Calculate the limit $\lim\limits_{n\to\infty}\frac{\sqrt[n]{|x-1|}+1}{(x+1)^n+1}$ How can I compute the following limit?

$$\lim_{n\to\infty}\frac{\sqrt[n]{|x-1|}+1}{(x+1)^n+1}$$

You have to look at different values of $x$. Look at $|x+1|<1$ and $|x+1|\geq 1$ separately. In each case, what can you say about the numerator? Does it go to a finite limit? What can you say about the denominator? Does it blow up? Stay constant? Tend to a finite limit?

You'll have to use that $$|r|<1\implies r^n\to 0$$ $$|r|> 1\implies |r^n|\to  \infty$$ $$r=1\implies r^n\to 1$$  $$\forall r>0\; ; \; r^{1/n}\to 1$$

$$\lim_{n\to\infty}\left({\sqrt[n]{|x-1|}+1\over (x+1)^n+1}\right)\to{\lim_{n\to\infty}\left({\sqrt[n]{|x-1|}+1}\right)\over\lim_{n\to\infty}\left({(x+1)^n+1}\right)}$$
$\lim_{n\to\infty}\left({(x+1)^n+1}\right)$ is only defined if x>-1, where it is infinity, so the limit is 0 becuase any number over infinity is 0;


Is there an infinite amount of primes in base n made from an equal amount of even and odd digits. Is there an infinite amount of primes in base n made from an equal amount of even and odd digits?
A list of primes that have this property is this sequence
$$23,29,41,43,47,61,67,83,89,1009,1021,1049,1061,\small\dots$$
I think this is true because if you pick a big random number the number of digits in that number with an n or an m is about equal. but I'm sure most numbers don't have an exact number of 2's as 1's just really close to equal. So I'm guessing that the number of primes like this is either less and less frequent or more and more frequent.
So my second question is what is the percentage of primes that have this property. is it almost 0% as the number of primes goes to infinity or does it go to 50%?

As twnly already mentioned in their answer, research by Mauduit and Rivat on a related problem proved (a generalized version of the result) that asymptotically half the primes have an even/odd sum of digits. I want to point out that this seemingly simple statement was an open problem for over 40 years, was proved less than a decade ago, and the proof was published in arguably the most prestigious mathematics journal in the world.
Questions about digits of primes are very hard!
The probability that a randomly chosen $2n$-digit integer has an equal number of odd and even digits is asymptotically $1/\sqrt{\pi n}$ (from asymptotics for central binomial coefficients). The probability that a randomly chosen $2n$-digit integer is prime is asymptotically $1/(2n\ln10)$ (from the prime number theorem).
The natural conjecture would be that these two events are asymptotically independent, so that the probability that a randomly chosen $2n$-digit integer both is prime and has the same number of odd/even digits should be asymptotically $1/(n^{3/2}\sqrt\pi\ln10)$. In particular, the probability that a randomly chosen $2n$-digit prime has the same number of odd/even digits should also be asymptotically $1/\sqrt{\pi n}$, which in particular tends to $0$ as $n\to\infty$. But also, in particular, there should be infinitely many primes with this property—this heuristic predicts that the number of such primes is rather larger than the number of twin primes, for example.

This is not an exact answer, but here are some results about digits of prime numbers that might be of interest.
In
Christian Mauduit and Joël Rivat. Sur un problème de Gelfond : la somme des chiffres des nombres premiers, they show that asymptotically, 50% of the primes have an even sum of digits and 50% have an odd sum of digits in base-10.
Also there is a result of Bourgain, where you can specify arbitrary fixed positions of binary digits and still get primes with the prescribed digits. And Maynard has shown that there are infinitely many primes with no digit 7.
All this is written up in a survey by Maynard that can be found here.


What is the runner's average acceleration during the first 4.1s? A world-class sprinter accelerates to his maximum speed in 4.1s . He then maintains this speed for the remainder of a 100-m race, finishing with a total time of 9.0s. What is the runner's average acceleration during the first 4.1s ?

I know that average velocity is $\frac{\Delta v}{\Delta t}$, where $\Delta t =4.1s$. I also know that the initial velocity is 0, but I don't know how to find the final velocity to find the average velocity. Can someone please help me?

Let denote the way the runner is passing while he's accelerating $s_1$, because we want to find the average acceleration we have:

$$s_1 = v_0t + \frac{at_1^2}{2}$$

The initial velocity is 0, so we left out the first term. Also we make a change $a = \frac{\Delta v}{t_1}$

$$s_1 = \frac{\Delta vt_1}{2}$$

We know that $\Delta v = v - v_0$, where $v$ is the maximum reached velocity. We know that $v_0 = 0$, so $\Delta v = v$. Now we get:

$$s_1 = \frac{vt_1}{2}$$

Know let's denote the second part of the way, when the velocity is constant as $s_2$. We have:

$$s_2 = v(t-t_1)$$

We know that the sum of the two distances is $100m$ so we have:

$$s_1 + s_2 = 100$$
$$\frac{vt_1}{2} + v(t-t_1) = 100$$
$$\frac{4.1 \cdot v}{2} + v(9-4.1) = 100$$
$$2.05 \cdot v + 4.9 \cdot v = 100$$
$$6.95 \cdot v = 100$$
$$v = \frac{100m}{6.95s}$$
$$v \approx 14.39m/s$$

Now we substitute back:

$$a = \frac{\Delta v}{t_1} = \frac{v}{t_1} = \frac{14.39}{4.1} \approx 3.51 \frac{m}{s^2}$$

$$\text{Average velocity} = \frac{\text{Total distance run}}{\text{Total time taken}} = \frac{100}{9} = 11.11\  m/ \sec$$


Conditional expectation, cartesian product Let $\Omega=X\times Y$ be a sample space of pairs $(x,y)$.

Let $\zeta$ be a random variable on $\Omega$.

How would you interpret notation $\mathbb{E}(\zeta \mid x)$? Does this have any sense? I've never come across this. Any ideas?

This is what I think they mean. The random variable $\zeta$ is a function $X\times Y\rightarrow \mathbb{R}$. Given that we choose a certain $x\in X$ we have the function $\zeta|_{x\times Y}:x\times Y\rightarrow \mathbb{R}$. I guess they mean the expectation of this random variable (You might need to rescale to actually make it a random variable).


If $\lim\limits_{x \to \infty} f(x) = A$ then prove $\lim\limits_{x \to \infty} \frac{1}{f(x)} =\frac{1}{A}$ This is a part of my homework and I was asked to prove that if $\lim\limits_{x\to \infty} f(x) = A$, then $\lim\limits_{x\to \infty} \frac{1}{f(x)} = \frac{1}{A}$.

In the previous question, I was asked to show for $\lim\limits_{x\to \infty} f(x) = A$, and $\lim\limits_{x\to \infty} g(x) = B$, $\lim\limits_{x\to \infty} f(x)+g(x) = A+B$

I solved that by showing $|f(x)+g(x) - (A+B)| \le |f(x)-A| + |g(x)-B| = \frac{\epsilon}{2} + \frac{\epsilon}{2}$ but I am not sure if this help on this question

Assume $A\neq0$.

Set $\epsilon>0$. We want $\left|\dfrac1{f(x)}-\dfrac1{A}\right|<\epsilon$ by making $|f(x)-A|$ as small as we please.

Now, $\left|\dfrac1{f(x)}-\dfrac1{A}\right|<\epsilon\implies \left|\dfrac{f(x)-A}{A\cdot f(x)}\right|<\epsilon$.

We can make $|f(x)-A|$ as small as we like, so that we only need to worry about $\left|\dfrac1{A\cdot f(x)}\right|$.

To bound $|A\cdot f(x)|$, make $|f(x)-A|<\dfrac{|A|}2$ which puts a lower bound on $|f(x)|$ of $\dfrac{|A|}{2}$.

Thus, $\left|\dfrac1{A\cdot f(x)}\right|<\dfrac1{|A|\cdot \left|\dfrac{A}{2}\right|}=\dfrac{2}{|A|^2}$.

Now, we can write this down formally.

Set $\epsilon>0$.

Since $\lim_{x\to\infty} f(x) = A$, there is some $N_0$ such that $|f(x)-A|<\min(\dfrac{|A|^2\epsilon}2,\dfrac{|A|}{2})$ for all $x>N_0$.

So, suppose $x>N_0$. Since, $|f(x)-A|<\dfrac {|A|}2$, $\dfrac{|A|}{2}<|f(x)|$.

Thus, $\left|\dfrac1{f(x)}-\dfrac1{A}\right|= \left|\dfrac{f(x)-A}{A\cdot f(x)}\right|=\dfrac{|f(x)-A|}{|A|\cdot |f(x)|}<\dfrac{\dfrac{{|A|}^2}{2}\epsilon}{|A|\cdot \dfrac{|A|}{2}}=\epsilon$.

The statement is only true if $A \neq 0$. If you believe that $\lim_{x\to\infty}\frac1x =0$, then put $g(x)=1/x$ and note that $\lim 1/f(x)=\lim g(f(x))= g(\lim f(x)) = g(A)=1/A$ since $g$ is continuous at $A\neq 0$.


Evaluating $\sqrt{1+\sqrt{2 - \sqrt{3 + \sqrt{4 - \cdots}}}}$ $$x =\sqrt{1+\sqrt{2 - \sqrt{3 + \sqrt{4 - \sqrt{5+ \sqrt{6 - \cdots}}}}}}$$
Find $x$.

I am not sure how to proceed.
Is this a sort of Arithmetico-Geometric Progression? Will this converge at a point?
Any help would be sincerely appreciated.

I doubt a closed form exists for this number, but I'll show convergence and give some bounds on the values.
Let $f_m^n = \sqrt{m\pm\sqrt{(m+1)\mp \sqrt{\cdots\pm \sqrt n}}} $
be the the version where we go from $m$ to $n$ (don't take the $\pm$'s too seriously). We have the recursive expression
$$
f_m^n = \begin{cases}
\sqrt{m + (-1)^{m+1}f_{m+1}^n} & m<n \\
\sqrt n & m=n
\end{cases}
$$
Realness and general bounds:
Let $3 \le m < n$ with $n$ large enough. I claim that
$$
f_m^n - \sqrt{m-1} \in \begin{cases}
[0.5, 0.7] & m \textrm{ odd} \\
[-0.4, -0.6] & m \textrm{ even}
\end{cases}
$$
These constants can surely be refined, and maybe the ranges can be made to depend on $m$ as well to sharpen it further. Furthermore, it shouldn't be hard to show that $f_m^\infty-\sqrt m \to \pm \frac12$ for $m\to \infty$.
In particular everything is real in these cases (which is easy to check). We can show it by backwards induction on $m$. You can check that the base case $m=n-1$ holds for $n\ge8$.
For $m<n-1$ we must check for odd $m$ that
$$
f_m^n \in \left[ \sqrt{m + \sqrt{m} - 0.4}, \sqrt{m + \sqrt{m} - 0.6} \right]
\subset \left[ \sqrt{m-1} + 0.5, \sqrt{m-1} + 0.7 \right]
$$
and for even $m$ that
$$
f_m^n\in \left[ \sqrt{m - \sqrt{m} - 0.7}, \sqrt{m - \sqrt{m} - 0.5} \right]
\subset \left[ \sqrt{m-1} - 0.6, \sqrt{m-1} - 0.4 \right]
$$
These are quadratic inequalities, so we can run the numbers and see that they all hold for $m\ge3$.
Bounds on $f_1$:
Assume for now that $f_m^n$ converges as $n\to\infty$, which I'll prove below. Then the same bounds as above hold for $f_m := \lim_{n\to\infty}f_m^n$. Let's get some better bounds on $f_3$. We have
$$
f_3 = \sqrt{3+\sqrt{4-f_5}}
\in \left[\sqrt{3+\sqrt{4-\sqrt{4}-0.7}}, \sqrt{3+\sqrt{4-\sqrt{4}-0.5}}\right]
\approx [2.035, 2.055]
$$
We could get increasingly better bounds by using a higher term than $f_5$. Assuming the principal squareroot for complex numbers, we get approximately
$$
f_2 = \sqrt{2-f_3}
\in [0.186i, 0.235i]
$$
This finally gives us bounds on $f_1$. If $f_1=x+iy$, then $f_1$ lies on the curve $x^2-y^2=1$ and between the values approximately $1.00430 + 0.0928i$ and $1.00681 + 0.117i$.
Convergence:
We want to use that
$$
\lvert \sqrt{a+x} - \sqrt a \rvert \le \frac{|x|}{2}
$$
or
$$
\sqrt{a+x} = \sqrt a + \varepsilon\left(\frac{|x|}{2}\right)
$$
where $\varepsilon(x)$ means some number that is smaller than $x$ in absolute value.
This holds for all $x$ as long as $a\ge4$. We only need to show convergence for one value of $m$, so choose $m_0$ large enough that $(f_m^n)^2 \ge 4$ for all $n \ge m \ge m_0$ (we can do this using the bounds above). We unravel the radical from the inside out and get inductively (with the base case $k=0$ evident):
$$ \begin{split}
f_{n-k}^n
&= \sqrt{(n-k) \pm f_{n-k+1}^{n}} \\
&= \sqrt{(n-k) \pm f_{n-k+1}^{n-1} + \varepsilon\left(\frac{\sqrt n}{2^{k-1}}\right)} \\
&= \sqrt{\left(f_{n-k}^{n-1}\right)^2 + \varepsilon\left(\frac{\sqrt n}{2^{k-1}}\right)} \\
&= f_{n-k}^{n-1} + \varepsilon\left(\frac{\sqrt n}{2^{k}}\right) \\
\end{split}
$$
In particular
$$
f_{m_0}^n = f_{m_0}^{n-1} + \varepsilon\left(\frac{\sqrt n}{2^{n-m_0}}\right)
$$
which means that
$$
f_{m_0} = \sum_{n=m_0}^\infty \varepsilon\left(\frac{\sqrt n}{2^{n-m_0}}\right)
$$
which is absolutely convergent.

$$x = \sqrt{1+\sqrt{2-\sqrt{3+\sqrt{4-\cdots }}}} $$
If I put write the recurring equation of this statement
$f[n] = \sqrt{n-(-1)^{n}\cdot f[n+1] }$
But the value of $x$ is a complex number so recurring equation can't reveal much, But instead if we mirror $x$, by a change of sign from $+ $ to $ - $
We get it's conjugate
$$ y = -\sqrt{1-\sqrt{2+\sqrt{3-\sqrt{4+\cdots }}}} $$
$y$ is also a complex number, but conjugate to $x$
$x$ and $y$ can therefore appear as the roots of some quadratic, so that $x + y $ and $ x \cdot y $ are real
What remains is to reduce the identities


Injective linear map between modules If you have an injective linear map between two free modules of equal dimension, is the determinant of the matrix representing the map necessarily nonzero?  If not is there an obvious counterexample?  (Everything is over a multivariate polynomial ring over a field.)

Thanks!

Over an integral domain $D$ this is straightforward as Jack Schmidt says in the comments. Suppose $T : D^n \to D^n$ is a morphism such that $\det(T) = 0$. Then $T \otimes \text{Frac}(D) : \text{Frac}(D)^n \to \text{Frac}(D)^n$ has the same property, where $\text{Frac}(D)$ is the fraction field. Since we are now over a field, $T \otimes \text{Frac}(D)$ is not injective, and so it annihilates some nonzero vector in $\text{Frac}(D)^n$. Scaling this vector suitably puts it in $D^n$, and so $T$ is not injective.

Over a general commutative $R$ I have a proof for $n = 2$ (consider the adjugate) but it doesn't seem to generalize. I'll get back to you.

Let $T: A^n \to A^n$ be a morphism and let's use the same notation for its matrix, $T=(t_{ij})$. Let us assume that $det(T)=0$ and I will prove that $T$ fails to be injective.

There must be an integer $k$ such that there exists a minor of rank $k$ with nonzero determinant and any minor of larger rank has determinant zero. Without loss of generality, we can assume that the minor sits in the first $k$ rows and first $k$ columns. If $k=0$, then $T=0$, which is trivial. Since our task is to find a non-zero vector $x=(x_1,\cdots,x_n)$ killed by $T$ and inparticular our vector can be chosen to have the form $(x_1,\cdots,x_{k+1},0,\cdots,0)$, it suffices to assume that n=k+1.

Having adopted these assumptions, since $TT^*=det(T)I=0$, we just have to find a vector $x\neq 0$ such that $T^*x=0$, where $T^*$ is the adjugate matrix of $T$. Now $T^*$ is obviously nonzero since the first $k\times k$ minor of $T$ has nonzero determinant. Hence, such an $x$ must exist(assuming that the ring in question has a unit) and we are home!


Dividing by terms that can potentially be equal to zero If I have the equality $(x + 1)(x - 1) = \dfrac{-n}{m}(x - 1)$, where $m \not = 0$ and $x \in \mathbb{Z}^+$, is it an invalid operation to divide both sides by $(x - 1)$? Since $(x - 1)$ has the potential to equal $0$ if $x = 1$, it seems like it should be an invalid operation?

I would greatly appreciate it if people could please take the time to clarify this.

@The Pointer

You have:

$$(x + 1)(x - 1) = \frac{(-n)}{(m)}(x - 1)$$

Subtracting

$$\frac{(-n)}{(m)}(x - 1)$$

On both sides:

$$(x + 1)(x - 1) - \frac{(-n)}{(m)}(x - 1) = 0$$

Now you put $(x - 1)$ on evidence:

$$(x - 1)[(x + 1) - \frac{(-n)}{(m)}] = 0$$
Now, this can happen only if :

1) $$(x - 1) = 0$$ which leads to $$x_1 = 1$$
2) $$(x + 1) - \frac{(-n)}{(m)} = 0$$
So, in this case, add $$\frac{(-n)}{(m)}$$ to both sides, which gives
$$(x + 1) = \frac{(-n)}{(m)}$$
Subtracting 1 on both sides leads to $$x_2 = \frac{(-n)}{(m)} - 1$$

If you divide by (x - 1) you lose one root. Instead, subtract it on both sides and factr it as :
(X-1)[(x+1)-(-n/m)] = 0

So now you have that (x-1) = 0
Or
(X+1) - (-n/m) = o

Which gives the roots x1 = 1 and x2 = (-n/m) - 1

If you have divided you would only find x2


how to understand the definition of continuity in analysis? Please have a look at the picture above.
This is about the continuity in analysis.
I don't really understand how to utilize this definition?

It says that is statement is equivalent to f is continuous at x.
How to relate it to neighbourbood and open ball?
Can anyone please help me out?

The italicized statement is a definition of "$f$ is continuous at $x$" where $f$ is a function from a metric space $X$ to a metric space $Y$.  Any neighbourhood of $f(x)$ in $Y$ contains an open ball $B_\epsilon(f(x)) = \{y: d_Y(y,f(x))<\epsilon\}$.  The statement says that for every such neighbourhood of $f(x)$
there is an open ball $B_\delta(x) = \{z: d_X(z,x) < \delta\}$ that $f$ maps into the neighbourhood.

Hint:

This definition simply states that $f$ is continuos at $x$ iff $$\lim_{z\rightarrow x}\, f(z) = f(x)$$

It can be proved that it is equivalent to the topological definition of coninuity i.e. $f$ is continuous iff the image of an open subset of $\mathbb{R}$ through $f$ is an open subset of the set $\mathrm{Range}(f)$.

An open ball is a particular kind of neighbourbood, since $\forall r >0$, defining $B(x,r)$ you can always find an open ball, like $B(x,r/2)$ which is an open set completely inside $B(x,r)$.


counter example of inverse function theorem Prove that in the inverse function theorem, the hypothesis that $f$ is $C^1$ cannot be weakened to the hypothesis that $f$ is differentiable. I read an example of my teacher, but I can't have any analysis argument for the fact that $f$ is not one to one in any neighborhood of $0$. Here it is,

$$f(0)=0, \qquad f(x)=x + 2x^2\sin \frac{1}{x}, x\neq 0.$$

Except at $0$, the function is $C^1$, so you can analyse it by examining its derivate on the interval $(0,\varepsilon)$. If $f$ was injective on $(0,\varepsilon)$, then it would have to be monotone and hence its derivative could not have a change of sign there. Show that it does.

Note: I had originally and erroneously looked at the function $f(x)=x + x^2\sin \frac{1}{x}$. That function is indeed bijective, but the set of zeros of $f'$ has an accumulation point at $0$, so it, too, does not yield a diffeomorphism when restricted to any neighbourhood of $0$.

It seems to me that your teacher is wrong. Look here  for a version of the inverse function theorem when the function is only everywhere differentiable but not $C^1$ (see for example this MO question).


Nilpotence criterion for solvable Lie algebras Let $\mathfrak{g}$ be solvable Lie algebra. Lie’s theorem states, that adjoint representation is a homomorphism $\operatorname{ad}:\mathfrak{g}\to \mathfrak{t}$, where $\mathfrak{t}$ is an algebra of upper-triangular matrices. Let $\mathfrak{d}\subset\mathfrak{t}$ be a subalgebra of diagonal matrices.

Is it true, that $\mathfrak{g}$ is nilpotent iff $\operatorname{ad}^{-1}(\mathfrak{d})=Z(\mathfrak{g})$? In one direction it is exactly the Engel’s theorem, but I cannot find counter examples or proof for the other direction.

UPD I want to rephrase my question in an equivalent way. Is it true, that all solvable, but not nilpotent subalgebras of upper-triangular algebra contain nonzero diagonal elements?

Consider the non commutative algebra of dimension $2$ generated by $x,y$ defined by $[x,y]=x$.

The matrix of $ad_x$ in the basis $(x,y)$ is $\pmatrix{0&1\cr 0&0}$ the matrix of $ad_y$ is $\pmatrix{-1&0\cr 0&0}$. $ad_y$ is diagonal, but this Lie algebra is solvable and not nilpotent and $y$ is not in the center.


Linear Programming with $ {L}_{1} $ Norm Constraint When solving a linear programming problem in MATLAB using linprog of the form

$$ \min c^T x $$

subject to

$$ Ax \leq b, \; \left\| x \right\|_{1} = 1 $$

if we need to enforce the additional constraint that $L_1$ norm of $x$ is $1$, or $\|x\|_1=1$, how can this be converted into a set of inequations of the above form so that the trivial solution $x=0$ can be avoided?

As Brian notes, the problem is not convex. There are two workarounds for this.

It seems like the right hand side of $||x||_1=1$ is arbitrary. Maybe you can replace it with $x_1=1$ (if you know that $x_1\geq  0$)? Or solve two problems: one with $x_1=1$ and one with $x_1=-1$? This only works if you know $x_1 \neq 0$.
You can use binary variables to deal with the nonconvexity:
$$ \sum y_i = 1$$
$$ y_i = x_i + s_i$$
$$ y_i = -x_i + t_i$$
$$ 0 \leq s_i \leq M b_i$$
$$ 0 \leq t_i \leq M (1-b_i)$$
$$ y_i \geq 0$$
$$ b_i \in \{0,1\}$$
The big-M constraints set either $s_i=0$ or $t_i=0$. Consequently, $y_i = x_i$ or $y_i = -x_i$. Since $y_i\geq 0$, we obtain $y_i = |x_i|$.

We have

$\|x\|_1= |x_1| + |x_2| + ... + |x_n|$

which is equivalent to

$ \max\{x_1,-x_1\}+\max\{x_2,-x_2\}+...+\max\{x_n,-x_n\}$

So $\|x\|_1=1$ can be substituted as

$y_i \geq x_i$

$y_i \geq -x_i$

$y_i \geq 0$

Then the minimum $L_1$ norm solution for $x$ is given by $y$.


Why set of natural numbers is infinite, while each natural number is finite? In his book Analysis Vol. 1, author Terence Tao argues that while each natural number is finite, the set of natural numbers is infinite (though has not defined what infinite means yet). Using Peano Axiom, if a property holds for P(0) and whenever P(n) is true, P(n+1) is also true, then it is true for all natural numbers. [See image attached at the end.]

However, he has not provided an argument/proof why the set of natural numbers in infinite.
If we go by the same argument, the set of natural numbers should also be finite.

Just like finiteness let’s say P is property called count associated with each natural number. Count can be defined as P(n) = n+1. (Intuitively count means number of elements in the set till n, or the number of elements in the set till n). Now P(0) = 1, which is finite. If P(n) is finite (i.e. n+1), then P(n+1) will also be finite. Hence, the number of elements in the set of natural numbers should also be finite.

Suppose for the sake of contradiction that the set of natural numbers is finite. Then there exists a maximum element $m$. But $m+1=n$ is also a natural number and $n>m$. This contradicts the maximality of $m$, so our original assumption was false, and hence set of natural numbers is not finite but rather infinite.

The induction argument fails because it shows $P(n)$ is finite for every natural number $n$, but it does not show that $P(\infty)$ is finite because $\infty$ is not a natural number.

I am reading the book and I had the same query. I think Tao's presentation of the idea is incorrect. I say that because the property of a natural number being finite cannot be deduced from the axioms Tao's provided. This is so for the axioms he's provided do not address the finite property of all natural numbers. In addition, his induction argument is unnecessarily confusing because one has to assume that 0 is finite and its successors are also finite leading to a circular argument.

I think Tao's made an error in the way he's presented the idea. I think he wants to say that a set that consists of only one natural number is finite and the set containing all natural numbers is infinite. The reason I say his argument is circular is because the finiteness or infiniteness of a set is dependent on the definition of counting (mapping the elements of a set to a set of natural numbers). In the chapter that Manish is referring to, Tao wanted to present an idea on the inductive property of natural numbers and made a mistake in choosing the example he's chosen (since the example depends on definitions he hasn't presented yet).


Solving a recurrence relation: can't figure out how to convert from summation I am really struggling to solve this recurrence.

$$
T(n) = T(\sqrt{n}) + n.
$$

I am asked to give asymptotic upper and lower bounds for $T(n)$. I am free to use any method to arrive at my answer, whether it be iteration method, substitution method, or master method. I'm not quite comfortable with substitution method (as I don't even have any idea as what to guess), and the master method doesn't apply here. I wrote out a recursion tree, and I now believe that the solution can be described as:

$$
\sum_{i=0}^{n} \frac{1}{c^{2^{i}}}
$$

However, I can't (for the life of me) figure out what it sums to. Any help would GREATLY be appreciated, or if someone could explain how to do it via substitution method would also be appreciated (either one). Thanks in advance!

Well ... if you're willing to believe that $T$ is an increasing function, then it's probably enough, for these estimates, to work with numbers of the form
$$
n_k = (((2^2)^2)\ldots)^2,
$$
where the $2$ is squared $k$ times. Then you get that
$$
T(n_k) = T(n_{k-1}) + n_k
$$
and unrolling this gives just the kind of expression you've got (except that I've used "2" instead of $c$).

I can't tell you what it sums to either, but that's not what you were asked. You were asked to give an asymptotic upper bound, which is just something that it's no greater than. Assuming that $c > 1$ (because it is), we have $c^{2^i} > c^i$, for instance, so
$$
\frac{1}{c^{2^i}} < \frac{1}{c^i}
$$
and you can (upper) bound your series with a simple geometric series.

There is the question of "what is $c$ in terms of $n$?", and the answer looks like something like an iterated logarithm, but I haven't worked through that.

Hint.

From

$$
T(n) = T(\sqrt n) + n
$$

we have

$$
\mathcal{T}(\ln n) - \mathcal{T}\left(\frac{\ln n}{2}\right) = n
$$

or making $z = \ln n$

$$
\mathcal{T}(z) - \mathcal{T}\left(\frac{z}{2}\right) = e^z
$$

and also

$$
\mathbb{T}(\log_2 z) - \mathbb{T}(\log_2 z-1) = e^z
$$

and now making $\log_2 z = m$

$$
\mathbb{T}(m) - \mathbb{T}(m-1) = e^{2^m}
$$

now this is a linear nonhomogeneous difference equation that can be solved as

$$
\mathbb{T}(m) = \mathbb{T}_h(m)+\mathbb{T}_p(m)
$$

or

$$
\mathbb{T}(m) =\mathbb{T}(0) +\sum_{k=1}^{m}e^{2^k}
$$

The return to $T(n)$ is left to the reader.


Question involving De Moivre's Theorem Hey guys,

I've been stuck on this question 1b in the image. I got 1a...writing it in general polar form but I'm really not sure of the procedure used to do 1b.

Could someone please give some step-by-step instructions to how I can get solve it?

Any help would be much appreciated.

Thanks!

Well, you need to convert the terms to polar form to adequately use De Moivre's Theorem.

$$ z^2 = i - 1  = \sqrt{2}e^{i(3\pi/4)}$$

You should be able to use De Moivre's Theorem from here.

a) $r = \sqrt {1^2 + 1^2} = \sqrt{2}$ and $\theta = \tan^{-1}\left(\frac {-1}{1} \right) =  \frac {3 \pi}{4}$

Therefore $w = \sqrt{2}~ \mathrm{cis} \left(\frac{3\pi}{4}+ 2k\pi\right)$

b) $$\begin{align} z + 1 - i = 0 \implies z &= -1 + i \\ &= \left (\sqrt{2} ~ \mathrm{cis} \left( \frac{3 \pi}{4} + 2k \pi \right)\right)^2 \\ &= 2~  \mathrm{cis}^2~ \left(\frac{3\pi}{4} + 2k\pi\right) \\ &= 2 ~ \mathrm{cis} \left(\frac{3 \pi}{2} + 4k\pi\right) \end{align}$$


Continuous function with non-negative second derivative in the weak sense is convex I am currently working through a section of Peter Petersen's Riemannian Geometry in which he talks about weak second derivatives of functions. I am trying to work through the details of why a function on a Riemannian manifold $(M,g)$ with non-negative Hessian in the weak sense is convex.
I understand how one can reduce the problem to the case where $M$ is the real line with the Euclidean metric by precomposing with unit speed geodesics, so I have been able to reduce the problem to the following:

Proposition: Let $f$ be a continuous function defined on an open interval in $\mathbf{R}$. We say that $f''(p)\geq 0$ in the weak sense if for every $\varepsilon>0$ there exists a smooth function $f_\varepsilon$ defined in a neighborhood of $p$ such that
(1) $f(p)=f_\varepsilon(p)$
(2) $f\geq f_\varepsilon$
(3) $f_\varepsilon''(p)\geq -\varepsilon$.
Then if $f''\geq 0$ everywhere in the weak sense, then $f$ is convex.

I have tried for a while to come up with a direction to move in from here. My ideas keep running into the problem that the definition of $f_\varepsilon$ only necessarily has nice properties in a small neighborhood of $p$.
This makes the naive approach of trying to directly apply the definition of convexity more difficult (compared to when we assume our function is actually twice differentiable). Given $x_1$, $x_2$ and $t\in(0,1)$, for example, we cannot guarantee the function $f_\varepsilon$ defined around $p=tx_1+(1-t)x_2$ is even defined at $x_1$ or $x_2$.
I have a feeling I am over-thinking this problem. I would appreciate any input/hints anyone would be willing to provide as to how I might proceed from here.

I got stuck exactly at the same point. Here is a quite simple argument to prove your Proposition.
First of all, you can reduce to prove this:

Lemma: Let $f$ be a continuous function defined on an open interval in $\mathbf{R}$. Suppose for every $p$ there exists a convex function $g_p$ defined in a neighborhood of $p$ such that
(1) $f(p)=g_p(p)$
(2) $f\geq g_p$.
Then $f$ is convex.

To deduce the Proposition from the Lemma, fix any $\epsilon>0$ and set $h_\epsilon(x):=\epsilon x^2$.
For every $p$ choose a function $f_\epsilon$ (defined on $U_p$) as in the hypothesis and observe that $f+h_\epsilon\ge f_\epsilon+h_\epsilon$ (on $U_p$), with equality at $p$.
Moreover $(f_\epsilon+h_\epsilon)''(p)\ge -\epsilon+2\epsilon>0$, so wlog $g_p:=f_\epsilon+h_\epsilon$ is convex on $U_p$.
Now the Lemma implies that $f+h_\epsilon$ is convex. As $f+h_\epsilon\to f$ (uniformly on compact sets) as $\epsilon\to 0$, we conclude that $f$ is convex too.
Proof of the Lemma: observe that $f$ is convex iff, for any linear function $\phi$, the sublevel sets $\{f-\phi\le \alpha\}$ are convex for any $\alpha\in\mathbb{R}$ (tell me if this is not clear).
So pick any linear function $\phi$ and put $F:=f-\phi$. Let $x_1,x_2\in\{F\le\alpha\}$; we have to prove that $[x_1,x_2]\subseteq\{F\le\alpha\}$.
Assume by contradiction that $M:=\max_{[x_1,x_2]}F>\alpha$ and call $p\in(x_1,x_2)$ the first point (of the interval) at which $f$ equals $M$ (i.e. $p:=\min f^{-1}(M)\cap [x_1,x_2]$). Now $F$ still satisfies the hypotheses (of the Lemma), so we can find a convex $G:U\to\mathbb{R}$ ($U\subseteq (x_1,x_2)$ being a neighborhood of $p$) such that $F\ge G$ with equality at $p$.
So $G$ reaches its maximum at $p$ (since $G\le F\le F(p)=G(p)$ on $U$); but convexity easily implies that $G\equiv M$ on all of $U$: for example we know that the slope $s(x):=\frac{G(x)-G(p)}{x-p}$ is an increasing function on $U\setminus\{p\}$, but $s(x)\ge 0$ if $x<p$ and $s(x)\le 0$ if $x>p$ since $G(p)$ is the maximum value. If you draw a picture everything becomes clear.
Thus $r\equiv 0$, so $G\equiv M$. Finally, since $F\ge G$, we get $F\equiv M$ on $U$; but this contradicts the fact that $F<M$ on $[x_1,p)$. $\blacksquare$

Here's an outline of what I believe to be a proof. Simpler ones may exist, but this is what I came up with. I haven't really been pedantic at all about details (as you'll see) but maybe this will give you some ideas.

Lemma. If $f'' \geq 0$ everywhere in the weak sense, then for every smooth, compactly supported function $\varphi$ on $\mathbb{R}$ with $\varphi \geq 0$,
$$
\int f \varphi'' \geq 0.
$$

Proof sketch. For any $\epsilon$, show that one can cover the support of $\varphi$ with finitely many disjoint half-open intervals $J_1, \dots, J_k$ such that there exists for each $1 \leq i \leq k$ a smooth function $f_i$ defined on $J_i$ and satisfying $f \geq f_i$ and $f_i'' > -\epsilon$ on $J_i$. Estimate the integral above by breaking it up into integrals over the $J_i$ and using integration by parts and the functions $f_i$.

Proposition. If $f'' \geq 0$ everywhere in the weak sense, then $f$ is convex.

Proof sketch. By adding a linear function to $f$, it suffices to show that for $a,b \in \mathbb{R}$ with $f(a) = f(b) = 0$, we have $f(t) \leq 0$ for $a < t < b$.

Suppose otherwise. Then there is $c$ with $a < c < b$ such that $f(c) = C > 0$. By the intermediate value theorem there exists $c_1, c_2$ with $a < c_1 < c < c_2 < b$ such that $f(c_1) = f(c_2) = C/4$. By continuity, the values of $f$ are close (say, within $\epsilon$) to $c$ or $c/4$ respectively within an $\eta$-neighborhood of the points $c$ and $c_1, c_2$ respectively.

One now constructs an appropriate smooth function $\psi$ supported on the $\eta$-neighborhoods of these three points, positive near $c_1$ and $c_2$ and negative near $c$, and sufficiently well-controlled so that (1) There exists a second antiderivative $\varphi$ of $\psi$ which is smooth and of compact support, and (2) The integral $\int f \varphi''$ can be estimated (and proved to be negative). Together with the lemma, this proves the proposition.


Compact refinement of a covering Suppose $X$ is compact. $A$,$B$ open sets which cover $X$. Can $X$ be covered by compact sets $C$,$D$ such that $C \subseteq A$ and $D \subseteq B$?

Let $X = \alpha(\mathbb{Q})$ denote the Alexandroff compactification of $\mathbb{Q}$. It is a compact non-Hausdorff $T_1$-space which is moreover a KC-space (which means that all compact subsets are closed). See the answers to https://math.stackexchange.com/q/2793610.

Now assume that $X$ satisfies the "compact refinement" property of your question. Then $X$ would be normal since it is a KC-space. But among compact $T_1$-spaces, Hausdorff is equivalent to normal. This is a contradiction.

We need normality of $X$ (so $X$ Hausdorff is sufficient too), because the fact that every two-set open cover has a two-set closed refinement is equivalent to normality. (easy proof by taking complements.)


Solving $\sqrt{a +\sqrt{a-x}}+\sqrt{a-\sqrt{a+x}}=2x, \ a\in\Bbb{R}$ Given
  $$\sqrt{a +\sqrt{a-x}}+\sqrt{a-\sqrt{a+x}}=2x$$
  and $a\in\Bbb{R}$, express $x$ in terms of $a$.

I rationalised the above expression and then again rationalised which gave me :

$$\sqrt{a+\sqrt{a-x}}-\sqrt{a-\sqrt{a+x}}=\frac{1}{\sqrt{a+x}-\sqrt{a-x}}$$

Now, what should I do?

If didn't rationalised the original Equation, or if the squared both sides twice then too bi-quadratic or higher degree polynomial will be obtained.

What should be done?

Here is a simplification, which leads to a parametrized solution.

Write the original equation as two equations
$$
\sqrt{a +\sqrt{a-x}} = x +q\\
\sqrt{a-\sqrt{a+x}}= x -q
$$
where $q$ is a function of $a$.
Double squaring both equations gives
$$
(x+q)^4-2a(x+q)^2+x+a^2-a = 0 \\
(x-q)^4-2a(x-q)^2-x+a^2-a = 0
$$
Subtracting and adding gives the two equations
$$
x^2+q^2-a -\frac{1}{4q}= 0 \\
x^4+ 6 q^2x^2 + q^4-2a(x^2+q^2)+a^2-a = 0
$$
The first equation gives the result
$$
x = \sqrt{-q^2+a +\frac{1}{4q}}= \sqrt{a + f(q)}
$$
with the offset $f(q)$. In here, $q$ has to be inserted, where $q = q(a)$ is obtained by replacing all terms with $x^2$ in  the second equation with the first equation, giving
$$
(-q^2+a +\frac{1}{4q})^2+ 6 q^2(-q^2+a +\frac{1}{4q}) + q^4-2a(a +\frac{1}{4q})+a^2-a = 0
$$
Expanding this expression gives an implicit formula for $q(a)$ as follows:
$$
- 64 q^6 + 64 a q^4 + 16 q^3 - 16 a q^2 + 1 = 0
$$
I wasn't able to solve this further for $q$. However, the equations directly lead to a parametrized solution. Let $q$ be a free parameter. The range where $q$ is to be chosen from is discussed below. Then for each $q$, we have $a$ and $x$ according to
$$
a  = \frac{64 q^6 - 16 q^3 - 1}{64 q^4 - 16 q^2} \\
x = \sqrt{-q^2+a +\frac{1}{4q}}
$$

The following observations can be made:

For large $a$, the equation for $a$ gives $q^2 \to \frac{1}{4}$. This can also be observed by regarding the very first set of two equations and treating $q$ as a small disturbance, which can in first order be neglected if all other terms grow exceedingly large. We then have
$$
\sqrt{a +\sqrt{a-x_1}} = x_1\\
\sqrt{a-\sqrt{a+x_2}}= x_2
$$
which are solved (see Solve the equation: $x=\sqrt{a-\sqrt{a+x}},(a\geq 1)$) by
$$
x_1 = \frac12 + \sqrt{a -3/4}\\
x_2 = -\frac12 + \sqrt{a -3/4}
$$
So indeed, the $x_{1,2}$ grow exceedingly large, and $q = -\frac12$ gives both the correct offset term $-\frac34$ under the previous root $
x = \sqrt{-q^2+a +\frac{1}{4q}}
$, as well as the two deviations $x - x_{1,2}$. For small $a$, correction terms  must be added to $q(a)$ (in the sense of an expansion).

Addendum:
1. As a side observation, one can use the two solutions $x_{1,2}$ to establish that the modified equation
$$
\sqrt{a +\sqrt{a-x -\frac12}}+\sqrt{a-\sqrt{a+x-\frac12}}=2x
$$
is solved by $x = \sqrt{a -3/4}$.

Note that in the general solution $x =\sqrt{a + f(q)}$, the offset $f(q) = -q^2 +\frac{1}{4q}$ actually has a maximum at $q = -0.5$. This can be seen by
$f'(q)  =  \frac{-1 - 8 q^3}{4 q^2}$ which is zero at $q = -0.5$. Hence, the offset has its maximum $f(q= -0.5) = -0.75$ for $\alpha \to \infty$. It will be lower for other $q$. The behavior of $q(a)$ is given in the following plot. For low $a=1.5$, for example, we have $q\simeq -0.81$ and $f(q) \simeq -0.965$. The plot $q(a)$, together with the offset behavior $f(q)$, establish that the offset is monotonously rising with $a$ in a small range of $f$-values.

Squaring both sides, $$a\require{\cancel}\cancel{+\sqrt{a-x}}+a\cancel{-\sqrt{a-x}}+2\sqrt{\left(a+\sqrt{a-x}\right)\left(a-\sqrt{a-x}\right)}=2\left(a+\sqrt{a^2-a+x}\right)=4x^2.$$ since for all $m,n$ one has that $(m+n)(m-n)=m^2-n^2$. Now, dividing both sides by $2$,$$\begin{align} a+\sqrt{a^2-a+x}&=2x^2 \\ \Leftrightarrow (2x^2-a)^2 &= 4x^4 + a^2 - 4ax^2 \\ &= a^2 - a + x.\end{align}$$ Cancel out the term $a^2$ from both sides and expanding,$$4x^2(x^2-a) = x-a.$$ Now, notice that $$\frac{x^2-a}{x-a} = \frac{x^2-a+x-x}{x-a} = \frac{x^2-x}{x-a}+1.$$ Therefore, $$\begin{align}4x^2\left(\frac{x^2-a}{x-a}\right)= 4x^2\left(\frac{x^2-x}{x-a}+1\right)&=1 \\ \Leftrightarrow \frac{4x^3(x-1)}{x-a}+4x^2&=1.\end{align}$$ Now, since $4x^2 = (2x)^2$ then converting it to that and subtracting $1$ from both sides, $$\begin{align} \frac{4x^3(x-1)}{x-a} + (2x+1)(2x-1)&=0 \\ \Leftrightarrow \frac{4x^3(x-1)}{x-a} + x(2x+1) + (x-1)(2x+1) &= 0 \\ \Leftrightarrow (x-1)\left(\frac{4x^3}{x-a} + (2x+1)(x+1)\right)&=0.\end{align}$$ Solve for $x$ and $a$. To start you off, $x=1$.


Finding all solutions for $3^c=2^a+2^b+1$ Given the equation:

$3^c=2^a+2^b+1$

Find all solutions for $a,b,c$ - given that they are positive integers and $b>a$.

Any ideas?

This equation can be solved using Baker's method. It is explained in the book Unsolved problems in number theory by Richard Guy. He writes "In fact, neither $3^a=1 +2^b+2^c$ nor $2^a+3^b=2^c+3^d$ can be reduced to a single congruence. But Baker's method makes it possible to solve these last two (exponential) Diophantine equations." Then he gives the relevant bibliography where this has been done.
In fact, the solution is attributed to Pillai. All integer solutions of $3^a=1 +2^b+2^c$ are
$$
(a,b,c)=(1,0,0),(2,2,2),(4,6,4),(4,4,6).
$$

S.S. Pillai: On the equation $2^x-3^y=2^z+3^w$, Bull. Calcutta Math. Soc. $37$, 1945.

working in $\pmod3$ we find that
$0=(-1)^a+(-1)^b+1 \pmod3$

$edit;$

it only shows that $a$ and $b$ must be even, in $\pmod4$ we see that $c$ also must be even


Differentiable curves that are not smooth We call a curve admitting a parameterization $t\to z(t)$, $t\in[0,1]$ differentiable if the vector function $z$ is differentiable. We call the curve smooth if it is differentiable and its derivative is non-zero.

What are examples of differentiable curves that are not smooth? What are the geometrical differences between the two types? Are there curves, which are not smooth at a point and still have a unique tangent at that point? Which are some differentiable curves do not have a tangent.

the smoothness that you mean is what we call geometric continuity of a curve , in other words a curve is sad to be geometric continuous if it has at each point a tangent vector and varie continuously along the curve.
I m interseting at such field of geometry,  i will be happy if you give me any open question in .


Prove subset of $\mathbb R^2$ is open Problem. Let $G = \{(x,y): x \ne y\}$. Prove $G$ is an open subset of $\mathbb R^2$.

What I am thinking: If I could show that $\mathbb R^2 \setminus G = \{(x,y): x = y\}$ is a closed set, then its complement $G$ is open. I might be totally off. Any suggestions?

Let the function $f\colon \mathbb R^2\rightarrow \mathbb R,$ $(x,y)\mapsto x-y$ then $f$ is continuous since it's a polynomial on $x$ and $y$. We have
$$G=f^{-1}(\mathbb R^*)$$
Do you know what theorem we should use to conclude?

HINT: $\Bbb{R}^2 - G$ results in a line. Can this complete $G$?


Bayes theorem, confusion between $P(X|Z)$ and $P(Z|X)$ In a problem I'm trying to solve it states:

There is a machine displays the inserted credit. Unfortunately, the display is broken. At time $k = 3$, we can infer from the display that the inserted credit is either $2$ or $3$, but certainly not $0$ or $1$.
Without taking into account the additional information provided by the display, the state x(3) has the probability distribution

$f(x_3=0) = 0.1,$

$f(x_3=1) = 0.2,$

$f(x_3=2) = 0.2,$

$f(x_3=3) = 0.5.$

What is the probability distribution of $x(3)$ given the information from the display?

Now my reasoing was the following: the sentence we can infer from the display that the inserted credit is either $2$ or $3$, but certainly not $0$ or $1$ is equivalent to say: given the state of the display, we can say there is $50\%$ probability that the coin is $2$ and $50\%$ probability that the coin is $3$.
So the probability distribution asked would be simply $50\%$ for $2$ and $50\%$ for $3$.

But from the solution of the exercise, it looks at the problem from another perspective. It states that if we can infer something from the display means that we can depict the conditional probability of $z$ given $x$ (and not the opposite as I thought) and then uses the prior information $f(x_3)$ and this likelihood ($f(z|x_3)$) with the bayes theorem to arrive to the solution.

Why is my reasoning wrong?

You're applying the principle of indifference where it doesn't apply. There are two requirements for that principle to apply. First, there must be symmetry among the alternatives (or, as the Wikipedia article puts it, they must be “indistinguishable except for their names”). This is not the case here – the credit states $2$ and $3$ are distinguishable, as you can readily convince yourself by replacing $3$ with $1000$: You probably wouldn't believe that the credit states $2$ and $1000$ are equiprobable. The second requirement is the one you state, that we don't have further information. In the present case, we do have further information, namely, the given a priori distribution, which again distinguishes between the credit states $2$ and $3$. Thus the principle of indifference doesn't apply.

OK this is an interesting problem. So we know that inserted credit equals $2$ or $3$. Let us try to calculate the probability that the inserted credit is $3$ given that we know it is $2$ or $3$. I will just call the inserted credit $x$. We use the formula $P(A|B)=P(A \wedge B)/P(B)$. Where $A$ is $x=3$ and $B$ is $x=2 \vee x=3$. This becomes $P(x=3 | (x=2 \vee x=3))$. The fact that $A$ implies $B$ and so $P(A \wedge B)=P(A)$ is important to keep in mind. We can calculate this as $0.5/0.7=0.71$. By a similar method, we can work out that the probability that the display read $2$ is $0.2/0.7=0.29$. These add up to $1$ which is good news. It must be one of them! Another way to have come at this would have been to say the answer must be $2$ or $3$ so the probabilities of these scores must split in the ration $0.2:0.5$ or $2:5$ or $0.29:0.71$.


Why is $p^{-1}(1)=\mathbb{Z}$? I encountered this proof of fundamental group of the circle by Allen Hatcher. There is a part where I have some doubt.

There is this fact that: For each path $f:I\to X$ starting at a point $x_0\in X$ and each $\tilde{x_0}\in p^{-1}(x_0)$ there is a unique lift $\tilde{f}:I\to\tilde{X}$ starting at $\tilde{x_0}$.

Let $f:I\to S^1$ be a loop at the basepoint $x_0=(0,1)$, representing a given element of $\pi_1(S^1,x_0)$. By the fact above there is a life $\tilde{f}$ starting at $0$. This path $\tilde{f}$ ends at some integer $n$ since $p\tilde{f}(1)=f(1)=x_0$ and $p^{-1}(x_0)=\mathbb{Z}\subset\mathbb{R}$.

My doubt is why $p^{-1}(x_0)=\mathbb{Z}$? How could I make sense of this? Could somebody help clarify my doubt?

Thanks.

The map $p : \Bbb R\to \mathcal S^1$ may be defined as $t\mapsto e^{2\text{i}\pi t}$. Thus, the fiber over a point $e^{2\text{i}\pi \theta}\in\mathcal S^1$ is $\{\theta+k : k\in\Bbb Z\}\simeq \Bbb Z$. That's all.


How to prove the divisibility rule for $3\, $ [casting out threes] The divisibility rule for $3$ is well-known: if you add up the digits of $n$ and the sum is divisible by $3$, then $n$ is divisible by three.  This is quite helpful for determining if really large numbers are multiples of three, because we can recursively apply this rule:

$$1212582439 \rightarrow 37 \rightarrow 10\rightarrow 1 \implies 3\not\mid 1212582439$$
$$124524 \rightarrow 18 \rightarrow 9 \implies 3\mid 124524$$

This works for as many numbers as I've tried.  However, I'm not sure how this may be proven.  Thus, my question is:

  Given a positive integer $n$ and that $3\mid\text{(the sum of the digits of $n$)}$, how may we prove that $3\mid n$?

HINT: Suppose that you have a four-digit number $n$ that is written $abcd$. Then

$$\begin{align*}
n&=10^3a+10^2b+10c+d\\
&=(999+1)a+(99+1)b+(9+1)c+d\\
&=(999a+99b+9c)+(a+b+c+d)\\
&=3(333a+33b+3c)+(a+b+c+d)\;,
\end{align*}$$

so when you divide $n$ by $3$, you’ll get

$$333a+33b+3c+\frac{a+b+c+d}3\;.$$

The remainder is clearly going to come from the division $\frac{a+b+c+d}3$, since $333a+33b+3c$ is an integer.

Now generalize: make a similar argument for any number of digits, not just four. (If you know about congruences and modular arithmetic, you can do it very compactly.)

This can be easily proven by "Digital Root" concept.

Digital root: A digit obtained by adding digits of number until a single digit is obtained.

All natural number is partitioned into 9 equivalence class by "Digital root".

Any number of Digital root 1 is represented by $1+9\times n$
 any number of Digital root 2 is represented by $2+9\times n$ and so on.

(the reason for writing like this is: 9 is identity in the case of finding digital root)

So a number whose sum is 3. i.e., Digital root is 3 can be written as $3+9*i$.
Which is divisible by 3.(It's clear from this representation).


Why can a quadratic equation have only 2 roots? It is commonly known that the quadratic equation $ax^2+bx+c=0$ has two solutions given by: $$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ But how do I prove that another root couldn't exist?

I think derivation of quadratic formula is not enough....

Suppose there are three distinct roots $x,y,z$. One has
$$\begin{cases}ax^2+bx+c=0\\ay^2+by+c=0\\az^2+bz+c=0\end{cases}\Rightarrow\begin{cases}a(x^2-y^2)+b(x-y)=0\\a(x^2-z^2)+b(x-z)=0\end{cases}\Rightarrow\begin{cases}a(x+y)+b=0\\a(x+z)+b=0\end{cases}$$ It follows $$a(z-y)=0\Rightarrow z=y$$ which is a contradiction.

If you wrote the theorem out, it would look like this

THEOREM: Let $a,b,$ and $c$ be real numbers with $a \ne 0$. Then $ax^2+bx+c=0$ if and only if $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

That means

if $ax^2+bx+c=0$, then $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

and

if $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, then $ax^2+bx+c=0$.


$a+bi$ is algebraic over $\mathbb{Q}$ iff $a$ and $b$ are algebraic over $\mathbb{Q}$. A complex number $a + bi$($a$,$b\in \mathbb{R}$,$i^2=-1$) is algebraic over $\mathbb{Q}$ if and only if $a$ and $b$ are algebraic over $\mathbb{Q}$.

Actually,I want to know how to get $a$ and $b$ are algebraic over $\mathbb{Q}$ from $a+bi$ is algebraic over $\mathbb{Q}$.

Consider $F=\mathbb{Q}(a,b,i)$, this is an finite extension of $\mathbb{Q}$ (prove it, it's not difficult, use that for finite extensions $L/H$ and $H/K$ you have that $[L:K]=[L:H][H:K]$, and a finite extension is in particular an Algebraic extension), so $F$ is algebraic over $\mathbb{Q}$. Therefore, every element of $F$ is an algebraic element over $\mathbb{Q}$, so $a+ib\in F$ is.

On the other hand, if $a+ib$ is algebraic over $\mathbb{Q}$, we have that if $b=0$ the result follows, so we suppose that $b\neq 0$. Let $f(x)\in\mathbb{Q}[x]$ the minimal monic polynomial associated with $a+ib$ (over $\mathbb{Q}$), so $a-ib$ is a root of $f(x)$, so $a-ib$ is algebraic over $\mathbb{Q}$, then $((a+ib)+(a-ib))/2=a$ is algebraic over $\mathbb{Q}$ (the set of algebraic elements over a field is a field). Now, in the same way $((a+ib)-a)/i=b$ is algebraic over $\mathbb{Q}$.

Hint: Investigate sums, products, and complex conjugates of algebraic numbers.


Find an equivalent to $\sum_{k=n}^\infty \frac{1}{k!}$ I would like to find an equivalent of $\sum_{k=n}^\infty \frac{1}{k!}$. Can I do as follow ? (since I always have doubt with those $o$ and $O$, I would like your opinion.

$$n!\sum_{k=n}^\infty \frac{1}{k!}=1+\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}+\cdots$$
$$=1+\frac{1}{n+1}+\mathcal O\left(\sum_{k=n+1}^\infty \frac{1}{k^2}\right)=1+o(1).$$
and thus the claim follow.

Q1) Is it correct ?

Q2)

I have that $\mathcal O\left(\sum_{k=n+1}^\infty \frac{1}{k^2}\right)$ because $$\frac{1}{(n+1)(n+2)}\leq \frac{1}{(n+1)^2},\quad \frac{1}{(n+1)(n+2)(n+3)}\leq \frac{1}{(n+2)^2},\quad \frac{1}{(n+1)(n+2)(n+3)(n+4)}\leq\frac{1}{(n+3)^2} \cdots$$

and so $\frac{1}{(n+1)(n+2)\cdots(n+k)}\leq \frac{1}{(n+k)^2}$. But suppose we could only prove that $$\frac{1}{(n+1)(n+2)\cdots(n+k)}\leq \frac{C_k}{(n+k)^2},$$
i.e. that the constant depend on $k$ and we can't have a uniform bound. Could I conclude that $$ \frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}+\cdots=\mathcal O\left(\sum_{k=n+1}^\infty \frac{1}{k^2}\right) \ \ ?$$

See this post: Calculate $\lim_{n \rightarrow \infty}$ ($n!e-[n!e]$)?

It follows that the integer part of $(n-1)!e$ is $(n-1)!\sum\limits_{k=1}^{n-1} \frac{1}{k!}$, so the fractional part  of $(n-1)!e$ is $(n-1)!\sum\limits_{k=n}^{\infty} \frac{1}{k!}$.
Hence, an equivalent expression is $\frac{\{(n-1)!e\}}{(n-1)!}$.


Let $F$ be a field. How do we show that maximal ideals of $F[x]$ are the principal ideals generated by the monic irreducible polynomials? Let $F$ be a field. How do we show that maximal ideals of $F[x]$ are the principal ideals generated by the monic irreducible polynomials?

In Algebra by Artin, he says this proposition is proven analogously to:

Here, he shows that if $n$ is prime, then $\mathbb Z/(n)$ is a field. Then we use the fact that $R/I$ is a field iff $I$ is maximal, and he concludes that $(n)$ is maximal.

The analogous proof would be that if $f(x)$ is monic irreducible, then $F[x]/(f)$ is a field. The only problem is that he has not proven that $F[x]$ modulo a monic irreducible polynomial is a field.

For any field $F$, $F[x]$ is a principal ideal domain; this is a very well-known and oft-quoted result, which I will accept here.
Now let
$M \subset F[x] \tag 1$
be a maximal ideal; since $F[x]$ is a principal ideal domain, we have
$M = (m(x)) \tag 2$
for some
$m(x) \in F[x]; \tag 3$
we may clearly take $m(x)$ to be monic, since the leading coefficient $\mu$ of $m(x)$, satisfying as it does $\mu \ne 0$, is a unit; thus $\mu^{-1} m(x)$ is monic and
$(\mu^{-1} m(x)) = (m(x)); \tag 4$
now if $m(x)$ were reducible in $F[x]$, we would have
$m(x) = p(x)q(x), \; p(x), q(x) \in F[x], \; \deg p(x), \deg q(x) \ge 1; \tag 5$
consider the ideal
$(p(x)) \subsetneq F[x]; \tag 6$
it is clearly proper:  since $\deg p(x) \ge 1$, $(p(x))$ contains no polynomials of degree $0$, that is, contains no elements of $F$ other than $0$. Also,
$(m(x)) = (p(x)q(x)) \subsetneq (p(x)), \tag 7$
for
$p(x) \notin (p(x)q(x)) \tag 8$
lest for some
$r(x) \in F[x] \tag 9$
we have
$p(x) = r(x)p(x)q(x), \tag{10}$
or
$p(x)(r(x)q(x) - 1) = 0, \tag{11}$
whence
$r(x)q(x) = 1, \tag{12}$
which yields
$\deg r(x) + \deg q(x) = \deg 1 = 0, \tag{13}$
impossible in light of the assumption $\deg q(x) \ge 1$; thus we have shown that
$(m(x)) = (p(x)q(x)) \subsetneq (p(x)) \subsetneq F[x] \tag{14}$
in the event that $m(x)$ is reducible, which further shows that $(m(x))$ is not a maximal ideal in $F[x]$; this contradiction implies that $m(x)$ is irreducible in $F[x]$.  Finis.

There's a classic result in commutative algebra that you can apply:

Let $B$ be an integral domain, $A$ be a subring such that $B$ is
integral over $A$. Then $B$ is a field if and only if $A$ is a field.


What exactly is calculus? I've researched this topic a lot, but couldn't find a proper answer to this, and I can't wait a year to learn it at school, so my question is:

  What exactly is calculus?

I know who invented it, the Leibniz controversy, etc., but I'm not exactly sure what it is. I think I heard it was used to calculate the area under a curve on a graph. If anyone can help me with this, I'd much appreciate it.

In a nutshell, Calculus (as seen in most basic undergraduate courses) is the study of change and behaviour of functions and sequences. The three main points are:

Limits: How sequences and functions behave when getting closer and closer to a desired point (geometrically, what happens when you "zoom in" near a point)
Derivatives: How functions change over a parameter (geometrically, the "slope of a graph at a given point")
Integrals: What's the cumulative effect of a function (geometrically, the "area under a graph")

And obviously (and maybe especially), how these relate to one another; the crowning jewel of Calculus is probably the Fundamental Theorem of Calculus, which truly lives up to its name and was developed by none other than Leibniz and Newton.

It's too simple. The derivative of a function gives its rate of change at a given point. The integral of a function gives the area under the curve at a given point. The fundamental theorem merely says that the rate of change of the area under the curve is the value of the function at a given point.

In other words, the derivative of the integral of a function is the function itself. Voila!


How can find this limit? $\lim_{x\to-∞}(\sqrt{x^2+6x}-\sqrt{x^2-2x})$ $$\lim_{x\to-∞}(\sqrt{x^2+6x}-\sqrt{x^2-2x})$$
plugging in infinity gives infinity - infinity, what kind of manipulation can I do to solve this?

To conclude this The answer is actually -4 , because you can do this step


Is there a name to this equation: $(y - a|x|^b)^2 + (cx)^2 = d$? While doing a survey of the various equations that generate the universal love symbol, a heart curve, I find that many fit into this parametrised form:

$$(y - a|x|^b)^2 + (cx)^2 = d $$

Where   a b c d   are parameters.

Is there an existing name to this equation?
Or is this equation being used in some fields of mathematics or physics?
If not, I'll name it Paul Ma's Heart Equation   :-)

Examples of a heart curve for the parameters   a b c d   are:

Set   a=0.75   b=1   c=0.75   d=1

Set   a=0.6   b=(2/3)   c=0.8   d=0.9

And if you set

    a=(4/5)   b=0.5   c=(4/5)   d=(16/25)

it is identical to the equation in the math.stackexchange forum:

An equation that generates a beautiful or unique shape for motivating students in mathematics

My Heart Equation

    $(y - a|x|^b)^2 + (cx)^2 = d$

can also be used to deform the heart curve.  Examples:

For boomerang: Set   a=0.5   b=1   c=0.13   d=1

For circle: Set   a=0   b=any   c=1   d=1

I have written a light hearted blog on this Heart Equation:

  http://onemanadreaming.blogspot.com.au/2014/08/mathematical-equation-of-love-heart.html

where the heart curves are graphed for the above mentioned parameters.

there are equation $a^2 (abs(x)-y)^2 + b^2 (abs(x)+y)^2 = 2(ab)^2$
its make love symbol

if ratio $abs(a/b)>1$ :its make the heart

if ratio $abs(a/b)$ too lagre : its make the boomerang

if ratio $abs(a/b) =1$ : its circle

if ratio $abs(a/b) < 1$ : making it upside down

where $a,b$ is in equation $x^2/a^2+y^2/b^2 =1$


Vieta's With Restrictions How many positive integers $n$ less than 100 have a corresponding integer $m$ divisible by 3 such that the roots of $x^2-nx+m=0$ are consecutive positive integers?

I'm thinking that this problem is Vieta's. Here's my reasoning. Let $a_1$ and $a_2$ be the roots of this polynomial. We have that $a_1+a_2=n$ and $a_1a_2=m$. We also have that $a_1+1=a_2$. We know that since $m$ is divisible by $3$, we have that $a_1$ or $a_2$ is divisible by $3$. In this case, we know that the other root will have to be even. We have that $n$ will be odd, and $m$ will be odd as well. Another key bit of information is that $n$ and $m$ are both integers. From here, I am not able to solve. Help is greatly appreciated.

Let the smallest solution be $a$. Then, by Vieta's, $m=a(a+1)$. Since m is divisible by 3, $a$ or $a+1$ is divisible by 3. This gives that $a$ can be $2$, $3$, $5$, $6$, etc. Remember that $n<100$, so $2a+1<100$, which gives $a<49.5$. This means that the largest value of $a$ is $48$. Note that for every multiple of $3$ $a$ is equal to, there is a value of $a$ one less than it. So the answer is the number of multiples from $3$ to $48$ times $2$, which is $16*2=\boxed{32}$. I am not sure that this is correct, so you may want to double check.

Hint:

If the roots are $a,a+1$

$m=a(a+1)\implies3|a(a
+1)$

If $a=3b,$ we need $0<2(3b)+1<100$

What if $3|(a+1),a+1=3c$(say)


need help finding the coordinates of AB Find AB if the coordinate of A is -5 and the coordinate of B is 17. i have been out of school for over 20 years and have little to no memory of this process. i examined my daughter's book and there is absolutely NO reference to this material at all. please help me understand how to solve this equation?

You want $B-A$, the distance from $-5$ to $17$

The question is a little vague but I will answer what I think it means.  I am presuming A is a vector with coordinates (0, -5) and B is a vector with coordinates (0,17); equivalently they are the complex numbers -5i and 17i.  If you multiply them together you get (-5)(17)($i^2$) = 5x17.

For polar coordinates the angle (0, -5) makes with the x-axis is $-\pi/2$, so the polar coordinates of the endpoint are 5$e^{-i\pi/2}$.  For B the angle of (0,17) is $\pi/2$ and the polar coordinates of the endpoint are 17$e^{i\pi/2}$.

To multiply these two numbers together you multiply the 5 and 17, and add the exponents on the e which come out to 0. So you get 5 x 17 x $e^0$ which is 5x17.

It's a little harder when the angle isn't $\pm 90^o$, but perhaps that is another subject.


Derivative of square root What would be the derivative of square roots? For example if I have $2 \sqrt{x}$ or $\sqrt{x}$.

I'm unsure how to find the derivative of these and include them especially in something like implicit.

Let $f(x) = \sqrt{x}$, then $$f'(x) = \lim_{h \to 0} \dfrac{\sqrt{x+h} - \sqrt{x}}{h} = \lim_{h \to 0} \dfrac{\sqrt{x+h} - \sqrt{x}}{h} \times \dfrac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}} = \lim_{x \to 0} \dfrac{x+h-x}{h (\sqrt{x+h} + \sqrt{x})}\\ = \lim_{h \to 0} \dfrac{h}{h (\sqrt{x+h} + \sqrt{x})} = \lim_{h \to 0} \dfrac1{(\sqrt{x+h} + \sqrt{x})} = \dfrac1{2\sqrt{x}}$$
In general, you can use the fact that if $f(x) = x^{t}$, then $f'(x) = tx^{t-1}$.

Taking $t=1/2$, gives us that $f'(x) = \dfrac12 x^{-1/2}$, which is the same as we obtained above.

Also, recall that $\dfrac{d (c f(x))}{dx} = c \dfrac{df(x)}{dx}$. Hence, you can pull out the constant and then differentiate it.

$\sqrt{x}$
Let $f(u)=u^{1/2}$ and $u=x$
That's
$\frac{df}{du}=\frac{1}{2}u^{-1/2}$ and $\frac{du}{dx}=1$
But, by the chain rule
$\frac{dy}{dx}=\frac{df}{du}•\frac{du}{dx}
=\frac{1}{2}u^{-1/2} •1
=\frac{d}{dx}\sqrt{x}$
Finally
$\frac{1}{2\sqrt{x}}$


What is minimum and maximum probability What is meant by minimum and maximum probability of an event. I came across a questions that asks minimum and maximum probability of three heads in three coins are flipped. We need to find minimum and maximum probability in two cases

When all three coins are not independent
All pairs of coins are mutually independent

The probability of head and tail on each individual coin is 0.5. I am more concerned on how to approach this problem rather than its solution. So far, I am unable to find any material on minimum and maximum probability.

Assign probability $1/4$ to each of the  outcomes TTT, HHT, THH, THH. (That is, spread the mass uniformly on the subset where there is an even number of heads.  One can do this by flipping the first two coins independently, and then using the parity of the first two to determine the output of the 3d flip.  At any rate, in this case, $P(HHH)=0$.  This strange probability law evidently minimizes $P(HHH)$.

Now assign probability $1/2$ to the outcomes HHH and TTT.  Now $P(HHH)=1/2$, which is maximal.  (If $P(HHH)>1/2$ we'd make the first coin unfair, giving it probability $>1/2$.)

That's the answer.  A general principle here is that the joint probability law is constrained by having certain marginals.  The constraints are linear (these probabilities must add up to that) and so the OP's problem boils down to a linear programming problem.

The probability of three heads with three FAIR coins is $({1\over 2})^3$. The min/max may relate to conditional probabilities where the chance of getting heads on the seconds coin depends on what happened with the first coin etc. I'm only guessing that it may mean this as I have never encountered min/max in probability questions such as this.


Large regime asymptotics of sum Can we give an asymptotic description of the behaviour of the sum

$$F(x):=\sum_{\{n_1,n_2 \in \mathbb Z: n_1^2+n_2^2\le x\}}   (n_1+n_2)^2 $$

for large $x$?

I was thinking of rewriting this as a Riemann sum

$$F(x):=\left(\sum_{\{n_1,n_2 \in \mathbb Z: n_1^2+n_2^2 \le x\}}  1 \right) \times \frac{1}{\sum_{\{n_1,n_2 \in \mathbb Z: n_1^2+n_2^2 \le x\}}1  }\sum_{\{n_1,n_2 \in \mathbb  Z: n_1^2+n_2^2 \le x\}}   (n_1+n_2)^2 $$

but did not get anywhere with this.

EDIT: My apologies, it is $n_1^2+n_2^2$ in the summation index not $(n_1+n_2)^2.$ Please let me know if you have any questions.

$F(x) = \frac{\pi}{2}x^2+O(x^{3/2})$.

$$F(x) = \sum_{(n_1,n_2) : n_1^2+n_2^2 \le x} n_1^2+n_2^2+2n_1n_2.$$ Note that $\sum_{(n_1,n_2) : n_1^2+n_2^2 \le x} n_1n_2 = 0$ since we can pair up the terms $(n_1,n_2)$ and $(n_1,-n_2)$. We're left with $2\sum_{(n_1,n_2) : n_1^2+n_2^2 \le x} n_1^2$. We get a factor of $4$ by restricting to $n_1,n_2 \ge 0$. So we have $$8\sum_{n_1 \le \sqrt{x}} n_1^2\sum_{n_2 \le \sqrt{x-n_1^2}} 1 = 8\sum_{n_1 \le \sqrt{x}} n_1^2\left[\sqrt{x-n_1^2}+O(1)\right],$$ where, e.g., "$n_1 \le x$" means "$1 \le n_1 \le x$". Now, using the general $\sum_{n \le t} n^2 \sim \frac{1}{3}t^3$, we have $$8\sum_{n_1 \le \sqrt{x}} n_1^2 = O(x^{3/2}),$$ so ignore the $O(1)$ term above. We're left with $$8x^2\sum_{n_1 \le \sqrt{x}} \left(\frac{n_1}{\sqrt{x}}\right)^2\sqrt{1-\left(\frac{n_1}{\sqrt{x}}\right)^2}\frac{1}{\sqrt{x}},$$ which I wrote in this way so that it is basically the Riemann integral $8x^2\int_0^1 t^2\sqrt{1-t^2}dt$, which gives $\frac{\pi}{2}x^2$. The convergence of the sum to the Riemann integral can of course be justified fairly easily.

The easiest way to get the desired estimate is
\begin{align*}
&\int_{-\sqrt{x}}^{\sqrt{x}} \; \int_{-\sqrt{x-n_1^2}}^{\sqrt{x-n_1^2}} \; (n_1+n_2)^2 \, \mathrm{d}n_2 \, \mathrm{d}n_1  \\
&= \int_{-\sqrt{x}}^{\sqrt{x}} \; \frac{2}{3} (x+2n_1^2) \sqrt{x-n_1^2} \,\mathrm{d}n_1  \\
&= \frac{\pi}{2} x^2  \text{.}
\end{align*}

We might wonder if this estimate is even close to the true value of $F$.

Looking at the second line in the display, a guess at the error term is $O(x^{3/2})$.  (This method of guessing is usually wildly pessimistic.)  Plotting residuals, this seems feasible on $x \in [0,1000]$, but seems pessimistic as we push out further, to $x = 10\,000$.  (The dashed orange lines are $\pm x^{3/2}$.)

Continuing to larger $x$, we see that $O(x^{3/2})$ continues to be pessimistic...


For complex $z$, find the roots $z^2 - 3z + (3 - i) = 0$ Find the roots of:

  $z^2 - 3z + (3 - i) = 0$

$(x + iy)^2 - 3(x + iy) + (3 - i) = 0$

$(x^2 - y^2 - 3x + 3) + i(2xy -3y - 1) = 0$

So, both the real and imaginary parts should = 0. This is where I got stuck since there are two unknowns for each equation. How do I proceed?

Use $(z-a)(z-b)=z^2-(a+b)+ab=0$ to get
$$
(a+b)=3 \tag{1}
$$
$$
ab=(3-i) \tag{2}
$$
From (1) you get $\Im(a)=-\Im(b)$. So $(a_r+ia_i)(b_r-ia_i)=(a_rb_r+a_i^2)+i(-a_r+b_r)a_i=(3-i)$, thus $a_rb_r+a_i^2=3$ and $(b_r-a_r)a_i=-1$. Now you can try a few values like ...

$a_i=1$ and figure out that $a_r=2$ and $b_r=1$.
So you finally rewrite it as $(z-(2+i))(z-(1-i))$.

You could also just use the standard way (as proposed by anon) to get:
$$
\frac{3\pm\sqrt{9-4(3-i)}}{2}=\frac{3\pm\sqrt{-3+4i}}{2}.
$$
Now note that $(i(2-i))^2=-3+4i$, so
$$ \frac{3\pm i(2-i)}{2}=\frac{3\pm (2i+1)}{2}=\frac{3\pm 1 }{2} \pm i $$


is continuous image of dense set dense? Here is the problem:

Let $f : X \to Y$ be a continuous map between topological spaces.  Let $E$ be a dense subset of $X$ (that is, $\operatorname{Cl} E = X$), where $\operatorname{Cl}E$ represents the closure of $E$.  Prove: $\operatorname{Cl} f(E) = f(X)$

I know continuity of $f$ implies $f(\operatorname{Cl} E) \subseteq \operatorname{Cl} f (E)$, but I have no idea to show the converse.

$f[X] = f[\operatorname{cl}(E)] \subset \operatorname{cl}(f[E])$. This implies that every point of $f[X]$ is in the closure of $f[E]$, i.e. $f[E]$ is dense in $f[X]$.

Alternatively from the above inclusion $f[X] \subseteq \operatorname{cl}(f[E])$ we see: $$f[X]\supseteq \operatorname{cl}_{f[X]}(f[E]) = \operatorname{cl}(f[E]) \cap f[X] \supseteq f[X] \cap f[X]=f[X]$$ and hence $\operatorname{cl}_{f[X]}(f[E]) = f[X]$ which says $f[E]$ is dense in $f[X]$.


LogSine Integrals $\int_0^{\pi/3}\theta \ln^2\big(2\sin\frac{\theta}{2}\big)d\theta$. Hi this will soon end my posts on Log Sine integrals, and we can progress into other classes of integrals.

Given the generalized log sine integral,

$$\rm{Ls}_n^{(k)}(\sigma) = -\int_0^{\sigma}\theta^k \Big(\ln\big(2\sin\tfrac{\theta}{2}\big)\Big)^{n-1-k}\,d\theta$$

The log sine integral I am trying to calculate is given by $k=1$, $n=4$,

$$
\rm{Ls}_4^{(1)}\big(\tfrac{\pi}{3}\big)=-\int_0^{\pi/3}\theta\, \ln^2\big(2\sin\tfrac{\theta}{2}\big)\,d\theta=-\,\frac{17\pi^4}{6480}.
$$
What a beautiful closed form!
This integral is known as one of many "Log Sine integrals at $\pi/3$."  It also seems that many people have not been responding to these log sine integral posts.  This is quite puzzling as these integrals are fundamental to mathematical analysis and are quite old..

If anybody wants literature on these integrals, please let me know.

I am not quite sure how to evaluate this.  Perhaps  a change of variables or logarithmic expansion may work.  Thanks for your help.

EDIT:  This is also of interest when discussing Mahler measures.  http://en.wikipedia.org/wiki/Mahler_measure


Degree of extensions and their composite Let $F<E$ and $F<K$ be finite extensions and assume that $EK$ is defined as composite of two fields. I need to show that $[EK:F] \leq [E:F][K:F]$, with equality if $[E:F]$ and $[K:F]$ are relatively prime.

I am stuck with even the less than or equal to part, and also I couldn't see the thing about the case being these two relatively prime.

Since $E$ and $K$ are finite extensions of $F$, they are algebraic and finitely generated. So, we can write $E = F(a_1,\dots,a_n)$ and $K = F(b_1,\dots,b_m)$ for some algebraic elements $a_1,\dots,a_n,b_1,\dots,b_m$. The composite field $EK$ is then equal to $F(a_1,\dots,a_n,b_1,\dots,b_m)$.

Now, $[EK : F] = [EK : E] [E : F]$, so it suffices to show that $[EK : E] \leq [K : F]$. Let $E_0 = E$ and $E_i = E_{i-1}(b_i)$ for each $1 \leq i \leq m$. Note that $E_m = EK$. Let $F_0 = F$ and $F_i = F_{i-1}(b_i)$ for each $1 \leq i \leq m$. In particular, $F_i \subseteq E_i$ for each $i$. Also note that $K = F_m$. Hence,
$$
\begin{align}
[EK : E] = [E_m : E_0] &= [E_m : E_{m-1}] \cdots [E_1 : E_0], \quad \text{and}\\
[K : F] = [F_m : F_0] &= [F_m : F_{m-1}] \cdots [F_1 : F_0].
\end{align}
$$
For each $1 \leq i \leq m$,
$$
\begin{align}
[E_i : E_{i-1}] = [E_{i-1}(b_i): E_{i-1}] &= \deg \min(E_{i-1},b_i)\\
&\leq \deg \min(F_{i-1},b_i)\\
&= [F_{i-1}(b_i) : F_{i-1}]\\
&= [F_i : F_{i-1}]
\end{align}
$$
Hence,
$$
[EK : E] = \prod_{i=1}^m [E_i : E_{i-1}] \leq \prod_{i=1}^m [F_i : F_{i-1}] = [K : F].
$$
Thus, $[EK:F] \leq [E:F] [K:F]$.

$$
[EK:F] = [EK:E][E:F] \implies [E:F] \mid [EK:F],\\
[EK:F] = [EK:K][K:F] \implies [K:F] \mid [EK:F].
$$
$$
\gcd([E:F],[K:F]) = 1 \implies [E:F][K:F] \mid [EK:F].
$$
$$
\therefore [E:F][K:F] \leq [EK:F].
$$
By the previous part,
$$
[EK:F] \leq [E:F] [K:F].
$$
Hence, $[EK:F] = [E:F] [K:F]$ if $[E:F]$ and $[K:F]$ are relatively prime.

Divide the inequality by one of the factors on the right; you should be able to subsequently simplify the left-hand side by using the product formula for towers. Use the PET for the numerator field on the right-hand side, so that both sides of the inequality turn into the degree of an algebraic element only the one on the left side is considered over a larger base field...

For the second part, remember your knowledge of arithmetic and elementary number theory: in particular recall that $a,b\mid n$ and $\gcd(a,b)=1$ imply $ab\mid n$. What are $a,b,n$ in this problem?


Krein-Rutman for cones with empty interior My question concerns the following theorem (a finite-dimensional version of Krein-Rutman):

  Let $V$ be a finite dimensional real normed space and $C \subseteq V$ a closed cone (i.e. a convex subset which is closed under multiplication with positive scalars). Suppose $C$ has nonempty interior. Then for every linear map $T : V \rightarrow V$ with $T(C) \subseteq C$ there exists an eigenvector $v \in C$ of $T$ whose corresponding eigenvalue is the spectral radius of $T$.

I was wondering if the cone really has to have nonempty interior for this to work. It is clear to me that I cannot hope to get an eigenvector corresponding to the spectral radius, but maybe still an eigenvector to some nonnegative eigenvalue?

My idea is as follows. If $C$ is any closed cone in $V$ and $T : V \rightarrow V$ leaves $C$ invariant, then it also leaves invariant the subspace $V_0 = C - C$  which is the subspace of $V$ generated by $C$. Considering $C$ as a cone of $V_0$, $C$ has nonempty interior, because $V_0$ has a basis $\{v_1,\dots,v_n\} \subseteq C$ and the convex hull of $\{0,v_1,\dots,v_n\}$ is an $n$-simplex contained in $C$ which definitely has nonempty interior in the $n$-dimensional space $V_0$. Applying the above theorem (with $V$ replaced by $V_0$) to the restriction $T|_{V_0} : V_0 \rightarrow V_0$ we get an eigenvector $v \in C$ with eigenvalue the spectral radius of $T|_{V_0}$.

Is this argument correct? I have seen this kind of theorem in the literature only with the condition of a nonempty interior, so I was wondering if I might have made some mistake here.

E.g. in $\mathbb R^3$ let $C = \{(x,0,0): x \ge 0\}$.  Any matrix of the form
$$T =  \pmatrix{a & 0 & 0\cr
            0 & b & c\cr
            0 & d & e\cr}$$
with $a > 0$ leaves $C$ invariant, but if the spectral radius is the spectral radius of $\pmatrix{b & c\cr d & e\cr}$ there is no eigenvector $v \in C$ for that spectral radius.


Find a combinatorial proof for $\binom{n+1}{k} = \binom{n}{k} + \binom{n-1}{k-1} + ... + \binom{n-k}{0}$ Let $n$ and $k$ be integers with $n \geq k \geq 0$. Find a combinatorial proof for
$$\binom{n+1}{k} = \binom{n}{k} + \binom{n-1}{k-1} + \cdots + \binom{n-k}{0} .$$

My approach:
I was thinking to use the binomial formula as in $$2^n = \sum{\binom{n}{k}1^k1^{n-k}} .$$
I also tried to use Pascal's Identity $\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}$.

I think you can try to interpret this identity using the following real-life problem:

  Imagine you have a plate of $n+1$ treats: $k$ of them are chocolate pieces, and the rest are Brussels sprouts. In how many ways can you eat all of them, one by one? Assume the chocolate pieces are indistinguishable from each other, and so are Brussels sprouts.

The answer is, of course, $\binom{n+1}{k}$. However, let's count differently, depending on what you eat first:

You bravely go straight on to a Brussels sprout, before eating any chocolate. There are $\binom{n}{k}$ ways to do that, as there will be $n$ treats left and $k$ of them are chocolate pieces;
You first eat one piece of chocolate, and then go on to eating a Brussels sprout: there are $\binom{n-1}{k-1}$ ways to do that, as there will be $n-1$ treats left, $k-1$ of them chocolate;
You first eat two pieces of chocolate, then go on to a Brussels sprout: similarly, it can be done in $\binom{n-2}{k-2}$ ways;
You first eat three pieces of chocolate...

etc. until:

You eat $k-1$ chocolate pieces first, then one Brussels sprout. Then, only one of the remaining $n-(k-1)$ treats is a chocolate, and the number of ways to do that is $\binom{n-(k-1)}{1}$;
You eat all the chocolate pieces first, and then you eat all of the Brussels sprouts: it can be done in only one way, which can also be written as $1=\binom{n-k}{0}$.

Generally, if you eat first $m$ pieces of chocolate ($0\le m \le k$) before getting on with your first Brussels sprout, there are $\binom{n-m}{k-m}$ ways to do it: after eating the initial chocolate pieces and the Brussels sprout, there will be $n-m$ treats left on the plate, $k-m$ of them being chocolate, so you can eat them in $\binom{n-m}{k-m}$ ways.

Altogether all those numbers must add to our original result $\binom{n+1}{k}$, i.e.

$$\binom{n+1}{k}=\binom{n}{k}+\binom{n-1}{k-1}+\binom{n-2}{k-2}+\cdots+\binom{n-(k-1)}{1}+\binom{n-k}{0}=\sum_{m=0}^k\binom{n-m}{k-m}$$

It is obvious,  upon reflection, that ${n \choose k} ={{n-1} \choose k} +{{n-1} \choose {k-1}}$.

Just apply repeatedly...


Cole-Hopf transformation for $u_{t}+c_{1}c_{2} (u_{x})^{2}=c_{1} u_{xx}$ I came across ,in some paper, that the
 $$u_{t}+c_{1}c_{2} (u_{x})^{2}=c_{1} u_{xx}$$

can be solved via the Cole-Hopf transformation. Is there a generalized Cole-Hopf transformation? I tried plugging in
$$u=2c_{1}\frac{f_{x}}{f}$$

and got

$$f^{3}(f_{t}-c_{1}f_{xx})_{x}+f^{2}f_{x}(c_{1}f_{xx}-f_{t})+2c_{1}^{2}c_{2}(f_{xx}f-(f_{x})^{2})^{2}=2c_{1}f_{x}f(f_{x}^{2}-ff_{xx})$$

Thanks

Also, what if $c_{1}(t),c_{2}(t)$ are functions of $t$ and not constants?

The Cole-Hopf transfomation for the PDE
$$
u_t - a \Delta u + b |D u|^2 = 0
$$
is $w = e^{-b u/a}$, which, in your case, is $ w = e^{-c_2 u}$. Performing this transformation on your equation, leads to
$$
w_t - c_1 w_{xx} = 0,
$$
a standard heat equation. You'll have to be careful with the boundary conditions, though.


What are the Differences Between a Matrix and a Tensor? What is the difference between a matrix and a tensor? Or, what makes a tensor, a tensor? I know that a matrix is a table of values, right? But, a tensor?

Maybe to see the difference between rank 2 tensors and matrices, it is probably best to see a concrete example. Actually this is something which back then confused me very much in the linear algebra course (where we didn't learn about tensors, only about matrices).
As you may know, you can specify a linear transformation $a$ between vectors by a matrix. Let's call that matrix $A$. Now if you do a basis transformation, this can also be written as a linear transformation, so that if the vector in the old basis is $v$, the vector in the new basis is $T^{-1}v$ (where $v$ is a column vector). Now you can ask what matrix describes the transformation $a$ in the new basis. Well, it's the matrix $T^{-1}AT$.
Well, so far, so good. What I memorized back then is that under basis change a matrix transforms as $T^{-1}AT$.
But then, we learned about quadratic forms. Those are calculated using a matrix $A$ as $u^TAv$. Still, no problem, until we learned about how to do basis changes. Now, suddenly the matrix did not transform as $T^{-1}AT$, but rather as $T^TAT$. Which confused me like hell: how could one and the same object transform differently when used in different contexts?
Well, the solution is: because we are actually talking about different objects! In the first case, we are talking about a tensor that takes vectors to vectors. In the second case, we are talking about a tensor that takes two vectors into a scalar, or equivalently, which takes a vector to a covector.
Now both tensors have $n^2$ components, and therefore it is possible to write those components in a $n\times n$ matrix. And since all operations are linear resp. bilinear, the normal matrix-matrix and matrix-vector products together with transposition can be used to write the operations of the tensor. Only when looking at basis transformations, you see that both are, indeed, not the same, and the course did us (well, at least me) a disservice by not telling us that we are really looking at two different objects, and not just at two different uses of the same object, the matrix.
Indeed, speaking of a rank-2 tensor is not really accurate. The rank of a tensor has to be given by two numbers. The vector to vector mapping is given by a rank-(1,1) tensor, while the quadratic form is given by a rank-(0,2) tensor. There's also the type (2,0) which also corresponds to a matrix, but which maps two covectors to a number, and which again transforms differently.
The bottom line of this is:

The components of a rank-2 tensor can be written in a matrix.
The tensor is not that matrix, because different types of tensors can correspond to the same matrix.
The differences between those tensor types are uncovered by the basis transformations (hence the physicist's definition: "A tensor is what transforms like a tensor").

Of course, another difference between matrices and tensors is that matrices are by definition two-index objects, while tensors can have any rank.

"Tensors are generalizations of scalars (that have no indices), vectors (that have exactly one index), and matrices (that have exactly two indices) to an arbitrary number of indices. " [1]

You can't index into a scalar $s\in \mathbb{R}$. You can index into a vector $v\in \mathbb{R}^d$ with a single index $i$ as $v_i$. You can index into a matrix $M\in \mathbb{R}^{n \times m}$ with two indices $(i, j)$ as $M_{ij}$.

If I continue adding indices, you might imagine a matrix-like object $T\in\mathbb{R}^{n\times m \times p}$ you can index into with three indices $(i, j, k)$ as $T_{ijk}$. How about $4, 5, 6, ...$ indices? A tensor is such an object that allows any number of indices you want.

In this way scalars, vectors and matrices are the special case of tensors with 0, 1 and 2 indices.

[1] http://mathworld.wolfram.com/Tensor.html


Shouldn't the graph of $y=1$ be the same as $y=\frac{\sqrt{4-x}}{\sqrt{4-x}}$ because $1=\frac{\sqrt{4-x}}{\sqrt{4-x}}$? Shouldn't the graph of $y=1$ be the same as $y=\frac{\sqrt{4-x}}{\sqrt{4-x}}$ because $1=\frac{\sqrt{4-x}}{\sqrt{4-x}}$ ?
So why is it that my calculator said otherwise? And I know that you can't square root a negative number, because it will gave an imaginary number. Is it because the calculator assume that if it there is an undefined value in the equation, the final result will be undefined too? Or is there a proof that I aren't aware of?
Unnecessary comment below:
And if graph of $y=1$ aren't the same as $y=\frac{\sqrt{4-x}}{\sqrt{4-x}}$ just like the calculator said. That would be great, because you don't have to clarify the domain outside the expression. You could just do $y=2x\cdot\frac{\sqrt{4-x}}{\sqrt{4-x}}$ instead of using a set notation such as $\{x\in| \mathbb{R},x>4\}$

If you define $y=1$ on all real numbers, including $4$, they cannot be the same as you cannot define the other function on the same domain, and for two functions to be the same (formally) their domain has to be the same.
On $\mathbb{C}$ the same holds, as for $x = 4$, the second function is not defined.
If you write these functions down formally you would have to say what there domain of definition is. If you define the first function $y=1$ on the same domain as the second they are the same. So it all depends on your choice of domain, which shows how important it is to keep in mind, that when defining a function you always have to state from where to where it maps, otherwise the definition is uncomplete.
I hope this also clarifies the question in your comment.

No
Your logic is sound in thinking that since $\frac{\sqrt{4-x}}{\sqrt{4-x}} = 1$, that $y=\frac{\sqrt{4-x}}{\sqrt{4-x}}$ could be equal  to $y=1$, but since your value of $x$ isn't defined, in this case, $y=\frac{\sqrt{4-x}}{\sqrt{4-x}}$ is not equal to $y=1$


Why the linear numerator for fractions with irreducible denominators and constant numerators for reducible denominators? For example:

$\Large{\frac{2x^3+5x+1}{(x^2+4)(x^2+x+2)}}$ breaks down to $\Large{\frac{ax+b}{x^2+4}+\frac{cx+d}{x^2+x+2}}$

I have been told that since the denominators are irreducible, the numerators will be either linear or constant.

Now my question is for something like $\Large{\frac{2x^3+5x+1}{x^2-4}}$ you would make it equal $\Large{\frac{A}{x+2}+\frac{B}{x-2}}$, why do we assume that the numerators are constant? Why couldn't they be linear??

First you divide the denominator into the numerator, getting a remainder.  So, for your example, $$\frac{2x^3+5x+1}{x^2-4}=2x+\frac{13x+1}{x^2-4}$$ and expand the last fraction in partial fractions.  As the denominator now has a higher degree than the numerator, the limit as $x \to \infty$ is $0$, so you don't want a constant term.  In fact, $$\frac{13x+1}{x^2-4}=\frac {25/4}{x+2}+\frac{27/4}{x-2}$$ and you don't want or need a linear term in the numerator.

That is actually incorrect... you need to do polynomial long division first.


Constructive proof of the Cauchy Schwarz inequality The famous CS inequality states

$$
\left| \left< x , y \right>\right|
  \le \left\| x \right\| \cdot \left\| y \right\|
$$

for $x,y$ in an inner product space $X$ over $\mathbb{K}$. Every proof I found involves some kind of case distinction; namely one may use w.l.o.g $\Vert x \Vert, \Vert y \Vert > 0$ since the inequality is trivial otherwise. However, I was looking for a constructive proof (i.e. without using the law of excluded middle) for the inequality.

I will add some remarks to the question.

1) Law of excluded middle: This axiom states that $A \vee \neg  A$ is true. This is not considered an axiom in constructive mathematics. One might interpret this in the following way: Indirect proofs are not allowed. However, I find this not completely accurate. Precisely the implication $A \rightarrow \neg \neg A$ is true in constructive mathematics; the implication $ \neg \neg A \rightarrow A$ not in gerneral.

2) The definition of an inner product is the same as in "classical" mathematics.

3) The norm $\Vert \cdot \Vert$ is given by $\Vert x \Vert = \sqrt{\langle x, x\rangle }$.

I don't think that you can actually prove Cauchy-Schwartz (CS) for an arbitrary inner product space (constructively). In Constructive Analysis by Bishop & Bridges they define inner product spaces so that CS holds:

However, for the special case of $\mathbb{R}^n$ it is possible to give a constructive proof, which can be found on page 83 of Constructive Analysis:

Breaking into cases doesn't necessarily rely on the law of excluded middle. It is true in a Hilbert space that either $x = 0$ or $x \neq 0$, we don't need to appeal to excluded middle for this fact. This constructively/intuitionistically reduces the problem to finding a proof for both the case $x=0$ and the case $x \neq 0$.

In other words, I agree that a case analysis uses excluded middle in some sense, but if you can constructively prove excluded middle for the cases you are considering, then you don't need to assume it in general.


What are the Laws of Rational Exponents? On Math SE, I've seen several questions which relate to the following. By abusing the laws of exponents for rational exponents, one can come up with any number of apparent paradoxes, in which a number seems to be shown as equal to its opposite (negative). Possibly the most concise example:

$-1 = (-1)^1 = (-1)^\frac{2}{2} = (-1)^{2 \cdot \frac{1}{2}} = ((-1)^2)^\frac{1}{2} = (1)^\frac{1}{2} = \sqrt{1} = 1$

Of the seven equalities in this statement, I'm embarrassed to say that I'm not totally sure which one is incorrect. Restricting the discussion to real numbers and rational exponents, we can look at some college algebra/precalculus books and find definitions like the following (here, Ratti & McWaters, Precalculus: a right triangle approach, section P.6):

The thing that looks the most suspect in my example above is the 4th equality, $(-1)^{2 \cdot \frac{1}{2}} = ((-1)^2)^\frac{1}{2}$, which seems to violate the spirit of Ratti's definition of rational exponents ("no common factors")... but technically, that translation from rational exponent to radical expression was not used at this point. Rather, we're still only manipulating rational exponents, which seems fully compliant with Ratti's 2nd property: $(a^r)^s = a^{rs}$, where indeed "all of the expressions used are defined". The rational-exponent-to-radical-expression switch (via the rational exponent definition) doesn't actually happen until the 6th equality, $(1)^\frac{1}{2} = \sqrt{1}$, and that seems to undeniably be a true statement. So I'm a bit stumped at exactly where the falsehood lies.

We can find effectively identical definitions in other books. For example, in Sullivan's College Algebra, his definition is (sec. R.8): "If $a$ is a real number and $m$ and $n$ are integers containing no common factors, with $n \ge 2$, then: $a^\frac{m}{n} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m$, provided that $\sqrt[n]{a}$ exists"; and he briefly states that "the Laws of Exponents hold for rational exponents", but all examples are restricted to positive variables only. OpenStax College Algebra does the same (sec. 1.3): "In these cases, the exponent must be a fraction in lowest terms... All of the properties of exponents that we learned for integer exponents also hold for rational exponents."

So what exactly are the restrictions on the Laws of Exponents in the real-number context, with rational exponents? As one example, is there a reason missing from the texts above why $(-1)^{2 \cdot \frac{1}{2}} = ((-1)^2)^\frac{1}{2}$ is a false statement, or is it one of the other equalities that fails?

Edit: Some literature that discusses this issue:

Goel, Sudhir K., and Michael S. Robillard. "The Equation: $-2 = (-8)^\frac{1}{3} = (-8)^\frac{2}{6} = [(-8)^2]^\frac{1}{6} = 2$." Educational Studies in Mathematics 33.3 (1997): 319-320.
Tirosh, Dina, and Ruhama Even. "To define or not to define: The case of $(-8)^\frac{1}{3}$." Educational Studies in Mathematics 33.3 (1997): 321-330.
Choi, Younggi, and Jonghoon Do. "Equality Involved in 0.999... and $(-8)^\frac{1}{3}$" For the Learning of Mathematics 25.3 (2005): 13-36.
Woo, Jeongho, and Jaehoon Yim. "Revisiting 0.999... and $(-8)^\frac{1}{3}$ in School Mathematics from the Perspective of the Algebraic Permanence Principle." For the Learning of Mathematics 28.2 (2008): 11-16.
Gómez, Bernardo, and Carmen Buhlea. "The ambiguity of the sign √." Proceedings of the Sixth Congress of the European Society for Research in Mathematics Education. 2009.
Gómez, Bernardo. "Historical conflicts and subtleties with the √ sign in textbooks." 6th European Summer University on the History and Epistemology in Mathematics Education. HPM: Vienna University of Technology, Vienna, Austria (2010).

You have put your finger precisely on the statement that is incorrect.

There are two competing conventions with regard to rational exponents.

The first convention is to define the symbol $a^x$ for $a > 0$ only. The symbol $\sqrt[n]{a}$ is defined for negative values of $a$ so long as $n$ is odd, but according to this convention, one wouldn't write $a^{1/n}$, for instance.

In defining $a^{p/q}$ to be $(\sqrt[q]{a})^p$, the author you quoted chose the fraction $p/q$ to be in lowest form so that the definition would be unambiguous. For example, $a^{10/15}$ is defined to be $(\sqrt[3]{a})^2$. However, it is preferable to define $a^{p/q}$ to be $(\sqrt[q]{a})^p$ in all cases and to prove that this definition is independent of the particular representation chosen for $p/q$; this is what more rigorous books tend to do. That is, you prove that if $p/q = r/s$, then $(\sqrt[q]{a})^p = (\sqrt[s]{a})^r$. There is no mention of lowest form.

The competing convention is to also allow $a^x$ to be defined for all $a \ne 0$ and all rational numbers $x = p/q$ that have at least one representation with an odd denominator. You then prove that $(\sqrt[q]{a})^p$ is independent of the particular representation $p/q$ chosen, so long as the denominator is odd. Thus you can write $a^{3/5} = (\sqrt[5]{a})^3 = (\sqrt[15]{a})^{9} = a^{9/15}$. All of that is fine. However, you cannot write $a^{6/10} = (\sqrt[10]{a})^6$, or even $a^{6/10} = \sqrt[10]{a^6}$. The number $a^{6/10}$ is well-defined, but to write down its definition, you must first select a fraction equivalent to $6/10$ that has an odd denominator, which could be $3/5$ or $9/15$ or something else. For $a^{1/2}$, this can't be done at all, so $a^{1/2}$ is undefined for $a < 0$.

The rules for exponents break down if you start allowing $a < 0$ and exponents that can't be written with an odd denominator. For example, the rule $a^{xy} = (a^x)^y$ is valid, but only so long as $x$ and $y$ are both rational numbers that can be written with an odd denominator. This is not the case if you write $a^1 = (a^2)^{1/2}$, despite the fact that both sides of the equation are defined since $a^2 > 0$.

Edit Reading the paper by Tirosh and Even, I was surprised to learn this matter has drawn serious attention from math educators.

A long time ago, I assumed that, apart from complex extensions, $a^x$ for non-integer $x$ should be defined only for $a > 0$. I reasoned that it made no sense to have a function $(-2)^x$ defined only for rational numbers $x$ with odd denominator. I objected strenuously to notations like $(-8)^{1/3}$.

But that was before I taught a calculus class, which is when I realized why some textbook authors are so happy to define $a^x$ for $a < 0$, following the second convention. The reason is that the formula $\frac{d}{dx}(x^r) = rx^{r-1}$ is perfectly valid for $x < 0$ and $r$ with odd denominator.

Edit: I'm now disavowing this answer, but keeping it here to maintain the comments. Consider my newer answer elsewhere for, I think, a better treatment.

I would currently argue that it's the 2nd equality in the example, $(-1)^1 = (-1)^\frac{2}{2}$, that contains the initial error. It's illegitimate to write $(-1)^\frac{2}{2}$ because that's an undefined expression, although the exact reason depends on which of two common definitions of $a^\frac{m}{n}$ is in use in a given textbook:

Some books at the college-algebra level define rational exponents for any real number $a$, but only for rational exponents written in lowest terms (i.e., $m$ and $n$ with no common factors). In this case, $(-1)^\frac{2}{2}$ in invalid because of the common factor in the exponent.
Other books, including those at the real analysis level, define rational exponents only for nonnegative real $a$ (i.e., $a \ge 0$), but permit any $m$ and $n$ regardless of common factors. In this case, $(-1)^\frac{2}{2}$ is undefined because of the negative base.

In summary: for $a > 0$, any $(-a)^\frac{n}{n}$ is a meaningless piece of writing. The way that this expression violates the definition of a rational exponent depends on the exact definition in use.


Deriving an equation for the orbital period of a satellite I'm combining three equations from my physics text book:

Newton's law of gravitation: $F = -\frac{GMm}{r^2}$
The centripetal force equation: $F = \frac{mv^2}{r}$
The equation for the speed of an object traveling in a circle: $v = \frac{2 \pi r}{T}$

I wanted to create an equation to find the Time period, $T$ and ended up with: $T = \frac{2 \pi r^2}{GM}$ Which is wrong...

EDIT

I've worked it out again, this is my working:

I put Newton's law of gravitation and the centripetal force equation equal to each other:

$\frac{GMm}{r^2} = \frac{mv^2}{r}$

Multiply both sides by $r$:

$\frac{GMm}{r} = mv^2$

Sub in $v = \frac{2 \pi r}{T}$ for $v$:

$\frac{GMm}{r} = m(\frac{2 \pi r}{T})^2$

Divide both sides by $m$:

$\frac{GM}{r} = (\frac{2 \pi r}{T})^2$

Root both sides:

$\sqrt{\frac{GM}{r}} = \frac{2 \pi r}{T}$

Flip both sides and divide by $2 \pi r$:

$T = \frac{2 \pi r}{\sqrt{\frac{GM}{r}}}$

EDIT 2
Which I can simplify:

Multiply both sides by $\sqrt{\frac{GM}{r}}$:

$T \times \sqrt{\frac{GM}{r}} = 2 \pi r$

Square both sides:

$T^2 \times \frac{GM}{r} = (2 \pi r)^2$

Divide both sides by $\frac{GM}{r}$:

$T^2 = \frac{(2 \pi r)^2}{\frac{GM}{r}}$

Clean it up:

$T^2 = \frac{(2 \pi r)^2 \times r}{GM}$

Take out $r$ to get the final answer:

$T^2 = \frac{(2 \pi)^2}{GM}r^3$

If you take out the constant you get Kepler's law (as Ross Millikan said):

$T^2 \propto r^3$

Is this correct? It's going in my A-Level Physics notes and I don't want to be learning the wrong stuff when it comes to the exam.
This is correct now, thanks guys!

If anybody's interested, I've open sourced the notes here

It is not correct.  Kepler's third lawstates:  The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.  Your solution has the square, not the $\frac 32$ power of the axis.  You are using the correct input, so if you show your work we may find the problem.  $v$ should be proportional to $\frac 1{\sqrt r}$

F(g)=GMm/(R+H)^2  (gravitational force) F(centripetal force) =Mv^2/R+H (here v is orbital velocity)
here centripetal force is provided  by gravitational force
therefore
GMm/(R+H)^2=Mv^2/R+H
by working out we get
v^2=Gm/R+H
H is comparatively small than R (radius of earth) and is negligible
therefore
v= squareroot of GM/R

but g=GM/R^2 (g is acceleration due to gravity)
or GM/R=Rg

atlast v =squareroot of Rg(where v is orbital velocity)


Is the area of a line = 1? I tried to teach my son multiplication using a rectangle.
(e.g. 3cm * 4cm = 12cm^2).  Now I have 12 little squares.  But how do I explain where the "little square" came from?

My best guess: If I cut a little square in two, I can get a rectangle 2cm long * 1/2cm deep.  Then I cut that rectangle in two, I get 4cm long * 1/4 cm deep.  I keep going and say I get a rectangle 100,000cm long by 1/100,000 cm deep.

This is very long and very thin.  And I deduce, that the limit would be a line.  And so I conclude that the area of a line = 1.  (units become irrelevant, infinity takes care of that).

I also like this to explain why the integral of y=0x+1 (for x from 0 to 1) is 1.
Note: this should be a 1 unit square, and integrals are an area but we say the answer is 1 not 1 unit^2.

My question: Do you agree? (if not, why not?)

The ‘little squares’ are simply the result of dividing each side into $1$ cm segments and using those to form a grid of $1\times 1$ squares within the $3\times 4$ rectangle. There’s no reason to look any further for an explanation of where they come from. We define the area of a $1\times 1$ square to be $1$ square unit (where unit is whatever linear unit was used to measure the sides of the square), so the obvious way to find the area of some other shape is to divide it up into $1\times 1$ squares $-$ if we can.

Most of the rest is simply wrong, I’m afraid. For starters, a straight line has no area. If you want to explain geometrically/pictorially why a rectangle of dimensions $\frac12\times 2$ has an area of $1$, divide it into $4$ squares of dimensions $\frac12\times\frac12$ and observe that they can be rearranged into a single $1\times 1$ square, which by definition has area $1$ square unit. The same reasoning applies to a rectangle of dimensions $\frac1{100}\times 100$: since $100=10000\cdot\frac1{100}$, it can be divided into $10000$ squares of dimensions $\frac1{100}\times\frac1{100}$, which can then be arranged in $100$ rows of $100$ squares to form a unit square. There is no limiting procedure involved here.

Finally, $\int_0^1 1~dx$ is $1$ square linear unit (e.g., sq. cm.) when it’s interpreted as the area between the $x$-axis and the line $y=1$ over the interval $0\le x\le 1$; it has other units when the integral is given other interpretations.

Lines simply have no area. I'm still a bit confused by the fact that you managed to conclude that the area of a line is 1, and I don't know why you mentioned infinity. Use basic logic. A line has only one spatial dimension, and a square has two. Let's say that the area of that square is 4. If so, then we know that each side must measure at 2. Since we know that to find the area of a square you must 'square' on of the side's measurements, we form the equation l x w = a. A line, by definition, can only have either a length or a width (whichever you prefer to call it). The other dimension has a value of 0, or none. So, no matter how long a line is, it will never have an area because one measurement is missing. Having 'no' measurement is different than just having '0', however. A line's missing measurement can not be 0, and here's why. If the length of a line is equal to infinity (meaning that it is a true line and not just a segment), then that length of infinity must be multiplied by 0. Since 0 multiplied by infinity is just infinity, then the area of the line would be infinity. We know this is not true, which is how we conclude that a line has no second dimension.


Galois group of $x^4 + 2$ over $\mathbb{Q}$ I have a problem finding Galois group of $x^4 + 2$ over $\mathbb{Q}$. Not sure whether to start. It's irreducible over $\mathbb{Q}$ and also have 4 different roots.

Note that $\varepsilon_8^i\sqrt[4]2$, for $i=1,3,5,7$, are roots of this polynomial, where $\varepsilon_8=\frac{\sqrt 2}{2}+i\frac{\sqrt 2}{2}$ is the eighth primitive root of unity. Hence splitting field is $K=\mathbb Q(\varepsilon_8,\sqrt[4]2)=\mathbb Q(\sqrt[4]2,i)$. You can easily see that $|K:\mathbb Q|=8$, hence $Gal(K/\mathbb Q)$ has $8$ automorphisms. Automorphism $f$ is determined by $f(i)$ and $f(\sqrt[4]2)$, and $f(i)\in\{i,-i\}$ (roots of minimal polynomial of $i$ over $\mathbb Q$) and $f(\sqrt[4]2)\in \{\sqrt[4]2,-\sqrt[4]2,i\sqrt[4]2,-i\sqrt[4]2\}$ (roots of minimal polynomial of $\sqrt[4]2$ over $\mathbb Q$; it is $X^4-2$). So, every combination gives an automorphism.

Now you can calculate that $Gal(K/\mathbb Q)\cong\mathbb D_4$.

Let $0<a\in \mathbb Q$ be an element which it not a square in $\mathbb Q$. The roots are $\pm\sqrt[4]{a}$ and $\pm i\sqrt[4]{a}$.

Now we determine $[L:\mathbb Q]$ where $L=\mathbb Q(i,\sqrt[4]{a})$ is the splitting field of $f$

We have $[L:\mathbb Q]=[L:\mathbb Q(\sqrt[4]{a})]\cdot [\mathbb Q(\sqrt[4]{a}):\mathbb Q]$

The first factor is two because the degree of the minimal polynomial of $i$ is bounded by 2. ($i$ is a root of $X^2+1$). However it can't be one because $\mathbb Q(\sqrt[4]{a})\subset \mathbb R$ and $L$ contains complex numbers, hence $[L:\mathbb Q]=8$.

The second one is $4$ because $a$ is not a square in $\mathbb Q$

There are 5 possibilities for $Gal(L/K)$. $3$ are abelian and $2$ not. But $1$ of the $2$ non-abelian groups has only normal subgroups (Q_8) in contrast to the other one (D_4).

So write down some elements of the Galois group and see if they generate a normal subgroup or not.

Note that the elements of the Galois group are determined by the images on the generators, for example:

Let $\phi_1:i\mapsto -i$, $\sqrt[4]{a}\mapsto \sqrt[4]{a}$. Then $<\phi_1>=\{\phi_1,id\}$

and $\phi_2:i\mapsto -i$, $\sqrt[4]{a}\mapsto i\sqrt[4]{a}$.

Now $<\phi_1>$ is not normal because $\phi_2\circ\phi_1\circ\phi_2^{-1}\notin<\phi_1>$, hence

$Gal(L/K)\cong D_4$

This list may also help.

Question: What is the Galois group if $a<0?$