and $B = \{ x|-5 \le x \le 5 \}$.

$$A \subset B$$

$$x \in A \Rightarrow x \in B.$$


Is taking the reciprocal of a differentiable strictly rigorous? I've read a proof of $\frac{d}{dx}(\ln x)=\frac{1}{x}$ in which the reciprocal of $\frac{dx}{dy}$ is taken and was wondering if this is strictly rigorous. The proof was: $$y=\ln x$$ $$e^y=x$$ then take the derivative with respect to $y$, so $$e^y=\frac{dx}{dy}$$ then take the reciprocal of the differentiable to give $$\frac{1}{e^y}=\frac{dy}{dx}$$ and since $e^y=x$, $$\frac{dy}{dx}=\frac{1}{x}$$

the question is simply: is taking the reciprocal of a differentiable rigorous? I know considering it as a variable to add or subtract to different sides of an equation, but I (obviously) am not sure about this case. Any help is appreciated.

This is a consequence of the chain rule.  For a function composition $F=f \circ g$, the rule states $$F'(x)=f'(g(x))g'(x)$$.  In other words, if you denote $y=g(x)$ and $z=f(y)$, then $$\frac{dz}{dx}=\frac{dz}{dy}\frac{dy}{dx}$$ Now if $g$ is an invertible function and $f$ is its inverse, then $$F=g^{-1}\circ g$$ is just the identity $F(x)=x$. In other words, $z=x$. Then $$\frac{dz}{dx}=1=\frac{dx}{dy}\frac{dy}{dx}$$ or
$$\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}$$ That's why you were okay to do it.

I cannot comment, yet. Therefore, I write an answer. It is not really rigorous as presented here. But if you use the right theorem (https://en.wikipedia.org/wiki/Inverse_function_theorem) it works.


$T(n) = 4T({n/2}) + \theta(n\log{n})$ using Master Theorem I am trying to solve the following recurrence relation using the master theorem:

$$T(n) = 4T({n/2}) + \theta(n\log{n})$$

So:

$a = 4$, $b = 2$, and $f(n) = n\log{n}$

So we are comparing:

$n^{log_b{a}}$ with $n\log{n}$

$n^{log_2{4}} = n^2$ so we are comparing $n^2$ with $n\log{n}$

Now i know that $n^2$ is larger but is it polynomial larger than $n\log{n}$?

Can I apply the master theorem to this problem, if so which case applies to this problem?

Any help would be appreciated.

Your recurrence relation falls into Case 1: $f(n) = n \log n$ is $O(n^{log_{b}{a}-\epsilon}) = O(n^{2-\epsilon})$.

To show why this is Case 1, as Louis says, logarithmic functions ($\log n$) are asymptotically bounded by  polynomial functions ($n^a$, where $a > 0$). This can be shown by taking the limit:
$$
\lim_{n \to \infty} \frac{\log n}{n^a} = 0
$$
through L'Hôpital's rule. In particular, $\log n \in O(n^{1-\epsilon})$ for small $\epsilon$. (We can go even further and say that $\log n \in o(n^{1-\epsilon})$.)

Then by multiplying both sides by $n$, (an allowed operation in big-O notation), $n \log n \in O(n^{2-\epsilon})$.

Therefore by the Master Theorem, $T(n)$ is $\Theta(n^2)$.

$$
T(n)=4T\left(\frac{n}{2}\right)+\theta(n\log n)
$$
Step 1:
$$
T(n)=4T\left(\frac{n}{2}\right)+\theta(n\log n) \iff T(n)=aT\left(\frac{n}{b}\right)+\theta(f(n))
$$
where $a=4,b=2,f(n)=n\log n$.
Whenever we are solving by using the master method, remove the polynomial, logarithmic, exponential part from the equation. Therefore , $f(n)=n$.
Step 2: $n^{\log_ba} = n^{\log_2 4} =n^2$.
Step 3: Compare $n_2$ v/s $f(n)$, i.e. $n^2$ v/s $n$. Since $n^2$ is greater than $n$ so, Case 1 : if $n^{\log_b a} > f(n)$, then $T(n)= θ(n^{\log_ba}) =θ(n^2)$ with
$$
\begin{split}
\epsilon &= \text{exponential power of }n^{\log_ba} - \text{ exponential power of }f(n)=n^2-n \\
&=\text{power of }n^2 =2-(\text{power of }n =1) = 2-1 =1
\end{split}
$$


Calculate limit without L'Hopital I need a some help with this.

Calculate:

$$\lim_{x\to\infty}\frac{\ln(x-1)}x$$

I know the answer is zero. But dont know how to handle the $\ln(x-1)$

You have that $$\log \sqrt{x -1} < \sqrt{x - 1} \Rightarrow \log (x-1) < 2\sqrt{x-1} \Rightarrow \frac{\log (x-1)}{x} < 2\frac{\sqrt{x-1}}{x}$$

Can you take it from here?

Well, if it's infinity over infinity you can still use L'Hopital's rule.

the derivative of both is:

( 1/(x - 1 ) ) / 1

invert and multiply and then plugin infinity in for X.

The denominator is getting infinitely large and therefore we can assume it's getting closer and closer to 0. Which is why it's 0 (or at least that's how my instructor taught me)


Can en eigen value be a vector? Given

$$b=\sqrt \frac{3}{22}\begin{pmatrix}2\\2\\-1
\end{pmatrix}\\
B=\frac 1 9 \begin{pmatrix}1 & 1 & 0\\
1 & 1 & 0\\
0 & 0 & 0
\end{pmatrix}$$

How can I deduce the eigenvalue $\lambda$ associated to $b$ from $b'Bb$?

Because

\begin{align}
b'Bb = \lambda b \Leftrightarrow \overbrace{\frac{3}{22}\begin{pmatrix}2 & 2 & -1\end{pmatrix}\frac 1 9 \begin{pmatrix}1 & 1 & 0\\
1 & 1 & 0\\
0 & 0 & 0
\end{pmatrix}}^{\mbox{isn't a scalar !!}}b=\lambda b\\
\end{align}

Presumably you are using $b'$ to be the transpose of $b$.  It should be $b'Bb=b'(Bb)=b'\lambda b=\lambda b'b$ which is a scalar.  But in your example $b$ is not an eigenvalue of $B$.  Leaving aside the constants we have $$Bb= \begin{pmatrix}1 & 1 & 0\\
1 & 1 & 0\\
0 & 0 & 0
\end{pmatrix}\begin{pmatrix}2\\2\\-1
\end{pmatrix} =\begin{pmatrix}4\\4\\0
\end{pmatrix}$$
which is not proportional to $b$

I believe you have the wrong formula. You want

$$b'Bb = \lambda b'b$$

This comes from,

$$Bb = \lambda b$$
$$b'Bb = b'\lambda b$$
$$b'Bb = \lambda b'b$$

and both sides of that equation are scalar.


Proving convergence of improper integral, g(t) I want to prove convergence of the improper integral,

$$\int_1^\infty g(t) dt$$
Where $g(t)$ is given by: $g(t)=(\frac{1-e^{-2t^2}}{t^2})$ , $t\neq0$

I have thought about using different tests, but I keep getting stuck at the fact that as t approaches zero from the right hand side, the function value increases towards infinity, so I can't figure out a limit from the right hand side.

I hope you can help.

Best regards,
Christoffer

$0<\frac{1−e^{−t^2}}{t^2} < \frac{1}{t^2}$ which is integrable on $[1,∞)$, so by comparison, your integral converges.

If you look at the integrand, it can be rewritten as
$$
f(t) = 2 \int_{0}^{1} e^{-2t^2x} dx
$$
So the whole expression (with $\varepsilon \to \infty$)
$$
L = 2 \int_{1}^{\varepsilon}\int_{0}^{1}e^{-2xt^2}dxdt = 2\int_{0}^{1}\int_{1}^{\varepsilon} e^{-2xt^2}dt dx
$$
Use the $\texttt{erf}$ function for the second integral:
$$
L = \sqrt{2} \int_{0}^{1}\frac{1}{\sqrt{x}}[\texttt{erf}(\varepsilon) - \texttt{erf}(1)]
$$
Now, compute the simple integral, and then take the limit as $\varepsilon \to \infty$. Can you handle from here?


limit of $\frac{xy-2y}{x^2+y^2-4x+4}$ as $(x,y)$ tends to $(2,0)$ Am I able to substitute $x$ by ($k$+$2$) with $k$ tending to $0$, then using polar coordinates to deduce its limit?!

\begin{equation*}
\lim_{(x,y)\to(2,
0)}{xy-2y\over x^2+y^2-4x+4}.
\end{equation*}

You can use Polar coordinates .And whenever you see  squares .Polar coordinates make it easier because of the  $cos^2x+sin^2x=1$.


Simplify $\tan^{-1} ( \frac{x-\sqrt{1-x^2}}{x+\sqrt{1-x^2}} )$ with trigonometric substitution I will explain my approach, help me with the last step please!
$$ \tan^{-1} {\left(\frac {x - \sqrt {1-x^2}}{x + \sqrt {1-x^2}}\right)}$$
substituting x = $\sin \theta$ (as learnt from book) and solving 1-$\sin^2 \theta$ = $\cos^2 \theta$
$$ \tan^{-1} {\left(\frac {\sin \theta - |\cos \theta|}{\sin \theta + |\cos \theta| }\right)}$$
For solving modulus, it was important to determine range of $\theta$ , therefore  I defined it (as it is my variable,i can define it my way) for [-$\pi$/2 , $\pi$/2] so that sine covers all values from $-1$ to $1$ (as ,  $ -1 \le x \le 1 \,$ , from domain  ) and $\cos \theta$ is positive , and hence $|\cos \theta| = \cos \theta$.
$$ \tan^{-1} {\left(\frac {\sin \theta - \cos \theta}{\sin \theta + \cos \theta }\right)}$$
= dividing by $\cos \theta$ $$ \tan^{-1} {\left(\frac {\tan \theta - 1}{\tan\theta + 1 }\right)}$$
= by formula of $\tan (\theta - \pi/4)$ $$ \tan^{-1}( \tan{\left(\theta - \pi/4\right)})$$
That's where I am stuck ,as according to the identity,$\quad$ $tan^{-1} ( \tan \alpha) = \alpha$ $\quad$ only when $\, -\pi/2 <\alpha < \pi/2$ . But here $$ -3\pi/4 \le \,(\theta-\pi/4) \, \le \pi/4 $$
Therefore, I am not going to get ($ \,\theta - \pi/4 $) out of the expression. What  i get will be based on that graph of $\bf {\tan^{-1} (\tan x)}$ .
$$ (\theta - \pi/4) +\pi \,$$  for  $\,-3\pi/4 \le \, (\theta -\pi/4) \, < -\pi/2 \,\,$ and
$$\theta -\pi/4$$
for $\,-\pi/2 < \, (\theta -\pi/4) \, \le \pi/4 \,\,$
My teacher just cancelled arctan and tan and wrote $\theta - \pi/4$ and he didn't even include that modulus function over $\cos \theta$.
So what will be the exact answer because if everyone decide $\theta$ as per they like then there will not be a finite answer. Everyone will have their own answers and in each answer they have multiple cases as I just discussed above.
So please help me, very hopefully I signed up in stackexchange!
Found Solution :-
I was confused because  I was thinking that there can be many solutions differing person to person, but even if you choose any value of $\theta$ , you are going to get two solutions which are in the asked question above. The problem resolves when we write $\theta$ in terms of $sin^{-1} x$  as then we would not simply write like $$ \theta = \sin^{-1} x $$
we would write an equation,$$ \sin^{-1} x = \sin^{-1} (\sin \theta)$$,
now if $\theta$ is not in range of $-\pi/2$ and $\,\pi/2$ , then there would be some constant in $\pi$ (like , $\pi/4 , 2\pi$ etc. we would have to add or subtract according to the graph of 'sin inverse sin' and when we would put that value of $\theta$ , we would end with the solutions as answered by people.
(I write the answer in this edit  to help anyone who will reach here after searching  web , thanks to everyone for answers)

The domain of the function is $x\in [-1,-\frac1{\sqrt2})\cup (-\frac1{\sqrt2},1]$. Then, with the substitution $\sin \theta =x$, we have $\theta \in[-\frac\pi2,-\frac\pi4)\cup (-\frac\pi4,\frac\pi2]$ and correspondingly
$$\tan^{-1} {\left(\frac {x - \sqrt {1-x^2}}{x + \sqrt {1-x^2}}\right)}
=\tan^{-1}\left[\tan{\left(\theta - \frac\pi4\right)}\right]$$
$$=
\begin{cases}
 \theta+\frac{3\pi}4 = \sin^{-1}x +\frac{3\pi}4  & x\in [-1,-\frac1{\sqrt2})\\
\theta -\frac\pi4 = \sin^{-1}x -\frac\pi4 & x\in (-\frac1{\sqrt2},1] \end{cases}
$$

In the case of $$-\pi/2 \lt \theta-\pi/4 \le \pi/4 $$ There is no problem, and we get $\theta -\pi/4$.
For the case $$-3\pi/4 \le \theta-\pi/4 \lt -\pi/2 $$ You need to adjust by adding $\pi$, so that $$\pi/4 \le \theta +3\pi/4 \lt \pi/2 $$ The answer should really be $$\begin{cases} \theta-\pi/4 , & -\pi/4 \lt \theta \le \pi/2 \\ \theta+3\pi/4, & -\pi/2\le \theta \lt \pi/4 \end{cases}$$  Your teacher is wrong.


evaluate integral the simplest way possible I want to evaluate the integration $\int \limits x \sqrt{\frac{x-1}{x+1}}dx$ ?

I tried putting $t = \sqrt{\frac{x-1}{x+1}}$ but that did not help !

Is there a quick and smart way to evaluate.

For the square root to be defined you need for instance $x\ge 1$, this suggest a substitution $x=\cosh(u)$.

The integral becomes $$I=\int\cosh(u)\sinh(u)\sqrt{\frac{\cosh(u)-1}{\cosh(u)+1}}\mathop{du}$$

Now using $\cosh(u)^2-\sinh(u)^2=1$ we get

$\displaystyle\int\frac{\cosh(u)\sinh(u)\sqrt{\cosh(u)^2-1}}{\cosh(u)+1}\mathop{du}=\int\frac{\cosh(u)\sinh(u)^2}{\cosh(u)+1}\mathop{du}=\int\cosh(u)^2-\cosh(u)\mathop{du}$

Just linearise the $\cosh(u)^2$ to get the result $$I=\frac u2-\sinh(u)+\frac 14\sinh(2u)$$

Finally use $\sinh(\cosh^{-1}(x))=\sqrt{x^2-1}\quad$ and $\quad\sinh(2\cosh^{-1}(x))=2x\sqrt{x^2-1}$

To get $$I=\frac 12\cosh^{-1}(x)+\sqrt{x^2-1}\left(\frac x2-1\right)+C$$

Inside the radical, multiply top and bottom by $x-1$ to get

$$\int \frac{x^2-x}{\sqrt{x^2-1}} \; dx$$

This wants the usual trig sub $x = \sec \theta$ which yields

$$\int \sec^3\theta - \sec^2 \theta \; d\theta.$$

The $\sec^2 \theta$ is easy.  The $\sec^3\theta$ is less easy, but is a standard integral.


How to Prove this Multinomial Distribution Inequality I have the following lemma, but there seems to be one (or two) mistakes in the proof found in this paper (lemma 3).
The lemma states that for $Multinomial(n,p_1,\ldots,p_k)$ distributed $(X_1,\ldots,X_k)$ the following inequality holds:

$P(\sum|X_i-np_i|\geq \epsilon) \leq 3\exp(-n\epsilon^2/25)$

However, in the proof he defines some new Variables, especially a $Poisson(n)$ variable N, independent from iid $U_1,U_2,\ldots$ with $P(U_1=i)=p_i$ for $i=1,\ldots,k$.

He then defines $X_1',\ldots,X_k'$ with $X_i'$ being the number of occurences of $i$ in $U_1,\ldots,U_N$, that is $X_i'=\sum_{j=1}^N 1_{U_j=i}$. Of course they are each $Poisson(np_i)$ distributed, but then he states, that they are independent as well, but that cannot be, can it? For example, take $k=2$, then $X_1'=N-X_2'$. Sadly he uses this independence like 10 lines later.

My question is - am I missing something obvious? If not, is there another way to prove this fact or a similar (by which I mean, a different right hand $R(n,\epsilon)$ of the inequality, for which at least $\sum_{n\geq 1} R(n,\epsilon) < \infty$)

(In the first step of equation (4) of the proof seems to be a mistake as well, but maybe I have not given it enough thought yet)

There is no error actually. I proved this to my computer science professor. First, I need to give you major components:

[1] Switching between Poisson distribution and multinomial. Suppose X Poisson. Let $x_i$ be number of times the value i appeared (ex. $x_5$ would be the number of times 5 was observed). Also, let $\lambda$ be the total number of observations. We see that if $\lambda$ is not fixed, X is Poisson. If $\lambda$ is fixed, we can think of $p_i$ = $x_i$/$\lambda$ as the probability of each value. $p_i$'s have to add up to 1. Thus, when $\lambda$ is fixed (known in advance), X become multinomial distribution.

[2] Equation (2): Using triangle inequality. Plop $X_i^{'}$ right in the middle of the inequality like this:
$$\sum_{i=1}^{k}(1/n)|X_i-np_i|$$ becomes
$$\sum_{i=1}^{k}(1/n)|X_i-X_i^{'}+X_i^{'}-np_i|$$ and apply triangle inequality to get right hand side.

[3] Between (2) and (3). Let's break open the expectation.
$$E(e^{t|U-\lambda|}) = \sum_{u=0}e^{t|u-\lambda|} = \sum_{u=0}^{u=\lambda}e^{t|u-\lambda|} + \sum_{u=\lambda + 1}e^{t|u-\lambda|}$$
We see that we can now get rid of absolute signs: if u is greater than $\lambda$ then remove the absolute signs. Otherwise, swap the locations. Doing this results in the following:
$$= \sum_{u=0}^{u=\lambda}e^{t(\lambda-u)} + \sum_{u=\lambda + 1}e^{t(u-\lambda)}$$
We see the following situation. The left hand side of the equation in the paper are both expectations. For ours, first one only goes up to $\lambda$ (and misses the remaining infinite number of terms) and the other is missing the first lambda number of terms, and every term is strictly positive (due to being e^). So our expression, which is equivalent to the left hand side, must be strictly smaller. But the author is lazy so he puts $\le$.

The next left hand side is just using characteristic function property.

[4] The first move is Chernoff bound rewritten to fit into one line. Then, use [2]. Next, the author uses a new trick. ln(1+x) where x is really close to 0 is approximately x. Then last part, he approximates (1+x) with 2 entirely for laziness.

[5] The final step (4). He uses (2) but surprisingly, N-n appears out of nowhere. We see that he gets this by realizing that summation in left hand side is actually Expectation written out. Thus N and n are means in expectation sense. N for expectation of $X_i$ (capitalized to symbolize that N is still a random variable due to being Poisson) and n for expectation of $X_i^{'}$ (lowercase, since n is fixed for multinomial).

Then everything else is straight forward.

Actually, he uses this obscure Taylor series somewhere but I forgot where. You must use this: $$\ln (z)  = \frac{(z-1)^1}{1} - \frac{(z-1)^2}{2} + \cdots$$.

Good luck!

The independence follows from the thinning property of a Poisson process. I'm not sure about the first step in equation (4).


If $E$ has measure zero, then does $E^2$ have measure zero? Currently I'm just working through measure theory just to see if its something I would like to take.

Unfortunately I am stuck on this problem from Carothers.

  If $m^*(E)=0$, then $m^*(E^2)=0$.

Where $m^*$ denotes outer measure and
$$E^2=\{x^2:x\in E\}.$$

I toyed with the idea that $I_k < 1 \Rightarrow I^2_k < I_k$.  However I am at a loss as to how to set up a chain of inequalities (which is what I am assuming I need).

Here’s one approach.

(1) First show that if $E\subseteq [0,M]$ for some real number $M>0$, and $m^*(E)=0$, then $m^*(E^2)=0$. HINT: You can cover $E$ with arbitrarily short intervals $[a,b]$ in such a way that for each of these intervals $m^*([a^2,b^2])=b^2-a^2<(M+1)(b-a)$.

(2) Use (1) to take care of the case $E\subseteq [M,0]$ for some $M<0$.

(3) Use (1) and (2) to show that for any $\epsilon > 0$ and any positive integer $n$, $m^*(E\cap [-n,n])<2^{-n}\epsilon$. Then $E = \bigcup_{n=1}^\infty(E\cap [-n,n])$ and $\epsilon = \sum_{n=1}^\infty 2^{-n}\epsilon$, so ... .

See that

any set of Outer Measure zero is measurable
any Lipshcitz continuous function takes measure zero sets to measure zero set.
$f:\mathbb{R}\rightarrow \mathbb{R}$ given by $x\mapsto x^2$ is Lipschitz
Image of $E$ under $f$ is $E^2$.

Done!

This is a bit more general question than you have asked.


Bijection example given set of integers to natural numbers with condition. State an example of a bijection $f:\mathbb{Z}\to\{ 2n \space|\space n\in\mathbb{N} \}$.
I know for a bijection the two sets must have the same cardinality. An example I came up with for this condition is:
$$
f(n)=2n^2, f:\{ ...,-3,-2,-1,0,1,2,3,... \}\to\{ 2,4,6,8,... \}
$$
But I'm not sure if this is valid or if I am missing something?

What about $f(n) =4(n+1)$ for $n\ge 0$ and $f(n)=-4n-2$ for $n \lt 0$?

To answer you question about whether your function $f$ is bijective, the anwer is no. In fact, $f(n)=2n^2$ is very well surjective, since $f(\mathbb{Z})\subset \{2,4,6,...\}$. How every, it is not injective, since, for example, $f(2)=f(-2)$ while $2\not=-2$.


Modified two child problem. Find the probability that both are girls, given that at least one is a girl born in March. A family has two children. Assume that birth month is independent of gender,
with boys and girls equally likely and all months equally likely, and assume that the
elder child’s characteristics are independent of the younger child’s characteristics).

What is the probability that both are girls, given that at least one is a girl who was
born in March.

Looks like we owe this variation to Gary Foshee.

In the simpler (no month information) formulation of the two-child problem we have BG, GB, GG as the possible genders of the children for a 1/3 likelihood that both are girls.  Here we can again enumerate the possibilities to find (counterintuitively) that the likelihood that both are girls has increased!

We have B[all months]G[march], G[march],B[all months], G[all months]G[march], and G[march]G[all months] as the possible outcomes - but have double-counted the case of G[march]G[march] - which gives us $\frac{2*12-1}{4*12-1} = \frac{23}{47}$ or just shy of a 50% chance that both children are girls.

The fact that one of the girls were born in March is irrelevant.  There are four equally likely possiblities: older=girl, younger=boy; older=girl,younger=girl; older=boy, younger=girl; and older=boy, younger=boy.  In 3 out of four cases, one of the children is a girl.  So the probability is $3/4$.


Non negative eigenvalues for a certain definition of hermitian matrix If a matrix A is defined by $(A)_{ij}=(X_i)^*X_j$, show that eigenvalues of A are non negative.

I was able to show that A is hermitian and an eigenvalue $\lambda$ can be written for an eigenvector Y as:
$\lambda=(Y)^*(AY)/((Y)^*Y)$.

However I am unable to show that this is non negative.

Thanks for the help.

Hint: Note that
$$
Y^*AY = \left(\sum_{i=1}^n Y_i X_i\right)^*\left(\sum_{i=1}^n Y_i X_i\right)
$$

Let's $x_i = (X_i)^*$, then $A = xx^*$. Now if $y$ is eigenvector of $A$ we have
$$
Ay = xx^*y = (x^*y)x = \lambda y,
$$
so $y = \alpha x$ for some number $\alpha$, which implies $\lambda = x^*x = \sum x_i^*x_i = \sum |x_i|^2 > 0$.


Does Gödel's Incompleteness Theorem really say anything about the limitations of theoretical physics? Stephen Hawking believes that Gödel's Incompleteness Theorem makes the search for a 'Theory of Everything' impossible. He reasons that because there exist mathematical results that cannot be proven, there exist physical results that cannot be proven as well. Exactly how valid is his reasoning? How can he apply a mathematical theorem to an empirical science?

Hawking's argument relies on several assumptions about a "Theory of Everything". For example, Hawking states that a Theory of Everything would have to not only predict what we think of as "physical" results, it would also have to predict mathematical results. Going further, he states that a Theory of Everything would be a finite set of rules which can be used effectively in order to provide the answer to any physical question including many purely mathematical questions such as the Goldbach conjecture. If we accept that characterization of a Theory of Everything, then we don't need to worry about the incompleteness theorem, because Church's and Turing's solutions to the Entscheidungsproblem also show that there is no such effective system.

But it is far from clear to me that a Theory of Everything would be able to provide answers to arbitrary mathematical questions. And it is not clear to me that a Theory of Everything would be effective. However, if we make the definition of what we mean by "Theory of Everything" strong enough then we will indeed set the goal so high that it is unattainable.

To his credit, Hawking does not talk about results being "unprovable" in some abstract sense. He assumes that a Theory of Everything would be a particular finite set of rules, and he presents an argument that no such set of rules would be sufficient.

Since "Theory of Everything" needs to be expressed in some kind of proof, which will be partially mathematical and partially observation, it is not clear, how such proof can be validated without the component of mathematics.

Quantum computing might solve this problem for us, as it is able to operate using mathematical constants that are not verifiable or provable after they are executed, just like some theories in mathematics, that we can observe.


Simulating a Stochastic Integral of OU process The stochastic integral I want to simulate is
$$\int_{0}^{1}J_c(s)dJ_c(s)$$
where $J_c(s) = \int_{0}^{s}e^{-c(s-r)}dB(r)$, is an OU process.

I simulate the data using Matlab and the sample codes are as follows. n is the sample size.

dB_x = mvnrnd(zeros(K, 1), Omega_xx/n, n); % n-by-m matrix
B_x = cumsum(dB_x,1); % n-by-m matrix
B_x = [zeros(1,K); B_x];
J_c = exp(-C*[1:n]/n)'.*cumsum(exp(C*[1:n]/n).*[zeros(K,1) dB_x(1:n-1,:)'])';
dJ_c=[J_c(1,:); diff(J_c)]; % m-by-n matrix
int_J =  1/n*J_c'*dJ_c;

I am not sure whether my calculation is correct or not, can anybody give me a correct way of simulating this integral?

Note that
\begin{align*}
dJ_c(s) = dB_s -cJ_c(s) ds.
\end{align*}
Moreover, $d\langle J_c, J_c\rangle_t = dt$.
Then, by Ito's lemma,
\begin{align*}
\int_0^1J_c(s) dJ_c(s) &= \int_0^1 d\left(J_c^2(s) -s \right)\\
&= J_c^2(1) - 1.
\end{align*}
Since
\begin{align*}
J_c(1) &= \int_0^1 e^{-c(1-s)}dB_s
\end{align*}
is normal with zero mean and the variance given by
\begin{align*}
E\left(J_c^2(1) \right) &= \int_0^1e^{-2c(1-s)} ds\\
&=\frac{1}{2c}(1-e^{-2c}).
\end{align*}
Then \begin{align*}
J_c(1) = \sqrt{\frac{1}{2c}(1-e^{-2c})}\, Z,
\end{align*}
and
\begin{align*}
\int_0^1J_c(s) dJ_c(s) &= \frac{1}{2c}(1-e^{-2c})\, Z^2 -1.
\end{align*}
where $Z$ is a standard normal random variable. The simulation becomes a draw of a sample from a standard normal random variable.

Finally, I found it wrong. The correct way is

dB_x = mvnrnd(zeros(K, 1), Omega_xx/n, n); % n-by-m matrix
B_x = cumsum(dB_x,1); % n-by-m matrix
B_x = [zeros(1,K); B_x];
J_c = exp(C*[1:n]/n)'.*cumsum(exp(-C*[1:n]/n).*[zeros(K,1) dB_x(1:n-1,:)'])';
dJ_c=[J_c(1,:); diff(J_c)]; % m-by-n matrix
int_J = J_c'*dJ_c;


Find all solutions to $f\left(x^2+xf(y)\right)=xf(x+y)$ Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that $$f\left(x^2+xf(y)\right)=xf(x+y)$$ for all $x,y\in\mathbb{R}$.

This is somewhat related to this question, but with an $xf(y)$ term instead of $yf(x)$. I've managed to get $f(0)=0$, $f\left(x^2\right)=xf(x)$ and $f$ is odd, but not much more. I think that $f(x)=x$ and $f(x)=0$ are the only solutions, but I'm not sure.

The function defined by $f(x) = 0$ for all $x \in \mathbb{R}$ is a solution to the functional equation.

Suppose that $f$ is some other solution. Then there is a real number $c$ such that $f(c) \neq 0$.

For any $r \in \mathbb{R}$, let $x = \frac{r}{f(c)}$, and let $y = c - x$. The functional equation then gives us that
$$
  f \left( x^2 + xf(y) \right) = xf(x + y) = xf(c) = \frac{r}{f(c)} \cdot f(c) = r
$$
showing that $r \in f(\mathbb{R})$. i.e. The function is surjective.

As you noted, we have that $f(0) = 0$, which follows by taking $x = y = 0$ in the functional equation.

Now let $x = -f(y)$ in the functional equation. We obtain that
$$
  0 = f(0) = f \left( f(y)^2 - f(y) f(y) \right) = -f(y) f(y - f(y))
$$
for all $y \in \mathbb{R}$.

Then for any $y \in \mathbb{R}$, we have that if $f(y) = 0$, then $f(y - f(y)) = f(y) = 0$, and if $f(y) \neq 0$, then $-f(y) f(y - f(y)) = 0$ implies that $f(y - f(y)) = 0$. We thus have that
$$
  f(y - f(y)) = 0
$$
for all $y \in \mathbb{R}$.

Also as noted by you, we have that $f(x^2) = xf(x)$ by taking $y = 0$ in the functional equation.

Now for any real number $x$, we showed earlier that there is some $y \in \mathbb{R}$ such that $f(y) = x$.

We then have that
$$
  xf(x) = f(y) f(f(y)) = f(f(y)^2) = f \left( f(y)^2 + f(y) f \left( y - f(y) \right) \right)
$$
using the fact that $f(y - f(y)) = 0$. The functional equation then gives us that
$$
  xf(x) = f(y) f \left( f(y) + y - f(y) \right) = f(y) f(y) = x^2
$$
holds for all $x \in \mathbb{R}$. For $x = 0$, we already have that $f(x) = 0 = x$. If $x \neq 0$, then we can divide the previous relation by $x$ to find again that $f(x) = x$. We thus conclude that $f(x) = x$ for all $x \in \mathbb{R}$.

Since you've allready know that, I'll continue from $f(0) = 0$ and $xf(x)=f(x^2)$

first, see that for every $x\ne0$: $$-xf(-x)=f\left(\left(-x\right)^2\right)=f(x^2)=xf(x)$$
$$f(x)=-f(-x)$$
Now, for every $a>0$:
$$f(a)=\sqrt{a}f(\sqrt{a})$$
which means:
$$f(a)=a^{\frac{1}{2}}f(a^{\frac{1}{2}})=a^{\frac{1}{2}}\cdot a^{\frac{1}{4}}f(a^{\frac{1}{4}})=a^{\frac{1}{2}}\cdot a^{\frac{1}{4}}\cdot a^{\frac{1}{8}}f(a^{\frac{1}{8}})=\dots=a^{\frac{2^n-1}{2^n}}f(a^{\frac{1}{2^n}})$$
therefore:
$$f(a)=\lim_{n\rightarrow\infty}{a^{\frac{2^n-1}{2^n}}f(a^{\frac{1}{2^n}})}
=
\lim_{n\rightarrow\infty}{a^{\frac{2^n-1}{2^n}}}\cdot\lim_{n\rightarrow\infty}{f(a^{\frac{1}{2^n}})}=af\left(\lim_{n\rightarrow\infty}{a^{\frac{1}{2^n}}}\right)=af(1)$$
which means basicly that theres $m\in\Bbb{R}$ such that for every $a>0$, $f(a)=ma$. But now, for every $a<0$: $$f(a)=-f(-a)=-m\cdot(-a)=ma$$
therefore $f(a)=ma$ for every $x\in\Bbb{R}$


"If everyone in front of you is bald, then you're bald." Does this logically mean that the first person is bald? Suppose we have a line of people that starts with person #1 and goes for a (finite or infinite) number of people behind him/her, and this property holds for every person in the line:

  If everyone in front of you is bald, then you are bald.

Without further assumptions, does this mean that the first person is necessarily bald? Does it say anything about the first person at all?

In my opinion, it means:

  If there exist anyone in front of you and they're all bald, then you're bald.

Generally, for a statement that consists of a subject and a predicate, if the subject doesn't exist, then does the statement have a truth value?

I think there's a convention in math that if the subject doesn't exist, then the statement is right.

I don't have a problem with this convention (in the same way that I don't have a problem with the meaning of 'or' in math). My question is whether it's a clear logical implication of the facts, or we have to define the truth value for these subject-less statements.

Addendum:

You can read up on this matter here (too).

You can see what's going on by reformulating the assumption in its equivalent contrapositive form:

  If I'm not bald, then there is someone in front of me who is not bald.

Now the first person in line finds himself thinking, "There is no one in front of me.  So it's not true that there is someone in front of me who is not bald.  So it's not true that I'm not bald.  So I must be bald!"

Let's try to isolate the problematic part of the question.  Imagine you are the first person in line.  Your cell phone rings; the man on the other end says to you, "My psychic powers tell me that everyone in front of you is bald.  Am I wrong?"  How do you answer?

From the standpoint of mathematical logic, the answer is unequivocally No, he is not wrong.  This is not a "convention", but rather an inevitable consequence of how logic works:  The person on the other end of the phone would only be wrong if there were a non-bald person in front of you; since there is no non-bald person in front of you, he is not wrong.

Now here is where it gets tricky:  If he is not wrong, is he right?

If we assume a logical system in which every statement has a truth value, and there are only two possible truth values, then if he is not wrong, he must be right.  So in such a logical system it must be true that everybody in front of the first person in line is bald.

For similar reasons, the largest natural number is odd, and the final digit of the decimal expansion of $\pi$ is a 4.  These statements are true because they are not false; they are not false because there is no largest natural number and $\pi$ has no final digit.

But human language does not operate according to the principles of formal logic.  In the real world, we would not tell the person on the cell phone "You are wrong" or "You are right"; what we would say is "There is nobody in front of me."  More precisely, we would infer the tacit additional premise "There is at least one person in front of you", and reject the entire construction because that premise is false.


Finding out missing observation $a$ and $b$ The mean of $a,b,8,5,10$ is $6$ and the variance is $6.8$. find out value of $a$ and $b$

I tried something like this:

$$\bar{x}=6=\frac{a+b+8+5+10}{5}$$
$$7=a+b$$

Then i have put the value in the formula of variance

$$variance\ \sigma^2={\frac{{\Sigma x_i^2}}{n}-(\bar{x}^2)}$$
$$6.8={\frac{{a^2+b^2+189}}{5}-(36)}$$
$$25=a^2+b^2$$

I don't this my second equation is correct.If it correct then can anyone one guide me what do further?

You wrote the correct equation for $a^2+b^2$, but I get $$6.8={\frac{{a^2+b^2+189}}{n}-(36)}\\34=a^2+b^2+189-180\\25=a^2+b^2$$  Coupled with your correct value for $a+b$, you should be able to get home.

Added: Now you have two equations in two unknowns.  You might be able to solve them by inspection.  Otherwise, you can write $b=7-a$ in substitute that into $25=a^2+b^2=a^2+(7-a)^2$, getting a quadratic in $a$.

Your second equation is wrong as well.

$$variance\ \sigma^2={\frac{{\Sigma (x_i - \bar{x})^2} }{n}}$$

That reduces to $$13 = (a-6)^2 + (b-6)^2$$

You can then substitute for either a or b from the first equation a + b = 7 to get a fairly simple quadtratic equation.


Analytic $F(z)$ has $f(z)$ as derivative $\implies$ $\int_\gamma f(z)\ dz = 0$ for $\gamma$ a closed curve Hypothesis: Suppose that $F(z)$ has $f(z)$ as a derivative.  Suppose further that $F(z)$ is analytic.  Now consider the complex line integral

$$
\tag{1}
\int_\gamma f(z)\ dz
$$

Question: Does this imply that $(1)$ is equal to zero if $\gamma$ is a closed curve?  If so, why?

Attempt:

There is a theorem that says that for $\gamma$ a closed curve, we have that

$$
\int_\gamma p\ dx + q\ dy = 0 \iff p\ dx + q\ dy \text{ is an exact differential}
$$
Then $f(z)\ dz = f(z)\ dx + i f(z)\ dy$ implies that

$$
f(z)\ dz = f(z)\ dx + i f(z)\ dy = \underbrace{{\partial F \over \partial x}\ dx + i \left(- i{\partial F \over \partial y}\right)\ dy}_{\text{applying CR-equations to $F(z)$}} = {\partial F \over \partial x}\ dx + \left({\partial F \over \partial y}\right)\ dy
$$
Then $f(z)\ dz = dF = {\partial F \over \partial x}dx + {\partial F \over \partial y} dy$
so that $f(z)\ dz$ is an exact differential as desired.
Then via $(1)$ we have that

$$
\int_\gamma f(z)\ dz = 0
$$

as desired.

Is my proof correct?

Perhaps simpler is, assuming the domain of $\gamma $ is $[a,b]$,

$$\int _\gamma f=\int \limits _a^b f\left(\gamma (t)\right)\gamma '(t)\,\mathrm dt=\int \limits_a^b (F\circ \gamma)'(t)\,\mathrm dt=F\left(\gamma (b)\right)-F(\gamma(a))=0.$$

Or you can write $\gamma=\gamma _1+\gamma_2$ and if $\gamma$ is a closed curve suppose that $\gamma_1$ is going anti-clockwise and $\gamma_2$ clockwise. Now we must use the independency of paths which states that the integral of an analytic function on any two curves from a point $a$ to a point $b$ is the same.

Thus $\int_{\gamma_1}f=\int_{-\gamma_2} f$ and $\int_{\gamma} f=\int_{\gamma_1} f+\int_{\gamma_2} f=0$.


integration about $\arcsin $ and $\arccos $ To integrate $\frac{1}{\sqrt{a^2-x^2}}$, we get $\arcsin(x/a)+C $

To integrate $\frac{-1}{\sqrt{a^2-x^2}}$, we get $\arccos(x/a) + C$

but what if I take out the $-1$?

So

$\displaystyle\int \frac{-1}{\sqrt{a^2-x^2}}\,dx$ =  -$\arcsin(x/a)+C $?

Because $-1$ is a constant and we can just take it out of the integral?

I am confused.

HINT:

$$\arcsin(x)+\arccos(x)=\pi/2 \tag 1$$

SPOILER ALERT:  Scroll over the highlighted area to reveal the solution

  Note that we have $\arcsin(x)=\int_0^x \frac{1}{\sqrt{1-t^2}}\,dt$ and therefore in terms of anti-derivatives, we can write $$\arcsin(x)=\int \frac{1}{\sqrt{1-x^2}}\,dx+C$$Using $(1)$, the arccosine function can be written as the anti-derivative $$\begin{align}\arccos(x)&=\pi/2-\arcsin(x)\\\\&=\pi/2 -\left(\int \frac{1}{\sqrt{1-x^2}}\,dx+C\right)\\\\&=-\int \frac{1}{\sqrt{1-x^2}}\,dx +C'  \end{align}$$where $C'=\pi/2-C$

arcsinx + arccosx = pi/2 (some constant).
They hold a linear relationship.
So yeah, it's somehow correct.
Try solving more differential equations of first order and first degree.


Stupid geometry question I'm doing a proof in geometry.

There is a line AC, with a point B also on the line, between A and C. In other words, the three points on the line are ABC, in that order.

Another line, let's call it DA, intersects with ABC at A. This creates an angle CAD.

CAD is the same angle obviously as BAD.

But what is the justification in a geometric proof (something about the nature of lines?) that says that these two are the same.

(NB BAD and CAD are separately important as part of different triangles).

Thanks!

This question can be answered from Hilbert's axioms standpoint.

Among primitive notions we have a ternary betweenness relation which describes exactly the situation "point $B$ lies between points $A$ and $C$ (on the same line)". Let's denote $A*B*C$ for such situation.

To answer this question in detail, I will need the precise definition of a halfline in terms of betweenness relation. First look at the theorem:

Theorem. For any line $l$ and any point $O\in l$ the binary relation $\sim$ on the set $l\setminus\{O\}$ defined by the formula:
$$A\sim B :\iff \left(O*A*B \vee A=B \vee O*B*A\right)$$
is an equivalence relation with exactly two equivalence classes.

We will call the set $r$ a halfline whenever $r$ is an equivalence class of relation $\sim$ for some line $l$ and some point $O\in l$. It can be proved that both the line $l$ and the point $O\in l$ are uniquely determined by a halfline $r$. $O$ is called the origin of $r$.

Also let's denote $\overrightarrow{OA}:=\{B: O*A*B \vee A=B \vee O*B*A\}$ for any distinct points $O$ and $A$. It is quite obvious that $\overrightarrow{OA}$ is a halfline with origin $O$ contained in the line $\overleftrightarrow{OA}$.

A very simple consequence of all above statements is the following property:

Property 1. If $O*A*B$ then $\overrightarrow{OA}=\overrightarrow{OB}$.

Basically if follows from the fact that $O*A*B$ implies that $A\sim B$ where $\sim$ is the relation introduced before for line $\overleftrightarrow{OA}=\overleftrightarrow{OB}$ and point $O$. The well known fact about equivalence relations is that if $A\sim B$ then the equivalence classes of $A$ and $B$ are equal.

Another definition needed here is the angle definition:

If $r,s$ are halflines with the same origin, then an unordered pair $\{r,s\}$ is called an angle.

Also whenever $A\neq O\neq B$ we denote $\angle AOB:=\{\overrightarrow{OA},\overrightarrow{OB}\}$

Answer to the question

From assumptions we have $A*B*C$.

By the property 1, $\overrightarrow{AB}=\overrightarrow{AC}$.

Next $\angle BAD=\{\overrightarrow{AB},\overrightarrow{AD}\}=\{\overrightarrow{AC},\overrightarrow{AD}\}=\angle CAD$.

An angle is between two lines. One line is $D\vee A$. The other line is (i) $A\vee C$ and (ii) $A\vee B$. We therefore have to prove that $A\vee C=A\vee B$. You have said that "$B$ is in the middle of $A$ and $C$". This implies that $B\in A\vee C$, but $B\ne A$. Now there is exactly one line containing $B$ and $A$, namely $A\vee B$. Since $B$ is also on the line $A\vee C$ we can conclude that in fact $A\vee B=A\vee C$.


Given the function $f:[0,1]→[0,1]$; $f(x)=x^2$, check which one(s) of the properties it has. This homework is past due, but I am still fiddling trying to figure this out.

question: I do not understand what the heck the notation of $f:[0,1] \to [0,1]$; means. I thought I did, but my repetition states otherwise.

Here is the image of the problem for better understanding

Don't even give me the answer to the question itself. Just please explain what f:[0,1] -> [0,1]; means. I thought it was all points $0 \le x \le1$. If it is, I have no idea what the heck the answer is.

Any feedback or direction is appreciated.

It means the domain and codomain of the function are the closed interval from $0$ to $1$. Given any $x$ such that $0 \le x \le 1$, you must have $0 \le f(x) \le 1$.  This is true for $f(x)=x^2$.  You don't have to have that every point in $[0,1]$ is the image of some point in the domain.  If that is true, the function is surjective.

It is injective and surjective - both increasing and strictly increasing.
Graph:


How to get a general rule from the recursive definition Assume we have an arithmetic sequence $f_n$ that has a non-recursive definition as follows:
$$f_n=2n+1$$
This sequence can also be defined recursively:
$$f_n=2+f_{n-1}$$
$$f_0=1$$
The same can be done if we assume f_n is a geometric sequence:
$$f_n=3^n$$
$$f_n=3f_{n-1}$$
$$f_0=1$$
Note that if either the non-recursive rule or the recursive definition of a geometric sequence or an arithmetic sequence is known, the other definition can be derived. This is possible, however, only when the sequence is known to be distinctly arithmetic or geometric. In other words, the recursive definition should be $f_n=a*f_{n-1}$ or $f_n=a+f_{n-1}$. My question is that is there a proof if it possible to derive a general rule for a sequence with the combined recursive definitions of an arithmetic and a geometric sequence which looks like:
$$f_n=af_{n-1}+b$$
$$f_0=c$$

If it is possible, what would be the way to derive the general rule ?

P.S. I am not that much informed in this area, and because of this, I would be very glad if you point out the unclear or wrong parts of the question. Constructive feedback is always useful :)

If $a=1$ we have an arithmetic progression with $f_n=bn+c$

Otherwise it would be easier if we could find a $k$ such that $g_n=f_n+k$ and $g_n=ag_{n-1}$ making $g_n=a^ng_0$

so we want $f_n+k = a(f_{n-1}+k)$ while we know  $f_n = af_{n-1}+b$

and this means we need $k=\frac{b}{a-1}$ and we do not have division by zero as $a \not=1$,
so $$f_n=g_n -\frac{b}{a-1} = a^n g_0 -\frac{b}{a-1}= a^n\left(c+\frac{b}{a-1}\right)-\frac{b}{a-1} $$

I'm honestly not what the answer is, but you may want to check out the well-written wikipedia post on recurrence relations.


Tree Diagrams Probability A box contains $20$ chocolates, of which $15$ have soft centres and $5$ have hard centres. Two chocolates are taken at random, one after the other. Calculate the probability that,
a.  both chocolates have soft centres, $[2]$
b.  one of each chocolate is taken, $[2]$
c.  both chocolates have hard centres, given that the second chocolate has a hard centre. $[4]$

a.
$P(SS)= (15/20)(14/19)= 21/38$

b.
$P(\text{sorting})= (SH)+(HS)= (15/20 x5/19)+(5/20 x15/19)= 15/38$

c.
P(Both H and second is H) divide by P(second is H)

Using formula: $P(BH | 2H)/(2H)$

How to tackle with it, i try all combination but do not get the answer $(20/83)$

The answer for (c) does not come out to be 20/83 according to my method either, so can you show what answer you get?

I get 4/19 as the required probability for (c) by

P(H ∩ H) / (P(H ∩ H) + P(S ∩ H))

where,
S is the event of choosing a soft centered chocolate and H is the event of choosing a hard centered chocolate.

Possibility Space = $\{HH,SS,SH,HS\}$

Sample Space $= 20P2 = 380$

$n(HH) = 5P2 = 20\\
n(SS) = 15P2 = 210\\
n(SH)+N(HS) = 380-210-20 = 150\\
n(HS) = 75\\
N(SH) = 75$

Given the second chocolate is HARD leaves $4$ possibility for the FIRST to be HARD.
therefore the number of possible combinations with the SOFT chocolates are
$4$ Hard $\cdot 15$ Soft $= 4\cdot15 = 60$
$\begin{align}n(HH)' &= \text{Possibility Space} - n(HH)\\
       &= 4-1\\
       &= 3\end{align}$

The probability of first and second being hard given the second is hard is

$\begin{align}P(HH) &= n(HH)/[(4\text{Hard}\cdot15\text{Soft}) +n(HH)+ n(HH)']\\
      &= 20/(60+20+3)\\
      &= 20/83\end{align}$


How To Prove This Is Continuous Or Discontinuous How do I prove this is continuous/discontinuous?
A function $f(x)$ (can be any function, in general) is defined on the interval $[3,7]$, where $f(3)=2$ and $f(7)=4$. The range of $f(x)$ on this interval is $$2\leq f(x)\leq 4$$ for all $x \in [3,7]$.
$f$ has unique values on $[2,4]$ (so if $b$ is in $[2,4]$ there is only one $c$ in $[3,7]$ such that $f(c)=b$)
How should I prove that this is continuous/not continuous? I think using the intermediate value theory, we can assume that f(x) is continuous first and then use say that there is a point $c$ and  $$f(c) = \alpha$$ and $f(a) < \alpha < f(b)$. What do I do with the point next? The function $f(x)$ can be anything so I cannot use a single value for it. Please can someone help.

If you can prove that for every alpha between f(a) and f(b) the function takes on a value, then you have proved the intermediate value theorem. In this cases, what I look for is a counter example. What you want is to look for all the f(c). That would be tedious. Just look for a counter example, such that f(d) = $\beta$ and $\beta$ does not lie in the interval between f(a) and f(b). If you cannot find such d, then from the intermediate value theorem, it is continuous.


Multiplying and dividing composite functions I'm given:

$$f(x) = \frac{1}{x-1}$$

$$g(x) = \sqrt{\log2x}$$

I'm asked to multiply $f(x)$ and $g(x)$ and to divide them, and then state the domain for both.

Domain is x>1 because radicals have to be positive, logs can't be 0, and x cannot equal 1 or else you divide by zero which is undefined. You can simply multiply f(x) and g(x) together...here is a link if you have forgotten.

http://www.purplemath.com/modules/fcnops.htm


Why if $f:A\rightarrow A_{1}$ is an $R$-module homomorphism then $f(B) \subset B_{1} $ and $f(C) \subset C_{1}.$? The question is:

If $R$ has an identity and $A$ is an $R$-module, then there are submodules $B$ and $C$ of $A$ such thatt $B$ is unitary $RC = 0$ and $A = B \oplus C.$
And if $A_{1}$is another $R$-module, with $A_{1} = B_{1} \oplus C_{1}$, $B_{1}$ unitary ,$RC_{1} = 0.$

Why if $f:A\rightarrow A_{1}$ is an $R$-module homomorphism then $f(B) \subset B_{1} $ and $f(C) \subset C_{1}.$?

I assume your modules are not necessarily unital. Let $M$ be a left $R$-module; define
$$
z(M)=\{x\in M:1x=0\}\qquad u(M)=\{x\in M: 1x=x\}
$$
and prove $z(M)$ and $u(M)$ are submodules of $M$, with $Rz(M)=0$.

Now let $x\in M$ and consider $x=(x-1x)+1x$; then
$$
1(x-1x)=1x-1x=0
\qquad
1(1x)=1x
$$
so $x-1x\in z(M)$ and $1x\in u(M)$, so $M=z(M)+u(M)$. Also, $x\in z(M)\cap u(M)$ implies $1x=0$ and $x=1x$, so the sum is direct.

Now prove that for a homomorphism $f\colon M\to N$ you have $f(z(M))\subseteq z(N)$ and $f(u(M))\subseteq u(N)$.

Hint:Assume that $f(B) \not\subset B_{1}$. We have some $0\neq b\in B$ such that that $f(b)\in C_{1}$ (i.e., $f(b)\notin B_1$). So, $f(b)=f(1_Rb)=1_Rf(b)\in C_1$.But, we know that $RC_1=0$, so $1_Rf(b)=f(b)=0$. This is a contradiction of $b\neq 0$. ($f(b)=0=f(0)$).
You can use above idea for $f(C) \subset C_{1}$.


Alternative Monty Hall Problem So the typical set up for Monty Hall problem, there are 3 doors where 2 have goats and 1 has a car. I, the contestant, get to randomly guess a door looking to get the one with the car, after this the host will open a door that will always be a goat. Thus out the two doors that are left, I have to choose to stay with the original door I chose or switch to the other door. As many analysis of this problem have been done, switching my choice gives me a higher probability of winning. This largely has to do with the fact that since the host always reveals a goat the asking of whether to stay or not, is the same as did you guess right or not, and you have $\frac{2}{3}$ of wrong so you should switch

Now it seems, this "strange" result largely has to do with the fact that the host always reveals a goat. But what if alternatively you had this situation

  You are given 3 doors, 2 with a goat and 1 with a car. You randomly choose a door (looking to get one with the car). The host will randomly choose to reveal what is behind one of the 2 doors you haven't chosen. Given that he reveals goat, what is the probability of getting car if you chose to stay with choice?

My analysis of this problem goes as follows:

Let $D$ be event the door I guessed has car and Let $G$ reprsent the event that host reveals a goat thus what I want to calculate is $P(D|G)$ with this I have
$$P(D|G)=\frac{P(D\cap G)}{P(G)}=\frac{P(G|D)P(D)}{P(G|D)P(D)+P(G|D^{c})P(D^{c})}=\frac{1\left(\frac{1}{3}\right)}{1\left(\frac{1}{3}\right)+\frac{1}{2}\left(\frac{2}{3}\right)}=\frac{1}{2}$$

So it seems it doesn't matter if I choose to switch or not, and this is the result most people come up with when first thinking of problem.

Question: First is my analysis correct for this problem? Second, is it true in general that if you guess out of $n$ doors and host reveals $k$ doors that all have goats, will the probability that car is behind door you choose just $\dfrac{1}{n-k}$?

UPDATE

So I ended up asking my statistics/probability professor about this question and he said the result I got was correct. He explained that the reasoning that the Monty Hall problem inherently causes confusion is because many don't notice the only randomness in the original problem is just in your choice while hosts choice of door is deterministic. Now the problem I asked now has two sets of randomness, your original choice of door and the hosts choice thus the problems are inherently different

Your analysis is correct. Suppose that there are $n$ doors, one of which has a car, the other have goats. The host randomly chooses $k$ doors and opens them. I will use your notation, so $D$ is the event you have chosen the car and $G$ is the event the host reveals $k$ goats.

Then we have
$$
\mathbb{P}(D|G) = \frac{\mathbb{P}(D \cap G)}{\mathbb{P}(G)} = \frac{\frac{1}{n}}{\frac{n-k}{n}} = \frac{1}{n-k}.
$$
This is because

the probability $\mathbb{P}(G)$ that the host only reveals goats is $\frac{n-k}{n}$ (as it is the probability that the car is among one of the other $n-k$ doors),
the probability $\mathbb{P}(D \cap G)$ that you have chosen the car and the host only reveals goat is $\frac{1}{n}$ as this is the same as the probability $\mathbb{P}(D)$ that you have chosen the car.

Your analysis of the intuition looks good. I think your right that people think his way.
But no. No matter the number of goats revealed, your odds of initially guessing right are always 1 in $n$.

It helps to imagine that there are 100 doors. And that the host reveals 98 goats.
It is easier to see that you should switch.


Coordinate Transformation on Local coordinate system I am having a point $P(x,y,z)$ in $3D$ with respect to global coordinate system. I want to create an another Local  Coordinate System by picking three points $N1, N2, N3$ in 3D. Now I want to know the new coordinate values of the point $P'(x', y', z')$ with respect to the Local Coordinate System.

Please suggest me to create correct transformation matrix, or any procedure.

Thanks in advance.


Frequency from probability Assume an event has a probability $p=1/100$ of happening, per trial. Here are three statements that I believe are true:

On average, the event is witnessed in a one-hundredth of the the
trials.
The average number of trials between consecutive events is
$100$.
It is more likeley than not to witness at least one event in a
set of $69$ trials. This is found by computing the probability of the complement, i.e. the probability of no event in $69$ trials, which is $(1-p)^{69}=0.4998...$.

I can't understand how statements 2 and 3 can both be true. Is there a flaw in the reasoning somewhere ? Can they both be true? What is the "frequency" of the event ?

Let $X$ be the number of trials before one event (without loss of generality, we can look at the first time an event happens, i.e. start at trial $0$). Then, the second item can be rewritten
$$
\mathbb{E}[X] = 100
$$
and if I am not mistaken, the third item is equivalent to saying that $$\mathbb{P}\{X > 69\} < \frac{1}{2}.$$
To see why this is not contradictory, you can for instance look at the formula for expectation
$$
\mathbb{E}[X] = \sum_{n=1}^\infty n\mathbb{P}\{X=n\}
 = \sum_{n=0}^\infty \mathbb{P}\{X > n\}.$$

Looking at the second expression, the knowledge that $\mathbb{P}\{X > 69\} < \frac{1}{2}$ may lead to think that "only the first $69$ or so terms of the sum matter (after, the terms must be very small, as they are decreasing), so that the sum cannot be much more that $69\cdot 1 = 69$." However, this is not quite true: indeed, the terms of the sum after $n=69$ will get smaller and smaller, but you have infinitely many of them, and summing them together will add a significant contribution to the sum.

This is not very intuitive, but said differently (and in a very handwavy way): look at the first expression for the expectation, $\sum_{n=1}^\infty n\mathbb{P}\{X=n\}$. Item 3 means/hints that for $n > 69$, the value $\mathbb{P}\{X=n\}$ will be small (and actually getting smaller and smaller very quickly). But you multiply this quantity by $n$, which makes the term slightly bigger; and you sum all these things. So even if the probability that you have to wait, say, $3000$ trials, is very very small, it eventually adds up (along with the probability you have to wait $10^6$ trials, even smaller, etc), biasing the expectation towards higher values than you would "expect."

You have not formulated statement 2 as a precise statement, i.e., as the result of a well-defined, unambiguously stated probability problem.

As for statement 3, you have implicitly assumed the outcome per trial to be independent across trials, yet nowhere is this assumption included in your assumptions.  Statement 3 can be false if the outcomes per trial are not independent.  The correctness of a precisely stated version of statement 2 can also be affected by independence or not of the outcomes per trial.

As for your paradox under your interpretation of statement 2 and your implicit independence assumption, the bottom line is that not all probability distributions are symmetric, and therefore the expected value of a random variable need not be equal to its median value.  In particular, the waiting time between occurrences of an event, when the event's occurrence is independent across trials, is not symmetric.

Edit:
Regarding statement 2, consider the following probability distribution, which is fully consistent with the original problem specification:
With probability 1/100, the event occurs on every trial, and with probability 99/100 it never occurs.
In such case, the expected number of trials between occurrences of the event, given that it occurs, is 1.  The probability that the event ever occurs is 1/100.

There are various ways to make statement 2 precisely defined and unambiguous.  One way would be to define it as the expected number of trials between occurrences of the event, given that the event occurs.  There are other possibilities as well, such as defining the expected number of trials between occurrences of the event as equal to expected number of trials between occurrences of the event if the probability that the event ever occurs equals 1, and as equal to infinity if the probability that the event ever occurs is less than 1.  In the example provided, the former definition results in the answer 1, whereas for the second definition, the result is infinity.


How do I give a good estimate for $3^{3.8}$ or $3^{308}$ or $3^{3.7}$ etc. quickly and without a calculator? How do I give a good estimate for $3^{3.8}$ or $3^{3.08}$ or $3^{3.7}$ etc. quickly and without a calculator?

Back of the envelope estimate of $3^{3.8}$ to follow:

We know $\log 3$ is slightly bigger than $1$.

Let $f(x) = 3^x$, so that $f'(x) = 3^x \log 3$.

If $x = 4$ and $dx = -.2$ then $y = 81$ and $dy = 3^4 \log 3 (-.2) = -16.2 (\log 3)$.

So a not-too-accurate guess using differentials could be $81 - 16.2(\log 3)$. Since $\log 3$ is slightly bigger than $1$ let's just call it $81 - 17 = 64$.

If you go a bit further and use the fact that $y = 3^x$ is convex, we know the initial estimate is going to be too low, so perhaps bump it up a bit to 65.

Do you know how these things are calculated in a calculator, they use the taylor approximation.

https://en.wikipedia.org/wiki/Taylor_series

So if you could memorize up the first few terms of the series of $a^x$

$a^x = 1 + x log(a) +\frac{1}{2} x^2 log^2(a) + \frac{1}{6}x^3 log^3(a)....$

https://www.wolframalpha.com/input/?i=expand+a%5Ex%5C

and if you know the $log_{10}$ values, this can be used.


Prove that the perpendicular from the origin upon the straight line Prove that the perpendicular drawn from the origin upon the straight line joining the points $(c\cos \alpha, c\sin \alpha)$ and $(c\cos \beta, c\sin \beta)$ bisects the distance between them.

My Attempt:
Equation of the line joining the points $(c\cos \alpha, c\sin \alpha)$ and $(c\cos \beta, c\sin \beta)$ is:
$$y-c\sin \beta=\dfrac {c\sin \alpha - c\sin \beta}{c\cos \alpha- c\cos \beta} (x-c\cos \beta)$$
$$y-c\sin \beta =\dfrac {\sin \alpha - \sin \beta}{\cos \alpha - \cos \beta} (x-c\cos \beta)$$
$$x(\sin \alpha - \sin \beta)-y(\cos \alpha - \cos \beta)=c \sin \alpha. \cos \beta - c \cos \alpha. \sin \beta$$
$$x(\sin \alpha - \sin \beta) - y (\cos \alpha - \cos \beta)= c\sin (\alpha - \beta)$$

The two points $A$ and $B$ lie on the circle of radius $c$ centred on the origin $O$, having arguments $\alpha$ and $\beta$ respectively. Now consider $\triangle AOB$; because $AO=OB$ the triangle is isosceles and the perpendicular to $AB$ through $O$ bisects $AB$. This completes the proof.

Prove that the perpendicular bisector  from origin to the line joining the points $(c\cos\alpha,c\sin\alpha)$ and $(c\cos\beta,c\sin\beta)$ passes through the center as well.


How to determine the N-smallest eigenvalues of a symmetric matrix using the Power Method? I was assigned to make a program that finds the largest, the N-largest, the smallest and the N-smallest eigenvalues of a symmetric matrix, using the Power Method. So far, I've been able to succesfully calculate the largest eigenvalue using the traditional Power Method, the N-largest using the Power Method with Deflation, and the smallest using the Inverse Iteration (the Inverse Iteration as described here in section 3-2: Iterative Methods).

But, right now I have no idea how to determine the N-smallest. I tried using the Inverse Iteration with Deflation to calculate the N-smallest but it is not working. When I calculate the second smallest (and so on...) I don't get the expected results, as if it's not possible to simply apply the Inverse Iteration with deflation. What am I missing? What should be the right way to calculate de N-smallest?

Your help is deeply appreciated.

I assume by "largest" and "smallest" you mean in absolute value.  The $n$'th smallest eigenvalue of $A$ is $\pm \sqrt{t - \mu}$ where $\mu$ is the $n$'th largest eigenvalue of $t I - A^2$ if $t$ is large enough.  See how this moves if you replace $A$ by $A-\epsilon I$ for small $\epsilon$ and you can tell whether it's $+$ or $-$.

Try this,

import numpy as np
from numpy.linalg import norm

from random import normalvariate
from math import sqrt
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

def randomUnitVector(n):
    unnormalized = [normalvariate(0, 1) for _ in range(n)]
    theNorm = sqrt(sum(x * x for x in unnormalized))
    return [x / theNorm for x in unnormalized]

def svd_1d(A, epsilon=1e-10,orth=[]):
    ''' The one-dimensional SVD, find smallest eigen '''
    n, m = A.shape
    x = randomUnitVector(min(n,m))
    lastV = None
    currentV = x

    if n > m:
        B = np.linalg.inv(np.dot(A.T, A))
    else:
        B = np.linalg.inv(np.dot(A, A.T))

    iterations = 0
    while True:
        iterations += 1
        lastV = currentV
        currentV = np.dot(B, lastV)
        for v in orth:
            currentV -= currentV.dot(v) * v     #ensure orthogonality with vectors in orth
        currentV /= np.linalg.norm(currentV)
        if abs(np.dot(currentV, lastV)) > 1 - epsilon:
            # print("converged in {} iterations!".format(iterations))
            return currentV

def svd(A, k=None, epsilon=1e-10):
    '''
        Compute the singular value decomposition of a matrix A
        using the power method. A is the input matrix, and k
        is the number of singular values you wish to compute.
        If k is None, this computes the full-rank decomposition.
    '''
    rng = np.random.RandomState(42)
    A = np.array(A, dtype=float)
    n, m = A.shape
    svdSoFar = []
    if k is None:
        k = min(n, m)

    for i in range(k):
        matrixFor1D = A.copy()

        orth = [np.array(v) for singularValue, u, v in svdSoFar[:i]]
        if n > m:
            v = svd_1d(matrixFor1D, epsilon=epsilon,orth=orth)  # next singular vector
            u_unnormalized = np.dot(A, v)
            sigma = norm(u_unnormalized)  # next singular value
            u = u_unnormalized / sigma
        else:
            u = svd_1d(matrixFor1D, epsilon=epsilon,orth=orth)  # next singular vector
            v_unnormalized = np.dot(A.T, u)
            sigma = norm(v_unnormalized)  # next singular value
            v = v_unnormalized / sigma

        svdSoFar.append((sigma, u, v))

    singularValues, us, vs = [np.array(x) for x in zip(*svdSoFar)]
    return singularValues, us.T, vs

def plot_all(A,v,ax):
    ax.scatter(A[:,0],A[:,1],A[:,2])
    x=[0,v[0]]
    y=[0,v[1]]
    z=[0,v[2]]
    ax.set_xlim(-1, 1)
    ax.set_ylim(-1, 1)
    ax.set_zlim(-1, 1)
    ax.set_xlabel('X Label')
    ax.set_ylabel('Y Label')
    ax.set_zlabel('Z Label')
    ax.plot(x,y,z)

if __name__ == "__main__":

    def funC(t,rng,r):
        return t +r*2*(rng.rand()-0.5)

    rng = np.random.RandomState(42)
    xaux=[-1+float(i)/float(50) for i in range(100)]
    yaux=[funC(i,rng,0.5) for i in xaux]
    zaux=[funC(i,rng,0.01) for i in xaux]
    movieRatings= np.array([[xaux[i],yaux[i],zaux[i]] for i in range(100)])
    # v1 = svd_1d(movieRatings,orth=[np.array([1,0,0])])
    S,_,V=(svd(movieRatings,k=2))

    # fig = plt.figure()
    # ax = fig.add_subplot(111, projection='3d')
    # plot_all(movieRatings,v1,ax)

    fig2 = plt.figure()
    ax2 = fig2.add_subplot(111, projection='3d')

    print (S)
    print (V)

    for v in V:
        plot_all(movieRatings,v,ax2)

    plt.show()

    # print(v1)
    #print(svd(movieRatings,k=1))

apologies for the messy code.


Unitary Central Character by Schur's Lemma Consider an irreducible smooth representation $\pi$ of the group $G=GL_n(\mathbb{Q}_p)$ with center $Z$. Does there exist a unitary central character for $\pi$?

More precisely, is there a (quasi-)character $\omega: G \to \mathbb{C}^{\times}$ such that $\pi \otimes \omega$ when restricted to the center $Z$ is a unitary character for $Z$?
I find this result casually stated in many references, where they say it follows from Schur's lemma. But I am unable to see it directly from Schur's lemma.

Recall the following from Garrett's Automorphic Representations and L-functions:

Definition: A central character is simply a continuous group homomorphism $\omega : Z_A \to $ $\mathbb C^\times$ where $Z$ is the center of a reductive linear group $G$ over a field $k$ and $Z_A$ denotes the adele points of $Z$.
Triviality of $\omega \in Z_k$ is often assumed, resulting in the mapping $\omega : Z_A \to Z_k \diagup Z_A \to $ $\mathbb C^\times$. Furthermore, $\omega$ is said to be unitary when, $\forall z \in Z_A$, $|\omega(z)| = 1$

The key theorems are as follows:

Theorem 4.1 (Schur's Lemma over $\mathbb C$). If $V$ is an irreducible complex $G$-representation, then every linear operator $\phi : V \to V$ that commutes with $G$ is a scalar.
Proof. Let $\lambda$ be an eigenvalue of $\phi$ and assume that the eigenspace $E_\lambda$ is $G$-invariant. Then $v \in E_\lambda \implies \phi(v) = \lambda v,$ whence $\phi(gv) = g\phi(v) = g(\lambda v) = \lambda * gv$, so $gv \in E_\lambda$ (g was arbitrary). However, by irreducibility, this implies that $E_\lambda = V$, thus $\phi = \lambda \text{Id}$.

So, consider an irreducible smooth $G$-representation $\pi$.

Theorem 3.8 (Schur's Lemma). For a td-group $G$, any irreducible smooth $G$-representation $\pi$ satisfies End$_G$($\pi$) = $\mathbb C$.

Since a unitary central character is, by definition, a continuous group homomorphism $\omega : Z_A \to \mathbb C^\times$ such that $|\omega(z)| = 1 \space \forall z \in Z_A$, and, by Theorem 3.8, any  irreducible smooth $G$-representation $\pi$ satisfies End$_G$($\pi$) = $\mathbb C$, we can associate a central character $\omega_\pi(z) : Z \to \mathbb C^\times$ such that, for all matrix coefficients $f$ of $\pi$, $f(zh) = \omega_\pi(z)f(h).$ (commuting with G as a scalar, cf. Thm 4.1)
A matrix coefficient defined in Definition 5.1 of 2.

There is a central character $\omega:Z \rightarrow \mathbb{C}^{\times}$ by Schur, but it need not be unitary. E.g., consider $∣\text{det}∣$.


Problem with Directed Angles which sum to ${\pi \over 2}$. I was solving a section in my book(EGMO Lemma 1.30) where the author discusses the uses of directed angles, when I came across-

Points $A, B, C$ lie on a circle with center $O$. Show that $\measuredangle$ $OAC$ = $90^\circ$ − $\measuredangle$ $CBA$.

Let me denote directed angle with $\measuredangle$.(everywhere)

Here is an attempt; the author talks about the directed angles in blue, and it is to be shown that they sum to half $\pi$ radians. Lines in red are my own construction.
By directed angles, we know that $\measuredangle$ $CBA$ = $\measuredangle$ $CXA$ = ${1\over 2}$ $\measuredangle$ $COA$ (the inscribed angle theorem).
And also that $\measuredangle$ $OAC$ = $\measuredangle$ $ACO$ (triangle $OAC$ is isosceles).
Now by a theorem of directed angles, $\measuredangle$ $OAC$ $+$ $\measuredangle$ $ACO$ $+$ $\measuredangle$ $COA=0$
But after this, as we're working modulo $\pi$ radians, it's unintelligible to multiply or divide by $2$, which I have to do, so my attempt failed.
Answers are thankfully welcome.

By directed angles, we know that $\measuredangle$ $CBA$ = $\measuredangle$ $CXA$ = ${1\over 2}$ $\measuredangle$ $COA$ (the inscribed angle theorem).
And also that $\measuredangle$ $OAC$ = $\measuredangle$ $ACO$ (triangle $OAC$ is isosceles).

Now by a theorem of directed angles, $\measuredangle$ $OAC$ $+$ $\measuredangle$ $ACO$ $+$ $\measuredangle$ $COA=0$

After this we can write $2\times \measuredangle$ $OAC$ $+$ $\measuredangle$ $COA=0$
and resubstitute $\measuredangle$ $COA$ as $2\times \measuredangle$ $CBA$
We get, $2\times \measuredangle$ $OAC$ $+$ $2\times \measuredangle$ $CBA=0^\circ (\text{mod}\ 180^\circ)$
which is equivalent to writing as $2\times \measuredangle$ $OAC$ $+$ $2\times \measuredangle$ $CBA=180^\circ (\text{mod}\ 180^\circ)$
Divide both sides by $2$, and proceed to get $\measuredangle$ $OAC$ + $\measuredangle$ $CBA$ = $90^\circ \ (\text{mod}\ 90^\circ)$
using : If a ≡ b (mod c) and gcd(c, d) = g then a/d ≡ b/d (mod c/g)
Hence proved.

$\angle CBA \ne \angle CXA$, but $\angle CBA +\angle CXA =\pi$.


Consider $T$ = $9 \times 99 \times 999 \times 9999 \times \cdots \times \underbrace{999....9}_{2015 \:nines}$ Consider $T$ = $9  \times 99 \times 999 \times 9999 \times
 \cdots \times \underbrace{999....9}_{2015
\:nines}$

Find the last 3 digits of $T$

Advise: I wrote it down wrong the first time, it should be a product of "2015" numbers, i apologize about that, i realized my fault when i was travelling and i couldn't repare it in my cellphone.

My try

I know the last digit, i found it easily, but the struggle is with the others. I tried this:

$9  \times 99 \times 999 \times \cdots \times \underbrace{999 \ldots 9}_{2015
\:nines}$  $= 9 \times 9(11) \times 9(111) \times \cdots \times 9(\underbrace{111 \ldots 1}_{2015
\:ones})$

So $T$ = $9(1+11+111+ \cdots +\underbrace{111 \ldots 1}_{2015
\:ones})$

But from here i found nothing, any hints?

Well, following from where you ended up:

$T = 9(1 \times 11 \times 111 \times ...)$

Notice that except for the first and second terms, the other 2013 terms are congruent to 111 (mod 1000).

So $T \equiv 9(1 \times 11\times 111^{2013}) \pmod {1000}$

We will use the property that $\varphi (n^m) = n^{m-1} \varphi (n)$

So $\varphi (10^3) = 10^2 \varphi (10) = 100 \times 4 = 400 $

Since $2013 = 400 \times 5 + 13$,

$9(1 \times 11\times 111^{2013} \equiv 9(11 \times 111^{13}) \equiv 9(11 \times 111^{10} \times 111^3) \equiv 9(11 \times (111^2 )^5 \times 1367631) \equiv 9(11 \times 12321^5 \times 631) \equiv 9(11 \times 321^5 \times 631) \equiv 9(11 \times 3408200705601 \times 631) \equiv 9(11 \times 601 \times 631) \equiv 9(4171541) \equiv 9(541) \equiv 4869 \equiv 869 \pmod {1000}$


Term for the graph of the cycles I am looking for the term which describes the graph G* of a graph G, composed by the same edges, but each cycle becomes the nodes of G*.

How to call G*? (I grew up calling it "Star Graph" but now reality has changed, and star graph is something completely different https://en.wikipedia.org/wiki/Star_(graph_theory) )

https://en.wikipedia.org/wiki/Dual_graph   => "In the mathematical discipline of graph theory, the dual graph of a plane graph G is a graph that has a vertex for each face of G. The dual graph has an edge whenever two faces of G are separated from each other by an edge, and a self-loop when the same face appears on both sides of an edge."


What is so special about Klein 4-group? This is my first course in abstract algebra and so far I am only learning about groups. So is there anyone who can explain to me why Klein 4-group is so special that it warrants a category of its own. Please explain as clearly as possible as I am still new to algebra. Thanks

The smallest anything is often of interest.  One reason is that it lets us think about why it works when nothing smaller does.  In this case, it hints that prime groups are always cyclic, and might provide a motivation to prove Lagrange's theorem, which is extremely useful.

It's the symmetry group of the rectangle. Don't you think rectangles are important enough to warrant a category of their own?


Is a r.v. supported on $\mathbb Q$ discret or continuous ? Same question for $\mathbb R\backslash \mathbb Q$. I know that a discret r.v. is a r.v. that is supported on a finite set or a countable set. But if $X$ is s.t. $\mathbb P\{X\in\mathbb Q\}=1$, how should I right the expected value :
$$\mathbb E[X]=\sum_{x\in\mathbb Q}x\mathbb P\{X=x\}$$
or $$\mathbb E[X]=\int_{\mathbb Q}x f_X(x)dx\ \ ?$$
The sum doesn't has sense for me, so I would say that the integral is the correct formulae. But if it is, is really $X$ a discrete r.v. ?
Same question if $\mathbb P\{X\in \mathbb R\backslash \mathbb Q\}=1$. To me s.t. a r.v. is not continuous, but is not continuous as well. So what can it be ?

There is nothing wrong with the sum. It looks strange because you don't usually think of $\mathbb{Q}$ as being well ordered, but that's just a small flaw. For the sum to make sense you will of course need it to converge absolutely and then it will converge independently of how you choose the well ordering of $\mathbb{Q}$.

(re.: acetone's comment) The definition of the expected value for a countably infinitely supported r.v. contains the requirement that the sum converge absolutely.

As to the second question in the OP. You don't point out what the probability space here is or what $\mathbb{P}$ is and for continuous random variables that is quite important.

But for example take $X$ a continuous random variable with a normal distribution on the reals. Then $\mathbb{P}(\{X\in\mathbb{R}\setminus\mathbb{Q}\})=1$ will be true since $\mathbb{Q}$ has Lebesgue measure $0$.

Again following from helpful comments by acetone:

Continuous random variable might be the part that's confusing you OP. Depending on what your definition of that term is you get very different results. In general you have discrete random variables (those are on probability spaces with finite or countably infinite many elements), continuous random variables those have a continuous probability distribution function (usually we think of them as having values in some subset of the reals with the corresponding borel $\sigma$-algebra as the measurable sets), and then there the "other" random variables.

Sometimes the "other" random variables get lumped up into the continuous random variables because they are not discrete and we think of the underlying sets (Reals, Complex numbers) as being continuous in some sense but that leads to confusion.


Given a light source and a triangle in space, find area of the triangle's projection onto a plane Problem

We have a light source in $L = (0, -2, 5)$, and we have a triangle $ABC$, given by the points $(0, 0, 2), \ (3, 0, 2), \ (0, 0, 3)$.
We are also given that the point $C$ projects to $C' = (0, 3, 0)$ on the $xy$ plane.
Find the area of the shadow created by the triangle on the $xy$ plane.

Thoughts
I don't see the importance of giving us the $C'$ point, seeing as we still need to find $A'$, $B'$ ourselves anyway, right?
The best method I can think of is finding these two points, and then we can simply calculate the area of the shadow using, for example, the cross products of two vectors that span it.
But the problem seems to be hinting at some generalized way of calculating the area of the projection of the triangle onto a plane.

If, sooner or later, you are asked the area of the projected  are, here is a way to do it.

Let us consider, for the sake of simplicity, that the center of central projection $L$ has coordinates $(0,0,h)$ (i.e., is situated on $z$ axis). If it is not the case, translate your figure in such a way that this is so.

It is not difficult to see that the image of point $(x,y,z)$ is point $(X,Y,0)$ with

$$X=\dfrac{hx}{h-z}=p(x,y), \ \ Y=\dfrac{hy}{h-z}=q(y,z)$$

Then, the area of the projected triangle is (see for example here) :

$$\dfrac12 \begin{vmatrix}p(x_1,y_1)&p(x_2,y_2)&p(x_3,y_3)\\q(y_1,z_1)&q(y_2,z_2)&q(y_3,z_3)\\1&1&1\end{vmatrix}$$

which can be written :

$$\dfrac{h^3}{2(h-z_1)(h-z_1)(h-z_1)} \begin{vmatrix}x_1&x_2&x_3\\y_1&y_2&y_3\\1-\tfrac{z_1}h&1-\tfrac{z_2}h&1-\tfrac{z_3}h\end{vmatrix}$$

Verification : if $h\to +\infty$, we tend to the situation of vertical projection, and we find back the area of triangle

Hint. Find the points $A'$ and $B'.$ We have that the line $LA$ lies in the $yz$ plane, so that its equation is given by $2z=4-3y,$ and on setting $z=0$ we find that $y=4/3.$

Now since the triangle $ABC$ lies in the $xz$ plane, and has orthogonal dimensions $AC=1$ and $AB=3,$ we see that $A'C'=5/3,$ so that the scale factor is $5/3.$ Hence the area of $A'B'C'$ is given by $$\left(\frac53\right)^2\left(\frac12×3×1\right).$$


Is this piecewise function continuous? Is the above function continuous? Intuitively, I think the answer is no because at points where the norm is equal to 1 and points where the norm is slightly larger than 1, it feels like there is a "jump discontinuity" but I'm not sure how to show that.

Yes, it is continuous. It is the product of two continuous functions: the identity function and the function$$\begin{array}{rccc}a\colon&\Bbb R^n&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}1&\text{ if }\|x\|<1\\\frac1{\|x\|}&\text{ otherwise.}\end{cases}&\end{array}$$The function $a$ is continuous because, for each $x\in\Bbb R^n\setminus\{0\}$,$$a(x)=\min\left\{1,\frac1{\|x\|}\right\}=\frac{1+\frac1{\|x\|}-\left|1-\frac1{\|x\|}\right|}2.$$


If $[G:H]=n$, then $g^{n!}\in H$ for all $g\in G$. I have the following question:

  Let $G$ be a group and let $H$ be a subgroup of finite index of $G$. Let $|G:H|=n$
  Then it holds: $g^{n!}\in H$ for all $g\in G$.

Why is this true?

I think, that's not very difficult, but I have no idea at the moment.

Thanks!

Let $G$ act on the left cosets of $H$ by left multiplication,
$$g: aH\longmapsto gaH.$$
This gives an action of $G$ on a set of $n$ elements, hence induces a homomorphism $G\to S_n$. In particular, for every $g\in G$, $g^{n!}$ must lie in the kernel, since $|S_n|=n!$. But the kernel is contained in $H$: if $g$ lies in the kernel, then $gaH = aH$ for all $a\in G$, hence in particular $gH=H$, so $g\in H$. Thus, for every $g\in G$, $g^{n!}\in H$.

This is an easy question if you start by showing that $g^n \in H$. Consequently, $g$ raised to the power of any multiple of $n$ is also in H (since $[G:H] = n$). Just an idea.


Estimating the total cost of purchasing every item in a grocery store My friend and I were arguing for way too long the other night about how much it would cost you to buy every single thing in a grocery store. Our first go at it went something like this:

Assume there are $N_{\text{items}}$ items per row in the grocery store, and let $p_{\text{avg}}$ be the average price for each item. Then say that there are $N_{\text{rows}}$ rows. Multiplying this out we get a total price $P_{\text{total}}$ as

$$ P_{\text{total}} = N_{\text{items}}p_{\text{avg}}N_{\text{rows}}$$

The only issue is, there is a vast range of difference prices for items, and vast ranges of items per row, depending on what row you're in. For instance, if you go down the aisle with all the spices, there's a ton of items at very low cost, but the coffee aisle has a lot of items at very high cost; the meat aisle has relatively average number of items at a much higher cost, as well as the kitchen-utensils/kitchenware aisle etc.

This got me thinking that there must be a better way to do an accurate estimation for a problem like this. Perhaps come up with some sort of intelligent distribution for prices (I was thinking maybe a log-normal distribution with a maximum around some arbitrary "most-probable" price, based on observation). And possibly do the same thing with the number of items per row? Estimating $N_{\text{rows}}$ is relatively straight forward since most grocery stores have somewhere between ten and twenty rows, so letting be $N_{\text{rows}}$ Gaussian centered at ten should take care of that, if we even want to get that fancy with that variable.

Anyway, I'm not that savy with probability/statistics in the first place, so I thought I would ask you brilliant people: how would you most intelligently try to take a stab at this estimation?

Your question is what's known as a Fermi Problem: a question about estimating a large or unwieldy number, such that answering the question calls for some rough estimates and back of the envelope calculations. I think that your attempt, and dxiv's and Producer of BS's responses are all pretty good (although theirs' would require more leg-work than yours. I'll give this a shot too, and I'll go with an approach similar to yours:

I want to pick a single number to be the average value of all the prices in the store. Now if the actual values of the prices follow Benford's Law as Joce suggests, then we could pick the mean of a random variable obeying Benford's law, which is $3.44$. But this only cleanly works if you're considering items priced between \$1.00 and \$9.99. There are items priced lower and higher than that. I'm willing to ignore the lower priced items because they won't matter as much, but there are enough items with price greater than \$10, like meats, kitchen utensils, and saffron, that I don't want to ignore those. So for the sake of not thinking too much harder about this, let's just bump that \$3.44 estimate up to \$4 even to account of the expensive items. In the end, this tiny estimate won't matter much anyways in comparison to the other estimates that have to be made.

I want to say there are twenty rows in a typical store. There are really closer to fifteen, but you've got to account for those outer areas that aren't rows, like the butcher counter and bakery and produce and floral too.

Now I want to break down the number of items in each row more precisely. Each row (besides maybe the outer two) has two sides, is about 70 feet long, has an average of 10 shelves, and each shelf is about 2 feet deep. Now we just need an estimate for the number of items on the typical square-foot of a single shelf. Thinking about Benford's Law again, and assuming that most products come in packages that have a roughly square base, I'd estimate there to be $3.44^2$ items per square-foot of shelf. Multiplying all of these we get that there are about

$$2 \times 70 \times 10 \times 2 \times 3.44^2 \approx 33,\!134 $$

items per row. Then we calculate the total price to be

$$\begin{align}
\mathrm{TotalPrice}
&=
\mathrm{AverageItemPrice}\times
\mathrm{NumberOfRows}\times
\mathrm{ItemsPerRow}
\\
&=
\$ 4\times
20\times
33,\!134
\\ &= \$ 2,\!650,\!720
\end{align}
$$

Enter the grocery shop and say "I want everything.".  The answer is the price they ask you to pay.


Linear Independence of Secant-squared or tangent-squared Due to a result from Chowla, we know that the set $\cot(2\pi k/n)$, such that $\gcd(k, n) = 1$, is linearly independent over the Rationals.

Do we have any similar results for $\sec^2(2\pi k/n)$ or $\tan^2(2\pi k/n)$ ?

Thanks

K. Wang, On a theorem of S. Chowla, J. Number of theory 15 (1982), 1-4
all the derivatives of cot(x) which includes sec-squared and hence tan-sqared.


If $f(x)=\lim_{n\to \infty }\left(1+\frac{x}{n}\right)^n$ prove that $f$ is continuous. I'm stuck on this problem. I proved that $f_n(x)=\left(1+\frac{x}{n}\right)^n$ converge for all $x\in\mathbb R$. Set $$f(x)=\lim_{n\to \infty }f_n(x).$$ I'm now trying to prove that $f$ is continuous. I tried to prove that $f_n\to f$ uniformly on compact but I failed. I tired as follow :

Set$$g_n(x)=f_n(x)-e^x.$$
Then $$g_n'(x)=\left(1+\frac{x}{n}\right)^{n-1}-e^x.$$
Now, how can I solve $g_n'(x)=0$ to find it's maximum ? Also, I tried to show that $g_n\to 0$ uniformly on all compact, but also, I failed. Any idea ?

Yes, that sequence converges uniformly on every compact subset of $\mathbb R$. That is sufficient to prove that the limit is continuous. Of course, it is enough to prove that the convergence is uniform on any interval $[a,b]$.

Note that\begin{align}e^x-\left(1+\frac xn\right)^n &=\sum_{k>n}\frac{x^k}{k!}+\sum_{k=0}^n\left(\frac1{k!}-\binom nk\frac1{k!}\right)x^k \\&=\sum_{k=0}^\infty a_{n,k}\frac1{k!}x^k,\end{align}where$$a_{n,k} = \begin{cases} 1-\left(1-\frac1n\right) \cdots\left(1-\frac{k-1}n\right)& \text{ if }k \leqslant n\\1&\text{ otherwise.}\end{cases}$$Note also that $0 \leqslant a_{n,k} \leqslant 1$ for all $k,n\in\mathbb N$ and, for any fixed $k$, $\lim_n a_{n,k}=0$.

Let $\varepsilon>0$ and choose $N\in\mathbb N$ such that, for $n\geqslant N$, $\sum_{k=n+1}^\infty\frac1{k!}b^k<\frac\varepsilon2$. Then we have\begin{align}
\left\lvert e^x-\left(1+\frac xn\right)^n\right\rvert&\leqslant\left\lvert\sum_{k=0}^N a_{n,k}\frac1{k!}x^k\right\rvert+\frac\varepsilon2 \\&\leqslant\sum_{k=0}^N a_{n,k}\max\left\{1,b^N\right\}+\frac\varepsilon2\end{align}
Now choose $N^\ast\geqslant N$ such that $\sum_{k=0}^N a_{n,k}\leqslant \frac\varepsilon{2\max\left\{1,b^N\right\}}$ for every $n\geqslant N^\ast$, in order to get$$(\forall x\in[a,b]):\left\lvert e^x-\left(1+\frac xn\right)^n\right\rvert\leqslant\varepsilon.$$

Hint $:$ Observe that $f(x) = \lim\limits_{n \rightarrow \infty} \left (1 + \frac x n \right )^n = e^x,$ for all $x \in \Bbb R.$