Showing independence of a set of functions $\{1,x,x^2,x^3,(x-3)^3_+,(x-5)^3_+\}$ I want to show that the set of functions $\{1,x,x^2,x^3,(x-3)^3_+,(x-5)^3_+\}$ are linearly independent.

I.e., I want to show that if $p(x)=c_1+c_2x+c_3x^2+c_4x^3+c_5(x-3)^3_+ + c_6(x-5)^3_+=0$ then $c_1=c_2=c_3=c_4=c_5=c_6=0$.

The last two functions is what is confusing me.

Since $p(x)=c_1+c_2x+c_3x^2+c_4x^3+c_5(x-3)^3_+ + c_6(x-5)^3_+=0$ means $p(x)=0$ for all $x$. Could I consider three different cases: $x<3, 3 \leq x <5, x \geq 5$, and for each case show how if $p(x)=0$ then all coefficients are $0$ using the argument posted in the answer to this question Prove that $1, x, x^2, \dots , x^n$ are linearly independent in $C[-1,1]$ ?

I was thinking along these lines, because for those three cases of $x$, you have a cubic polynomial.

If this is not a correct approach, how could I solve this question?

EDIT:

$(x-3)^3_+$ is $(x-3)^3$ if $x \geq 3$ and $0$ otherwise

If I understand your argument correctly you are saying the following: On each of the three regions, the spline is a cubic polynomial.  Since cubic polynomials can have only three roots, unless the polynomial is zero, the only way we can linearly combine the basis functions to be the zero function is if we set all the $ci=0$.  Is this right?

But then consider applying the argument to the following linear combination of basis functions

$c_1 + c_2x + c_3x^2 + c_4x^3 + c_5(x + 2x^2 + 3x^3)$

Then we can claim that the above is a cubic polynomial, hence it has at most three roots, and hence the assumption that the linear combination is the zero function forces c$1 = c2 = .. = c5 = 0.$   But in this case that is not true because we can get the zero function with $c1=0, c2=-1, c3=-2, c4=-4$ and $c5=1.$    The problem with the argument is the assumption that the zero function forces all ci=0.

Your idea is correct. To add more to it, consider eliminating coefficients as you go. Start with $x<3$. Then if $p(x)=0$ for all $x\in \mathbb{R}$, then when $x<3$,
$$
c_1+c_2x+c_3x^2+c_4x^3=0
$$
If $c_j \neq 0$ for some $j$, then a nonzero polynomial has a root at every element in $(-\infty,3)$, a contradiction. Hence, $c_1=c_2=c_3=c_4=0$.

Then move on to $3\leq x <5$ and show $c_5=0$. Finally, finish with $x\geq 5.$

To show that a set of functions $\{f_{1}, f_{2}, \cdots, f_{n}\}$ is linearly independent you must proove that
$$P(x) = c_{1}f_{1} + c_{2}f_{2} + \cdots c_{n}f_{n} = 0$$
Only for
$$c_{1} = c_{2} = \cdots = c_{n} = 0 $$

Which can also be proven by using the Wronskian determinant. This is if $W\{f_{1},f_{2},\cdots,f_{n}\} \neq 0 \implies \{f_{1},f_{2},\cdots,f_{n}\} \text{ is LI} $

This particular set is actually linearly dependant for any $x$:
$$ (x-5)^{3} =  x^{3} - 15x^{2} + 75x - 125$$
$$ (x-3)^{3} =  x^{3} - 9x^{2} + 27x -27$$
$$ (x-5)^{3} - \frac{125}{27}(x-3)^3 = \frac{-98x^{3}}{27} + \frac{80x^2}{3} - 50x$$
Then you can see easily that:
$$ (x-5)^{3} - \frac{125}{27}((x-3)^3) + \frac{98}{27}(x^{3}) - \frac{80}{3}(x^{2}) + 50(x) + 0(1) = 0$$

For the cases when the $(x-n)^3$ terms nullify it is trivial to prove that if there is a zero in the set then such set is linearly dependant.


How do I integrate these functions? I deleted my last question, since some of you wanted me to rewrite the question properly. I feel sorry for inconvenience, but please understand that this is the first time I use $\texttt{MathJax}$.

Up to now, I tried every method I know in integration, like substituition, partial fractions, uv-method, etc. But seems like nothing works. I would appreciate to have your help. Thanks.

$$
\int_{0}^{1}\frac{\mathrm{d}x}{\left(x + 1\right)
\left[x^{2}\left(1 - x\right)\right]^{1/3}}\,,
\qquad\qquad\int_{0}^{1}\frac{\mathrm{d}x}{\left(x^{2} + 1\right)
\left[x^{2}\left(1 - x\right)\right]^{1/3}}
$$

First integral can be rewrite in the following form:
$$I=\int _0^1\frac{(x/(1-x))^{(1/3)}}{\left(x+1\right)x}\:dx$$
This view suggests the following change of variabel $u=x/(1-x)$, $dx=\dfrac{du}{(1+u)^2}$
$$I=\int _0^\infty\frac{ u^{-2/3}du}{(1+2u)}=\frac{1}{2^{1/3}}\int _0^\infty\frac{ u^{-2/3}du}{(1+u)}$$
The latter integral is the Beta function
$$\frac{1}{2^{1/3}}B(1/3,2/3)=\frac{1}{2^{1/3}} \Gamma(1/3)\Gamma(2/3)=\frac{2^{2/3}\pi}{\sqrt{3}}\,.$$
Here I use the identity $\Gamma(x)\Gamma(1-x)=\dfrac{\pi}{\sin(\pi x)}$

To evaluate the secon integral I decompose the fraction $\dfrac{1}{1+x^2}=\frac{1}{2}\left(\dfrac{1}{1+i x}+\dfrac{1}{1-ix}\right)$
The integral
$$I_a=\int _0^1\frac{(x/(1-x))^{(1/3)}}{\left(1+a x\right)x}\:dx=\frac{2\pi}{\sqrt{3}(1+a)^{1/3}}$$
The integral $I_a$ is calculated absolutely analogously to the integral $I$.
Note that this identity is valid not only for real positive $a$ but also for complex one. It is related wit the fact that the integral

$$I=\int _0^\infty\frac{ u^{-2/3}du}{(1+(1+a)u)}$$ can be evaluated for complex a using contour integration technique, but here I don't want touch it.

Thus we have
$$ I_2=\int _0^1\frac{(x/(1-x))^{(1/3)}}{\left(1+x^2\right)x}\:dx=\frac{2\pi}{\sqrt{3}}\frac{1}{2}\left(\frac{1}{(1-i)^{1/3}} + \frac{1}{(1+i)^{1/3}}\right)=\frac{2\pi}{\sqrt{3}} \frac{cos(\pi/12)}{2^{1/6}}=\frac{\pi}{2^{2/3}} \frac{\sqrt3+1}{\sqrt3}
$$

EDIT:
Let us show that
$$I_a=\int _0^\infty\frac{ u^{-2/3}du}{(1+(1+a)u)}=\frac{2\pi}{\sqrt{3}(1+a)^{1/3}}\,.$$
The integrant has a cut in the complex plane from zero to infinity. We set this cut along posinive real axis.
We will consider integral
$$\int_\gamma \frac{ z^{-2/3}dz}{(1+(1+a)z)}\,,$$
contour $\gamma$ is consists of two parts:

1) $z$ from $-i \epsilon +\infty$ to $-i\epsilon$ (against positive real axis a littel bit below cut

2) $z=\epsilon e^{-i \phi}$ $\phi$ from $\phi=\pi/2$ to $\phi=3\pi/2$ (semicircle)

3) $z$ from $i\epsilon$ to $i \epsilon +\infty$ (along positive real axis a littel bit higer cut

$$\int_\gamma \frac{ z^{-2/3}dz}{(1+(1+a)z)}=I_a(1-e^{2\pi i/3 })$$
In the other side this function has a pole in the poin $z=-1/(1+a)$ and we can calculate this integral by residue
$$\int_\gamma \frac{ z^{-2/3}dz}{(1+(1+a)z)}=\frac{-2\pi i e^{i \pi/3}}{(1+a)^{1/3}}\,. $$
Thus we obtain
$$ I_a=\frac{\pi }{\sin(\pi/3)(1+a)^{1/3}}=\frac{2\pi}{\sqrt{3}(1+a)^{1/3}}\,.$$


Let A be a set of all infinite sequences consisting of 0's and 1's. Prove that A is not countable. Sequences such as {010101010101...., 10100100100...., etc}

if i am not wrong these sequences can represent all the real numbers in the binary format, so a such a set will not be countable. but i am not sure how to go about proving it and how  cantor's diagonalization fits in this case.

Assume it is countable, i.e. there exists an enumeration $(a_n)$ of $A$.
Let $a_i:=a_i^1\cdots a_i^n\cdots$, $i=1,2,\cdots$. The element
$$b=b_1\cdots b_n\cdots$$
such that $b_i\neq a_i^i$ does not coincide with any $a_i$.

Suppose $\{0,1\}^{\mathbb{N}}$ is countable. Then we can write $\{0,1\}^{\mathbb{N}}$ as a sequence which elements are sequences of $0$s and $1$s,  $(x^1, x^2, \dots ) $. Because this has every sequence of $0$s and $1$s in fact we should have the sequence $(y_n)_{n\geq 1}$ such that $y_k = 1 -x_k^k$ that is $1$ minus the $k$-th element of the $k$-th sequence (note that this is well definided, that is $y_n$ is a sequence of $0$s and $1$s), now it follows that $(y_n)_{n\geq 1}$  must be an element of $(x^1, x^2, \dots ) $ for some $j$ that is $(y_n) = (x^j_n)$, we have arrived at a contradiction because:

$$y_j = x_j^j \quad \text{  and } \quad y_j = 1 - x_j^j$$


Finding $\lim_{n\to \infty}\sqrt n \int_0^1 \frac{\,dx}{(1+x^2)^n}$ $$
\lim_{n\to\infty} n^{1/2}
\int_{0}^{1} \frac{1}{(1+x^2)^n}\mathrm{d}x=0
$$

Is my answer correct?
But I am not sure of method by which I have done.

No. With the change of variable $ t = \sqrt{n} x $ you get $$ \sqrt{n} \int_0^1 \frac{1}{(1+x^2)^n} \, dx = \int_0^\sqrt{n} \frac{1}{\left( 1 + \frac{t^2}{n} \right)^n} \, dt = \int_0^{+\infty} \frac{1}{\left( 1 + \frac{t^2}{n} \right)^n} \chi_{[0,\sqrt{n}]}(t) \, dt. $$Observe that $$ \lim_{n \to \infty} \left( 1 + \frac{t^2}{n} \right)^n \chi_{[0,\sqrt{n}]}(t) = e^{t^2}$$pointwise everywhere, say for $t>0$; also the sequence $ n \mapsto (1 + t^2 /n)^n$ is increasing for all $t \in \mathbb{R}$ which implies that $$\frac{1}{1+t^2} \ge \frac{1}{\left( 1 + \frac{t^2}{n} \right)^n} \ge \frac{1}{\left( 1 + \frac{t^2}{n} \right)^n} \chi_{[0,\sqrt{n}]}(t) $$ for all $n \in \mathbb{N}$ and for all $t \in \mathbb{R}$. By the Dominated Convergence Theorem, since $1/(1+t^2) \in L^1([0,+\infty))$, we have that $$\lim_{n \to \infty} \int_0^{+\infty} \frac{1}{\left( 1 + \frac{t^2}{n} \right)^n} \chi_{[0,\sqrt{n}]}(t) \, dt = \int_0^{+\infty} e^{-t^2} \, dt = \frac{\sqrt{\pi}}{2}. $$

$$I(n)=\sqrt{n}\int_0^1\frac{dx}{(1+x^2)^n}$$
and since $$(1+x^2)^{-n}=\sum_{k=0}^\infty{{n+k-1}\choose k}(-1)^kx^{2k}$$
$$I(n)=\sqrt{n}\int_0^1\sum_{k=0}^\infty{{n+k-1}\choose k}(-1)^kx^{2k}dx=\sqrt{n}\sum_{k=0}^\infty{{n+k-1}\choose k}\frac{(-1)^k}{2k+1}$$
and we know that:
$${{n+k-1}\choose k}=\frac{(n+k-1)!}{k!(n-1)!}$$
and as $n\to\infty$ this will $\to1$
so we are left with:
$$I(n)\approx\sqrt{n}\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}$$
now take the limit


Finding the volume using double integrals Find the volume  of the wedge sliced from the cylinder $x^2+y^2 =1$ by the planes $z=a(2-x)$ and $z=a(x-2)$  $a>0$

I am confused because $x^2+y^2 = 1$ is a unit circle not a cylinder. and the other two are lines in the $zx$ plane where $y=0$. I don't see how they are planes, are they planes or are they just lines in the $zx$ plane (when $y=0$). I tried drawing the 3d picture but it made no sense since a cylinder wasn't formed.

In a figure showing the $(x,z)$-plane the two given planes appear as lines intersecting the $x$-axis at $x=2$ and the $z$-axis at $z=\pm2a$. Looking at the figure one realizes that because of symmetry the volume in question is $4a\pi$, whereby $\pi$ stands for the area of the circle $x^2+y^2\leq1$..

It is clear that the problem is given in $\mathbb{R}^3$. Therefore when someone says "the cylinder $x^2+y^2=1$" they mean to indicate the following set:

$$\{(x,y,z)\in\mathbb{R}^3: x^2+y^2=1\}$$

which is an infinite cylinder obtained by sweeping the circle $x^2+y^2=1$ up and down vertically in the $z$-direction, that is, perpendicular to the plane $z=0$.

Now the general equation of a plane in $\mathbb{R}^3$ is this:
$$Ax+By+Cz = D$$
for some $A,B,C,D\in \mathbb{R}$. Choosing $A=-a,B=0,C=1,D=2a$, gives the plane $z=a(2-x)$. It is a plane perpendicular to the vector $(-a,0,1)$ that lies a distance of $2a$ from the origin. If you take $A=a, B=0, C=1, D=-2a$, you get the other plane, namely $z=a(x-2)$, which is perpendicular to the vector $(a,0,1)$ and also lies at distance of $2a$ from the origin.

So you got the entire geometry mixed up. Try it now.


Solve for $z$ in $z^3=8i$ I only have a question about the end results. I answered the question fully but my professor knocked off 1 point for my $w_1^0$ result, but I don't know why. He circled the $i\pi /6$ in my answer but I can't figure out what I did wrong. Does anyone know what might be wrong here?

$$z^3=8i$$
$$z=(8i)^{1/3}$$
Converting to polar form and letting $k=0,1,2$ and $-\pi \leq \theta \leq \pi$, we get

$$w_1^0 =2e^{i\pi /6}$$
$$w_2^1=2e^{i5\pi /6}$$
$$w_3^2=2e^{9i \pi /6} = 2e^{-i \pi /2}$$

What's wrong with the $\frac{i\pi}{6}$ in $2e^{i\pi /6}$??

Perhaps your teacher expected a little more details, in particular on how you obtained that angle:

$$z^3=8i=8e^{\frac{\pi i}2+2k\pi i}=8e^{\frac{\pi i}2\left(4k+1\right)}\implies$$

$$z_k=(8i)^{1/3}=\sqrt[3]8e^{\frac{\pi i}6\left(4k+1\right)}=2e^{\frac{\pi i}6\left(4k+1\right)}\;,\;\;k=0,1,2\implies$$

$$\begin{cases}z_0=2e^{\frac{\pi i}6}=2\left(\frac{\sqrt3}2+\frac12i\right)=\sqrt3+i\\{}\\z_1=2e^{\frac{5\pi i}6}=2\left(-\frac{\sqrt3}2+\frac12i\right)=-\sqrt3+i\\{}\\z_2=2e^{\frac{3\pi i}2}=-2i\end{cases}$$

Hint use $$a^3-b^3=(a-b)(a^2+ab+b^2)$$ solve them and get the roots .


Which simple puzzles have fooled professional mathematicians? Although I'm not a professional mathematician by training, I felt I should have easily been able to answer straight away the following puzzle:

Three men go to a shop to buy a TV and the only one they can afford is £30 so they all chip in £10. Just as they are leaving, the manager comes back and tells  the assisitant that the TV was only £25. The assistant thinks quickly and decides  to make a quick profit, realising that he can give them all £1 back and keep £2.
So the question is this: If he gives them all £1 back which means that they all  paid £9 each and he kept £2, wheres the missing £1?
3 x £9 = £27 + £2 = £29...??

Well, it took me over an hour of thinking before I finally knew what the correct answer to this puzzle was and, I'm embarrassed.
It reminds me of the embarrassement some professional mathematicians must have felt in not being able to give the correct answer to the famous Monty Hall problem answered by Marilyn Vos Savant:
http://www.marilynvossavant.com/articles/gameshow.html

Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what's behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?
Yes; you should switch.

It's also mentioned in the book: The Man Who Only loved Numbers, that Paul Erdos was not convinced the first time either when presented by his friend with the solution to the Monty Hall problem.
So what other simple puzzles are there which the general public can understand yet can fool professional mathematicians?

OK, here's one which had me stumped and when I knew the answer, I was pretty embarrassed. Even more so when I told my mother and she immediately saw the solution. And she's no mathematician.

  A music lover lives in a town which has two concerts every weekend. One is in the North of the town, the other in the South. The music lover lives in the centre of town and decides to take the bus to the concert. There's a bus to the North and one to the South every hour. The music lover just decides to take the first bus he sees and go to that concert. At the end of the year, he notices that he only went to the concert in the South. How come?

Here's the answer:

   The clue is that there's a bus every hour, but I didn't specify at what time in the hour the buses leave for the North and South. Imagine that the bus for the South leaves only 5 minutes before the one for the North. Then, it is more likely that if the music lover arrives at a random time, he will take the bus to the South. The odds are 1 to 11 that he takes the one to the North. If they leave even closer to one another, the probability will be even smaller.


Let $z, w \in \mathbb{C}$. Prove that if $zw$ and $z + w \in \mathbb{R}$, then $z = \bar{w}$ Let $z, w \in \mathbb{C}$. Prove that if $zw$ and $z + w \in \mathbb{R}$, then $z = \bar{w}$.So I have started to see complex numbers in class and by reading I find this problem. So I know that $$Re(z)= \frac{z + \bar{z}}{2},$$ but I don't seem to get to the answer. I also tried using other basic properties of $\mathbb{C}$.

Any help/hints?

The quadratic equation $t^2 - (z+u) t + zu =0$ has solutions $z$ and $u$.

Since $z+u, zu \in \mathbb{R}$, we have $\overline{t^2 - (z+u) t + zu} = \overline{t}^2 - (z+u) \overline{t} + zu = \overline{0} = 0$, which implies that if $t$ is a solution then $\overline{t}$ also is. This lead us to $z = \overline{u}$.

You can prove much general thing: If $z$ is a zero of some real-coeffienct polynomial $p(t)$, $\overline{z}$ is another zero of $p$.

I'll just explain this with barely any algebra because this just shows how beautiful complex numbers are.

First of all, when you multiply two complex numbers, you rotate by adding the angles and dialate by multiplying their moduluses:
$$r_1e^{i\theta_1}\cdot r_2e^{i\theta_2}=(r_1r_2)e^{i(\theta_1+\theta_2)}$$
So, for a complex number to be real, it has to have $\theta_1+\theta_2=\pi n$ for any integer $n$, but taking the principal we'll just look at $0$ and $\pi$, so we either need
$\theta_1=-\theta_2$ or $\theta_1=\pi-\theta_2$. The first equation reflects on the $x$-axis while the second reflects on the $y$-axis. (with greater or smaller magnitude)

Now when you add two complex numbers, you translate one by a vector corresponding to the other. Now, if you noticed, if you tried to translate a complex number by it's reflection on the $y$-axis, it would always be away from the $x$-axis, and with some trivial geometry you can find out why, so we need the reflection to be on the $x$-axis.

For such two complex numbers, the translation to end up on the $x$-axis and be a real number  (considering the discussion of the angles above), you have to have the other complex number be of equal length, and hence the conclusion. (a conjugate complex number is a reflection on the $x$-axis with equal length)

I suggest you study closely how complex numbers work, it will help you visualize them and understand how much they're useful at the core before just wanting to do the algebra.


Cantor's diagonal argument applied to rationals How to prove that the new number produced by the Cantor's diagonalization process applied to $\Bbb Q$ is not a rational number ?

Suppose, someone claims that there is a flaw in the Cantor's diagonalization process by applying it to the set of rational numbers. I want to prove that the claim is false by showing that the new number produced by this process is not rational. How to prove this ? Hope I have made my question clear.

This has very little to do with rational numbers themselves. You apply the diagonal argument to construct a number that is not on the list. Now the only reason that you must have produced an irrational number is that all the rational numbers are on the list, so it cannot be any of them. There is nothing intrinsic the construction that favours irrational numbers: if just one rational number were missing from the list, it might be that you have constructed just that number.

Added. Actually, now that I think about that, it is not true: in order for given some rational number to be the result of the diagonal procedure, there are a great many (other) rational numbers that must be absent from the list. Assuming decimal notation and for simplicity a rational number without terminating decimal representation, there are for every digit position $9$ rational numbers that differ from it only in that digit position. Of those, only one can be on the list, and only if it occurs at the exactly right position. So an infinite number of those "neigbours" need to be absent from the list. And even if one assumes binary notation (with only one neighbour per digit position), there must be one neighbour absent for every non-neighbour that is on the list (because the position of the former is taken), still making for a requirement of infinitely many rational numbers from the list. All this is related to the fact that modifying a rational number in a single position will not make it irrational. Curious!

Despite replying to an old post, I would like to supplement things that that I missed out after reading all replies.
Fact:

$\pi$ is irrational

Cantor's technique for finding a contradiction has 1 property

The generated new number must be different

Cantor's contradiction makes sense because the new number is

Real number
Resulted from the aforementioned technique, which means its different from all enumerated real numbers

The contradiction will makes sense to $\mathbb{Q}$ if the new number is

Rational number
Resulted from the aforementioned technique, which means its different from all enumerated rationals

Here's the key. The very technique can give this number--$\pi$,

Rational number (No, it isn't)
Resulted from the method (Yes, it is different from all since its irrational)

This is how you conclude why is the generated number not necessarily rational, because the method gives a lot of freedom in picking the number


Differentiability of the distance function Suppose that $d:X \times X \to \mathbb{R}$ is a geodesic distance function on a smooth Riemannian manifold $X$ ($d$ is determined by metric tensor) and $x \in X$ is fixed. What can be said about the points in which $f:=d(x,\cdot)$ is differentiable? In particular:
-is it possible that $f$ is everywhere differentiable?
-how big the set of points in which $f$ isn't differentiable can be?
-what can be said about the structure of the set of point in which $f$ isn't differentiable (can it be a submanifold?)
I would be grateful if anybody could shed some light on these issues.

Travis's examples have a general explanation - on any complete, connected $n$-manifold, the distance function $d(p,\cdot)$ is smooth everywhere except at $p$ and on the cut locus of $p$. (You can make this conclusion by studying the exponential map - see e.g. Chapter 5.9 of Petersen's Riemannian Geometry.)

Thus your question amounts to studying the structure of the cut locus, which has been the subject of a lot of research over the last few decades; so while I'm not too familiar with it myself, you should be able to find many relevant papers online. While the picture I have in my head of the cut locus is a submanifold, I'm pretty sure there are degenerate cases where you at least lose differentiability.

If I'm interpreting the main theorem of this article correctly, you can conclude that the cut locus has Hausdorff dimension at most $n-1$, and thus that the distance function is almost everywhere differentiable, which answers the "how big" question.

For example, we can take a point $(x_0,\frac{1}{2})$ on $\mathbb{R}^2/\mathbb{Z}^2$  and geodesics $y=\pm\frac{x_0}{2}x$. Obviously, for $x_0\leqslant \frac{1}{2}$ these two lines are shortest geodesics. And if we move the point upper or lower, then the distance has a positive one-sided derivative and therefore is not differentiable at any point of $\{y=\frac{1}{2}\}$. The same for $\{x=\frac{1}{2}\}$.

Subset of singular points is not a subvariety.


Does this limit exist (ln)? Does this limit exist and is finite?
(It goes to one from the left

$\lim_{x\to 1^-}{{(\ln(x) \times \ln(ln(x))}})$

I'm not sure about what the $\lim_{x\to 1^-}{{(\ln(ln(x))}})$ returns here. What do I do in this case?

$\ln(\ln(x))$ is only defined for $x>1$, because if $x<1$, then $\ln(x)<0$ which means that $\ln(\ln(x))$ is not defined.

As noted by others if $x \to 1^{-}$ then the expression $\log\log x$ is not real and hence the limit does not exist as far as we are dealing with real functions of a real variable. On the other hand if $x \to 1^{+}$ then putting $t = \log x$ we see that $t \to 0^{+}$ and then $\log x \cdot \log \log x = t\log t \to 0$.


Intuitive reasoning behind $\pi$'s appearance in bouncing balls. This video is about an interesting math/physics problem that when cranked out churns out digits of $\pi$.

Is there an intuitive reason that $\pi$ is showing up instead of some other funky number like $e$ or $\phi$, and how can one predict without cranking it out that the solution has something to do with the mysterious $\pi$?

Edit to answer jay:

You need not watch all of it, but I'll attempt a quick summary.

Let the mass of two balls be $M$ and $m$ respectively. Assume that $M=16 \times 100^n m$. Now, we will roll the ball with mass $M$ towards the lighter ball which is near a wall. How many times do the balls touch each other before the larger ball changes direction?
(The large ball hits the small ball which bounces off the wall)

The solution is the first $n+1$ digits of $\pi$.

What an interesting problem! I will try to explain only the intuition, as much as possible, and leave the discussion of the math to  Galperin's original translated paper which, aside from a few typos, is quite accessible. Also, I will not follow the exact method in the video, but rather follow Galperin's original method, as I believe it is easier to understand that way. The main difference is that in Galperin's original method, all bounces in the system are counted, not just those between the two balls. This also results in a different relationship between the mass of the large ball and the mass of the small one, but it's the same basic idea. In particular, $\pi$ shows up for the same reason.

First, let's set up the problem as shown below:

Place the large ball of mass $M$ on the line $\ell$, and place the small ball of mass $m$ between the large ball and the wall. Let the distance of the small ball from the wall at time $t$ be given by the function $x(t)$, and let the distance of the large ball from the wall be given by the function $y(t)$. Now this gives us a way of representing the position of both balls as a point $P(t)=(x(t),y(t))$ in the Cartesian plane, $\mathbb R ^2$.

Play around with this a bit to recognize what various parts of the plane correspond to. First, since the little ball is always between the large ball and the wall, $x(t)\le y(t)$ and so the point $P(t)$ will be between the $y$-axis and the line $y=x$. For instance, when the little ball hits the wall, then $x(t)$ will be zero, and so the point $P(t)$ will be on the $y$-axis. When the large ball hits the little ball, then $y(t)=x(t)$ so the point will be on the line $y=x$. This gives us a good way of visualizing all activity in the system as the movement of a point in the two dimensional plane.

Now let's set up a relative criteria for the masses $M$ and $m$ (and here is where I deviate from the video). Let $M=100^Nm$, roll the large ball toward the small ball with some velocity $V$, and let's count the total number of bounces in the system. If the masses were equal, when $N=0$, we would get three total bounces in the system:

The "large" ball would hit the "small" ball and, similar to Newton's cradle, all the velocity of the moving ball would be transferred to the second with the first bounce.
The "small" ball would bounce against the wall and, conserving kinetic energy, move in the opposite direction with the same speed as going in.
The "small" ball would bounce against the "large" ball, transferring all it's momentum into the "large" ball.

This is 3 bounces, which happens to be the first digit of $\pi$. The point $P(t)$ will follow the dotted path outlined in Figure A. Since the masses were equal, note that we have a very nice reflective property of the path of $P(t)$, that is, the angle of incidence is equal to the angle of reflection when it "bounces" against $y=x$ and the $y$-axis. If the masses are unequal, however, this reflective property does not hold because the "large" ball will keep moving with some velocity toward the wall (see Figure B.).

To explain why this reflective property is desirable, let's "fold-out" the angle between the $y$-axis and the line $y=x$ for the case when the masses are equal, as shown in Figure C. (Think of flipping the angle over on itself in a clockwise manner.)

The dotted line still represents the movement of $P(t)$, but every time it would have bounced off of an object, this time it passes into the next folded-out section of the angle instead. Because the angle of incidence is equal to the angle of refraction, we have that the path of $P(t)$ is just a nice straight line, and it is very easy to see that it bounces three times. If only we could get the reflective property to hold when the masses were different sizes...

Well, as it turns out (and this does require some math to verify), we can get the reflective property to hold when the masses are different sizes by scaling the graph differently in the $y$ and $x$ directions. First, verify yourself that the ball bouncing against the $y$-axis always has the reflective property, and note that it doesn't matter how you scale the graph (stretch it in the $y$ or $x$ direction, and the bounce of $P(t)$ against the $y$-axis will still be reflective). Thus we only have to scale it in a way so that the bounce off the $y=x$ line is reflective. If you think about it, scaling the graph with different scales for the $x$ and $y$ axis will be akin to changing the angle of the $y=x$ line, and so what we want to do is "rotate" the $y=x$ line to a point where the reflective property holds. As it turns out, and again this does require some math to verify, the proper choice of scaling is $Y=\sqrt M y$ and $X=\sqrt m x$.

This scaling changes the angle between the $y$-axis and the line that represents a bounce between the two balls, which is now $Y=\sqrt{\frac M m}X$. Take the point $Y=\sqrt M$ and $X=\sqrt m$ which is on this line, and note that the triangle thus created by the points $(0,0)$, $(0,\sqrt M)$, and $(\sqrt m, \sqrt M)$ yields that $\tan(\theta)=\sqrt {\frac m M}$. Now, let's "fold-out" the angle, which now has the reflective property:

It's no matter that this angle doesn't perfectly divide $\pi$ (Ahah! There it is!), we do note however that the resulting "folding-out" of an angle will make $\left \lceil \frac \pi \theta \right \rceil$ copies of the angle, and therefore the line $P(t)$ will intersect $\left \lfloor \frac \pi \theta \right \rfloor$ lines. This is why the number that comes out is related to $\pi$!

Now the punchline: we have that $\tan \theta = \sqrt \frac m M$, so $\theta = \arctan \sqrt {\frac m M}$. When we set $M = 100^N m$ for some $n\in \mathbb N$, we have that

$$\theta = \arctan \sqrt {\frac m {100^N m}} = \arctan \sqrt {100^{-N}}=\arctan {10^{-N}}$$

And therefore $P(t)$ will intersect $\left \lfloor \frac \pi {\arctan 10^{-N}} \right \rfloor$ lines, which is exactly the same as saying the system will have exactly $\left \lfloor \frac \pi {\arctan 10^{-N}} \right \rfloor$ collisions. Now Galperin argues that

$$\left \lfloor \frac \pi {\arctan 10^{-N}} \right \rfloor \approx \left \lfloor \frac \pi {10^{-N}} \right \rfloor= \left \lfloor \pi 10^N \right \rfloor$$

and proves that it is exactly equal for values of $N<100,000,000$. He conjectures that it also holds more generally, but this is not proven. Clearly, $\left \lfloor \pi 10^N \right \rfloor$ is precisely the first $N+1$ digits of $\pi$!

This approach seems to make $\pi$ naturally come out, but (unless I'm mistaken, which is certainly possible) you can make the number of bounces line up with any number you'd like. For instance, let $M=\frac {\pi^2} {e^2} 100^N m$. Then we would end up with the number of bounces equal to $\left \lfloor e 10^N \right \rfloor$, that is, the first $N+1$ digits of $e$. Of course the interesting thing about the "natural" choice of $M=100^N m$ is that it does not depend on already knowing the value of $\pi$ in order to calculate it. On the other hand, if you had a need to create a physical system which bounced a particular number of times, you now can do it.

In fact, if you're willing to accept an uglier mass relation between the two balls, you can get it exact without resorting to Galperin's approximation. Just let $M=\cot^2(10^{-N})m$.

An easier way to see this is as follows.

Conservation of energy implies that the point (u, v) where u and v are the velocities of the balls, lies on an ellipse at all times. Rescaling u and/or v results in a circle (hence the connection to pi). More explicitly, conservation of momentum shows that the point on this circle, after each collision between the balls, moves by equal increments. The number of increments before u changes sign (or equivalently, the point on the circle has rotated through an angle of pi/2), allows the calculation of pi. I won't do the calculation here, but it is quite straightforward.

I have just discovered that my proof was already known--I might have guessed as it is the obvious approach.

Note that, by replacing 10 by b, it is clear that this phenomenon is true in base b.

A minor point, that probably IS original, is that the Nth digit after the decimal point of pi is not guaranteed correct--it could be too small by 1. This is because the approximation is based on replacing the arctangent with the angle itself. However the digit will be incorrect only if the next 2N + log_10(48) digits are zero. I doubt if the such a sequence occurs in the decimal expansion of pi. If pi behaves like a random sequnce of digits, the conditional probability that this ever happens, given that it has not occurred in the first billion digits is extremely small (of the order of 10^-2billion).

There could be some base, other than 10, for which the Nth digit after the point is incorrect. If it does happen, it would almost certainly only happen when N is small.


Showing that the limit of $\sin (n) $ as $n $ approaches infinity doesn't exist In order to do that I want to find two subsequences $a_n$ and $b_n$ of $\sin (n) $ such that $\lim_{n \to +\infty} a_n $ and $\lim_{n \to +\infty} b_n$ are not the same number.
Hints?

Well if it approached some limit $L$, then $\sin^2 n$ approaches $L^2$ and $\cos^2 n$ approaches $1 - L^2$. For any integer $a$, we use the identity
$$\sin(n + a) = \sin n \cos a + \cos n \sin a$$
This gives
$$(\sin(n + a) - \sin n \cos a)^2 = \cos^2 n \sin^2 a$$
Now take the limit as $n$ goes to infinity, and you get
$$(L - L \cos a)^2 = (1 - L^2) \sin^2 a$$
Equivalently,
$${L^2 \over 1 - L^2} = \bigg({\sin a  \over 1 - \cos a}\bigg)^2$$
(Take reciprocals of this if $L^2 = 1$).

However, ${\displaystyle \bigg({\sin a  \over 1 - \cos a}\bigg)^2}$ is different for $a = 1$ and $2$ for example. Hence the limit cannot exist.

Hint:

1) $a_n=n\pi$, $n\in\mathbb{Z}$

2) $b_n=(2n+1)\frac{\pi}{2}$, $n\in\mathbb{Z}$

What can you say about the value of $\sin$ evaluated at each term of the sequences $a_n,b_n$?


Binary long division for polynomials in CRC computation I am trying to learn binary long division, and I am confused. An example in my book gives that $10011010000/1101 = 11111001$ plus a remainder of $101$, which doesn't make sense, since $1001$ is not divisible by $1101$, and so the first digit should be $0$. When I use an online calculator, it gives me a result of $01011110$, which makes more sense. Could anybody enlighten me, please?

To add some context, this is the polynomial long division involved in CRC computation. Maybe the numbers, which represent the coefficients of polynomials, aren't supposed to be converted to decimal? The number being divided had already been extended by the degree of the divisor.

$1232 = 94 * 13 + 10$

$10011010000 = 1011110 * 1101 + 1010$

The example is wrong and your online calculator is calculating correctly.


Speculating on the stock exchange Imagine you model each stock as a random walk (fractal) and also that you can buy and sell at any price. Suppose also that it 'walks' with the pace of 1.

If you buy, for example, 1000 shares of several companies, for \$100, and you sell every falling position that hits \$95, but keep every company until it reaches \$110. Wouldn't that be a winning strategy?

If by "'walks' with a pace of 1" means that in every time unit there is a 50% chance that the price rises by 1 and a 50% chance that it falls by 1, then this the classic gambler's ruin with a fair coin problem.

Imagine that for a particular stock you are playing a game where you win $\$1$ or lose $\$1$ with equal probability and that you start with $\$5$ and your opponent starts with $\$10$. If you play until one of you is ruined (i.e., wind up with nothing) then the probability that you will be ruined is $10(5+10)=2/3$ and consequently the probability that your opponent will be ruined is $5/(5+10)=1/3$. This means that your expected winnings will be
$$
E=\frac{2}{3}(-$5)+\frac{1}{3}($10) = $0
$$
Your expectation is zero and the cost of doing the trades guarantees that you'll lose money in the long run.

The particular numbers are unimportant here: no matter how much money you and your opponent have, the expectation will always be zero if the "coin" is fair.

As a stock investor, I do not think it is possible to model the stock prices. There are systems that work but even then, you may be looking at one good winner to cover 7-8 losers. The problem is that the stock exchange is not logical, it is emotional and all sorts of things, mainly unsubstantiated rumours, have a big instant effect.


Advantage of multi-objective optimization over single objective What are the advantages of multi-objective optimization over single objective?

I am specifically thinking about MO and SO in Genetic Algorithm.

I have surfed the net and found many articles talking about advantages of multi-objective optimization. Most of them are wordy and make no sense. They just keep writing about optimization of a system with multiple desirable criteria while I do not consider them as advantages of multiple objective optimization.

In multiple objective optimization we find a pareto-optimal solution set. And at the end, we apply weights to make a trade off between the criteria. This is exactly what single objective does from the beginning. So, what is the advantage of multi-objective optimization over single objective optimization.

Myth: Multi-objective optimization is for problem with multiple objectives while single objective optimization is for problems with single objective.

Truth: Both single and multiple objective optimizations can handle problems with multiple objectives.

If you are the decisionmaker yourself and know which objective is more important than others then MO does not give you any advantage over SO.However, I know two situations where MO does make sense:

You are the person that has to solve the problem but are not the decision maker. In this case, you might compute a (small) set of solutions and present all of them to the decision maker who then chooses one of these to implement.
Solving your problem takes prohibitive long time but the specific instance to solve is not known long before a decision has to be made. In this case, you can input all parameters which are known long enough and treat the remaining parameters as objectives. Then you compute a (possible large) set of solutions upfront and once all remaining parameters are fixed, you only need to search through this set for a best possible solution.

Both MOO and SOO may lead to the same result but there are conditions that MOO may be a better tool. If the functions of the optimization problem is not sensible to the decision maker, it is not clear how to chose the weighing factors. in these conditions, MOO performs much better which doesn't require weighing factors.


Are there proofs in math which can't be verified by a single person in their life span (say $80$ years)? Are there any proofs in math that can't reasonably be verified in one human life span? Say $80$ years?

How about the four color theorem? Or Kepler conjecture?

Can a person theoretically prove any of these by hand in the course of their life time? Where we can assume they live for at least $80$ years? If these results can be proven by hand by one person in less than $80$ years, then does anyone know of other theorems which can't?

Obviously all this is rather vague and depends on the particular person etc. Also one could trivially create problems that require one to spend an enormous amount of time, for example verify that the $46576575476547665^{\text{th}}$ digit of $\pi$ in base $82$ is equal to $53$. So I'm asking in regards to more well known problems, not ones that were arbitrarily constructed like this souly to make it hard for someone to do by hand.

Again I'll iterate I know this is all still some what vague, but I'm interested in to what extent if any automated proof checking  may become applicable in modern mathematics towards the near future. As I know for example automated proof checkers were used to verify the four color theorem and Kepler's conjecture, as no human was willing to go through and check them by hand (again I'm even not sure if its possible for a single human to verify either of these theorems all by hand).

I'll assume that various things such as "what a person can do in 80 years" and "what are allowed as proofs" are precisely defined and fixed.

With tongue in cheek, here is a way to get lots of such proofs. Since only finitely many proofs can be verified by someone within an 80 year period and there are infinitely many proofs, it follows that almost all proofs have the property that you're looking for.

Incidentally, I'd like to say that all you have to do is randomly choose a proof, because the probability will be $1$ that the proof you chose will be one of the proofs you are seeking. Unfortunately, there does not exist a uniform probability measure on a countably infinite sample space (the set of proofs).

Shinichi Mochizuki's proof for abc conjecture.

In August 2012, Shinichi Mochizuki released a series of four preprints on Inter-universal Teichmuller Theory which is then applied to prove several famous conjectures in number theory, including Szpiro's conjecture, the hyperbolic Vojta's conjecture and the abc conjecture.Mochizuki calls the theory on which this proof is based "Inter-universal Teichmüller theory (IUT)". The theory is radically different from any standard theory and goes well outside the scope of arithmetic geometry. It was developed over two decades with the last four IUT papers occupying the space of over 500 pages and using many of his prior published papers.


Shorthand notation for "increases" and "decreases" I want to write out something like:

"As $x$ increases, $y$ decreases."

Is there a standard symbolic notation for this, such as an up arrow and a down arrow?  (And if you can tell me how to write it in latex, that would be awesome, too).

Thanks!

In fact, the issue @Angada is referring has nothing to do with proportionality and $\propto$ symbol.

It appears that there is no enough widely accepted mathematical notation for that, although some complex notations are developed for fuzzy logic and fuzzy reasoning that unfortunately collide wit logic symbols.

Anyway I think an easy convention for this is a necessity if we want to do operative reasoning based on qualitative rules. I found this lack of mathematical notation so irritating that a couple of years ago, I began  to develop a convention that today use successfully for my own .

The model of expression I was run after was  that of Bachmann-Landau O notation, so the only thing I would had to do is to choose a symbol "S" with the meaning: "S(x) is any arbitrary monotonic increasing function of x".

Finding an intuitive significant for a increasing function is not an easy issue, and perhaps this is the reason why this convention has (as long as I know) been never established. If not,  try yourself to imagine a particular shape associated with increasing.

My first idea was  to use a crescent moon  as an icon for such functions, but unfortunately this wouldn't stand for people living in the southern hemisphere as moon appearance is reversed from there.

Then I abandoned the onomatopoeic approach   and ended up with an hybrid of the symbol of Derivative ("D") and the positive sign ("+"),  leading to "Đ" (crossed D ,D with stroke  or dyet) that also reminds for Directly (but not necessarily proporcional) related.

The symbols Đ and its lowercase: đ are part of the croatian and vietnamese alphabets, and have unicode in latin U+0110 , and U+0111 respectively.

Now I use;

$f(x)=Đ(x)$ or $f(x)\in \{Đ(x)\}$ for strictly monotonically increasing funcions,
and $f(x)=đ(x)$ or $f(x)\in \{đ(x)\}$ for (relaxed) monotonically increasing functions.

That is:

$f(x)=Đ(x) \equiv a>b \Longleftrightarrow f(a) >  f(b)$
$f(x)=đ(x) \equiv a>b \Longleftrightarrow f(a)\geq  f(b)$

Monotonic decreasing funcions are then esaily denoted as $f(x)=-đ(x)$, or (strictly) $f(x)=-Đ(x)$

Some properties are:

$Đ(Đ(x))=Đ(x)$
$Đ(đ(x))=đ(x)$
$f(x)=Đ(x)\Rightarrow f(x)=đ(x)$
$Đ(-x)=-Đ(x)$
$đ(-x)=-đ(x)$
$đ(k x^n)=Sign(k)  đ(x)$ , for any $n \ge 0$  $x=0$
$đ(k x^n)=Sign(k) Sign(n) đ(x)$ , for any interval that does'nt contain $x=0$
etc.

Of course, this could be combined with $\forall$ in order to restrict domains as for instance:
$$f(x)=\left\{\begin{matrix} Đ(x) & \forall x \in (-\infty,2) \\
-đ(x)) & \forall x \in [2,4] \\Đ(x) & \forall x \in (4,\infty)
\end{matrix}\right.$$

To respond to your question: in order to express: "As x increases, y decreases.", you could simply write   $y=-Đ(x)$.

From the point of view of verbalization, $y=Đ(x)$ would be read as: "$y$ is directly related to $x$" (but not: "$y$ is directly proportional to $x$"); and $y=- Đ(x)$ as "$y$ is inversely related to $x$".

Of course this is only a proposal, and by now, it works well only for my own use. If you find it useful, I encourage you to use it. If so, I would appreciate a lot if You'd let me know.

Otherwise: if you know or find a simple notation more established, shorter, or more intuitive, please let me know and I  will be very happy to adopt it.

This is an inverse proportionality and physicists (?) are fond of writing them as $x \propto \dfrac{1}{y}$.

Note that this is merely suggestive of the inverse variation and should not be interpreted as a proportionality.

As for $\LaTeX$, here is the code \varpropto for $\varpropto$ or $\propto$ for $\propto$.


For what values of $r$, $x^r$ has infinite slope at $x=0$? I'm learning calculus form MIT OCW 18.01SC. In session 23 (it's about linear approximation), prof computes linear approximation near $0$ of some basic functions.
$$\sin{x}, \cos{x}, e^x, \ln{(1+x)}, (1+x)^r$$
Why two of them are shifted by $1$? I can see why we can't compute linear approximation of $\ln{x}$ near $0$, but why not $x^r$? Prof says "If you try to graph $x^r$, you'll discover that sometimes the slope is infinite, and so forth".

$\frac{d}{dx}x^r=rx^{r-1}$, how this is sometimes infinite (for not infinite $r$ and $x$)?

Think of the domain of this function for various values of r, and their corresponding graphs. For instance, r=-1.

http://www.sagemath.org/calctut/pix-calctut/onesided05.png

Consider negative values of r. Let r=-1. Then $d \over dx$$x^{-1}$= $-x^{-2}$=$-1 \over x^2$.

What happens when x=0?


What is the Fourier transform of $f(t)=1$ or simply a constant? Wolfram alpha gives the answer to be

$$F(\omega)=\sqrt{2\pi}\delta(\omega)$$

Does that mean that the function is valued $\sqrt{2\pi}$ at all points in the frequency domain? I think this is reasonable because such function i.e. $f(t)=1$ in the time domain would be sum of all the harmonics of a sinusoid and hence would contain all the frequencies. Maybe no, the function isn't varying at all and hence the frequency is $0$. But then the Fourier transform should have been $\delta(0)$ instead of $\delta(\omega)$.

Someone please shed some light on this!

I think the clearest way to see this is by noting that we have (depending on your convention for the placement of $2 \pi$ in Fourier transforms) that
$$\mathcal{F}(\mathcal{F}(f(x))) = 2 \pi f(-x)$$
Taking the convention that
$$\tilde{f}(k) = \int_{-\infty}^\infty e^{ikx} f(x) \; dx$$
so
$$f(x) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{ikx} \tilde{f}(k) \; dk$$
we get
$$f(-x) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{-ikx} \tilde{f}(k) \; dk = \frac{1}{2\pi} \mathcal{F(\tilde{f}(k))} = \frac{1}{2\pi}\mathcal{F}(\mathcal{F}(f(x)))$$
Note we have
$$\mathcal{F}(\delta(x)) = \int_{-\infty}^\infty e^{ikx} \delta(x) \; dx = 1$$
So then
$$\mathcal{F}(1) = \mathcal{F}(\mathcal{F}(\delta(x))) = 2 \pi \delta(-x) = 2 \pi \delta(x)$$
For other constants, note by linearity we have
$$\mathcal{F}(c) = c \mathcal{F}(1) = 2 \pi c \delta(x)$$

Just use the definition. $F(w) = \int_{-\infty}^\infty e^{-iwt}dt =\delta(w)$ by how the dirac delta is defined. This is normalized to 1at zero and 0 otherwise.


Geometric Interpretation of Image What is the geometric Interpretation of the image of the following linear mapping?
               $f: \bf R^3 \to\bf R^3$  with $f(x,y,z)=f(x-y+z,-x+2y+z,y+2z)$

If you mean the entire image, it is a subspace of $\mathbb{R}^3$, which means it could be a point, a line through the origin, a plane through the origin, or the whole space. The answer is given by the rank of the matrix $\begin{bmatrix} 1 & -1 & 1 \\ -1 & 2 & 1 \\ 0 & 1 & 2 \end{bmatrix}$. If it is zero, the image is a point, if it is 1, the image is a line, etc. You can compute the rank by row reduction.

HINT:

Since z* = x* + y* , we have sphere mapped to a cylinder etc... (forgot its name)


Show that $2^{\nu(n)} \leq d(n) \leq 2^{\omega(n)}$ I trying to prove $2^{\nu(n)} \leq d(n) \leq 2^{\omega(n)}$, where $n \geq 1$,  $\nu(n) =$ number of distinct prime factors of $n$, $d(n)=$ total number of divisors of $n$ and $\omega(n)=$ number of prime factors counted with multiplicity.

$\textbf{My attempt:}$ Let $P$ be the set of all distinct prime factors of $n$. Then each divisor $m$ of $n$ can be identified with the subset of $P$ containing the primes in $m$. So, for each subset of $P$, there exists at least one divisor of $n$. So, the number of subsets of $P\leq d(n)$. That is, $2^{\nu(n)} \leq d(n)$. On the other hand, let $Q$ be the set of all prime factors of $n$ counted with proper multiplicities. (in fact, the repeated primes can be labelled differently to form the set $Q$). Then each divisor of $n$ corresponds to a subset of $Q$. Thus $d(n) \leq 2^{\omega(n)}$.

However, I think this proof is more of intuitive in nature and not so rigorous. Also, may be I am unable to put it in proper form. Is there a simpler or more rigorous way of proving this?

Hint: Consider the prime factorization of $n$, $n = {p_1}^{k_1}$${p_2}^{k_2}$...${p_j}^{k_j}$, where $p_1$, $p_2$,...$p_j$ are the distinct prime factors of $n$ & $k_i$ $\geq 1$ for all $i=1,2,...,j$

Can you figure out $ω(n)$, $d(n)$ & $ν(n)$ ? (hint: use the fact that these are multiplicative functions).

I'm sorry in advance that I don't know how to make certain things look neat such as subscripts and exponents. Can somebody please comment and let me know and how to. Thank you.

Let $n=({p_1}^{k_1})({p_2}^{k_2})\dots ({p_x}^{k_x})$  where $p_1,p_2,p_3\dots$ are primes. Clearly, $k_1,k_2 \dots$ are greater than or equal to $1$.

$d(n)=(k_1+1)(k_2+2)...(k_x+1)$

$v(n)=x$

$w(n)=k_1+k_2+k_3...k_x$

Also, it is helpful to realize that if the inequality is expanded then each part has the same number of terms.

Now we can look at the first part of the inequality.

$\underbrace{2 \cdot 2 \cdots 2 }_\text {x times} \leq (k_1+1)(k_2+2)...(k_x+1)$

Since each $k_a$ is equal to or greater than $1$ every term on the second part of the inequality is greater than or equal to two, and since there are $x$ terms on both sides the second part of the inequality must be equal to or greater than the first part.

Before we look at the second inequality it is worth realizing that $2^k \geq k+1$ for all $k$.

Now lets look at the second part of the inequality.

$(k_1+1)(k_2+2)\dots (k_x+1) \leq (2^k_1)(2^k_2)\dots(2^k_x)$

If you compare both sides term by term and use the fact that $2^k>=k+1$ for all k then it is evident that each term on the right side is greater than the left side.

The original inequality is now proven.


Find all coefficients of $n$ degree polynomial whose roots are $1, 2, 3, ..........n$ Given a polynomial of degree $n$, i.e:
$$p(x) = \prod_{i=1}^n x-i=(x-1)\cdot(x-2)\cdot\ldots\cdot(x-n)$$

What will be the coefficient of each power of $x$?

I figured out the coefficient of $x^{n-k}$ is the sum of the product of all possible combinations of $k$ integers from $1$ to $n$, but I can't find a simple formula or method to compute each of these coefficients efficiently.

Can we use Inverse FFT in this case?

The roots of this polynomial are $\alpha_1 = 1, \alpha_2 = 2, ..., \alpha_n = n$. Let polynomial $P$ to be like this: $P(x) = x^n + a_{n-1}x^{n-1}+...+a_1x + a_0$ (it is clear that the coefficient $a_n = 1$).

By the Vieta's formulas we know that $\sigma_k = (-1)^k \cdot \frac{a_{n-k}}{a_n}$ (for $1 \leq k \leq n$), where $\sigma_k$ is the sum of all $k-$permutations of roots $\alpha_1,...,\alpha_n$. So, since $a_n = 1$ we get that $a_k = (-1)^{n-k}\cdot\sigma_{n-k}$ for $1\leq k \leq n -1$

I think that this is a simpler approach than Inverse FFT (if it's possible to do it that way).


Using trigonometric identities to prove following equation... I'm stuck on this problem:

  If
  $$a\sin^2x + b\cos^2x = c$$
  $$b\sin^2y + a\cos^2y = d$$
  $$a\tan x=b\tan y$$
  then prove that
  $$\frac{a^2}{b^2} = \frac{(d-a)(c-a)}{(b-c)(b-d)}$$

I think it's wrong.

Try $a=b=c=d=x=y=1.$


What is the range of convergence of $\sum_{n=0}^{\infty} {(-1)}^n\binom{1/2}{n}\frac{1}{2n+3}.$ I was fiddling with the integral
$$\int_0^1 x^2\sqrt{1-x^2} \ dx $$
and I expanded the term under square root using a binomial series. Integrating, I got the result
$$\sum_{n=0}^{\infty} {(-1)}^n\binom{1/2}{n}\frac{x^{2n+3}}{2n+3}\Biggr|_0^1.$$

I would like to know if evaluating this series at the upper limit 1 would make it converge, since binomial series has a convergence of $|x|<1$? Also if it does converge what is the range of convergence?

Since the coefficient of $x^{2n+3}$ is asymptotic to a constant times $n^{-3/2}$, and $\sum_n n^{-3/2}$ converges, this does converge for $|x| \le 1$.  The answer, btw, is $\pi/16$.

From $\int_0^1 x^2\sqrt{1-x^2} \ dx$

Using the following substitution:

$t=x^2, \frac {dx}{dt}=\frac{1}{2}\frac{1}{\sqrt(t)}$ , at $x=0 \rightarrow t=0$ and at $x=1 \rightarrow t=1$

Applying the definition and identity of $\beta$ function:

$\frac{1}{2}\int\limits_0^1 t^\frac{1}{2}(1-t)^\frac{1}{2}dt= \frac{1}{2} \beta(\frac{3}{2},\frac{3}{2})=\frac{1}{2}\frac{\Gamma(\frac{3}{2})^2}{\Gamma(3)}= \frac{\pi}{16}$


Showing the inequality $f(x)=x^{2(1-x)}+(1-x)^{2x}\leq 1$ for $0<x<1$ My proof is not really natural but I think it works.
We want to show 1:

Let $0<x<1$ such that  then we have :
$$f(x)=x^{2(1-x)}+(1-x)^{2x}\leq 1$$

Case $0<x\leq 0.25$
The proof of this case is due to user Batominovski:
By Bernoulli's inequality we have:
$$(1-x)^{2x}\leq 1-2x^2\quad (1)$$
We have also:
$$x^{2-2x}\leq 2x^2\quad (2)$$
Summing $(1)$ and $(2)$ we get the desired inequality.
Case $0.25\leq x\leq 0.49$:
I shall prove it later but we have :

Let $0.25\leq x\leq 0.49$ then we have :

$$p(x)=2^{2x}(1-x)x^{2}2\geq x^{2(1-x)}$$
And
$$h(x)=\cos^2\Big(x\frac{\pi}{2}\Big)(1+\frac{195}{100}(1-x)(0.5-x)x^2)\geq (1-x)^{2x}\quad (3)$$
With my work we have :
$$f^2(x)\leq \Big(\cos^2\Big(x\frac{\pi}{2}\Big)(1+\frac{195}{100}(1-x)(0.5-x)x^2)\Big)^2+\Big(2^{2x}(1-x)x^{2}2\Big)^2+4^{1.95}(x(1-x))^{2.95}2\leq 1$$
To show it we can use power series see here
Case $0.49\leq x \leq 0.5$
On the domain $[0.49,0.51]$ the function $g(x)=x^{2(1-x)}$ is concave or:
$$g''(x)\leq 0\quad (4)$$
So we have by Jensen's inequality:
$$g(x)+g(1-x)\leq 2g(0.5)=1$$
As $f(x)=f(1-x)$ it's proved for $0.5\leq x<1$
Question:
How to show $(3)$?
Thanks in advance!
1 Vasile Cirtoaje, "Proofs of three open inequalities with power-exponential functions",
The Journal of Nonlinear Sciences and its Applications (2011), Volume: 4, Issue: 2, page 130-137.
https://eudml.org/doc/223938

Actually there is an easier way. Lets search for the maximum of that function. If its outside of $0$ and $1$ than it should be at the boundaries. (But this is not the case)

Differentiate the function.
Set the derivative equal to zero.
Solve for $x$.


tensor product of Hopf algebras and coalgebraic structure on the dual algebra I am learning Hopf algebras, and there are two questions as follows:

Is the tensor product of two Hopf algebras still a Hopf algebra?

Let $A$ be an infinite dimensional algebra. Is the dual $A^*$ a coalgebra? (In the case of the dimension of $A$ is finite the answer is positive.)

If they are, where can I find the proofs?
Thanks for your help.

Given two Hopf algebras $\big(H,\mu_H, \eta_H, \Delta_H, \varepsilon_H, S_H\big)$, $\big(G,\mu_G, \eta_G, \Delta_G,\varepsilon_G, S_G\big)$ over a field $k$, we can form the tensor product Hopf algebra $\big(H\otimes G,\mu_{H\otimes G}, \eta_{H\otimes G}, \Delta_{H\otimes G},\varepsilon_{H\otimes G},S_{H\otimes G}\big)$ through:
$$\mu_{H\otimes G}=(\mu_H\otimes\mu_G)\circ(Id\otimes\tau\otimes Id) :H\otimes G \otimes H \otimes G \rightarrow H\otimes G
\\ \\ \\
\eta_{H\otimes G}=(\eta_H\otimes\eta_G)\circ\phi^{-1}:k\rightarrow H\otimes G
\\ \\ \\
\Delta_{H\otimes G}=(Id\otimes\tau\otimes Id)\circ(\Delta_H\otimes\Delta_G):H\otimes G\rightarrow H\otimes G \otimes H \otimes G , \\ \\ \\
 \varepsilon_{H\otimes G}=\phi \circ (\varepsilon_H\otimes\varepsilon_G):H\otimes G\rightarrow k \\ \\ \\
S_{H\otimes G}=S_H\otimes S_G:H\otimes G\rightarrow H\otimes G
$$
where $Id$ is the identity map and $\phi:k\otimes k\stackrel{\cong}{\rightarrow} k$ the natural isomorphism. In other words, the standard tensor product of algebras is combined with the tensor product coalgebra structure and the antipode is given by the tensor product of antipodes. This is a standard construction and can be found in almost any introductory Hopf algebra text.

Now, regarding your second question, as you correctly mention in your post, it holds if the algebra $(A,\mu,\eta)$ is finite dimensional: then $(A^*,\lambda^{-1}\circ\mu^*,\psi^{-1}\circ\eta^*)$ is a finite dimensional coalgebra. Here we denote with $\psi$, the natural v.s. isomorphism
$k\cong k^*$ defined by $\psi(\kappa)=f$ with $f(1)=\kappa$ and $\psi^{-1}(f)=f(1)$ and with $\lambda:A^*\otimes A^*\rightarrow(A\otimes A)^*$ the linear v.s. injection, defined by $\lambda(f\otimes g)(u\otimes v)=f(u)g(v)\in k$. Not that, in case $A$ is of finite $k$-dimension then $\lambda$ is a vector space isomorphism.

However, if $A$ is infinite dimensional, then the above construction meets the following problem: Since $\lambda$ is injective but not an isomorphism, $\lambda^{-1}$ is not well defined: $A^*\otimes A^*$ is a proper subspace of $(A\otimes A)^*$ and thus the image of $\mu^*:A^*\rightarrow(A\otimes A)^*$ may not lie inside $A^*\otimes A^*$. The problem is treated in the following manner: Instead of considering the whole of the (linear) dual space $A^*$, which seems too "big" to carry the construction, we consider its linear subspace $A^\circ$ defined as
$$
A^\circ=\{f\in A^*\big|\ker f \textrm{ contains an ideal of finite codimension} \}=\{f\in A^*\big|f(I)=0 \textrm{ for some ideal } I \textrm{ of } A \textrm{ such that } \dim A/I<+\infty\}
$$
$A^\circ$ and its elements  can be characterized by the equivalent conditions of the following proposition:
Proposition: Let $A$ be a $k$-algebra and $f\in A^*$. Then the following conditions are equivalent:

$f\in A^\circ$
$\mu^*(f)\in\lambda(A^*\otimes A^*)$
$\mu^*(f)\in\lambda(A^\circ\otimes A^\circ)$
$A\rightharpoonup f$ is finite dimensional
$f\leftharpoonup A$ is finite dimensional
$A\rightharpoonup f\leftharpoonup A$ is finite dimensional

$A^\circ$ is called the restricted dual of $A$ and in fact it is the largest subspace $V$ of $A^*$ such that $\mu^*(V)\subseteq V\otimes V$.

Now, it can be shown that $A^\circ$ together with the maps  $(\lambda^{-1}\circ\mu^*)$, $(\psi^{-1}\circ\eta^*)$ as before, is a $k$-coalgebra. More details on the proofs of this construction can be found in various sources. See for example,

Dascalescu, Nastasescu, Raianu, "Hopf algebras, an introduction"
S. Montgomery, "Hopf algebras and their actions on rings"
Moss E. Sweedler, "Hopf algebras"
Kassel, "Quantum groups"
etc

Related question: You can see darij grinberg's answer for 1).

The answer for 2) is negative. For example, you can see here

  while the K-dual of every coalgebra is an algebra, the K-dual of an infinite-dimensional algebra need not be a coalgebra.

ON THE APPLICATIONS OF COALGEBRAS TO GROUP ALGEBRAS

  Unfortunately, the dual of an infinite dimensional algebra is too large to canonically take a coalgebra structure.


Find the parametric equation for the curve of intersection of two surfaces I need to parametrize the curve of intersection of 2 surfaces, where the surfaces are:
$$x^2+y^2 = z$$  $$z^3 = 5x+y$$
I follow the standard step of substitution as below $$(x^2+y^2)^3 = 5x+y$$ However, this will give me a mass after I open the cube on the left side. Could anyone tell me what is the right approach for this problem?

substitute $x = r\cos(\theta) \ and \ y = r\sin(\theta) $
which yeilds $$ r^3 = 5\cos(\theta)+\sin(\theta) $$
the parametric equation of the curve is $$ \Big(\cos(\theta)\big(5\cos(\theta)+\sin(\theta)\big)^\frac{1}{3},\sin(\theta)\big(5\cos(\theta)+\sin(\theta)\big)^{\frac{1}{3}},\big(5\cos(\theta)+\sin(\theta)\big)^\frac{2}{3}\Big) $$


When doing separable differential equations, why does Integrating a constant give a variable? ie. $\int -3 d = -3y$ I get that it is abit handwavy at this point in my math journey, but why does integrating a constant like this: $\int -3 dy$ result in -3y?

I get that when integrating a variable like y gives $\frac {y^2}{2}$ (below), but don't understand what to do when we are not integrating a variable like the above.

Okay, so if we have a function $y=ax$, calculating the derivative we get: $\frac{dy}{dx}=a$ so the derivative of a linear function is a constant. Now since integrating is the "opposite operation" of taking the derivative it makes sense to say that:
$$\int\frac{dy}{dx}dx=y+C=ax+C$$
Notice how we have an integration constant here though

Remember the power rule for integration:  $$\int 3x^a\, dx = \frac{3(x^{a+1})}{a+1} +C$$
In your case, $a=0$:  $$\int 3x^0\,dx = \frac {3x^1}{1} +C = 3x + C.$$


Conditional Probability Equivalence I am reading this paper here: http://www.utm.utoronto.ca/~weisber3/articles/SobervBJPS3.SP.pdf which claims on page 10 that $p(E \wedge [H_x \wedge \sim R]) + p(E \wedge [H_x \wedge E])= p(H_x \wedge E)$. I am having trouble seeing this equivalence.

Any help is appreciated; thank you in advance.

It doesn't.  That step is erroneous, as is the one prior (the sum-rule is only valid when the two events are mutually exclusive, which they clearly are not).

However, both are also unnecessary, as: $[E\wedge [H_x\wedge \neg R]] \vee [E\wedge [H_x\wedge E]]) = E\wedge H_x$ , thus you immediately have the desired result.

$$p([E\wedge [H_x\wedge \neg R]] \vee [E\wedge [H_x\wedge E]])
\\[2ex] = p([[E\wedge H_x]\wedge \neg R]\vee [[E\wedge H_x]\wedge E])
\\[2ex] = p([[E\wedge H_x]\wedge \neg R]\vee [E\wedge H_x])
\\[2ex] = p([E\wedge H_x])
$$


Show $\bigcup _{i\varepsilon \mathbb{Z}}\left[ i,i+1\right) =\mathbb{R}$ My proof idea. $(\Rightarrow)$. Let $x\in\bigcup _{i\in \mathbb{Z}}\left[ i,i+1\right)$. We need to show that $x\in\left( -\infty ,\infty \right)$. Actually, is it clear?

$\left( \Leftarrow \right)$. Let $x\in\left( -\infty ,\infty \right)$. We need to show that $x\in\bigcup _{i\in \mathbb{Z}}\left[ i,i+1\right)$. How can I show?

I think the left inclusion is trivial since $[i, i+1) \subset \Bbb R \ \forall i \in \Bbb Z$. Now, for the other inclusion, pick $x \in \Bbb R$, and assume $x > 0$. Then by the Archimedean principle there exists some $n \in \Bbb N : n > x$. Furthermore, since $\Bbb N$ is well-ordered, we can pick the least such $n$. Then $n > x \ge n - 1 \implies x \in [n-1, n)$, and obviously $n-1 \in \Bbb Z$. For $x < 0$ a very similar argument can be made taking $n > - x$ (be careful with half open intervals extremes, though). The case $x=0$ is trivial.

$\left( \Leftarrow \right)$. $\forall x \in (-\infty, +\infty)$, always $\exists i \in \mathbb Z$, such that $i \le x < i+1$, thus $x \in [i, i+1)$, thus $x \in \cup_{i\in \mathbb Z}[i,i+1)$. Thus $(-\infty, +\infty) \subset \cup_{i\in \mathbb Z}[i,i+1)$


Representing a Plane as a Vector Equation A general way of representing a plane in $3D$ is, $ax + by+ cd = d$. But in my Calculus 3 course my instructor represented a plane with a vector equation $ a\vec{i} + b\vec{j} +c\vec{k} = d$
Is that a valid representation of a Plane? I mean, of course a plane can represented with vectors like in,  $\vec{r}.\vec{d} = p$ where $\vec{d}$ is a unit normal vector perpendicular to the plane and $p$ is the perpendicular distance of the plane from the origin. In this case, both the left hand side as well as the right hand side is a scalar whereas in the representation used by my instructor the left hand side is a vector while the right hand side is a scalar. It appears a bit disturbing to me.
I can't reach my instructor now. In all the quiz questions, she represented planes as $ a\vec{i} + b\vec{j} +c\vec{k} = d$. Can anyone confirm is it a correct representation? If yes, How one side of the equation is a vector and the other is a scalar?
Thanks a lot in advance

I think the equation your instructor used is incorrect, for the sum of vectors is necessarily a vector.
However we can say the following:
Let $\mathbf{u}$ and $\mathbf{v}$ be any two non-zero, non-collinear vectors in $\mathbb{R}^3$. Then the equation of the plane determined by $\mathbf{u}$ and $\mathbf{v}$ is
$$
\mathbf{x} = \lambda \mathbf{u} + \nu \mathbf{v},
$$
where $\lambda, \mu \in \mathbb{R}$ are the variables.


find the period of a trigonometric function I've found the period of this trigonometric function,
$$y=\sin^n(x)+\cos^n(x)$$.

when n （$n\neq2$）is odd, the period is $2\pi$,

when n is even, the period is $\frac{\pi}{2}$.

but how to proof it?

Consider the periods of both terms, the period of the sum is the least common multiple of both.

You could express the sine and cosine using complex exponentials, often that makes trigonometric identities trivial.


A way to directly see that the interior angles of triangle sum to $180^\circ$? I'm looking for a way to look at a triangle, and perhaps visualize a few extra lines, and be able to see that the interior angles sum to $180^\circ$.

I can visualize that supplementary angles sum to $180^\circ$. I'd like to be able to see the interior angle sum similarly...

I can see that the exterior angles must sum to $360^\circ$, because if you walked around the perimeter, you would turn around exactly once (though I can tell this is true, I don't really see it). I also saw a proof on KA, where the exterior angles were superimposed, to show they summed to $360^{\circ}$ (though I'm not 100% comfortable with this one).

Finally, for $a$, $b$, and $c$ exterior angles $a+b+c=360$:

\begin{align}
 (180-a) + (180-b) + (180-c) & = 3\times 180 - (a+b+c) \\
                             & = 3\times 180 - 360 \\
                             & = 180 \\
\end{align}

But I find this algebra hard to see visually/geometrically. Is there proof that enables one to directly see that the interior angles of triangle sum to $180^\circ$?

A couple of secondary questions:

am I visually deficient in my ability to imagine?
or, am I asking too much of a proof, that I be able to see it, and that beimg able to tell that it is true should be enough...?

Since that fact about the angle sum is equivalent to the parallel postulate, any visualization is likely to include a pair of parallel lines. Here's one from wikipedia:

Imagine a triangle $ABC$ with two of the sides, say $AB$ and $BC$, rigid but meeting at the joint $B$, which is movable in the plane of the triangle, like a hinge sort of (also, you may fix one of these rigid sides, say $BC$, and rotate the other). Let the side $AC$ be made of a sufficiently elastic material (at least so that $AC\leq AB+BC$ ).

If you play with this imaginary object, you should soon see that as you increase $A \hat B C$, the other two angles eventually diminish (certainly for $A \hat B C>π/2$). Indeed, you may extend $\angle ABC$ as wide as you please close to $\pi$, so that the two rigid sides approach a single rectilinear segment; then the angle between them becomes arbitrarily close to $π$ and the other two angles are vanishing; in the limiting (degenerate) position the side $AC$ coincides with the segment formed by the rigid sides.

If you can already see that any point in this sequence of transformations of $\triangle ABC$ the sum of angles is conserved, then it is clear why for any triangle this sum is always $π$.


Closure of a connected subset of $\mathbb{R}$ is connected? Prove or disprove: The closure of a connected set in $\mathbb{R}$ is always connected.

Response: I don't really have a grasp on this conceptually, but here's an attempt.
Proof: Let $X$ be a connected subset of $R$, and let $Z$ be the closure of $X$. Suppose (to get a contradiction) that $Z$ is disconnected. That means there are two nonempty open subsets of $R$ covering $Z$, call them $U$ and $V$, such that $U\cap Z$ and $V\cap Z$ are nonempty while $U\cap V\cap Z$ is empty.

Then $U$ and $V$ cover $X$ also. And since $U\cap V\cap X \subseteq U\cap V\cap Z$, then $U\cap V\cap X$ is empty. Pick points $u\in U\cap Z$ and $v\in V\cap Z$ (which exist since those sets are nonempty). That means $u$ is in the closure of $X$ and $U$ is an open neighborhood of $u$, so $U\cap X$ is nonempty. Similarly, $V\cap X$ is nonempty.

But then the previous paragraph shows that $X$ is disconnected, which is a contradiction, so $Z$ must have actually been connected.

Do you know any definition or property of connectivity? A good one is that if $X$ is connected then any continuous $f : X \to \{0, 1\}$ must be constant. Say $\forall x \in X$, $f(x) = 1$. How can you extend this to the closure of $X$ without destroying continuity? Note that this statement is not restricted to $\mathbb{R}$.

You can actually show that a connected subset in R must be singleton or an interval. In both the cases, closure is also connected.


Probability of selecting positive integer with odd number of digits (Grinstead and Snell Ch 1.2 Q 28 (d)) If $A$ is the set of all positive integers with an odd number of digits, let $A\,(N)$ be the number of elements of $A$ which are less than or equal to $N$. Show that $P\,(A) = \lim_{N\to\infty} A\,(N)/N$ does not exist.

My rusty knowledge of limits says $P\,(A) = 0$, since the denominator dominates. Any help?

If you look at the beginning of the even set with $N_k=10^{2k+1}$ where k is an Integer then the last $10^{2k+1}-10^{2k}$ have been elements with an odd number of digits. Therefore we get $P(N_k) = A(N_k)/N_k \geq \frac{10^{2k+1}-10^{2k}}{10^{2k+1}}  $ which goes to 0.9 for k to infinity. But we can also look at the $M_k=10^{2k}$. For this we can show that $P(M_k)$ goes to 0.1 . Therefore the limit can‘t converge since we found 2 different  accumulation points.


Notion of divisibility in a proof about number fields I don't understand a point in the proof of the theorem 9.4 in Anthony Scholl's Number Fields lecture notes.

Theorem: Let $K$ be a number field, $\mathcal{O}_K$ its ring of integers and $\alpha \in \mathcal{O}_K$. Then $\text{Tr}_{K/\mathbb{Q}}(\alpha^p) \equiv \text{Tr}_{K/\mathbb{Q}}(\alpha) \pmod p$.
Proof: By Fermat's little theorem $\text{Tr}_{K/\mathbb{Q}}(\alpha)^p \equiv \text{Tr}_{K/\mathbb{Q}}(\alpha) \pmod p$ and
$$\text{Tr}_{K/\mathbb{Q}}(\alpha)^p - \text{Tr}_{K/\mathbb{Q}}(\alpha^p) = \left( \sum_{i=1}^n \sigma_i(\alpha) \right)^p-\left( \sum_{i=1}^n \sigma_i(\alpha)^p \right) = \sum_{0 <k_1  \ldots  k_n<p \\ k_1+ \cdots + k_n=p} \frac{p!}{k_1! \cdots k_n!}\sigma_1(\alpha)^{k_1} \cdots \sigma_n(\alpha)^{k_n}$$
by the generalised binomial theorem where $\sigma_i$ are the complex embedings of $K$. Each of the terms is divisible by $p$, so the expression is $0 \pmod p$.

What I don't understand is how this can possibly be a valid argument when the conjugates $\sigma_i(\alpha)$ are complex numbers and divisibility is a notion in $\mathbb{Z}$. What does dividing mean here? Is it that they belong to the ideal $(p)$?

You're right in that one has to be a bit more precise here. Hint: Write down the division of the RHS by $p$ in ${\mathbb C}$ explicitly, and convince yourself that it's (a) invariant under all complex embeddings (or rather their extensions to ${\mathbb C}$), and (b) still an algebraic integer. Deduce.

Divisibility is a notion in any ring $(A,+,\cdot)$. For $a,b\in A$, we say $a\mid b$ iff there exists $d\in A$ with $b=da$. You know this for $A=\Bbb Z$, here it is used in $\mathcal O_K$. (Of course, in a field such as $K$, the concept would we almost worthless, as we can ake $d=\frac ab$ except when $b=0$).


How to prove that $\nexists x\in\mathbb{N}_0(0>x)$ This question is about ZFC set theory. Here I'll define a few things (sets) and then I have a question related to these sets I couldn't prove. Here are the definitions:

Let say that $0$ is symbol for $\{\}$ (empty set), then $1$ is symbol for $\{0\}$, then $2$ is symbol for $\{1\}$, etc. Now you can write every non-negative integer as a set.

Here is the first definition. Let $\mathbb{N}_0$ be the set for which

$$\forall a\left(a\in\mathbb{N}_0\Leftrightarrow a=0\lor\exists b\in\mathbb{N}_0\left(a=\{b\}\right)\right)$$

It may not be obvious, but it can be written as $\mathbb{N}_0=\{0,1,2,\dots\}$. However, the above definition is more formal. Now, lets define relation between non-negative integers:

$$\forall a,b\in\mathbb{N}_0\left(a>b\Leftrightarrow a=\{b\}\lor\exists c\in\mathbb{N}_0\left(a>c\land c>b\right)\right)$$

So, basically, it says that for all non-negative integers $a$ and $b$ we have that $a>b$ iff $a$ is successor of $b$ or there exists some $c$ which is between $a$ and $b$ (and less than $a$).

These was definitions. Now, lets try to prove some statements using these definitions (and of course ZFC axioms). The first statement I came up with is

  Prove that $\exists x\in\mathbb{N}_0(x>0)$.

Here is my solution (I'm posting this just to show an example what kind of detailed proof I'm looking for the final question):

  Lets expand the definition of non-negative integers relation symbol $>$
  $$\exists x\in\mathbb{N}_0\left(x=\{0\}\lor\exists a\in\mathbb{N}_0(x>a\land a>0)\right)$$
  Now, notice that at least one of the stamenets $x=\{0\}$ and $\exists a\in\mathbb{N}_0(x>a\land a>0)$ must be true iff the main statement is true. So, lets consider the first statement $x=\{0\}$. Because $\{0\}=1$, the statement we want to prove is true for $x=1$. However, we need to show that $1\in\mathbb{N}_0$. So, using the definition:
  $$1\in\mathbb{N}_0\Leftrightarrow1=0\lor\exists a\in\mathbb{N}_0\left(1=\{a\}\right)$$
  The $1=0$ is obviously false. Now, we need to show that there is some $a$ for which $1=\{a\}$. Because $1$ is non-empty set, then there must be some set $a$ for which $1=\{a\}$ (in this case $a=0$). The proof is completed.

This was not too hard. However, I came up with the following statement (which I couldn't prove yet):

  Prove that $\nexists x\in\mathbb{N}_0(0>x)$.

How to prove it? I tried to apply similar technique like in the previous proof, but I ended up in a recursive chain. I also tried to apply contradiction approach. I supposed that the statement is false, but I again ended up in a recursive chain. Any ideas?

Sorry for bad english. Thank you in advance.

P.S. Please notice that this question is not like standard questions, so maybe not everyone will find it helpful. This is more about recreational math or art of proofs. It is related to core math, ZFC axioms and if you don't find this question helpful, then just move on. Thanks.

Note that given $a,b \in \mathbb{N}_0$, with $\mathbb{N}_0$ as defined, you have $a>b$ if and only if $b \in \mathrm{trcl}(a)$, where $\mathrm{trcl}(a)$ denotes the transitive closure of $a$. This can be proved easily by $\in$-induction, after which point the result is immediate, since then $0>a$ if and only if $a \in \mathrm{trcl}(\varnothing)=\varnothing$.

Note $2=\{0,1\}$, not $2=\{1\}$.

If you assume there is indeed an $x$ such that $x<0$, well the definition of $<$ is "element of".  For example, $1<2$ because $1$ is an element of $\{0,1\}$.  Also, 0<2 because 0 is an element of {0,1}.

Note $2<2$ is false because of the axiom of foundation.

Back to x<0 now.  Note: $0$ is the empty set and $x<0$ iff $x$ is an element of $0$.  The latter proposition is false for all $x$.  Contradiction implying it is not the case that $x<0$.


Why does $1+2+3+\cdots = -\frac{1}{12}$? $\displaystyle\sum_{n=1}^\infty \frac{1}{n^s}$ only converges to $\zeta(s)$ if $\text{Re}(s) > 1$.

Why should analytically continuing to $\zeta(-1)$ give the right answer?

there are many ways to see that your result is the right one. What does the right one mean?
It means that whenever such a sum appears anywhere in physics - I explicitly emphasize that not just in string theory, also in experimentally doable measurements of the Casimir force (between parallel metals resulting from quantized standing electromagnetic waves in between) - and one knows that the result is finite, the only possible finite part of the result that may be consistent with other symmetries of the problem (and that is actually confirmed experimentally whenever it is possible) is equal to $-1/12$.
It's another widespread misconception (see all the incorrect comments right below your question) that the zeta-function regularization is the only way how to calculate the proper value. Let me show a completely different calculation - one that is a homework exercise in Joe Polchinski's "String Theory" textbook.
Exponential regulator method
Add an exponentially decreasing regulator to make the sum convergent - so that the sum becomes
$$ S = \sum_{n=1}^{\infty} n e^{-\epsilon n} $$
Note that this is not equivalent to generalizing the sum to the zeta-function. In the zeta-function, the $n$ is the base that is exponentiated to the $s$th power. Here, the regulator has $n$ in the exponent. Obviously, the original sum of natural numbers is obtained in the $\epsilon\to 0$ limit of the formula for $S$. In physics, $\epsilon$ would be viewed as a kind of "minimum distance" that can be resolved.
The sum above may be exactly evaluated and the result is (use Mathematica if you don't want to do it yourself, but you can do it yourself)
$$ S = \frac{e^\epsilon}{(e^\epsilon-1)^2} $$
We will only need some Laurent expansion around $\epsilon = 0$.
$$ S = \frac{1+\epsilon+\epsilon^2/2 + O(\epsilon^3)}{(\epsilon+\epsilon^2/2+\epsilon^3/6+O(\epsilon^4))^2} $$
We have
$$ S = \frac{1}{\epsilon^2} \frac{1+\epsilon+\epsilon^2/2+O(\epsilon^3)}{(1+\epsilon/2+\epsilon^2/6+O(\epsilon^3))^2} $$
You see that the $1/\epsilon^2$ leading divergence survives and the next subleading term cancels. The resulting expansion may be calculated with this Mathematica command
1/epsilon^2 * Series[epsilon^2 Sum[n Exp[-n epsilon], {n, 1, Infinity}], {epsilon, 0, 5}]
and the result is
$$ \frac{1}{\epsilon^2} - \frac{1}{12} + \frac{\epsilon^2}{240} + O(\epsilon^4) $$
In the $\epsilon\to 0$ limit we were interested in, the $\epsilon^2/240$ term as well as the smaller ones go to zero and may be erased. The leading divergence $1/\epsilon^2$ may be and must be canceled by a local counterterm - a vacuum energy term. This is true for the Casimir effect in electromagnetism (in this case, the cancelled pole may be interpreted as the sum of the zero-point energies in the case that no metals were bounding the region), zero-point energies in string theory, and everywhere else. The cancellation of the leading divergence is needed for physics to be finite - but one may guarantee that the counterterm won't affect the finite term, $-1/12$, which is the correct result of the sum.
In physics applications, $\epsilon$ would be dimensionful and its different powers are sharply separated and may be treated individually. That's why the local counterterms may eliminate the leading divergence but don't affect the finite part. That's also why you couldn't have used a more complex regulator, like $\exp(-(\epsilon+\epsilon^2)n)$.
There are many other, apparently inequivalent ways to compute the right value of the sum. It is not just the zeta function.
Euler's method
Let me present one more, slightly less modern, method that was used by Leonhard Euler to calculate that the sum of natural numbers is $-1/12$. It's of course a bit more heuristic but his heuristic approach showed that he had a good intuition and the derivation could be turned into a modern physics derivation, too.
We will work with two sums,
$$ S = 1+2+3+4+5+\dots, \quad T = 1-2+3-4+5-\dots $$
Extrapolating the geometric and similar sums to the divergent (and, in this case, marginally divergent) domain of values of $x$, the expression $T$ may be summed according to the Taylor expansion
$$ \frac{1}{(1+x)^2} = 1 - 2x + 3x^2 -4x^3 + \dots $$
Substitute $x=1$ to see that $T=+1/4$. The value of $S$ is easily calculated now:
$$ T = (1+2+3+\dots) - 2\times (2+4+6+\dots) = (1+2+3+\dots) (1 - 4) = -3S$$
so $S=-T/3=-1/12$.
A zeta-function calculation
A somewhat unusual calculation of $\zeta(-1)=-1/12$ of mine may be found in the Pictures of Yellows Roses, a Czech student journal. The website no longer works, although a working snapshot of the original website is still available through the WebArchive (see this link). A 2014 English text with the same evaluation at the end can be found at The Reference Frame.
The comments were in Czech but the equations represent bulk of the language that really matters, so the Czech comments shouldn't be a problem. A new argument (subscript) $s$ is added to the zeta function. The new function is the old zeta function for $s=0$ and for $s=1$, it only differs by one. We Taylor expand around $s=0$ to get to $s=1$ and we find out that only a finite number of terms survives if the main argument $x$ is a non-positive integer. The resulting recursive relations for the zeta function allow us to compute the values of the zeta-function at integers smaller than $1$, and prove that the function vanishes at negative even values of $x$.

Here is a useful place for Euler's transform and the Dirichlet eta function:

$$\zeta(s)=\sum_{n=1}^\infty\frac1{n^s}$$

$$\eta(s)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^s}$$

$$\zeta(s)-\eta(s)=\sum_{n=1}^\infty\frac{1+(-1)^n}{n^s}=\sum_{n=1}^\infty\frac2{(2n)^s}=2^{1-s}\sum_{n=1}^\infty\frac1{n^s}=2^{1-s}\zeta(s)$$

$$\zeta(s)-\eta(s)=2^{1-s}\zeta(s)\implies (1-2^{1-s})\zeta(s)=\eta(s)$$

  $$\zeta(s)=\frac1{1-2^{1-s}}\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^s}$$

After a quick application of Euler's transform, we get a nice analytic continuation to the entire complex plane.

$${\small E_1}\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^s}=\sum_{n=0}^\infty\left[\frac1{2^{n+1}}\sum_{k=0}^n\binom nk\frac{(-1)^k}{(k+1)^s}\right]$$

Finally giving us

  $$\zeta(s)=\frac1{1-2^{1-s}}\sum_{n=0}^\infty\left[\frac1{2^{n+1}}\sum_{k=0}^n\binom nk\frac{(-1)^k}{(k+1)^s}\right]$$

And at $s=-1$,

$$\zeta(-1)=-\frac1{12}$$


What is the difference between $E[X\mid Y]$ vs $E[X\mid Y=y]$ and some of the properties of $E[X \mid Y]$? I was trying to understand both intuitively and rigorously what the difference between $E[X\mid Y]$ vs $E[X\mid Y=y]$.

Let me tell you first the things that do make sense to me. $E[X\mid Y=y]$ makes sense to me (I think). For me it means the value that we expect $X$ to have on average given that the event $Y=y$ was observed and it has deterministic value:

$$E[X\mid Y=y] = \sum_x{xp_{X\mid Y}(x\mid y)} = \mu_{X\mid Y=y}$$

that can be computed like any other expectation (where the notation $\mu_{X\mid Y=y}$ denotes the actual real number that $E[X\mid Y=y]$ takes). $E[X\mid Y=y]$ is a function of y. Given $Y=y$, the conditional expectation will always be $\mu_{X\mid Y=y}$, which is governed by what specific y was observed.

However, $E[X\mid Y]$ makes less sense to me. I have read something similar to the following:

  Notice now that Y is a random variable this time. Therefore, $E[X\mid Y]$ is a random variable.

That sort of makes sense to me because, if $Y$ is random then the expected value of $X$ has to be random too. In other words, $Y$ being random, then consequently $E[X\mid Y]$ is random too. The statement $E[X|Y]$ is a r.v. and is the exact statement I would like to understand more precisely. I feel I understand less and I would like to address it.

I feel if I really understood this concept of what $E[X\mid Y]$ really means, I should be able to answer the following questions:

1) If $E[X\mid Y]$ is random, then, what are the possible values it can take? Can it only take the values $\mu_{X\mid Y=y}$ for $y \in Y$? Say if $Y=\{1,2,3\}$ and $U_{X,Y} = \{\mu_{X\mid Y=1}= 11, \mu_{X\mid Y=2} = 22, \mu_{X\mid Y=3} = 33 \}$. Then is there any chance that $E[X\mid Y] = 123$?

2) Is the distribution of $E[X|Y]$ over $U_{X,Y}$ or over $Y$?

3) What is the probability distribution of $E[X\mid Y]$ if we have all the information we need about the distributions of $p_X(x)$ and $p_Y(y)$? Is it just the same as $p_Y(y)$? Is there a closed/specific formula for it?

4) When one is asked to find the distribution of $E[X\mid Y]$ are we asked to find $Pr[E[X\mid Y=y]]$ or $Pr[E[X\mid Y] = \mu]$? Is there a difference between the two? Is one nonsense while the other one is a valid probability distribution?

5) If we were to sketch the probability density function for $E[X \mid Y]$, would the horizontal axis be $y$ or $\mu$ ? i.e. would the probability density be a function of $y$ or of $E[X\mid Y] = \mu$? i.e. would $p_{E[X\mid Y]}(k)$ be a function of $y=Y$ or $E[X\mid Y]$?

6) Related to the above two question, it seems to me that writing an explicit formula for $E[X\mid Y=y]$ is easy, while for $E[X\mid Y]$ it is not (or at least for me). Is the formula $E[X\mid Y] = \sum_x{xp_{X\mid Y}(x\mid Y)}$? I would guess it is but for me its a very strange equation because we are conditioning on a random variable, or we are saying given $Y$, but $Y$ is random so its really not given. Therefore, I can't seem to find an expression for it that makes sense to me.

Basically, it is not clear to me what $E[X\mid Y]$ means, because I don't know what its valid outcomes are, what its distribution is (in relation to $p_X(x), p_Y(y)$ or $p_{X,Y}(x,y)$ or anything) nor can I write an explicit formula for it that makes sense to me. I can't even decide if $p_{E[X \mid Y]}(k)$ is a function of y or $\mu=E[X|Y]$.

If $\varphi(y)=E[X\mid Y=y\,]$ is the non-random function of $y$, then $E[X\mid Y\,]$ is defined to be the random variable $\varphi(Y)$.

The easiest way is to write it as follows:

$$f(y) = E[X \mid Y = y] = \sum_{x \in \mathcal X} p_{X \mid Y}(x \mid y) x $$

$y$ is simply just a value. $f$ is a function that can be evaluated. It's usually an integral or a sum usually. Given a value of $y$ (any that you want ) it produces a real number. It is simply a deterministic function. It is parametrized by the distributions of random variables $X,Y$ but there is no sampling going on. Once we know the value of $y$ we want we can compute the conditional expectation in question.


Show that $2 - \sqrt{2}$ is irrational I suppose $2 - \sqrt{2} $ is rational.
so $$2- \sqrt{2} = {a/b} $$
where a,b are integers and gcd(a,b) = 1.

$$\text{Step 1. } 2 = (a/b)^2 \text{ //squared both sides }$$
$$\text{Step 2. } 2b^2 = a^2 \text{ //We see $a^2$ is even }$$
$$\text{Step 3. }2b^2 = (2k)^2$$
Step 3 since $a^2$ is even and so $a$ is even too. Also, k is an integer.
$$\text{Step 4. }b^2 = (2k)^2 \text{ //We see $b^2$ is even }$$
Step 4 since $b^2$ is even and so $b$ is even too.

We see $ a$ and $b$ are even. We see $gcd(a,b) \neq 1$

Is this proof correct ?

For the sake of Contradiction let us assume that  $2-\sqrt{2} \in \Bbb Q$ , say $2-\sqrt{2}=r \in \Bbb Q \implies \sqrt{2}= 2-r$ but since both $2,r \in \Bbb Q$ and $(\Bbb Q,+)$ is a subgroup of $(\Bbb R,+)$ (i.e. basically addition of two rational numbers is rational) it follows that $\sqrt{2} \in \Bbb Q$ , a contradiction!

Yet another proof: $0<2-\sqrt{2}<1$, so $(2-\sqrt{2})^n\to 0$.  Note that $(2-\sqrt{2})^n=a_n-b_n\sqrt{2}$ for some integers $a_n,b_n$ by binomial theorem, so $$0<(2-\sqrt{2})^n=(a_n-2b_n)+b_n(2-\sqrt{2}).\tag{1}$$
If $2-\sqrt{2}$ were rational, say $=p/q$, $p,q\in\mathbb{N}$, then by equation (1) $(2-\sqrt{2})^n$ must be at least $1/q$, contradicting $(2-\sqrt{2})^n\to 0$.


"Unique Limit" involving continuous functions and Hausdorff space I want to prove the following:

  Suppose $(X,\tau)$ is a topological space. $X$ has the property that whenever $f:(Y,\sigma)\to (X,\tau)$ is a continuous function defined on a topological space $Y$, and $b$ a limit point of $Y$, and $f(x)\to l$ as $x\to b$, then $l$ is unique. Show that $X$ is Hausdorff. Show also that the converse is true.

I find this rather difficult because it has a universal quantifier on $Y$ and $f$, which are not elements of any topological spaces. I end up partially proving the converse only.

Partial Proof of converse. Suppose $f(x)\to l,f(x)\to m, l\neq m$, but $X$ is Hausdorff. Then there exists open neighborhoods $U,V$ of $l,m$ such that $U\cap V=\emptyset$. Since $f$ is continuous, the preimages $f^{-1}(U), f^{-1}(V)$ are open sets. I guess, but cannot prove that $f^{-1}(U), f^{-1}(V)$ are disjoint neighborhoods of $b$, which is not possible since $b$ is a limit point.

But that logic doesn't seem to work for the other direction of the argument.

So how can I prove the whole statement?

EDIT I have now proved the converse part. How to prove the forwards direction?

So formally, $X(,\tau)$ has the property "unique function limits":

  UFL: For all spaces $(Y, \sigma)$ and all continuous $f:(Y, \sigma) \to (Y,\tau)$, and all $b \in Y$ that are limit points of $Y$ and all $l, l' \in X$  we have that $$(\lim_{x \to b} f(x) = l) \land (\lim_{x \to b} f(x)=l') \implies l=l'$$

And you want to show $$\textbf{UFL} \iff X \text{ Hausdorff }$$

You already started on the reverse: supposing the Hausdorff property and the setting as needed for UFL, so we have $Y$, $b$ and $\lim_{x \to b} f(x) = l$ and $\lim_{x \to b} f(x)=l'$ for some $l,l' \in Y$; Assume for a contradiction that $l \neq l'$ and take disjoint neighbourhoods $U$ of $l$ and $V$ of $l'$ such that $U \cap V = \emptyset$. Applying the limit to $l$ property we get a neighbourhood $W_1$ of $b$ such that $$f[W_1 \setminus \{b\}] \subseteq U$$ and from the limit with value $l'$ we get a neighbourhood $W_2$ of $b$ such that
$$f[W_2 \setminus \{b\}] \subseteq V$$

which implies that $W_1 \cap W_2$ is an open neighbourhood of $b$ such that

$$f[(W_1 \cap W_2) \setminus \{b\}] \subseteq U \cap V=\emptyset \text{ so }W_1 \cap W_2=\{b\} $$ so that $b \notin Y'$, a contradiction to the setup for UFL, so indeed $l=l'$ and so function limits are unique if they exist. Note that continuity of $f$ is not really needed, just the limits must exist.

For the forward direction: suppose UFL holds and we want to show $X$ is Hausdorff. So suppose not: then there are two distinct points $p \neq q $ of $X$ such that

$$\forall U \in \mathcal{O}(p): \forall V \in \mathcal{O}(q): U \cap V \neq \emptyset\tag{1}$$ where $\mathcal{O}(x)$ is the set of open neighbourhoods of $x$ for $x \in X$. Now try to define a "custom made" $Y$ and $f$ to contradict UFL, using the structure of neighbourhoods intersecting in $X$ (filters are an option, or nets/directed sets, if you know them), so that $\lim_{x \to b} f(x)=p$ and also $\lim_{x \to b} f(x)=q$ for some limit point  $b$ in that $Y$. This would be the required contradiction and $X$ must be Hausdorff after all.

You've almost got the converse, you need only observe if $f^{-1}(U) \cap f^{-1}(V) \neq \emptyset$, then there exists $y \in (f^{-1}(U) \cap f^{-1}(V))\subset Y$. What is $f(y)$? It must reside in $U$ and $V$, but that isn't possible by construction of $U$ and $V$, as we constructed them to be disjoint, hence $U \cap V = \emptyset$. (We are using the definition of function to conclude that $f^{-1}(U) \cap f^{-1}(V) = \emptyset$, that the inverse image of disjoint sets is disjoint is a property of functions themselves; try proving it yourself, it's not actually that hard).

Edit:
I'm having trouble with the forwards direction, but one thing that's really useful is that it is enough to show that $(X, \tau )$ has unique limits (when they exist) and that it's first countable. Those two properties together imply Hausdorff. Showing uniqueness of limits, when they exist, is easy. Let $(Y,\sigma)=(X,\tau)$ and $f$ be the identity map on $X$. Now if we have a limit point $b$ and sequence $x_k\rightarrow b$, then we have $x_k=f(x_k)\rightarrow l=b$. f is trivially continuous in this case, and the hypothesis implies that the limit must be unique. (I'm abusing the fact that the hypothesis is universally quantified, to pick on a helpful example of a $Y$ and $f$ that is continuous). This works, because the map is trivial, any sequence in $Y$ that has a limit is also a sequence in $X$ (namely itself), that also has a limit (which is the same point), and the hypothesis says that with these sequences and continuous f, we can conclude that the limit is unique. This leaves only establishing that $X$ is first countable.


Why can't we just say that $\infty-\infty$ equals zero? Let be $\lim\limits_{x\to \infty}x=A$  and   $\lim\limits_{y\to
 \infty}y=B$.

Can be $A-B=0$?

If the answer is "no" , why?

And my other example:

  $\displaystyle\int_{-\infty}^{\infty} \dfrac{x}{1+x^2}dx$

Why does this not equal zero?

I know this is an improper integral, and I know how solve this. But I don't think this makes sense.

We have a positive area A and a negative area B (with $A=-B$).

Why can't we just say that

  $\displaystyle\int_{-\infty}^{\infty} \dfrac{x}{1+x^2}dx=0$

And my last example:

  $\displaystyle\sum_{n=0}^\infty n^2=A, \quad \displaystyle\sum_{b=0}^\infty b^2=B$

Here can we say that $A-B=0$? If "no", then why?

Hint. Two examples.

Observe that, as $x \to \infty$,
$$
(x^2+x)-x^2=x \to \infty
$$ whereas, as $x \to \infty$,
$$
\left(x+\frac1x\right)-x=\frac1x \to 0.
$$ In each case you have an indeterminate form $\infty-\infty$.

The simplest answer to your question, without either sequences or limits, is that there exist distinct infinite numbers. The most commonly used extension of the real numbers to an ordered field is the field of hyperreals. If $H$ is an infinite hyperreal, then so is $2H$ or for example $H^2$ or $\sqrt{H}$, not to speak of $H+1$ :-)


Calculus problem with negative infinity $$\lim_{x\to-\infty} x(\sqrt{x²+1}-x)$$

Can someone simplify this and explain the steps. I'm having problems getting rid of sqrt.

No need to get rid of the square root.  As $x\to -\infty$, both $\sqrt{x^2+1}$ and $-x$ go to $+\infty$, and thus the product goes to $-\infty$.

At infinity sqrt( x^2+1) tends to x. So we have for the limitation process something like Lim x (x - x) as x tends to zero. I think it is an undefined condition and says nothing ...


A gamma function inequality I would like to prove $$\frac{\Gamma(n+\frac{1}{2})}{\Gamma(n+1)} \le \frac{1}{\sqrt{n}}$$ for all natural $n \ge 1$. The inequality does seem to be true numerically, but the proof eludes me.

Here's a direct proof:

The gamma function satisfies $\Gamma(n+1) = n \Gamma(n)$,
so the function $g(x) = \ln \Gamma(x)$ satisfies
$g(n+1) = \ln n + g(n)$.
In other words, the line segment between the points $(n,g(n))$
and $(n+1,g(n+1))$ has slope $\ln n$.

Moreover, $g$ is known to be a convex function for $x>0$
(cf. the Bohr–Mollerup theorem),
so the point $(n+1/2, g(n+1/2))$ lies below that line segment:
$g(n+1/2) < g(n+1) - \frac12 \ln n$.
Exponentiating both sides gives $\Gamma(n+1/2) < \Gamma(n+1) / \sqrt{n}$,
as desired.

\begin{eqnarray*}
\Gamma(x+1)>\frac{\frac{(900\gamma^2+73\pi^2)x}{900\gamma}+\frac{73}{100}}{\frac{\pi^2 x}{9\gamma}+1}\quad \textbf{[1]}
\end{eqnarray*}
i want to prove this inquality on $(0,1) is increasing
i defined it as

i defined it as

\begin{eqnarray*}
w(x)=\frac{\log[\Gamma(x+1)(\frac{\pi^2 x}{9\gamma}+1)]}{\log[\frac{(900\gamma^2+73\pi^2)x}{900\gamma}+\frac{73}{100}]} \quad [2]
\end{eqnarray*}
but [2] not increasing to prove it !
what  it need to get it increasing


How to solve the system $x_{t+1}=-x_t-2y_t+3t-2$, $y_{t+1}=-2x_t+2y_t+t+1$ I have the following system of recurrence equations:

  $$x_{t+1}=-x_t-2y_t+3t-2\qquad y_{t+1}=-2x_t+2y_t+t+1$$

I write this in matrix-vector form:
$$r_{t}=Ar_{t-1}+b\cdot (t-1)+c$$
I repeatedly apply the recurrence to get:
$$r_t=A^tr_0+\left (\sum_{i=1}^t(t-i)A^{i-1}\right)b+\left (\sum_{i=1}^tA^{i-1}\right)c$$

  Does this count as a "solution"?
  Is there a way to further simplify it?

$$\begin{bmatrix}x_{t+1} \\ y_{t+1} \end{bmatrix}=\begin{bmatrix}-1 & -2 \\ -2 & 2 \end{bmatrix}\begin{bmatrix}x_{t} \\ y_{t} \end{bmatrix}+\begin{bmatrix}3t-2 \\ t+1 \end{bmatrix}$$
Now
$$\begin{bmatrix}-1 & -2 \\ -2 & 2 \end{bmatrix}=RJR^{-1}=\begin{bmatrix}2 & -1 \\ 1 & 2 \end{bmatrix}\begin{bmatrix}-2 & 0 \\ 0 & 3 \end{bmatrix}\begin{bmatrix}\frac{2}{5} & \frac{1}{5} \\ -\frac{1}{5} & \frac{2}{5} \end{bmatrix}$$
Let
$$\begin{bmatrix}x_{t} \\ y_{t} \end{bmatrix}=\begin{bmatrix}2 & -1 \\ 1 & 2 \end{bmatrix}\begin{bmatrix}a_{t} \\ b_{t} \end{bmatrix}, \ \forall{t}$$
Thus
$$\begin{bmatrix}2 & -1 \\ 1 & 2 \end{bmatrix}\begin{bmatrix}a_{t+1} \\ b_{t+1} \end{bmatrix}=\begin{bmatrix}2 & -1 \\ 1 & 2 \end{bmatrix}\begin{bmatrix}-2 & 0 \\ 0 & 3 \end{bmatrix}\begin{bmatrix}a_{t} \\ b_{t} \end{bmatrix}+\begin{bmatrix}3t-2 \\ t+1 \end{bmatrix}$$
Now multiply by
$$\begin{bmatrix}\frac{2}{5} & \frac{1}{5} \\ -\frac{1}{5} & \frac{2}{5} \end{bmatrix}$$
To give
$$\begin{bmatrix}a_{t+1} \\ b_{t+1} \end{bmatrix}=\begin{bmatrix}-2 & 0 \\ 0 & 3 \end{bmatrix}\begin{bmatrix}a_{t} \\ b_{t} \end{bmatrix}+\begin{bmatrix}\frac{6}{5}t-\frac{3}{5} \\ -\frac{t}{5}+\frac{4}{5} \end{bmatrix}$$
So you have decoupled the equations, they have a solution
$$\begin{bmatrix}a_{t} \\ b_{t} \end{bmatrix}=\begin{bmatrix}-c_{1}2^{t} \\ c_{2}3^{t} \end{bmatrix}+\begin{bmatrix}\frac{2}{5}t-\frac{1}{3} \\ \frac{1}{4}t-\frac{9}{40} \end{bmatrix}$$
Thus
$$\begin{bmatrix}x_{t} \\ y_{t} \end{bmatrix}=\begin{bmatrix}2 & -1 \\ 1 & 2 \end{bmatrix}\begin{bmatrix}a_{t} \\ b_{t} \end{bmatrix}=\begin{bmatrix}\frac{9t}{20}-\frac{107}{120} \\ \frac{9t}{10}-\frac{47}{40} \end{bmatrix}-\begin{bmatrix}c_{1}2^{t+1}+c_{2}3^{t} \\ c_{1}2^{t} -2c_{2}3^{t} \end{bmatrix}$$
Should be correct If i haven't messed the multiplications)


Find the follwing limit if $f$ is differentiable If $f$ is differentiable at $x=0$ and $f'(0) =2$, then find $$\lim_{x\to 0} \frac{2f(x) - 2f(2x)}{x}.$$

I don't know how to start, a hint will be appreciated.