Can the integral $\int_0^\infty\ \frac{1}{\sqrt{x^2+a^2}} \cdot \frac{1}{1+e^\sqrt{x^2+a^2}}\,dx$ be solved analytically? I'm trying to solve the following integrals:

$$I_1(a)=\int_0^\infty\ \frac{1}{\sqrt{x^2+a^2}} \cdot \frac{1}{1+e^\sqrt{x^2+a^2}}\,dx$$
and
$$I_2(a)=\int_0^\infty\ \frac{1}{1+e^\sqrt{x^2+a^2}}\,dx.$$

They seem similar to the two integrals solved in this post, so I wonder if there is something in the Bessel function family that will do the trick. But so far, substitutions like $x=a \sinh(t)$ and $u=\sqrt{x^2+a^2}$ are not giving me integrals that can be manipulated into one of the Bessel function forms on Wikipedia or Wolfram.

The $I_2(a)$ integral can't be reduced to a say, a Bessel function with an argument that is a function of $a.$ I say this because I worked out the asymptotics to be
$$  I_2(a) \sim \sqrt{\pi a/2} \  \frac{1}{1+e^a}, \quad a \to \infty $$
$$ I_2(a) \sim \log(2) + \frac{a^2}{4}(\kappa + \log{a}) +..., a \to 0$$
where
$$ \kappa =  \gamma - 1/2 + \frac{5}{6} \  \log{2} + 12\ \zeta'(-1)  = -1.3302154...$$

Putting the constant question aside, $I_2(a)$ has a functional form of $a \ K_1(a) $ for $a \to 0$ and $a K_{1/2}(a) $ for $a \to \infty,$ where $a$ is the McDonald function, a fundamental obstacle to seeking a simple form.  It may be possible to get something as an infinite series of Bessel functions, but the integral is nice, so why bother.

Added later, sketch of how to find asymptotic formulas:

For $a \to \infty, \exp(\sqrt{x^2+a^2)} \sim \exp(a) \ \exp(x^2/(2a)). $ Then

$$ I_2(a) \sim \int_0^\infty dx \sum_{k=1}^\infty (-1)^{k-1}\Big(e^{-a} e^{-x^2/(2a)} \Big)^k =  -\sum_{k=1}^\infty \big(-e^{-a} \big)^k  \int_0^\infty dx \  e^{-x^2 k /(2a)} $$
$$ \quad = -\sqrt{\pi a/2} \sum_{k=1}^\infty \frac{ \big(-e^{-a} \big)^k}{\sqrt{k}}  = -\sqrt{\pi a/2} \, \text{Li}_{1/2}(-e^{-a}) $$
The extra complexity of the polylogarithm is probably not worth the effort; in fact $e^{-a}$ is sufficient, but numerically a better appx to  $ -\text{Li}_{1/2}(-e^{-a})$ is $1/(1+e^{-a}).$  By happenstance, the $e^{-a}$ form is better than the polylogarithm form for $1<a<3.$

For $a \to 0, \exp(\sqrt{x^2+a^2)} \sim \exp(x) \ \exp(a^2/(2x))  $ Then

$$ I_2(a) \sim \int_0^\infty dx \sum_{k=1}^\infty (-1)^{k-1}\Big(e^{-x}
 \ \exp(-a^2/(2x)) \Big)^k =\sqrt{2}a \sum_{k=1}^\infty (-1)^{k-1}K_1(\sqrt{2} k a)$$ where again $\Sigma$ and $\int$ has been interchanged and the integral done in closed form in terms of the McDonald function.  Now expand the McDonald function as $a \to 0.$ Then

$$ \sqrt{2} a K_1(\sqrt{2}ka) \sim \frac{1}{k} + k a^2\Big(\gamma - 1/2 +
\big(\log{a} + \log{(k/\sqrt{2}}) \big) \Big) + ... $$
The first sum, $\sum_{k=1}^\infty (-1)^{k-1}/k = \log{2},$ is easy. The second sum does not converge, but it can be made to have meaning within the context of zeta function regularization.  I don't want to spend the time trying to justify the arguments.  Instead, I numerically calculated to see that the result gave good agreement.

yes and very easy
let u=sqrt(x^2+a^2)
then the integral becomes
int(2/(1+exp(u))
multiply numerator and denominator by exp(-u).
the derivates of denominator appear in numerator. and so on


Proof by strong induction $F(0) = 0$

$F(1) = 0$

$F(n) = 1 + F(n-2)$ , when $n > 1$

Prove by strong induction that $F(n)=n$ div $2$ for $n\geq0$

$a$ div $b$ is a integer division. So $10$ div $2 = 5$, $11$ div $2 = 5$, $12$ div $2=6$ and so on.

Did the base cases and they hold? Got stuck in the stepcase. I don't know if I am doing it right. This is what I have done so far.

$F(n+1)= 1 + F(n+1-2)$

$F(n+1) = (n+1)$ div $2$

$1 + (n+1)$ div $2 - 2= n$ div $2$

Which we had to prove

Have another idea.

$F(n+1) = 1 + F(n+1 -2)$

$F(n+1-2) = F(n) (I.H) k>1$

$1+F(n) = 1+n div 2$

Which we had to prove.
Does my proof hold or is it wrong? I don't want the whole answer, just a nod in the right direction. Any tips would be appriciated

This is what I have done so far.

  $F(n+1)= 1 + F(n+1-2)$

  $F(n+1) = (n+1)$ div $2$

  $1 + (n+1)$ div $2 - 2= n$ div $2$

  Which we had to prove.

The problem with this is that it's really not a proof. A proof consists of a sequence of sentences, each of which is true and includes (typically) some justification for its truth. The "statement/reason" format can help with this: on the left, you write statements, on the right, you write reasons. So converting what you've provided to that format gives something like what follows, where I've used "indenting" instead of left/right to provide statements and reasons.

$F(n+1)= 1 + F(n+1-2)$

i. replacement of $n$ by $n+1$ in the third hypothesis.
$F(n+1) = (n+1)$ div $2$

ii. [No justification given]
$1 + (n+1)$ div $2 - 2= n$ div $2$

iii. This statement is actually false for $n = 2$, so ...

As you can see, when you start having to give reasons for each statement, proofs get a little trickier. Even the very first statement is incorrect, because in the third hypothesis, there's a restriction, which says that the formula works for $n > 1$. You have to include this in your statement, and then, having substituted $n+1$ for $n$, you have to modify the condition $n > 1$ to $n+1 > 1$, i.e., $n > 0$. So your proof would begin

$F(n+1)= 1 + F(n+1-2)$ for $n > 0$.

i. replacement of $n$ by $n+1$ in the third hypothesis.

The second statement is what you're trying to prove, so it's no surprise that after a single statement, you cannot yet justify it.

In short: that first proof needs a lot of work.

Using the induction step you get that $F(n-1)=(n-1)$div$2$. Then use commun fraction properties to compute $F(n+1)$. As I see it, your proof doesn't make sence.


For 2nd order linear homogenous ODE, what is the effect of changing the value of the IC for $\mathrm{d}y/\mathrm{d}x$? Given a linear homogenous second-order ODE

$$f_2(x)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + f_1(x)\frac{\mathrm{d}y}{\mathrm{d}x} + f_0(x)y = 0$$

with initial conditions

$$y(x_0)=0$$
$$\frac{\mathrm{d}y}{\mathrm{d}x}(x_0) = c$$

for some $x_0$, what is the effect of changing $c$?

The background for this is that I have been told that all solutions may be obtained from the solution for $c=1$. (In a context where multiplying the solution by a constant wouldn't make any difference, in case that is relevant.)

I can see that we would want to exclude $c=0$, because this would lead to the trivial solution $\forall x, y(x)=0$, but what does it mean to say that all solutions may be obtained with $y'(x_0)=1$?

Let $y$ be a solution of the differential equation
$$f_2(x)\frac{d^2 y}{dx^2}+f_1(x)\frac{dy}{dx}+f_0(x)y=0,$$ satisfying the initial conditions $y(x_0)=0$ and $\frac{dy}{dx}(x_0)=c\ne 0$.

Let $z=\frac{y}{c}$. Then $\frac{dz}{dx}=\frac{1}{c}\frac{dy}{dx}$, and $\frac{dz^2}{dx^2}=\frac{1}{c}\frac{dy^2}{dx^2}$, and therefore $f_2(x)\frac{d^2 z}{dx^2}+f_1(x)\frac{dz}{dx} +f_0(x)z=0$, and $z(x_0)=0$, $\frac{dz}{dx}(x_0)=1$.

Conversely, if
$$f_2(x)\frac{d^2 z}{dx^2}+f_1(x)\frac{dz}{dx} +f_0(x)z=0,$$
 with $z(x_0)=0$ and $\frac{dz}{dx}(x_0)=1$, then $y=cz$ is a solution of your original equation, with $y(x_0)=0$ and $\frac{dy}{dx}(x_0)=c$.

The "conversely" part says that all solutions with $c\ne 0$ can be obtained by scaling a solution which has $c=1$.

A shift in initial conditions sometimes represents a certain translation of our solution. For instance, increasing $c$ may represent a horizontal shift to the left and vice-versa-


Maximum number of path for acyclic graph with start and end node Say given an acyclic graph with n nodes,
which includes a starting node s0 and ending node e0,
what is the maximum number of path from s0 to e0?

If you want the maximum number for any graph (of some size), that's easy. Take the maximal graph with N vertices $v_1...v_n$ where $v_1$ is $s_o$ and $v_n$ is $e_o$, with edges $v_i \rightarrow v_j$ for all $1 \le i < j \le n$. Now any sequence $v_1v_{p_1}...v_{p_k}v_n$ where $1 < p_1 < p_2 < ... < p_k < n$ is a path from $s_0$ to $e_0$. Or to put it another way, any subset of the vertices which includes both $s_0$ and $e_0$ uniquely defines a path (by sorting the vertices into index order); there are $2^{n-2}$ such subsets.

If you have a specific graph, then you can use the following procedure to compute the number of paths:

1) Topologically sort the vertices. The first vertex in the topological sort must be $s_0$ and the last one must be $e_0$ (unless I misunderstand your question; if so, just use the portion of the topological sort between the start and end vertex.)

2) Associate a count with each vertex. Set the count associated with $e_0$ to be 1.

3) For each vertex in the topological sort in reverse order, starting with the vertex just before $e_0$, set its count to the sum the counts of all of its neighbour vertices.

4) The count associated with $s_0$ is the total number of possible paths.

You don't actually have to do the topological sort. You can just depth-first-search the tree starting with $s_0$, computing the counts recursively.

I suppose the graph is connected, otherwise there might be no path at all.
In general there are infinitely many paths starting from s0 and ending in e0 if you not require them to be simple. If you require them to be simple there is only one.


An intuitive understanding of the equation $Var(X)=E(X^2)-E(X)^2$ I know the equation $Var(X)=E(X^2)-E(X)^2$ and its proof. After reading the textbook of mine, I found that this equation has been used in a lot of place. I want to know whether there is an intuitive understanding of this equation?

If you add $E[X]^2$ to both sides it looks like the Pythagorean theorem. In fact, $|\mathrm{E}[x]|$ and $\sqrt{\mathrm{Var}[X]}$ are the lengths of two legs of a right triangle.

The right triangle exists in an abstract space called a Hilbert Space. The Hilbert space in question is the set of all variables with finite second moment ($\mathrm{E}[X^2]$) on a probability space. To be specific, let the space be $\mathcal{H}= \{X: \int_\Omega X^2 \ d\mathbb{P} < \infty\}$ and let us extend the notion of length on $\mathbb{R}^2$ to $\mathcal{H}$ by defining the norm of $X\in\mathcal{H}$ by $\|X\|=\left(\int_\Omega X^2 \ d\mathbb{P}\right)^{1/2}$. We can also extend our notion of orthogonality in $\mathbb{R}^2$ by saying that $X$ is orthogonal to $Y$ in $\mathcal{H}$ if $\int_\Omega X  Y \ d\mathbb{P}=0$.

Although this space is very abstract, there are theorems that allow us to extend our geometric intuition about the Euclidean plane to $\mathcal{H}$.
The important theorem here is Parseval's Identity which generalizes the Pythagorean Theorem. If $X$ and $Y$ are orthogonal in $\mathcal{H}$ and $Z=X+Y$, then $\|X\|^2+\|Y\|^2 = \|Z\|^2$.

In the context of the question, if we define $Y=X-E[X]$ (more about this choice of $Y$ at the end) and $Z=E[X]$, that is, $Z$ is the function on $\Omega$ that maps every element to $E[X]$, then the norm of $Y$ is
\begin{align}
 \|Y\| = \left(\int_\Omega (X-E[X])(X-E[X]) \ d\mathbb{P}\right)^{1/2} = \sqrt{\mathrm{Var}[X]},
\end{align} (which by assumption is finite and verifies that $Y\in\mathcal{H}$). If it exists, then the norm of $Z$ is
\begin{align}
  \|Z\|=\left(\int_\Omega E[X] \ d\mathbb{P}\right)^{1/2},
\end{align}
and since $E[X]$ is a constant,
\begin{align}
  \|Z\|&=\left(E[X]^2 \ \int_\Omega 1_\Omega \ d\mathbb{P} \right)^{1/2} \\
   &= |E[X]|\cdot 1 \\
   &= |E[X]|.
\end{align}
Thus, $\|Z\|=|E[X]|$ and $Z\in\mathcal{H}$.

In this case $Y+Z=X$, so if we can verify that $Y$ and $Z$ are orthogonal then we can apply Parseval's Theorem. The calculation is simple:
\begin{align}
   \int_\Omega Y Z \ d\mathbb{P} &=
        \int_\Omega (X-E[X])E[X] \ d\mathbb{P} \\
          &= E[X]\int_\Omega X \ d\mathbb{P} - E[X]^2\int_\Omega \ d\mathbb{P} \\
         &= E[X]^2-E[X]^2 \\
         &= 0.
\end{align}

Plugging these in yields
\begin{align*}
   \|X\|^2 &= \|Y\|^2 + \|Z\|^2 \\
   \mathrm{E}[X^2] &= \mathrm{Var}[X] + \mathrm{E}[X]^2.
\end{align*}

To add a little more intuition to the choice of $Y=X-E[X]$, $Y$ can be thought of as a projection of $X$ onto the subspace of $\mathcal{H}$ consisting of all variables with 0 mean. It is analogous to projecting a point in $\mathbb{R}^2$ onto a line. In Hilbert Space this line is more like a very high dimensional plane cutting through $\mathcal{H}$.

So in summary, there is an intuitive understanding and it comes from thinking of the set of variables as a geometric space.

This has a physical interpretation if $X$ is the speed of a moving particle parametrized by the interval $[0,1]$.  Here the expected value of $X$ is the length of the trajectory while the expected value of $X^2$ is the energy.

IF the trajectory of a uniformly moving particle is reparametrized, its length will remain the same but the energy will increase.  The difference is measured in terms of the variance of the changing speed.


Does there exist a function such that $f(a)f(b)=f(a^2b^2)?$ Given $S=\{2,3,4,5,6,7,\cdots,n,\cdots,\} = \Bbb N_{>1}$, prove whether there exists a function $f:S\to S$, such that for any positive $a,b$:
  $$f(a)f(b)=f(a^2b^2),a\neq b?$$

This is 2015 APMO problem 2 (this event ended yesterday: see APMO), maybe I think didn't exist such a function, but I can't prove it.

My attempt:

Consider $p_{i}$ is $i^{th}$ prime,such $f(2)=2^2,f(3)=5^2,f(5)=11^2,\cdots?$

Let $f:S\mapsto S $ be $ f(x)=y$ where $y$ is the smallest number in $S$ so that $y^k=x$ with $k\in\{1,2,3...\}. $ This function "drops" exponents.
It would work like this $f(36)=f(216)=f(6)=6$

Suppose $a=c^d$ and $b=e^f$ where $c\neq h^i,  e\neq j^k $with $h,j \in S$ and $i,k \in \{1,2,3...\}$ $. Let $g=gcd(d,f)$

Then $f(a^2b^2)=f((ab)^2)=f(ab)=f(c^de^f)=f((c^\frac{d}{g}d^\frac{f}{g})^g)=f(c^\frac{d}{g}d^\frac{f}{g})=c^\frac{d}{g}d^\frac{f}{g}$

$=f(c^{\frac{d}{g} g})f(d^{\frac{f}{g}g})=f(c^d)f(d^f)=f(a)f(b)$

showing that the function has the desired property.


Graph theory- Deleting a vertex from a graph Given an undirected connected graph $G=(V,E)$ and two vertices $s,t \in V$ . We know that the length of the shortest path from s to t is bigger than $V / 2  $.

I would love your help with proving that there must be a vertex which if we would delete it, along with the edges going out of it, there won't be a path between $s$ and $t$.

sadly I didn't come with any smart conclusions.

Thanks!

Hint: Prove that $s$ and $t$ do not belong to the same biconnected component of $G$.

I do not have enough reputation to post something in a comment so i would post it here. Thank you for your cooperation .

So my confusion is:

if we delete a vertex v:

v ∈ V - {s,t}

from an undirected connected graph G = (V,E).

Can we prove that the deletion of v disconnects G?


Why does L'Hopital's rule fail in calculating $\lim_{x \to \infty} \frac{x}{x+\sin(x)}$? $$\lim_{x \to \infty} \frac{x}{x+\sin(x)}$$

This is of the indeterminate form of type $\frac{\infty}{\infty}$, so we can apply l'Hopital's rule:

$$\lim_{x\to\infty}\frac{x}{x+\sin(x)}=\lim_{x\to\infty}\frac{(x)'}{(x+\sin(x))'}=\lim_{x\to\infty}\frac{1}{1+\cos(x)}$$

This limit doesn't exist, but the initial limit clearly approaches $1$. Where am I wrong?

Your only error -- and it's a common one -- is in a subtle misreading of L'Hopital's rule.  What the rule says is that IF the limit of $f'$ over $g'$ exists then the limit of $f$ over $g$ also exists and the two limits are the same.  It doesn't say anything if the limit of $f'$ over $g'$ doesn't exist.

There is another useful rule, which I don't seem to have seen written down explicitly:

  Let $f, g, r$ and $s$ be functions such that $g\to\infty$ and $r, s$ are bounded.
  Then the limit of $\dfrac{f}{g}$ and the limit of $\dfrac{f + r}{g + s}$ gives the same result.

Applied here, since $\sin x$ is bounded, the limit is the same as the limit of $\dfrac{x}{x}$.


Cancelling 3-handles in Kirby diagrams Recently been trying to understand the proofs of Gompf and Akbulut that certain 4-manifolds are $S^4$ (these 2 papers: Gompfs paper in Topology Vol. 30 Issue. 1, Akbulut). In which they use a clever 2 and 3 handle cancelling pair to reduce the Kirby diagrams associated to these manifolds.

The part of the proof that I don't quite understand is the way the 2 handle is added.

In an unpublished book on Akbulut's web page Book.pdf he states (on p.14/15),

"Notice that since 3-handles are attached uniquely, introducing a canceling 2- and 3- handle pair is much simpler operation. We just draw the 2-handle as 0-framed unknot, which is $S^2 × B^2$, and then declare that there is a canceling 3-handle on top of it. In a handle picture of 4-manifold, no other handles should go through this 0-framed unknot to be able to cancel it with a 3-handle."

A similar definition for cancelling a 2-3 handle pair is given on p.146 of 4 Manifolds and Kirby Calculus by Gompf and Stipsicz.

My problem is that in both the papers above they add in a 2-3 handle pair but the 2 handle is drawn in such a way that is goes through a 1 handle, which seems to contradict the above cancelling criteria. If anyone could explain this to me it would be greatly appreciated.

Continued from the comments, this response was too long to fit there.

Something like that. If you believe that the knot is the unknot, then you may attach a 2-handle and immediately cancel it with a 3-handle to either $\alpha$ or to $\beta$, and get the diffeomorphism indicated on the 6th-to-last line of page 2. Then the claim is that if you attach the 2-handle to $\beta$, and then do a handle slide, you untwist one of the $m$-twists and wind up with $\alpha$ with a 2-handle, which can then be cancelled by the previous statement, and therefore $\Sigma_m\simeq\Sigma_{m-1}$. You might need to go digging through the references to see why some of those claims are true (the ones I'd want to verify are that that thing is the unknot, and the handleslide untwisting $m$).

See LNM 722, Springer
Pag 16- 30

Proceedings  of Low Dimensional Manifolds

A  Link  Calculus for 4 Manifolds


Can any counting situation that works for one side of an identity work for the other (combinatorial proof) If I come up with a situation that works for one side of a combinatorial proof, does some interpretation always exist for how the other side counts that same situation?  Or is it possible that one side cannot count a situation that the other side can count, even though they are numerically equivalent?

Take for example the following identity.
$$\frac{(2n)!}{n!(n+1)!}=4\binom{2n-1}{n}-\binom{2n+1}{n}.$$
The left side has a smooth interpretation.

How to understand this interpretation from the right side?


Well, $E[X|Y]$ is a random variable and $E[X]$ is a number, so no...

But it is true that $E[X]=E[E[X|Y]]$ and   $E[Y]=E[E[Y|X]]$; a formula which holds more generally for jointly distributed random variables $X$ and $Y$.

Let's verify this for the random variables $X$ and $Y$ in your question. Here, $X$ refers to the first coordinate of your joint mass function and $Y$ refers to the second coordinate.

First let's  figure out what values $X$ takes and it's marginal distribution.

Looking at your joint mass function $p(x,y)$, we see that $X$ takes the values $0$, $1$, and $2$.

To find the marginal distribution of $X$ we need to calculate the probability that $X$ takes on a specific value:
$$\eqalign{
p_X(0)=P[X=0]&=\textstyle P[X=0 ,Y=0]+P[X=0,Y=1]= {1\over 18}+{3\over18}={4\over 18}\cr
p_X(1)=P[X=1]&=\textstyle P[X=1 ,Y=0]+P[X=1,Y=1]= {4\over 18}+{3\over18}={7\over 18}\cr
p_X(2)=P[X=2]&=\textstyle P[X=2 ,Y=0]+P[X=2,Y=1]= {6\over 18}+{1\over18}={7\over 18}.
}
$$

As a check of our work, note $p_X(0)+p_X(1)+p_X(2)=1$, as it should.

While we have everything in hand let's find $E[X]$:
$$\tag{1}
E[X]=0\cdot{4\over18}+1\cdot{7\over18}+2\cdot{7\over18}={14\over18}.
$$

Now on to $Y$:

$Y$ takes the values $0$, and $1$.

The marginal distribution of $Y$ is:
$$\eqalign{
p_Y(0)=P[Y=0]&=\textstyle P[Y=0 ,X=0]+P[Y=0,X=1]+P[Y=0,X=2]\cr &= {1\over 18}+{4\over18}+ {6\over 18}\cr & ={11\over18}\cr
p_Y(1)=P[Y=1]&=\textstyle P[Y=1 ,X=0]+P[Y=1,X=1]+P[Y=0,X=2]\cr &=  {3\over 18}+{3\over18}+ {1\over 18}\cr & ={7\over18}.\cr }
$$

Note $p_Y(0)+p_Y(1) =1$, as it should.

The expectation of $Y$ is
$$
E[Y]=0\cdot{11\over18}+1\cdot{7\over18} ={7\over18}.
$$

Note that if you wrote the joint mass function in tabular form
$$
\matrix{ X\backslash Y & 0&1  \cr  0\ \ \ \ \ &{1\over18}&{3\over18}  \cr
                        1\ \ \ \ \ &{4\over18} &{3\over18} \cr
\cr  2\ \ \ \ \  &{6\over18} &{1\over18}   }
$$
Then the marginal distribution of $X$ is given by row sums of the table and the marginal distribution of $Y$ is given by the column sums of the table.

Now on to the random variable $E[X|Y]$. Note that  $E[X|Y]$ is a function of the random variable $Y$. The value of  $E[X|Y]$  at $y$ is by definition $E[X|Y=y]$. The quantity $E[X|Y=y]$ is called the conditional expectation of $X$ given $Y$ and its value at $Y=y$ is defined to be
$$
E[X|Y=y] =\sum_x xP[X=x|Y=y]=\sum_x x{p(x,y)\over p_Y(y)}.
$$
Note also that to find, for example $P[X=0|Y=1]$, you can go to the table and take the $(0,1)$ entry (which is $p(0,1)$) and divide by the column sum for column 1 (this gives $p_Y(1)$).

We have

$$\eqalign{
\Bbb E[X|Y=0]&= 0\cdot P[X=0|Y=0]+1\cdot P[X=1|Y=0]+2\cdot P[X=2|Y=0]\cr
             &=0+{p(1,0)\over p_Y(0)}+{p(2,0)\over p_Y(0)}\cr
             &={4/18\over11/18}+{6/18\over11/18}\cr
             &={10\over11}.
} $$

$$\eqalign{
\Bbb E[X|Y=1]&= 0\cdot P[X=0|Y=1]+1\cdot P[X=1|Y=1]+2\cdot P[X=2|Y=1]\cr
             &=0+{p(1,1)\over p_Y(1)}+{p(2,1)\over p_Y(1)}\cr
             &={3/18\over7/18}+{1/18\over7/18}\cr
             &={4\over7}.
} $$

Now, since  $E[X|Y]$ is a function of the random variable $Y$:
$$\eqalign{
E[E[X|Y]]&= \sum_y E[X|Y=y]\cdot P[Y=y]\cr
 &=      E[X|Y=0]\cdot P[Y=0]+ E[X|Y=1]\cdot P[Y=1]                      \cr
&={10\over11}\cdot {11\over18}+{4\over7}\cdot{7\over18}={14\over18};
}
$$
the same value as found in $(1)$, as claimed.

I'll leave the verification that  $E[Y]=E[E[Y|X]]$ as an exercise.

Use the fact that $E[X] = E(E[X|Y])$ and $E[Y] = E(E[Y|X])$.


calculation of median of grouped data While calculating the median of grouped data of total frequency $N$, in order to find the median class which value should be taken into consideration to match against cumulative frequency : $\frac N2$ or $\frac{N+1}{2}$ (it seems both are used)? I think $\frac{N+1}{2}$ should be taken since in case of list of values (i.e. ungrouped data), its fractional value indicates that the average of $\frac N2 th$ and $(\frac N2 + 1) th$ values should give the median.

And then comes the second part of my question -- while calculating the median of grouped data, if the value of $\frac{N+1}{2}$ ( or $\frac N2$) is a fraction, say 50.5, and there is a cumulative frequency 50, then what should we do? Should we take two median classes, one having cumulative frequency 50 and another coming next to it, and calculate two medians considering each of the median class using the formula: $L + \frac {\frac N2 - C}{f} \times w$ and take their average as the ultimate median? Or do something else? I mean what is the correct procedure in this kind of situation?

EDIT:

So, here is a specific problem regarding the second part of my question-

We have to find out the median score from the following frequency distribution table:

Score                :  0-10    10-20    20-30    30-40    40-50
Number of students   :   4        3        5        6        7
Cumulative frequency :   4        7        12       18       25

Here intervals are of type (,] .

Now, $N=25 \implies  \frac N2 = 12.5$, which means that we have to look for the interval which covers 12th item and 13th item. Looking at the cumulative frequencies, we see that the 3rd interval(i.e. 20-30) covers the 12th item,while 4th interval(i.e. 30-40) covers the 13th item. If we are supposed to take both the intervals as median class for the sake of using the formula:
$median=L + \frac {\frac N2 - C}{f} \times w$, then we will end up with two medians. We can take the average of these as the required median, though. I want to know the correct procedure here.

Note 1:

I am only concerned with using the above formula and not any other method of finding median of grouped data. There is a variation of the above formula where $\frac{N+1}{2}$ is used instead of $\frac N2$, the first part of my question refers to this confusion as well.

Note 2:

In the formula,

L = lower boundary of the median class
N = total frequency
C = cumulative frequency of the class preceding the median class
f = frequency of the median class
w = width of the median class i.e. upper boundary - lower boundary

Note 3:

If we consider the interval 20-30 as the median class and use the above formula, then the median will be

$20 + \frac{\frac{25}{2} - 7}{5} \times 10 = 31$

Interestingly, considering the interval 30-40 as the median class, we would get the same median using the above formula. Though, I am not sure if this will be the case for every problem of this type. In that case we can take any of the two interval as the median class.

Note 4:

I don't know whether there is any rule for such kind of situation saying that we have to select that cumulative frequency (and hence the corresponding interval as the median class) which is nearer to the value of $\frac N2$, in that case we have to take the interval 20-30 in this example as median class. It will be great and enough if anyone can confirm such a rule.

Because this is essentially a duplicate, I address a few issues
that are do not explicitly overlap the related question or answer:

If a class has cumulative frequency .5, then the median is at the boundary of that class and the next larger one.

If $N$ is large (really the only case where this method is
generally successful), there is little difference between $N/2$
and $(N+1)/2$ in the formula. All references I checked use $N/2$.

Before computers were widely available, large datasets were
customarily reduced to categories (classes) and plotted as histograms.
Then the histograms were used to approximate the mean, variance,
median, and other descriptive measures. Nowadays, it is best
just to use a statistical computer package to find exact values
of all measures.

One remaining application is to try to re-claim the descriptive
measures from grouped data or from a histogram published in a
journal. These are cases in which the original data are no longer
available.

This procedure to approximate the sample median from grouped
data $assumes$ that data are distributed in roughly a uniform
fashion throughout the median interval. Then it uses interpolation
to approximate the median. (By contrast, methods to approximate
the sample mean and sample variance from grouped data one assumes
that all obseervations are concentrated at their class midpoints.)

I thought the formula of $(N+1)/2$ is used in situation where the sum of frequency has got even number e.g. if sum of $f = 50$. But if the sum of frequency is odd number, then use formula $N/2$, i.e. if $f = 61$, then use formula $N/2$.


Determining all the elements of S4? What is an easy way to determine the elements of $S_{4}$?

While going through my revision process in an attempt to stem out the nitty gritty areas that I am unsure of, I chanced upon this.
I tried to search for an answer via Google but there are no satisfying responses.

Any help is appreciated.

List all possible 4-letter words in the letters $1,2,3,4$ without repetition.

A systematic way is to list them in lexicographical order:

1234
1243
1324
1342
...
4321

A word $a_1 a_2 a_3 a_4$ corresponds to the permutation $i \mapsto a_i$ in $S_4$.

We know that every element of $S_4$ is an automorphism over ${1,2,3,4}$.
We want to count the number of elements in $S_4$ then lets start with $1$, it must be mapped to an element in ${1,2,3,4}$ - so it has 4 options, now for 2 - he has only 3 options for mapping (since $1$ already occupied an element in ${1,2,3,4}$), in the same way $3$ has $2$ options, and the mapping of $4$ is determined uniquely after determining all the other mappings.

Therefore, we finally get $4\cdot 3\cdot 2\cdot 1 = 4! = 24$ such elements.


Is $f'(x)=0$ true at a cusp? If I am given a graph that has a cusp and I am asked to find every point $x$ where $f'(x)=0$ is satisfied, does this include the cusp? I know that when the derivative of a function equals zero this means that there is a horizontal tangent at that point, I also know that the derivative does not exist at a cusp. However, a cusp does have a horizontal tangent but is not differentiable at that point, so do we include the point $x$ where the cusp is when the question is asking us where $f'(x)=0$?

By cusp I mean when
$$\lim_{x \to a^{+}} f'(x)=+ \infty \text{ and } \lim_{x \to a^{-}} f'(x)=- \infty$$

if the derivative does not exist it is, in particular not $=0$

Regards Jotam. If a derivative does not exist at the cusp, say at point $x_o$, then it means that $f'(x_o)$ does not exist, it does not have any unique value. The limits in both direction are different :
$$  \lim_{h \rightarrow 0} \frac{ f(x_o + h) - f(x_o) }{h} \ne  lim_{h \rightarrow 0^{-}} \frac{ f(x_o + h) - f(x_o) }{h}$$

A tangent line at a point $ x $ is a straight line with gradient $f'(x)$. Since there is no derivative at the cusp, there is no tangent at the cusp.

To be spesific for your question, $f'(x_o)$ cannot be zero because it can not have any unique value.

A tangent line represents the $\textbf{continuous} $ rate of change at the considered point. But you can observe that there is a discontinuity of change at the cusp. $ \textbf{There is a sudden change at the cusp}.$

A tangent line has a direction.

Thanks.


Complex and Conformal structure on a trivial tangent bundle of a higher dimensional manifold Let $M$ be an even dimensional Reimannian  manifold with trivial tangent bundle $TM$. Then there is a global orthonormal basis $\{e_1,e_2,...,e_{2n-1},e_{2n}\}$ of $TM$.

Define a bundle map $J :TM  \to TM$ which sends each $e_{2k-1}$ to $e_{2k}$ and
 $e_{2k}$ to $-e_{2k-1}$ for $k=1,...,n$. Then, does this define a complex structure on $M$? Is it well-defined?

Ref

Complex structure v.s. conformal structure in more than 1 complex dimension

Proof of equivalence of conformal and complex structures on a Riemann surface.

Edit for understanding the answer.

First, we show that $M = (S^2\times S^2)\#(S^1\times S^3)\#(S^1\times S^3)$  has a trivial tangent bundle and to do so, it suffices to show that

(I)  the Euler numbr of  $M$  is zero and

(II)  $M$ is a spin manifold.

Consider the (I).

For any triangularizable toplogical space $X$ and $Y$, there are simple algebraic relations between their Euler numbers.

$\chi(X \#Y) = \chi(X ) +\chi( Y)  -2,$

$\chi(X \times Y) = \chi(X ) \times \chi( Y). $

Using these formulas, we can calculate the Euler number of the manifold  $M = (S^2\times S^2)\#(S^1\times S^3)\#(S^1\times S^3)$ as follows.

$\chi(M) = \chi (S^2)\chi( S^2)+\chi(S^1)\chi( S^3)+\chi(S^1)\chi( S^3) -4.$

Because $\chi (S^2)=2$ and $\chi (S^1)=0$ and $\chi (S^3)=1-0+0-1=0$, we obtain

$\chi(M) = 2 \times 2 +0 \times \chi( S^3)+0 \times \chi( S^3) -4=0.$

Thus, $\chi(M)=0$ and thus (I) is O.K.

Next, consider the (II).

According to Albanese's comment, the connected sum of spin manifolds is spin.

Thus we show that the spaces $S^2\times S^2$, $S^1\times S^3$ and $S^1\times S^3$ are spin. To poove this, for example, we show $w_2(S^1\times S^3)=0$.

According to problem 4-A of page 54 in Milnor-Stasheff,

$w_2(S^1\times S^3)=w_2(S^1) \times  w_0(S^3) + w_1(S^1) \times  w_1(S^3)   + w_0(S^1) \times  w_2(S^3) =0$

where we use the total Stiefel-Whitney class of sphere is 1, namely, $w(s^d)=1$ (page 42 example 1 in Milnor Stasheff).

First of all, the fact that $M$ is Riemannian plays no role here.

The endomorphism $J$ is indeed well-defined and as $J\circ J = -\operatorname{id}_{TM}$, it is an almost complex structure on $M$. However, it need not be an integrable almost complex structure. That is, it may not be an almost complex structure induced by a complex manifold. For example, $M = (S^2\times S^2)\#(S^1\times S^3)\#(S^1\times S^3)$ has trivial tangent bundle and hence has an almost complex structure as you describe, but it does not admit a complex structure.

In real dimension two, every almost complex structure is integrable, see this question. In real dimension four and above, there are examples of almost complex structures which are not integrable. In real dimension four, there are manifolds (such as $M$) which admit almost complex structures, none of which are integrable. It is an open problem whether such manifolds exist in real dimension six and above.

I would estimate yes in the sense each 2n is mapped one-to-one in the sense each +2n may be mapped onto the pure reals for the real part of the complex mapping and each of the -2n may be mapped onto the pure imaginary numbers of the imaginary part of the complex system.  I am sorry here if will not do.  I am an economist, not a mathematician.  Sincerely,  Richard Anthony Baum


Similarity by SSA (side-side-angle) in obtuse triangles Source: Challenge and Thrills of Pre-College Mathematics, Page 74, Problem 54:

"In two obtuse triangles, an acute angle of the one is equal to an angle of the other, and sides about the other acute angles are proportional. Prove that the triangles are similar."

I tried this first by cosine rule, and the question reduced to:
Assume $\triangle ABC$ and $\triangle DEF$ with $\angle A$ and $\angle D$ being obtuse, $\angle B=\angle E$ and $AC/BC=DF/EF$. We may take $DF=k\cdot AC$, $EF=k\cdot BC$ and apply $\cos B= \cos E$, obtaining $DE=k\cdot AB$ and the problem is solved. EDIT: Even this however is difficult to prove, as pointed out by cosmo5 in his answer.
What is interesting and challenging is to prove the same without the cosine law (probably the expected solution, as cosine law doesn't seem to help), as the law hasn't been introduced in the book until later chapters, while this problem is taken from chapter 3.
My attempt:
Fix $\triangle ABC$ as well as points $E$ and $F$ of $\triangle DEF$. Assuming $DF=k\cdot AC$, $EF=k\cdot BC$, the locus of $D$ will be a circle with centre $F$ and radius $k\cdot AC$.
After this however, I couldn't go further with the same concept. Either a hint or a solution related to my thought process, or any other thought process for that matter, would be greatly appreciated.

Hint: note that there are at most two points on ray $BA$ which is of a certain distance from point $C$ (because you are intersecting a ray with a circle).

They are congruent if the known congruent angles are obtuse. If the known congruent angles are acute, then the triangles are congruent if the triangles are both acute or both obtuse.  Why this is not more widely distributed is above my paygrade!


Is there a 40 face single polygon toroid? Solutions for 24 and 32 face single polygon toroid have been given.  Is there a solution for the next step of a 40 face single polygon toroid?

Is there a simpler single polygon toroid?

The opening on the 24 face toroid is an equilateral triangle.  The opening on the 32 face toroid is a regular square.  It seems there should be a 40 face single polygon toroid with a regular pentagonal opening.

A simple google search with the title of this post brings up this very post on first place and a solution with 48 triangles on the second.

--- rk


Giving a formal proof of p ⇒(q ⇒ r) $\vdash$ (p ⇒ r)∨(q ⇒ r) using the rules of inference. I can prove this with semantic equivalences and truth tables but I'm struggling on the formal proof using rules of inference front.

Given its format I would assume it must finish on V-introduction which means using if-introduction at some point on p⇒r or q⇒r?

I've made 3 or 4 attempts now and I can't seem to get there!

Any advice or pointers in the right direction would be very much appreciated,
Thanks!

Edit: We've been using Γ ├ B to mean ‘B can be formally proved from the set of Γ of assumptions’.

Edit: The rules at my disposal are:

&-I
&-E
V-I
V-E
⇔-I
⇔-E
⇒-I
⇒-E
$\lnot$-I
$\lnot$-E
true-I
false-E

Your strategy of ending with a $\lor$ Introduction on either $p \rightarrow r$ or $q \rightarrow r$ is not going to work

If $p$ is true, but $q$ and $r$ is false, then $p \rightarrow (q \rightarrow r)$ is true, but $p \rightarrow r$ is false, so you will not be able to derive $p \rightarrow r$ by itself.

If $q$ is true, but $p$ and $r$ is false, then $p \rightarrow (q \rightarrow r)$ is true, but $q \rightarrow r$ is false, so you will not be able to derive $q \rightarrow r$ by itself.

Indeed, here is a general 'tip': if you ever have to prove a disjunction as your ultimate conclusion, then most likely you will not be able to prove either of then disjuncts by themselves!

So, what to do instead?  I would recommend either:

Doing this by a proof by contradiction: if $(p \rightarrow r) \lor (q \rightarrow r)$ is false, then both $p \rightarrow r$ and $q \rightarrow r$ are false, hence $p$ and $q$ are true but $r$ is false, contradicting $p \rightarrow (q \rightarrow r)$

or:

Start by proving $p \lor \neg p$. And then: if $p$, then $q \rightarrow r$ using Modus Ponens ($\rightarrow$ Elim I suspect) between $p$ and the premise. And if $\neg p$, then you can show $p \rightarrow r$. So, you can show $(p \rightarrow r) \lor (q \rightarrow r)$ in either case (so now you do $\lor$ Intro within each subproof .. that is something that will frequently occur in formal proofs). So, using proof by cases (presumably your $\lor$ Elim), you definitely have $(p \rightarrow r) \lor (q \rightarrow r)$

Since $(a \Rightarrow b) \Leftrightarrow (\lnot a \vee (a \wedge b)) \Leftrightarrow (\lnot a \vee b   ) $ for all $a,b$ by the material implication rule.

$$
[ p \Rightarrow (q \Rightarrow r) ] \Leftrightarrow [ \lnot p \vee (q \Rightarrow r) ] \Leftrightarrow [ \lnot p \vee ( \lnot q \vee r) ]
\Leftrightarrow [ \lnot p \vee \lnot q \vee r ]
$$
By the same rule,
$$
[ (p \Rightarrow r ) \vee  (q \Rightarrow r) ] \Leftrightarrow
[ (\lnot p \vee r  ) \vee ( \lnot q \vee r) ] \Leftrightarrow
[ \lnot p \vee  \lnot q \vee r]
$$
Hence the result.


hint

$$T_n (\cos (\theta))=\cos (n\theta) $$
where $T_n $ is the Tchebychev polynomial.


Improper integral calculation - limit at infinity Will you please help me prove the following limit is zero ?

  $$\lim_{x \to \infty} \int_0^{\infty} \frac{1-e^{-u^4}}{u^2} \cos(x u) du.
$$

Thanks in advance

If you don't want to appeal to Riemann-Lebesgue, you could also try integrating by parts (being careful to justify the differentiability at 0 of $u\mapsto (1-{\rm e}^{-u^4})/u^2$).


What is the order of the midpoint rule? If the Trapezoidal-Rule has the order $n=1$, and Simpson's has order $n=2$, what is the order $n$ of the midpoint rule?

And if the weights of the Trapeziumrule are ($1/2, 1/2$) and those of the Simpson-Rule are ($1/6, 4/6, 1/6$), what are the weights of the midpoint-rule?

Thank you in advance!

The error of the Midpoint rule is

$Error_m \leq \frac{k(b-a)^3}{24n^2 } = k(b-a)h^2/24$. So it equals to $O(h^2)$, where $k = max|f"|$.

The order of the midpoint rule is $n=2$. For examples, you can see Why does Trapezoidal Rule have potential error greater than Midpoint?


82% of an event in a persons life, what is the daily chance? If there is an 82% chance that within your average lifetime, lets assume 70 years (25567 days), that "E" event will happen: what is the percent chance that it will happen on any given day?
I'm assuming that it'll be something like .02347% or something small. My problem is I've only taken up through Advanced Algebra and that was six years ago. I would have searched for this on google to solve it myself, but I'm not even sure what to call this beyond "math I don't know".
So if you could help me identify the solution, how to solve it, and what this is (i.e. I imagine terms like 'graphing polonomials' 'deductive statistics' or some such name that would give me an idea of what exactly it is that I should be learning to handle similar questions on my own). Thanks for the help in advance.

If the probability that it happens in a specific day is $p$, then the probability that it does not happen is $(1-p)$ (complimentary events). Thus the probability that it will not happen in $25567$ days is $(1-p)^{25567}$ (product rule, independent events). Therefore the probability that it will happen in a lifetime is $1 - (1 - p)^{25567}$ (again, complimentary events), and that is your $82\%$.

So, we have that
\begin{align}
0.82 &= 1 - (1-p)^{25567} \\
0.18 &= (1-p)^{25567}\\
\sqrt[25567]{0.18} &= 1 - p\\
0.999933 &= 1-p\\
0.000067 &= p
\end{align}
so the probability of it happening in a given day (assuming it doesn't change with day of the week, time of the year, age, and so on) is $0.0067\%$.

This is a bit backwards from the standard textbook example (it would give you the daily probability and ask you to calculate the lifespan probability), but it's more or less elementary probability, as they should teach in any high school. The two main concepts used in setting up the first equation was complimentary probability and the product rule. Once the equation was set up, however, we basically forget all about the probability, and focus on the algebra of isolating and calculating $p$. When we have that, we go back to the probability interpretation, and say what it means.

A statistic that said if you live to be 70, your chance of being the victim of violent crime is 82% would be better expressed as "your chance of being the victim of one or more violent crimes" and couldn't really be used to calculate the daily probability.


Can someone give me an example of a challenging proof by induction? I'm about to give my first consultation and the topic is proof by induction. The students are new to calculus and I assume they don't know anything besides precalculus. Can anyone give me a proof by induction which is a bit different, challenging, maybe foreshadows other areas of calculus (derivation or whatever) because the prof who teaches them as well already have shown them a lot of easy ones. I'm glad even if you can give an example where it is hard to see that it can be proved by induction.

Here is one that my math professor showed us: show that for $n\ge 1$ a $2^n\times 2^n$ chessboard with one square removed can always be tiled by "L-shaped" pieces. That is, pieces formed by removing a corner from a $2\times2$ square.

Construct the natural numbers $\mathbb{N}$ by using the Peano Axioms. Denote by $\varphi:\mathbb{N}\rightarrow\mathbb{N}$ the map defining the injective map associating each number to the "next one".

Show that the map $+:\mathbb{N}\times\mathbb{N}\rightarrow\mathbb{N}$ defined by
$$
n+1=\varphi(n),\qquad
n+\varphi(m)=\varphi(n+m)
$$
ís a commutative associative operation.


What are some examples of classes that are not sets? After reading about Russell's paradox, I see that the set of all sets does not exist, so instead it is called a class.

What other commonly known classes exist that are not sets? I know the class of all singleton sets is not a set, because you can unite that class to get the class of all sets. This seems like the class of all sets of a fixed finite size is not a set either, correct?

What are some examples of other 'things' that can't be gathered up and put into a set?

First, any description which doesn't uniquely specify the elements of a set tends to give a proper class.

For example, a group is an set together with operations satisfying a certain collection of axioms.  Since these axioms don't pin down the elements of the group, one might expect the collection of all groups to not be a set.

This holds in a similar fasion for topological spaces, rings, vector spaces, fields, manifolds, metric spaces, and many of the other commonly studied objects.

This also applies to your "collection of all singletons" example, and similarly to any sort of "collection of all sets of a fixed size".

But just because a description DOES pin down the elements doesn't mean you have a set.  For example, the collection of all ordinal numbers is not a set, despite the fact that if there is an order preserving bijection between two ordinals $\alpha$ and $\beta$, then one must have $\alpha = \beta$ as sets.

Similarly, the collection of all cardinals doesn't form a set.

In two-sorted first order set theory with proper classes (such as NBG) it is an axiom:
$$
\forall x\, \exists X\, \forall z\, [ z \in x \Leftrightarrow z \in X ]
$$
where lower-case letters denote type-0 objects (sets) and upper-case letters denote type-1 objects (classes).

This would mean that there are coclasses for every set:
$$
\forall x\, \exists \bar X \, \forall z\, [ z \in x \Leftrightarrow z \notin \bar X\, ]
$$
Thus, the elements NOT in a given set, forms a proper class.


If $\int_0^\infty f\text{d}x$ exists, does $\lim_{x\to\infty}f(x)=0$? Are there examples of functions $f$ such that $\int_0^\infty f\text{d}x$ exists, but $\lim_{x\to\infty}f(x)\neq 0$?

I curious because I know for infinite series, if $a_n\not\to 0$, then $\sum a_n$ diverges. I'm wondering if there is something similar for improper integrals.

It depends on what $\int$ means and what else you know about $f$.
Here is a continuous example with limits of Riemann integrals:

Let $h(x)=\max\{1-|x|,0\}$ and set $f(x)=\sum_{n=1}^\infty (-1)^n h(nx-n^2)$.
This function has up and down "bumps" around integers that become smaller and smaller in area, but have fixed height.

If $\lim_{x\to+\infty}f(x)=l>0$, then $\exists M>0:l-\varepsilon<f(x)<l+\varepsilon\quad \forall x>M$, and so

$$
\int_M^{+\infty}f(x)dx>\int_M^{+\infty}(l-\varepsilon)dx=+\infty
$$

if $\varepsilon$ is sufficiently small.


Solving $\begin{cases}x' = t \sin^2(\frac 1 t) - x^2 \\ x(0) = 0 \end{cases}$ Is there a way to compute the solution of the ode:

$\begin{cases}x' = t \sin^2(\frac 1 t) - x^2 \\ x(0) = 0 \end{cases}$

where in $t = 0$ we define the field as $-x^2$.

I need for an application of Guiding the solution of ODE with curves. In particular to compute the derivative of  the following at $0$:

$\begin{cases}x' = f(t,x) + \epsilon (\gamma(t) - 2x) \\ x(0) = 0\end{cases}$

Hint:

Let $x=\dfrac{1}{u}\dfrac{du}{dt}$ ,

Then $\dfrac{dx}{dt}=\dfrac{1}{u}\dfrac{d^2u}{dt^2}-\dfrac{1}{u^2}\left(\dfrac{du}{dt}\right)^2$

$\therefore\dfrac{1}{u}\dfrac{d^2u}{dt^2}-\dfrac{1}{u^2}\left(\dfrac{du}{dt}\right)^2=t\sin^2\dfrac{1}{t}-\dfrac{1}{u^2}\left(\dfrac{du}{dt}\right)^2$

$\dfrac{d^2u}{dt^2}-\left(t\sin^2\dfrac{1}{t}\right)u=0$

Let $r=\dfrac{1}{t}$ ,

Then $\dfrac{du}{dt}=\dfrac{du}{dr}\dfrac{dr}{dt}=-\dfrac{1}{t^2}\dfrac{du}{dr}=-r^2\dfrac{du}{dr}$

$\dfrac{d^2u}{dt^2}=\dfrac{d}{dt}\left(-r^2\dfrac{du}{dr}\right)=\dfrac{d}{dr}\left(-r^2\dfrac{du}{dr}\right)\dfrac{dr}{dt}=\left(-r^2\dfrac{d^2u}{dr^2}-2r\dfrac{du}{dr}\right)(-r^2)=r^4\dfrac{d^2u}{dr^2}+2r^3\dfrac{du}{dr}$

$\therefore r^4\dfrac{d^2u}{dr^2}+2r^3\dfrac{du}{dr}-\dfrac{\sin^2r}{r}u=0$

$r\dfrac{d^2u}{dr^2}+2\dfrac{du}{dr}-\dfrac{\sin^2r}{r^4}u=0$

Let $v=ru$ ,

Then $\dfrac{dv}{dr}=r\dfrac{du}{dr}+u$

$\dfrac{d^2v}{dr^2}=r\dfrac{d^2u}{dr^2}+\dfrac{du}{dr}+\dfrac{du}{dr}=r\dfrac{d^2u}{dr^2}+2\dfrac{du}{dr}$

$\therefore\dfrac{d^2v}{dr^2}-\dfrac{\sin^2r}{r^5}v=0$


How to carry out arithmetics with Bachmann–Landau notations? Here's something from my class.

As $n \to \infty$, $|\mathcal{F'}| =
(1-o(1))\frac{n \choose t}{k \choose t}
+ o(1){n \choose t}
= (1 + o(1))\frac{n \choose t}{k \choose t}$

I can sort of see why this makes sense, and I guess writing out the expression by the definitions would verify it. But is there an intuitive way to interpret this calculation and others similar to it?

You can write the following estimate : $o(1){n \choose t} = o(1)\frac{{n \choose t}}{{k \choose t}}$, since $k$ and $t$ are fixed. Thus $|F'| = (1+o(1))\frac{{n \choose t}}{{k \choose t}} + o(1)\frac{{n \choose t}}{{k \choose t}} = \frac{{n \choose t}}{{k \choose t}} (1+o(1)+o(1)) = (1+o(1))\frac{{n \choose t}}{{k \choose t}}.$
You can interpret the notations $o$ and $O$ as : these are quantities whose names and exact values are not relevant, only their asymptotic behaviour.

Taking, for example, non negative case, we have following definitions:
$O(g) =  \left\lbrace f:\exists C > 0, \exists N \in \mathbb{N}, \forall n (n >  N \& n \in \mathbb{N}) (f(n) \leqslant C \cdot g(n)) \right\rbrace $
$o(f)=\{g: \exists \epsilon, lim_{n \to \infty}\epsilon=0,  \exists N, n>N, g=\epsilon \cdot f   \}$
Firstly let me notice, that equality type $f=O(g)$ is quite different from type $O(f)=O(g)$. First means "$\in$", while under second  many authors understand "$\subset$", though it should be said, that for most well known equalities we can understand it as "$\subset \land \supset$". Well known are following properties:
$$C \cdot o(f)=o(f \cdot C)=o(f)$$
$$o(f)+o(f)=o(f)$$
First gives you possibility to change "minus" to "plus" in your first member and by  second you obtain last equality.
As formal, so intuitive way, probably best, is think about brought above properties, as on both sides of the equality there are the same sets, simply written in different ways.
In the case of the expressed interest, I can write proof, that the equality holds exactly as in the case of sets i.e. right to left and vice versa "$\subset \land \supset$".


Simple puzzle from The Moscow Puzzles with wrong solution? I have a book of mathematical puzzles -- The Moscow Puzzles, edited by Martin Gardner -- and I'm struggling to make sense of the following puzzle. It seems utterly simple, yet the solution given seems completely wrong. Here's the puzzle and solution verbatim.

42. Tell "At a Glance"

Here are two columns of numbers:

123456789          1
12345678          21
1234567          321
123456          4321
12345          54321
1234          654321
123          7654321
12          87654321
1          987654321

Look closely: the numbers on the right are the same as on the left, but reversed, and in reverse order.

Which column has the higher total? (First answer "at a glance," then check by adding.)

Solution

The columns don't look like they have the same sums, but look closely: comparing digits, nine 1s match one 9; comparing tens, eight 2s match two 8s, and so on. Check by adding -- the sums are equal.

This solution seem totally wrong. The sum of the left column is 137174205, which is less than the value of the last number alone in the right column. We can also  tell this without adding simply by flipping the order of left column: 1 matches 1, but 12 is smaller than 21, and ditto all the way down the remaining numbers. Also, when the solution says "comparing tens, eight 2s match two 8s", the eight 2s are indeed in the tens column, but the two eights are not.

Can anyone interpret the puzzle in a way that makes any sense? It's a pretty well-know puzzle book, for such a glaring error to get through.

The left column's entries should be interpreted with padded zeroes on the right, so that e.g. the last summand is 100,000,000.

The numbers are individual not a total #. For example 123456789 is not 123,456,789 it is in fact the numbers 1, 2, 3 etc.....That is where they try to trick you. I came up with the same thing you did when I first did the puzzle. It was only after I read the solution that the trick became clear.


Limit of $\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)/n$ For $n\in\mathbb{N}^*$, define $a_n=\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)/n$. Does the sequence $(a_n)$ converge and how to prove it?

Let $a_n=\frac{1}{n}\sum\limits_{k=1}^n\frac{1}{k}$.

Thus, $$0<a_n<\frac{1}{n}\int\limits_1^{n+1}\frac{1}{x}dx=\frac{\ln(n+1)}{n}\rightarrow0,$$
which says that
$$\lim_{n\rightarrow+\infty}a_n=0.$$


What denotes the essence of a function object in mathematics? In other words, when does something become a function, and why? Take this, for example:

x = y (x + z) = 350

Is anything enclosed within the brackets considered an input to the function?

What if expressed this way:

x = y + xz = 350

Where is the function? What purpose does a function serve to measure an input and output when an input and output do not need a function? You would just employ the equivalent operator, a few variables/constants, and express a result.

What I'm asking is, why employ a function, and what is a function overall?

What you are describing is an equality.  For it to be a function you would have to specify what inputs are possible and what outputs you are looking for.  In your equality above you might say: if x is 2 and y is 3, what is z?  Now you are headed towards having a function.  If you say, z is defined by this equality where x and y are real numbers, you do have a function.  It takes certain inputs (real numbers) and spits out an output which is the real number z (whether you are actually able to compute z or not, which is a different issue).

By thinking of functions as input - operation -output we gain an immense amount -- almost impossible to discuss briefly.  We use mathematical notation like f(x,y) = z to express that x and y are inputs, f defines the operation and z is the output.  There are plenty of other ways to notate functions, and many of them are in constant use.  But this particular way of notating them has proved very fruitful.

There are several purposes to notation.  One is to express thoughts in a succint way, so that you don't need a 100 word essay to describe what you are talking about.  Compare f(x,y) = z with  "x and y are inputs, f defines the operation and z is the output".

Another feature of "fruitful" is that the notation should make it easier to understand what is being supposed, or discussed or is to be proved.

Yet another is that it organizes the information.  It's easy enough to make mistakes, and it can be very difficult to track them down if you have 18 pages of general disorganized junk (actually it is hard enough when things are organized).

Another is that it proves to be a useful way to write things in a very large number of disparate problems.

Most fruitful of all is when the notation not only accomplishes the above but is suggestive of what you might look for, or how you might prove things. It is so good it puts you onto truths that you might never suspect had not the notation pointed that way.

The notation y = f(x) (and its generalizations) certainly meet all the criteria above.


Why is $D_4/Z(D_4)$ "just like" $V$, Klein's 4-group? What makes the two similar? $D_4/Z(D_4)$ here is the quotient group of $D_4$ by its center.

There are a few computational details missing, but I will leave those for you to do.  After some computation, you will find that
$$ Z(D_4) = \{ e, r_2 \}, $$
where $r_2$ is the rotation of the square by $\frac{\pi}{2}$ radians.  Now $|Z(D_4)| = 2$, and so (recall the center of a group is always normal)
$$ |D_4 / Z(D_4)| = 4. $$
At this point, we may say that $D_4 / Z(D_4) \cong \mathbb{Z} / 4 \mathbb{Z}$ or $D_4 / Z(D_4) \cong V_4$ (the Klein $4$ group) because these are the only two groups of order $4$ up to isomorphism. Check that each element of $D_4 / Z(D_4)$ has order $2$, which implies $D_4 / Z(D_4) \cong V_4$.

Not only is D4/Z(D4) "like" the Klein 4 group, for all practical purposes it "is" the Klein 4 group, since it is isomorphic to the Klein 4 group. To see this fairly straightforwardly:

We know that |Z(D4)| = 2, and so |D4/Z(D4)| = 8/2 = 4 by Lagrange's Theorem (We can consider D4/Z(D4), since the center is always normal in a group). Also notice that 4 = 2^2.

Burnsides Theorem yields that any group of order p^2 where p is a prime is abelian.

Thus, D4/Z(D4) is an abelian group of order 4.

Now, by the fundamental theorem of finite abelain groups, D4/Z(D4) is either isomorphic to Z4 or Z2 X Z2. But Z4 is cyclic, and if D4/Z(D4) was isomorphic to Z4, then this would imply that D4/Z(D4) is also cyclic, which would imply D4 is abelian (by elementary theorem). This is clearly not true.

Thus, D4/Z(D4) is isomorphic to Z2 X Z2, which is precisely the Klein 4 group.


Finding a particular term of a binomial expansion How do I find the term involving $x^{10}$ in the expansion of:

$$
(3+2x^2)^7
$$

I know from the binomial theorem that:

$$
u_{n+1} = {^nC_r a^{n-r}x^r}
$$

and that $n=7, a=3, x=2x^2, r=10$ in this case. But that leads to:

$$
^7C_{10}\cdot3^{-3}\cdot{x^{10}}
$$

Which doesn't match the answer listed ($^7C_{5}\cdot{3^{2}}\cdot{2^5x^{10}}$). Could somebody point me in the right direction?

The binomial theorem for $n=7$ says:

$(a + b)^7 = {7 \choose 0}a^0b^7 + {7 \choose 1}a^1b^6 + {7 \choose 2}a^2b^5 \ldots {7 \choose 7}a^7b^0$

Take $a = 3$, $b = 2x^2$. We need $x^{10}$, so we take the term with ${7 \choose 2}$, which is ${7 \choose 2} \cdot 3^2 \cdot (2x^2)^5$, which is the required answer, as $(2x^2)^5 = 2^5 x^{10}$ and ${7 \choose 2} = {7 \choose 5}$.

$r$ must be equal to $5$, so the term that will have the exponent $10$ is the $6_{th}$ term..

since to get $r$,

(term desired) - 1

or

$t-1=r$


Combining the effects of two moduli If I have:

$((x \bmod m_1) \bmod m_2) = z$

With $x,m_1,m_2$ given positive integers, is there some $y$ such that

$x \bmod y = z$

$x\geq z$.

If $x=z$, then $m_1,m_2>z$, so you can take $y=m_1$.

If $x>z$, then you can take $y=x-z$.


Given $a=r^2\pi h,c=2r\pi h+r^2\pi$, how to find $r$ and $h$ in terms of $a$ and $c$? Given the values of $a$ and $c$ and that
  $$a=r^2\pi h,\quad
c=2r\pi h+r^2\pi,$$
  what are the values of $r$ and $h$?

So recently on a math test, I got this question which I couldn't seem to figure out. It was really boggling my mind and I later found out that I just had interpreted the question wrong, but I was still set on finding out the solution to the problem. I can't seem to do it and wonder if anyone could help.

From the first equation we have that
$$
2\pi r h = 2\frac{a}{r}
$$
so we can substitute to the second one and then
$$
c=2\frac{a}{r}+r^2 \pi\qquad\text{so}\qquad cr=2a+r^3\pi
$$
and now the only thing you have to do is to solve this equation using Cardano formula and get that
$$
r=\frac{\left(\sqrt{81 \pi  a^2-3 c^3}-9 \sqrt{\pi } a\right)^{2/3}+\sqrt[3]{3}
   c}{3^{2/3} \sqrt{\pi } \sqrt[3]{\sqrt{81 \pi  a^2-3 c^3}-9 \sqrt{\pi } a}}
$$
and now by substituting to the first formula you get
$$
h=\frac{a}{r^2 \pi}=\frac{3 \sqrt[3]{3} a \left(\sqrt{81 \pi  a^2-3 c^3}-9 \sqrt{\pi }
   a\right)^{2/3}}{\left(\left(\sqrt{81 \pi  a^2-3 c^3}-9 \sqrt{\pi }
   a\right)^{2/3}+\sqrt[3]{3} c\right)^2}
$$


Radius of Convergence of $\sum_{n=0}^{\infty} e^{-z^{2}\sqrt{n}}$ Find all $z \in \mathbb{C}$ such that the series

$$ F(z)=\sum_{n=0}^{\infty} e^{-z^{2}\sqrt{n}}$$

converges.

I tried the ratio test and root test, but both were inconclusive. I considered the Cauchy-Hadamard formula, but could not see how to apply it to this problem.

Any guidance would be much appreciated.

Let $z=x+iy$ then $z^2=(x^2-y^2)+i(2xy)$ and
$$e^{-z^{2}\sqrt{n}}=e^{(y^2-x^2)\sqrt{n}}\cdot e^{-2xyi\sqrt{n}}.$$
Note that $|e^{-z^{2}\sqrt{n}}|=e^{(y^2-x^2)\sqrt{n}}\to 0$ iff $|x|>|y|$ which implies that the series does not converge when $|x|\leq |y|$.

Moreover if $|x|>|y|$ then
$$n^2|e^{-z^{2}\sqrt{n}}|=n^2e^{(y^2-x^2)\sqrt{n}}=e^{(y^2-x^2)\sqrt{n}+2\ln n}\to 0$$
because $(y^2-x^2)\sqrt{n}+2\ln n\to -\infty$ as $n\to+\infty$.
Hence eventually $|e^{-z^{2}\sqrt{n}}|\leq \frac{1}{n^2}$. Since the series $\sum\frac{1}{n^2}$ is convergent, we conclude that that series converges iff $|x|>|y|$.

I will look at the convergence of $\sum\limits_{n=0}^{\infty}{e^{r\sqrt{n}}}$ for $r$ real. Can you see how it's related to your problem?

For $r\ge 0$, we have $e^{r\sqrt{n}}\ge 1$ for all $n$, and hence the series diverges.

For $r<0$, note that for large enough $n$ we have $r\sqrt{n} < -2\log n$, so for large enough $n$ we have $e^{r\sqrt{n}}< n^{-2}$. Thus, by the comparison test the series converges.


What is an image of empty set? Let $f: M \rightarrow N$ be a function. We always can define image of custom set $A \subset M$ like $f(A) = \{y\in{N}:(f(x) = y) \wedge(x\in A) \}$. As known, empty set is a subset of each set, so my question is: Is it proper to write $f(\varnothing) = \varnothing$?

Well, $y\in f(A)$ if and only if $y=f(a)$ for some $a\in A$.

However there are no elements in the empty set, so there is no $y\in f(\varnothing)$, therefore $\varnothing=f(\varnothing)$.

The notation {x∈X:P(x)} denotes the set of all elements x belonging to X for which the proposition P is true.

The truth table for "∧" shows (f(x)=y)∧(x∈A) is true only when both (f(x)=y) and (x∈A) are both T.

But if A=∅, then (x∈A) is always F.

As such, the proposition P in your definition [i.e. (f(x)=y)∧(x∈A) ] is always F ... and so the set of all elements y belonging to N for which the proposition P is true is empty.


divisibility of N from 1,2,3,...k Good day! I encountered this problem yesterday and I eventually answered the problem correctly.
The problem is this:If N is divisible by 1,2,3,. . . 13, then N must also be divisible by 14 and 15.

Using this same idea, what is the smallest integer  such that the following statement is true?

If N is divisible by 1,2,3,. . . M, then N must also be divisible by M + 1, M + 2, M + 3 and M + 4.
But I am wondering if there is a general rule/ proof/ algorithm for this? I used rough way to solve this one,,

Your statement is true if and only if each of $M+1,M+2,M+3,M+4$ is not a prime power.

I doubt there is a good way to find $M$ other than brute force. $M=53$ give $54,55,56,57$ without any prime powers, so that $M$ works, but I haven't found if it is the smallest.

More generally, for $M+1,M+2,\dots,M+k$, you'd need to know something about prime power gaps. I'm not sure what is known about that, but I'm assuming they are arbitrarily large.

For general $k$, I think you can use $M=(k+1)!^2+1$. Then $M+1,M+2,\dots,M+k$ are not prime powers, but this $M$ is not the smallest such $M$ by far.

A smaller $M$ for a general $k$ is:

$$M=1+\prod_{p\leq k+1} p^{1+\lfloor \log_p (k+1)\rfloor}$$

This gives $M=13$ for $k=2$ and $M=1+8\cdot 9=73$ for $k=3$.

hint

If one of the four numbers $M+1, M+2, M+3 , M+4$ is prime, the statement is not true.

the primes are
$2,3,5,7,11,13,17,19,23,29,31,37... $

So, the possible values of $M $ are

$$23,31,32,47,53$$
you check them one by one until 53 which works.


when is the particle speeding up and when is it slowing down Based on this graph i have to figuere out when the particle is speeding up and when it is slowing down.
My understanding is that when velocity and accelaration have the same sign then we are speeding up. however if velocity and accelaration have different signs then we are slowing down. Using this concept i applied it to this problem and i came up
speeding up using interval notation.
$(0,1)and(3,4) $
I'm getting the answer wrong can someone please help me understand the missing piece to this puzzle.
Thanks
Miguel

You've got the right idea, but not the right execution. Let's call the direction of positive $x$ "up" and the opposed direction down.

It is moving upward--has a positive velocity--on $(0,1)$ and $(3,4)$. It is moving downward--has a negative velocity--on $(1,3).$ Put another way, the graph's height is increasing on $(0,1)$ as we move to the right--likewise in $(3,4)$--and decreasing on $(1,3)$ as we move to the right.

It is accelerating upward--has a positive acceleration--on $(2,4)$. It is accelerating downward--has a negative acceleration--on $(0,2)$. Put another way, the graph's slope is increasing on $(2,4)$ as we move to the right, and the graph's slope is decreasing on $(0,2)$ as we move to the right.

Now check where the signs match.

A particle usually speeds up when the velocity and the acceleration have the same signs. It slows down when the acceleration and velocity signs are different. My advice (even though it might take a while) would be to sketch the curve from the given function. Then, you can find where the particle speeds up and slows down.


Inverse element in $\mathbb Z_{493}$ I am trying to find the inverse for $348$ in $\mathbb Z/493$.

So since $x \cdot x^{-1} = 1$, I have tried to solve it by using the extended euclidean algorithm. $\gcd(493, 348):$
\begin{align}
493 &= 348 \cdot 1 + 145\\
348 &= 145 \cdot 2 + 58\\
145 &= 58 \cdot 2 + 29\\
58 &= 29 \cdot 2 + 0
\end{align}
Then the linear combination:
\begin{align}
29 &= 145 - 52 \cdot 2\\
29 &= 145 - (348 - 145 \cdot 2) \cdot 2\\
29 &= -2 \cdot 348 + 5 \cdot 145\\
29 &= -2 \cdot 348 + 5 \cdot (493 - 348)\\
29 &= -7 \cdot 348 + 5 \cdot 493
\end{align}
Question: If there is an inverse, is it $-7$ here? Is that the correct way to find an inverse ?

I appreciate every hint.

Quite simply, this inverse does not exist.  Given an $x \in \mathbb{Z}$, the equivalence class to which $x$ belongs is invertible in $\mathbb{Z}_m \iff \gcd(x, m) = 1$.

In such a case, you would proceed exactly as you tried here.  Using  the extended Euclidean algorithm, we can find integers $a$ and $b$ such that $ax + bm = 1$.  Modding out both sides of this equation by $m$ yields $ax \equiv 1 \pmod{m}$; thus, the equivalence class to which $a$ belongs is the inverse of that to which $x$ belongs.

The inverse here is found from the continued fraction, not the residue.

               0       A   B     C
  493          1     493         1
  348    1     1     349   1     1
  145    2     3     144   2     3
   58    2     7      61   2     7
   29    2    17      22   2    17
    0                 17   1    24
                       5   3    89
                       2   2   202
                       1   2   493

This actually finds that the LCM of 493 and 348 is 29.  Therefore, this number does not have an inverse, relative to 493.  Here we find 202*493 is +/- 1, there are 8 numbers in the continued fraction, so 202 * 493 = -1, and we take 493-202=291 as the required answer.

The columns A start with the modulus (493) and the number for which the reciprocal is required (349).  Column A continues with remainders or the previous two entries.

Column B consists of the whole parts, eg 493 = 1*349+144.

Column C derives from column B, in C(n) = B(n).C(n-1)+C(n-2), where C(0)=0, C(1)=1.  The last entry should be the modulus itself (493).  The second last entry is + 1/x if there is an odd number of rows, and -1/x if there is an even number of rows.


Why is convexity more important than quasi-convexity in optimization? In the mathematical optimization literature it is common to distinguish problems according to whether or not they are convex. The reason seems to be that convex problems are guaranteed to have globally optimal solutions, so one can use methods such as gradient (steepest) descent to find such solutions, but I am not convinced.

For example, the function $|x|^{0.001}$  is not convex (see the yellow-shaded area in the picture below) but it has a single global minimizer.

So why is convexity so important in optimization? Why is e.g. quasiconvexity often not enough?

There are many reasons why convexity is more important than quasi-convexity in optimization theory.  I'd like to mention one that the other answers so far haven't covered in detail.  It is related to Rahul Narain's comment that the class of quasi-convex functions is not closed under addition.

Duality theory makes heavy use of optimizing functions of the form $f+L$ over all linear functions $L$.  If a function $f$ is convex, then for any linear $L$ the function $f+L$ is convex, and hence quasi-convex.  I recommend proving the converse as an exercise: $f+L$ is quasi-convex for all linear functions $L$ if and only if $f$ is convex.

Thus, for every quasi-convex but non-convex function $f$ there is a linear function $L$ such that $f+L$ is not quasi-convex.  I encourage you to construct an example of a quasi-convex function $f$ and a linear function $L$ such that $f+L$ has local minima which are not global minima.

Thus, in some sense convex functions are the class of functions for which the techniques used in duality theory apply.

I'm not an expert, but just from eyeballing your function you can see that gradient descent would not converge to the global minimizer.  Gradient descent looks for places where the gradient (which is just the derivative for 1-dimensional functions like the one you provide) is zero.  In a convex function, as you approach the minimum, the gradient approaches zero (and vice versa).  But you can see that, as you approach the minimum in your function, the derivative is getting further and further from zero.

If you try to do gradient descent on that function, you'll actually move further and further away from the minimum, since the gradient is minimized as you approach (positive or negative) infinity.

So at least one reason convexity is so important in optimization is that the global minimum is also the unique critical point (place where the gradient is zero), which allows you to search for one by searching for the other.  In some sense, you pick a point, look around a little bit, and move in the direction that makes your gradient smaller.  Repeat this enough and you'll get to a really small gradient, which corresponds to being close to the minimum.


Calculating the limit $\lim\limits_{x \to 0^+} \frac{\sqrt{\sin x}-\sin\sqrt{ x}}{x\sqrt{x}}$ Calculate $$\lim\limits_{x \to 0^+} \dfrac{\sqrt{\sin x}-\sin \sqrt{x}}{x\sqrt{x}}$$
  without use Taylor serie and L'Hôpital.

$$\lim\limits_{x \to 0^+} \dfrac{\sqrt{\sin x}-\sin \sqrt{x}}{x\sqrt{x}}\cdot\dfrac{\sqrt{\sin x}+\sin \sqrt{x}}{\sqrt{\sin x}+\sin \sqrt{x}}=\lim\limits_{x \to 0^+} \dfrac{\sin x-\sin^2\sqrt{x}}{x\sqrt{x}(\sqrt{\sin x}+\sin \sqrt{x})}$$

now what ?

$$\mathrm L =\lim\limits_{x \to 0^+} \dfrac{\sqrt{\sin x}-\sin \sqrt{x}}{x\sqrt{x}}= \lim\limits_{x \to 0^+} \dfrac{(\sqrt{\sin x}- \sqrt{x}) -(\sin \sqrt{x} - \sqrt{x})}{x\sqrt{x}} \\=  \overbrace{\lim\limits_{x \to 0^+}\dfrac{(\sqrt{\sin x}- \sqrt{x})}{x\sqrt{x}}}^{\large \rm L^\prime} - \overbrace{\lim\limits_{x \to 0^+}\dfrac{(\sin \sqrt{x} -  \sqrt{x})}{x\sqrt{x}}}^{\large \rm L^{\prime\prime}}$$
Let $y = \sqrt{x}$
Then $${\rm L^{\prime\prime}} = \lim\limits_{x \to 0^+}\dfrac{(\sin \sqrt{x} -  \sqrt{x})}{x\sqrt{x}} = \lim\limits_{x \to 0^+}\dfrac{(\sin y -  y)}{y^3} = -\dfrac 16$$
For the other limit,
$${\rm L^{\prime}}=\lim\limits_{x \to 0^+}\dfrac{(\sqrt{\sin x}- \sqrt{x})}{x\sqrt{x}} =\lim\limits_{x \to 0^+}\dfrac{(\sin x- x)}{x^3}\dfrac{x\sqrt{x}}{(\sqrt{\sin x} + \sqrt{x})}= \dfrac {-1}6\lim\limits_{x \to 0^+}\dfrac{x}{\left(\sqrt{\dfrac{\sin x}{x}} + 1\right)} = 0$$
Hence our limit is $\mathrm L = \dfrac {1}{6}$.

I used $$\lim_{x \to 0} \dfrac{\sin x - x}{x^3} = \dfrac{-1}6$$
Are all limits solvable without L'Hôpital Rule or Series Expansion ?


What is the limit $x\to\infty$ of $x^n \cdot e^{-x}$? What is the limit $x\to+\infty$ of $x^n \cdot e^{-x}$ ?  My guess is that it is $0$ because the exponential will go faster to $0$ than the $n$th power go to $+\infty$, but I don't see which theorem to apply to reach the result.

Thanks for any help !

This can be shown without using L'Hospital as follows:

Let $$\exp:\mathbb R\rightarrow\mathbb R,~x\mapsto\sum\limits_{n=0}^\infty \frac{1}{n!}x^n$$ and $P(x)$ a polynomial. We show that $$\lim\limits_{x\to\infty}\frac{P(x)}{\exp(x)}=0.$$

Let $n$ be the degree of $P$, then there exists $c>0,R>0$ with $|P(x)|\leq c|x|^n$ for alle $|x|\geq R$. From the series representation we directly obtain $$\exp(x)>\frac{1}{(n+1)!}x^{n+1}$$ for all $x\geq 0$. Using this we get $$\left|\frac{P(x)}{\exp(x)}\right|\leq \frac{cx^n}{\frac{1}{(n+1)!}x^{n+1}}=\frac{c(n+1)!}{x}$$ for all $x\geq R$. As $$\lim\limits_{x\to\infty}\frac{c(n+1)!}{x} =0$$ we are done.

The answer is $n!$. Bring $e^{-x}$ in denominator and differentiate $n$ times


How can I produce $\sqrt{4+4t^2+t^4}=2+t^2$ $\sqrt{4+4t^2+t^4}=2+t^2$

That is the solution my calculator produces but I don't understand how it gets that solution.

If I factor it I get:

$\sqrt{4(1+t^2)+t^4} = 2\sqrt{1+t^2+t^4}$

$2\sqrt{1+t^2(1+t^2)}$

So how do I get from there to $2+t^2$

Since $(2+t^2)^2=4+4t^2+t^4$ then you can deduce your relation.

You can factor it a different way. Factor the entire thing, not just the first two terms.


Prove that $d(x,y)=\sum_{i=1}^\infty \frac{|x_i - y_i|}{2^i}$ converges Question: Let S be the set of sequences of $0$s and $1$s. For $x = (x_1, x_2, x_3, ...)$ and $y = (y_1, y_2, y_3, ...)$. Define

$d(x,y)=\sum_{i=1}^\infty \dfrac{|x_i - y_i|}{2^i}$

Proof the infinite sum in the definition of $d(x,y)$ converges for all $x$ and $y$.

Incomplete answer: Since the max of $d(x,y)$ happens when all elements of one of $x$ and $y$ is $1$ and the other is $0$, and the min of $d(x,y)$ happens when $x_n=y_n$ for all indices $n$, then

$0=\sum_{i=1}^\infty \dfrac{0}{2^i}\leq d(x,y)=\sum_{i=1}^\infty \dfrac{|x_i - y_i|}{2^i}\leq \sum_{i=1}^\infty \dfrac{1}{2^i}=1 $

But how to prove that $d(x,y)$ converges to some point when $x$ and $y$ have fixed arbitrary elements?

Thank you.

The sequence $a_n = \sum_{k=1}^n \frac{|x_k-y_k|}{2^k}$ is monotonic and bounded from above, thus converges.

$$a_n \le \sum_{k=1}^n \frac{1}{2^k} \le  \sum_{k=1}^\infty \frac{1}{2^k} = 1$$

You have the right idea of bounding the the infinite sum. Now consider what the bounds would be if we started the series at the $n^{th}$ term. We can get that

$\sum_{i=n}^\infty \dfrac{|x_i-y_i|}{2^i}\leq \dfrac{1}{2^{n-1}}\to0$

as $n \to\infty$, which shows that the series converges.


Sum of a series given the formula for the nth term So I have a series $1+1+3+9+23+\cdots$ and the formula given to find the $n^{th}$ term $R_n=1-2n+2^n$. The first part is to verify that the term $115$ exists in this series. I knew that if I just solve for $n$ in $1-2n+2^n=115$ and if $n$ is an integer solution then the term $115$ exists. In fact I could't solve it and I just guessed and checked that $n=7$. So can someone guide me to solve the following equation?
\begin{align}
1-2n+2^n &= 115\\
-2n+2^n &= 114
\end{align}
Second part of the question is, how can I find the sum of the first 25 terms of this series?

Since

$$ 1-2\cdot n+2^n\ =\ 115 $$
then
$$ 2^n > 115 $$

hence

$$ n > \log_2(115) > 6 $$

Also, by induction,

$$ \forall_{n\ge 4}\quad 2^{n-1} > 2\cdot n-1 $$

Thus,
$$ \forall_{n\ge 4}\quad 2^{n-1}\ < 2^n-2\cdot n + 1
       \ =\ 115 $$

hence (for $\ n\ge 4$)
$$ n-1\ <\ \log_2 115 < 7 $$
i.e.
$$ n\ <\ 8 $$

It follows that IF there is any integer solution $n$
it has t be $\ n=7.\ $ But is there any? The calculation

$$ 2^7-2\cdot 7 + 1\ =\ 115 $$

shows that indeed, $\ n:=7\ $ is a solution, and it is the ONLY integer solution.

REMARK But is there any other REAL solution? -- it would still have to satisfy inequality

$$ 6 < n < 8 $$

But looking at the derivative of $\ 2^x-2\cdot x-1,\ $
which is strictly positive when $\ x>6,\ $ we see that there is no real solution $\ n\ $ but $\ n=7.$

Your guess and check is a good approach when you know $n$ must be an integer.  $2^n$ changes quickly compared with $2n$, so just ignore the $2n$ and find the $n$ which is close.  You can prove that $2^n-2n$ is increasing with $n$ for $n \gt 2$, so if you miss you can just increment $n$ by $1$ and have a proof that there is no $n$.

The $1-2n$ part is an arithmetic series and the $2^n$ part is geometric.  Sum each and add them together.


limit of n sin (pi/n) as n --> infinity? This question has been asked in:

Does the sequence converge or diverge? $an+ n[\sin(\pi/n)]$

That post/title is messy to understand, so please refer to the problem as stated in my title.

I don't understand the answers and am unable to comment, so I had to re-ask the question here. In the first answer, it is stated:

"It diverges, but for a different reason. sinx < x for x∈(0,∞)"
Why is this true?

It then states, "sinx≥0 for x∈(0,π/2]"
I understand this is true, but do not see how this is applied once the inequality is set up for the squeeze theorem.

My own personal work is as follows:

-1 <= sin (pi/n) <= 1
-n <= n sin(pi/n) <= n
using the range of sin(x), then multiplying by n.

The limits of both sides as n--> inf are not the same, so I'm unable to proceed from here.

Thanks!

NOTE:
The textbook answer is given as "pi."

Rewrite as $$\lim_{n\rightarrow \infty}\pi \frac{\sin(\pi/n)}{\pi/n}$$

We know that $\lim_{x\rightarrow 0}\frac{\sin(x)}{x} = 1$ so that is enough to determine that the limit exists.

We will use the following 2 facts:

For $x<10^\circ, \sin x\approx x$
As $n\to\infty$, $\frac{\pi}{n}\to0$

Fact 2 implies that, for large $n$, $\sin \frac{\pi}{n} \approx \frac{\pi}{n}$.  Then
$$\lim_{n\to\infty} (n\cdot\sin(\frac{\pi}{n})) = \lim_{n\to\infty} (n\cdot\frac{\pi}{n}) = \lim_{n\to\infty} \pi = \pi$$


Why can't the quadratic formula be simplified to $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-b\pm(b-2)\sqrt{ac}}{2a}$? I am currently taking Algebra 1 (the school year's almost over ), and we just learned the quadratic formula, another method to solve quadratic equations:

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

However, this strikes me as not being simplified. Isn't it more proper to write it like this?
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
$$x=\frac{-b\pm(\sqrt{b^2}-\sqrt{4ac})}{2a}$$
$$x=\frac{-b\pm((b-2)\sqrt{ac})}{2a}$$
$$x=\frac{-b\pm(b-2)\sqrt{ac}}{2a}$$

Why isn't $x=\frac{-b\pm(b-2)\sqrt{ac}}{2a}$ more commonly used as the quadratic formula??

I'm sorry for my typo, I have edited it.

I have now edited in my steps, per request of commenters.

Because $\sqrt{b^2 - 4ac}\neq 2b\sqrt{ac}$. Say for example that $c = 0$ and $b\neq 0$. Then you have
\begin{align*}
\sqrt{b^2 - 4ac} &= \sqrt{b^2}\\
&= \left|b\right|\\
&\neq 0,
\end{align*}
so you can see that this simplification cannot be correct.

It seems that in your proposed simplification, you have completely disregarded the subtraction occurring in the radical. Moreover, $\sqrt{b^2} = \left|b\right|$, not just $b$. To see this with an example, take $b = -1$. Then $\sqrt{(-1)^2} = \sqrt{1} = 1 = \left|-1\right|\neq -1$.

EDIT:

Again, the simplification is incorrect. While it is true in general that $\sqrt{a^2b} = \left|a\right|\sqrt{b}$, this is not the situation you are in here:
$$
b^2 - 4ac\neq (b - 2)^2 ac = (b^2 - 4b + 4)ac = b^2 ac - 4abc + 4ac.
$$
You seem to have made a few mistakes here (if I'm to take a stab at the reasoning behind the simplification): first you've incorrectly simplified $b^2 - 4ac$ as $(b^2 - 4)ac$ (which is not true, because the first term in the former has no $ac$), and then you've simplified $b^2 - 4$ as $(b - 2)^2$, which is also not true (take $b = 0$ to see why). In general, $(x + y)^n\neq x^n + y^n$: this is a common mistake algebra learners make! Remember that when expanding $(x + y)^2$, we need to use the distributive property, and not simply regard squaring as linear:
\begin{align*}
(x + y)^2 &= (x + y)(x + y)\\
&= x^2 + yx + xy + y^2\\
&= x^2 + 2xy + y^2.
\end{align*}

I think a better way to parse the quadratic equation is as:

$$ x^{*} = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} $$

The left term ($-\frac{b}{2a}$) is the x-coordinate of the axis of symmetry (often taught in Algebra as a formula, but easily recognized to students of calculus as the solution to the first-order condition of a quadratic).

The right-hand term is the "margin" around the axis of symmetry giving how far to the left/right of the axis we find the zeroes.

The numerator of the right-hand term is the discriminant, the sign of which determines the number of zeroes there are.

The denominator of the right-hand term shows that as $a$ increases, ceteris paribus, the zeroes will move closer to the axis of symmetry -- intuitively, $a$ is the "stretch" factor, and higher $a$ means the parabola will be more "squeezed" (relatively speaking). The more "squeezed" the parabola, the closer its zeros will be to the "middle".


Solutions of exponential diophantine equation How would I go about finding the solutions to the exponential diophantine equation  $18n+10=2^k$ ?

Hint: You want $2^k \equiv 10 \mod 18$.  Certainly $2^k \equiv 0 \equiv 10 \mod 2$ for $k \ge 1$, so it is sufficient for $2^k \equiv 1 \equiv 10 \equiv \mod 9$.  So try a few powers of $2$ until you find one congruent to $1$ mod $9$...


Elementary question about a fake-proof and greatest common divisors I have a question to an excercise - for which I have a wrong solution - and I wanted to ask you to help me understand my thinking error. The excercise was as follows:

Let $a, b, n \in \mathbb{N}$. Show that $\gcd(a,b) = \gcd(a,b+na)$.

My solution was (in a nutshell):

Let $d := \gcd(a,b)$.
Then I showed that $d$ is a common divisor of both $a$ and $b + na$
Then I showed, that any common divisor $c$ of $a$ and $b+na$ is smaller-or-equal than $d$
After that, I concluded that by definition $d = \gcd(a,b+na)$, and with the definition of $d$ I would have $\gcd(a,b) = \gcd(a,b+na)$, so my proof was completed.

However, my tutor did not accept the proof. He showed me the 'correct' proof, in which I woould have $d := \gcd(a,b)$ and $e := \gcd(a,b+na)$ and then demonstrate $d \leq e \leq d$.

Anyway, I do understand his proof. But I still do not understand where my thinking error is. After all, if my starting line would be, let's say, $d := 9$ and I would have been able to show the steps afterwards, then I would be able to conclude $9 = \gcd(a,b+na)$, no?

Any comments would be welcome, thank you very much in advance!

EDIT
Wow, you guys answered fast! Thank you very much. I will ask my tutor on thursday again. For completeness sake, I will formulate my complete proof (my original proof is in german, so maybe there will be something lost in translation, the [Reference] are references to our script).

Let $a,b,n \in \mathbb{N}$. Let $d := \gcd(a,b)$. By definition it means $d \mid a$ and $d \mid b$, so with [Reference] the following holds true: $d \mid (b + na)$. Therefor, $d$ is a common divisor of both $a$ and $b + na$. We will show now, that $d$ is the greatest common divisor.

Let $c$ be any common divisor of $a$ and $b + na$. Therefor $c$ divides $a$, and as such $c \mid na$ und therefor $c \mid ((b + na) - na) = b$ (again because of [Reference]). So $c$ divides $b$, and therefor $c$ is a common divisor of both $a$ and $b$. Since $d$ was the greatest common divisor of $a$ and $b$, we have $c \leq d$.

By definition of the greatest common divisor we get $d = \gcd(a,b+na)$.

End of Proof. My tutor wrote, that I only showed $\gcd(a,b+na) \leq \gcd(a,b)$ and also need to show the other direction. Then he showed me the 'correct'/'ideal' proof today. But I will ask him again on thursday!

Your proof is perfectly correct.  Either you have misunderstood what your tutor said, you communicated your proof to your tutor incorrectly, or your tutor is just wrong.

End of Proof. My tutor wrote, that I only showed $\gcd(a,b+na) \leq gcd(a,b)$ and also need to show the other direction.

I am pretty sure, that $\gcd(a,b+na) \geq gcd(a,b)$ follows from $d:=\gcd(a,b)$ and $d\mid(b+na)$ (plus the definition of $\gcd$).


Why is Cantor's diagonal argument usually gets applied to the interval $[0,1)$ and not $[0,1]$? I noted that many formulations of the Cantor's diagonal argument, using base 10, consider only the interval $[0,1)$.

It's mostly convenience, since you don't want to be worried about the integral part of the real number, and in $[0,1)$ it is always $0$.

Generally the diagonal argument when applied to the decimal base, uses digits other than $9$ and $0$ for the "digit change" anyway, so you're not in the situation where you usually run into $0.999\ldots$ as your diagonal number.

Moreover, note that if $[0,1)$ is uncountable, then certainly $[0,1]$ is uncountable. So that really doesn't make much of a difference. (And besides the point, there is a bijection between the two sets anyway.)

Because they never read Cantor's actual diagonalization proof. It was his second proof of the proposition "there are uncountable sets," and the reason he gave for publishing it was that it was both simpler than his first proof, which used the real numbers as the example set, and that it didn't use that set as the example.

Let me repeat that: Geoff Cantor DID NOT apply diagonalization to the real numbers. He used infinite-length strings of the two characters 'm' and 'w'. For example, "mwmwmw...".

Now, if you replace 'm' and 'w' with '0' and '1', you can think of his strings as the binary representations of every real number in [0,1]. And you are right that it should properly include 1=0.111111... .

The reason that it is usually taught using the real numbers, is because this elegantly-simple proof is "dumbed-down" to be taught to High Schoolers, and they already understand infinite decimal representations. Simplifications include (1) using decimals fractions instead of strings, (2) sometimes ignoring that you need to avoid decimals that end in all 9s, (3) treating the diagonalization process as a sequence of steps instead of as a complete, infinite string, and (4) calling it a Proof by Contradiction when Cantor didn't, and it in fact fails as one (the proof is still valid, but the alleged contradiction does not follow from all of what is assumed, just part of it).

Unfortunately, most of these simplifications have the negative effect of making inquisitive students doubt it. One issue is the one you raised, that 1=0.99999... . This can be gotten around if you never use '9' as the replacement character, but that is unnecessary if you use Cantor's strings.


How do I interpret this with logic notation? "not A doesn't mean not B" $$
\neg A \rightarrow \neg B \equiv B \rightarrow A
$$
Is this correct?

As negation have more high priority, then implication, we have
$$\neg A \Rightarrow \neg B \equiv \\
(\neg A) \Rightarrow (\neg B) \equiv \\
\neg(\neg A) \lor (\neg B) \equiv\\
A \lor (\neg B) \equiv\\
B \Rightarrow A$$
So, yes, you are correct.

This is an implication so if A is true and b is true it is true
if not A implies not B then A is False implies B is False so the implication is still true.
A implies B is the converse of the statement and the truth of a statement does not imply the truth of the converse


What is this proof syntax (Hoare 1974)? I'm reading the seminal "Monitors" paper by Hoare. On page 4 he proceeds with a logical proof using syntax I've never seen before, and neither know what it's called or how to properly read it.

Excerpts:

  ... This gives the proof rule for waits:

  g {b.wait} g & B

  ... Thus the proof rule for a signal is:

  g & B{b.signal}g

How does one interpret this syntax? When I think of proofs I think of quantifiers and implications, of which I see none here.

As Zhen Lin points out, this Hoare logic, a logical calculus. It is designed to model pre- and post-conditions of programs:

{ P }
statement;
{ Q }

This means that if property P (always) holds before the statement, then Q (always) holds after its execution.

You can think of it as program annotation; you would start with some preconditions, write properties between lines and get postconditions of your programs. If these imply the desired property (given by a formal specification) your program is correct. Note that you can not annotate arbitrary properties, of course. You have to conform to rules like

$\qquad\displaystyle \frac{\{P\}S_1\{Q_1\} \quad \{Q_2\}S_2\{R\}}{\{P\}S_1;S_2\{R\}}$

or

$\qquad\displaystyle \frac{}{\{P[c/x]\}x:=c\{P\}}$

that effectively define your language's semantics.

In proving programs correct ,some assertions are made of the format {P} Code {Q} where P and Q are Per and post conditions respectively.

For example:

{ x is even }

x = x+1

{ x is even }

Here pre condition and post condition are  shown, by using these we can easily check our programs correctness.


Understanding Solving Simple Equations I'm having trouble explaining/understanding how the theorem

(1) $\forall x,y,z\in\mathbb R(x=y\iff x+z=y+z)$

applies to solving simple equations. Suppose we are working with the equation $p_1(x): x+1=0$ and we wish to determine the values of $x$ for which $p_1(x)$ is true. Rather than carrying out the usual "add $-1$ to both sides..." I want to understand why (1) says that $p_1(x)$ is equivalent to $p_2(x): x=-1.$

The theorems says:

$x +1 = 0 \iff x + 1 + z = 0 + z$ for all $z$.

In particular $x+1 = 0 \iff x+1 + (-1) = 0 + (-1)$

So $x+1 = 0 \iff x = -1$.

The theorem says if you ever come across $x + a = c$ you can solve it by adding $-a$ to both sides because $x + a = c \iff x + a +(-a) = c +(-a)$.

Well, let's use your example. Say we got the formula $p_1(x) : x + 1 = 0$. We state $x = y$, Just like you said. Since we got $x + 1$ on the left hand side, it must be that $z=1$. Now given your definition of $x$, $y$ and $z$, $x = y$ and $x+z=y+z$ are interchangeable. That means that if we get rid of $z$ on both sides, it states $x=y$. $$x + z = 0$$
$$y + z = 0$$
$$y = -z$$
$$x = y = -z$$
Since $z=1$ $x=y=-1$.
Just as clarification, there is no real use in using the way we just did. We better could've said $x=0-1=-1$ at the beginning. One of the explanations of this is that, not only is the formula clearly just another way of writing this, it also only has 1 unknown (or 2 of $z$ is unknown too). So I hope I made clear what the theorem does, but basically it's still the same.


If $\cos 25^\circ + \sin 25^\circ = k,$ then what is $\cos 20^\circ$? Question:

If $$\cos 25^\circ + \sin 25^\circ = k,$$ then what is $\cos 20^\circ$?

What I did:

I tried to square both sides, and obtained that $\sin 50 = k^2 -1$, however, this didn't get me anywhere. Then I tried splitting 25 into 20 + 5 but that didn't get me anywhere either. Can someone just point me in the right direction?

Hint

Take into account that $20 = 45 -25$. Develop $\displaystyle\cos(45^\circ-25^\circ)$ and remember that $\displaystyle\cos(45^\circ)=\sin(45^\circ)=\frac{\sqrt 2}2$ and see what happens.

I am sure that you can take from here.

$cos 25^\circ+ sin 25^\circ=k$

$\implies cos (45-20) + sin (45-20) =k$

$\implies cos 45\cdot cos 20+ sin 45\cdot sin 20 + sin45\cdot cos 20 -cos 45\cdot sin 20=k$

$2\cdot(\frac{1}{\sqrt2})cos 20^\circ=k$

$cos 20^\circ=k\cdot(\frac{\sqrt{2}}{2})$


Evaluate $\sum_{n=1}^\infty \frac{(-1)^{n+1}n^2}{n^4+1}$ Evaluate
  $$\sum_{n=1}^\infty \frac{(-1)^{n+1}n^2}{n^4+1}$$

Does anyone have any smart ideas how to evaluate such a sum? I know one solution with complex numbers and complex analysis but I'm looking for some more smart or sophisticated methods.

I would not say that it is elegant, but:

The form $n^4+1$ in the denominator suggests that one should be able to get this series by expanding a combination of a hyperbolic and trigonometric function in a Fourier series.

Indeed, after some trial and error, the following function seems to work:

$$
\begin{gathered}
\left(\cos \left(\frac{\pi }{\sqrt{2}}\right) \sinh \left(\frac{\pi }{\sqrt{2}}\right)-\sin
   \left(\frac{\pi }{\sqrt{2}}\right) \cosh \left(\frac{\pi }{\sqrt{2}}\right)\right)\cos \left(\frac{x}{\sqrt{2}}\right) \cosh \left(\frac{x}{\sqrt{2}}\right) \\
+ \left(\cos \left(\frac{\pi }{\sqrt{2}}\right) \sinh \left(\frac{\pi }{\sqrt{2}}\right)+\sin \left(\frac{\pi }{\sqrt{2}}\right) \cosh
   \left(\frac{\pi }{\sqrt{2}}\right)\right)\sin \left(\frac{x}{\sqrt{2}}\right) \sinh
   \left(\frac{x}{\sqrt{2}}\right)
\end{gathered}
$$

It is even, and its cosine coefficients are
$$
\frac{\sqrt{2}\bigl(\cos(\sqrt{2}\pi)-\cosh(\sqrt{2}\pi)\bigr)(-1)^{n+1} n^2}{\pi(1+n^4)},\quad n\geq 1.
$$
(The zero:th coefficient is also zero). Evaluating at $x=0$ (the series converges pointwise there) gives
$$
\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}n^2}{1+n^4}=
\frac{\pi\left(\sin
   \left(\frac{\pi }{\sqrt{2}}\right) \cosh \left(\frac{\pi }{\sqrt{2}}\right)-\cos \left(\frac{\pi }{\sqrt{2}}\right) \sinh \left(\frac{\pi }{\sqrt{2}}\right)\right)}{\sqrt{2}\bigl(\cosh(\sqrt{2}\pi)-\cos(\sqrt{2}\pi)\bigr)}\approx 0.336.
$$

Partial elements methods  should works but are boring.

Let's try smthg else : having $n^4$ and $n$ of the same parity

$$ (-1)^{n^4}=(-1)^n  $$ so
$$ u_n= \int_0^1 n^2(-t)^{n^4} $$

So you can begin to have fun with it (I'm searching as I'm writting )


Prove that if $|x|<1$ then $x^{6}<1$ I am trying to prove that if $|x|<1$ then $x^{6}<1$ and that if $x^{6}<1$ then $|x|<1$.

For the first part, I thought of first considering $0<x<1$. Multiplying by $x$ (which is positive) I then get $0<x^{2}<x<1$. And repeating the multiplication I would end up with $x^{6}<1$.

Then I would consider $-1<x<0$. But now multiplying by $x$ (which is negative) gives $-x<x^{2}$ and I dont see how to get $x^{2}<1$ from here and then $x^{6}<1$.
There is probably a more elegant way of going about this?

For the second part I thought of using the contrapositive and prove $x\leq -1$ or $x\geq 1$ then $x^{6}\geq 1$, and then I would try to use the previous proof to break this into two cases.

$x\leq -1$ then $x^{6}\geq 1$ and $x\geq 1$ then $x^{6}\geq 1$.

The argument that dxiv made in a comment is, I think, the most elegant argument.  As comments are ephemeral, I'll reiterate it here (with some additional details):  note that
$$ \frac{|x|^6 - 1}{|x|-1} = |x|^5 + |x|^4 + |x|^3 + |x|^2 + |x| + 1. \tag{1}$$
This is actually a specific case of a much more general result: for any natural number $n$,
$$ \frac{t^n-1}{t-1} = \sum_{j=0}^{n-1} t^j. $$
This can be proved by an induction argument after noticing that
$$t^{n+1} - 1 = t^{n+1} - t^n + t^n -1 = t^n(t-1) + t^n - 1.$$  In this particular case, we take $t = |x|$.  In any event, since all of the terms on the right-hand side of (1) are nonnegative (and $1>0$, i.e. $1$ is strictly positive), it follows that
$$ \frac{|x|^6 - 1}{|x|-1} > 0. \tag{2}$$
Note that
$$ \frac{a}{b} > 0 \iff (a>0 \land b > 0) \lor (a< 0\land b < 0). $$
That is, a fraction is positive if and only if both the numerator and denominator have the same sign.  Applying this to (2), either
$$ |x|^6 - 1 < 0 \qquad\text{and}\qquad |x| - 1 < 0, \tag{3} $$
or
$$ |x|^6 - 1 > 0 \qquad\text{and}\qquad |x| - 1 > 0. $$
Therefore if we assume that $|x|<1$ it immediately follows from (3) that $|x|^6<1$, and vice versa.

Alternatively, we can make your approach work.  The first part of your argument is basically an induction argument.  We can actually strengthen that result a bit and get the following:

Proposition: If $|x| < 1$ then $|x|^n < 1$ for any natural number $n$.

Proof:  The proof is by induction.  Assume that $|x|< 1$, and note that $|x|^1 < 1$, providing a base for the induction.  Now suppose that $|x|^k < 1$.   Then
$$ |x|^k < 1 \implies |x|^k \cdot |x| < 1 \cdot |x| \implies |x|^{k+1} < |x|. $$
But $|x|<1$, which proves the result.$\tag*{$\blacksquare$}$
Taking $n = 6$ gives gives the result in the original question.  Then, as you note, the converse requires an additional argument.  Following your instinct of attempting contraposition, you might consider the following proposition:

Proposition: Let $n$ be a natural number.

If $|x| < 1$ then $|x|^n < 1$,
if $|x| = 1$ then $|x|^n = 1$, and
if $|x| > 1$ then $|x|^n > 1$.

There may be more elegant arguments, but the induction proof above can be used, mutatis mutandis, to prove the two new statements.  This gives a kind of trichotomy, from which you can conclude that if $|x|^n < 1$, then $|x| < 1$.

suppose $0<x<1$ and $x^6\ge1$

$x^6=x\cdot x^5$ and since $0<x<1$ multiplying by $x$ is like scaling a value making it $100\cdot x$% of the original value, which is smaller than $<1\cdot100$%.

So $x^6=x\cdot x^5<x^5<...<x\cdot x<x<1$ and that's impossible


Understanding Linear Algebra Geometrically - Reference Request I know geometry and I know linear algebra but when I understand a linear algebraic concept geometrically, my head just explodes and things just become so much clearer and easier to understand...not to mention easier to remember or figure out its properties and explain them to others.

Here are a few examples.

Orthogonal matrices - If you think of an orthogonal matrix as a rotation then some of its properties are obvious. Orthogonal matrices are always invertible because rotations can simply be reversed. They always preserve the Euclidean norm because rotating a vector doesn't change its length. Orthogonal matrices forming a group is also easy to see because it is easy to see them satisfying the group axioms.
Determinant - The determinant of a linear transformation can be understood as follows. Start with the (chosen) basis of your domain. It forms a parallelepiped. Call it $P$. It has a certain volume $V(P)$. Now apply your linear transformation $T$ to the chosen basis. A new parallelepiped $T(P)$ is formed and its volume in the range space (embedded in the codomain) is now $V(T(P))$. The determinant (in absolute value) is the ratio of the new volume to the old one. This intuitively explains, for example, why the determinant is zero for non-invertible transformations. The dimension of such a transformation will always be strictly less than the dimension of the domain/codomain so the volume of the transformed parallelepiped will always be zero. I always imagine a parallelepiped in $\mathbb{R}^3$ collapsing onto a plane. This also explains why the determinant of an orthogonal matrix is always $\pm1$ because rotating a parallelepiped won't change its volume. In addition, it kind of helps with the Jacobian determinant and why is the Jacobian "necessary" when transforming variables.
Singular value decomposition - Every matrix having an SVD says the fantastical fact that any linear transformation can be considered a rotation, then a dilation (different directions by different factors), and then a rotation again.
Projection matrices - Imagine an arbitrary vector's shadow onto a line or a plane. I imagine a vector collapsing onto its shadow and properties like $P^2=P$ are immediate for any projector operator $P$. Take this and run with it.

  My question is, can anyone point to some good reading material where a geometric interpretation of various linear algebra concepts is offered?

This could be anyone's class/teaching notes, published papers, something from recreational mathematics, or just a good book.

It might be slightly off-topic since you are searching for reading material, but the 3blue1brown YouTube channel is dedicated to this sort of geometric intuitions. More specifically, it presents abstract concept with geometric animations, spanning from neural networks to Fourier transform.

Although it is not restricted to such matter, it has an excellent playlist called Essence of linear algebra which adresses what you seek.

The vector $y-x$ connects the end of $x$ to the end of $y$.