How to find the value of Grandi's series using Ramanujan's summation I can't figure out how to solve the infinite sum of

$\sum ^{\infty }_{n=0}\left[( -1)^{n}\right]$

I know that Srinivasa Ramanujan solved it and I couldn't figure it out with Ramanujan's summation. Here's what I tried.

Note: I'm only interested in solving this divergent sum by means of Ramanujan's summation. I know about the other methods, but am unable to solve it with Ramanujan's summation. So here's what I tried

$ \begin{array}{l}
\sum ^{\infty }_{n=1}\left[( -1)^{n}\right]\overset{\Re }{=} -\frac{1}{2} +i\int\limits ^{\infty }_{0}\frac{e^{-\pi x} -e^{\pi x}}{e^{2\pi x} -1} dx=-\frac{1}{2} -i\int\limits ^{\infty }_{0}\frac{e^{-\pi x}\left( e^{2\pi x} -1\right)}{e^{2\pi x} -1} dx=-\frac{1}{2} -\frac{i}{\pi }\\
\Longrightarrow \\
\sum ^{\infty }_{n=0}\left[( -1)^{n}\right] =1-\frac{1}{2} -\frac{i}{\pi } =\frac{1}{2} -\frac{i}{\pi }
\end{array}$

Which is obviously wrong.

Here are some ideas that were tried though

$\sum ^{\infty }_{n=0}\left[( -1)^{n}\right]$ Has no imaginary part--therefore

$\sum ^{\infty }_{n=0}\left[( -1)^{n}\right] =\frac{1}{2}$

The other idea is

$ \begin{array}{l}
\sum\limits ^{\infty }_{n=1} f( n)\overset{\Re }{=} -\frac{f( 0)}{2} +i\int\limits ^{\infty }_{0}\frac{f( ix) -f( -ix)}{e^{2\pi x} -1} dx\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathfrak{Grandi's\ series}\\
\sum ^{\infty }_{n=0}( -1)^{n} =1+\sum\limits ^{\infty }_{n=1}( -1)^{n}\overset{\Re }{=}\\
1-\frac{1}{2} +i\int\limits ^{\infty }_{0}\frac{( -1)^{ix} -( -1)^{-ix}}{e^{2\pi x} -1} dx=\frac{1}{2} +i\int\limits ^{\infty }_{0}\frac{( -1)^{ix} -( -1)^{-ix}}{e^{2\pi x} -1} dx\\
\ ( -1)^{a} =( -1)^{-a} \Longrightarrow \\
\frac{1}{2} +i\int\limits ^{\infty }_{0}\frac{( -1)^{ix} -( -1)^{ix}}{e^{2\pi x} -1} dx=\frac{1}{2} +i\int\limits ^{\infty }_{0} 0dx=\frac{1}{2} \Re
\end{array}$

But this shouldn't work because the thing $( -1)^{a} =( -1)^{-a} ,\ a\in \mathbb{Z}$

And $x$ is not an integer , Right? Well, I can't figure this out. Any help is appreciated! Thanks!

Edit:

Actually, I realize that Srinivasa Ramanujan probably did it using the derichlet eta function$\Longrightarrow$

$ \begin{array}{l}
\sum ^{\infty }_{n=0}( -1)^{n} =1+\sum ^{\infty }_{n=1}( -1)^{n} =1-\eta ( 0) ,\ \eta ( 0) =-\zeta ( 0) =\frac{1}{2}\\
\boxed{\therefore \sum ^{\infty }_{n=0}( -1)^{n} =\frac{1}{2}}
\end{array}$

Instead of writing the terms as $f(n)=(-1)^n$ try using just the real part $f(n)=\cos(\pi n)$.
Edit: Let me elaborate.  Ramanujan uses the Euler–Maclaurin formula in the form
$$ \sum_{m=1}^n f(m) \sim c + \int_a^n f(x)\,dx+\tfrac{1}{2}f(n)+\sum_{r=1}^\infty \frac{B_{2r}}{(2r)!}f^{(2r-1)}(n) $$
but if you set $f(n)=(-1)^n$ then the infinite series on the right is purely imaginary while the finite sum on the left is purely real—that is why the constant $c$ that you obtained had an imaginary component.  This problem is circumvented if you choose $f$ to be purely real rather than a complex-valued function.


Calculate center of mass multiple integrals Can you help me with this problem?

Find the center of mass of a lamina whose region R is given by the inequality:

and the density in the point (x,y) is :

The region r is this one:

Is this the proper way to set up the integral for m:

$$\int_{-1}^{1}\int_{-x-1}^{x+1} \ e^{x+y} \ dy \ dx$$

Any help?  Thanks

The integral for total mass is $$\int_{xmin}^{xmax}\int_{ymin}^{ymax} \delta(x,y) \ dy \ dx$$ where the range $(xmin,xmax)$ is the whole range of the shape and the range $(ymin,ymax)$ is the range in $y$ at a given $x$.  Can you put the correct limits on?  Can you do the integral?

Rotate the entire thingy by angle $\frac{\pi}{4}$. Integrations will be
$$\int_{-1/\sqrt{2}}^{+1/\sqrt{2}}dx\int_{-1/\sqrt{2}}^{+1/\sqrt{2}}dy \;\;\delta'(x,y)$$
etc. where
$$\delta'(x,y) = \delta\left(\frac{x+y}{\sqrt{2}}, \frac{-x+y}{\sqrt{2}}\right)=e^{y\sqrt{2}}$$


Construct a compact set of real numbers whose limit points form a countably infinite set. I have seen examples of sets that have these properties, like:

$$A=\left\{\frac1n+\frac1 m:m,n\in\Bbb N\right\}\cup\{0\}$$
And it is clear that 0 and all 1/n are limit points. However, how does one show that there are no other limit points?

I am completely stuck here. I have found many examples of sets that have these properties, but always run into trouble showing there are no other limit points. For example, I also tried:

$$A=\{0\}\cup\left\{\frac1n:n\in\Bbb N\right\}\cup\left\{\frac{n}{kn+1}:k,n\in\Bbb N\right\}$$

And showed that 0 and all 1/n are limit points, but I am lacking in how I can show that the $n/(kn+1)$ terms are not. To me, this means showing that there is some deleted ball around each of them which contains no element of A.

Thank you so much for your help!

Consider the first set $A$ in your question. Pick a sequence of points of $A$ that converges to some point in $\mathbb R$. Either the sequence has infinite intersection with some interval $[1/n, 1/(n+1))$, or it does not.

If the first option holds, check that the sequence converges to $1/n$, because a subsequence of it does (namely, the subsequence whose values are in the interval).

The other option is that each of these intervals contains only finitely many points of the sequence. Check then that the sequence converges to $0$.

By the way, much more complicated countable compact subsets of $\mathbb R$ are possible. The typical examples are modeled after the countable successor ordinals. See also this answer.

Take a sequence $(a_n)$ in $A$ that converges. For every $\epsilon$ there are only finitely many elements $a\in A$ with $|a|>\epsilon$, so either the sequence becomes eventually constant, or $a_n \to 0$.


Find the recurrence formula! I have a sequence defined by recursion as follows:

$$\begin{cases}x_0=a\\ x_{n+1}=x_n\cdot B^{x_n} \end{cases}$$
where $a,B$ are fix natural numbers. Does anyone know how to find a recurrence formula for this?

I tried to write in a different way, and I figure out that another equivalent definition of the sequence could be

$$\begin{cases}x_0=a\\ x_{n+1}=a\cdot B^{x_0+\cdots+x_n}\end{cases}$$

Then I tried to use logarithms and differences, but really couldn't get to anything good.

You wanted to say : an explicit formula for $x_n$ I suppose ?
If you calculate the firsts terms :
$x_0 = a$
$x_1 = a B^a$
$x_2 = a B^a B^{aB^a}$
$x_3 = a B^a B^{aB^a} B^{a B^a B^{aB^a}}$

You can guess $x_n$ can be written in the form :
$x_n = B_0 B_1 .. B_n$
[EDIT : what I have written just after this was false]

EDIT : You have :
$x_n = B_0 B_1 .. B_n$
With $B_0 = a, B_n = B^{B_0 .. B_{n-1}}$

So you can understand the terms of $x_n$ using decreasing sequences of integers with first terms $\leq n$.

For instance :
n = 2.

$x_2 = a B^a B^{aB^a}$

To (0) will be associated the first $a$ at the left.

To (1) is associated the tower $B^a$. To (1,0) is associated the $a$ in the tower $B^a$

To (2) is associated the tower $B^{aB^a}$. To $(2,0)$, the first $a$ at the left in the exponent of the tower $aB^a$. To $(2,1)$ the term $B^a$ in $aB^a$. To  (2,1,0) the last exponent $a$.

In the general case, you have an algorithm to generate your expression of $x_n$ in function of $a$ and $B$.

Put the $a$ on the first floor, associated to $(0)$, put the $n$ $B$ on the first floor, the base of the towers associated to the sequences of 1 term (1), (2), .. , (n).
To fill the second floor, you look at the sequences $(k, l)$ with $0 \leq l < k$. To each of these sequence, you put a B on the $k$-th tower if $l\neq 0$, $a$ if $l=0$
You continue like this, until the $n$-th floor ; then you have finished.

If you index $B$ by the sequences I said, you can also have a recursion formula.

$B_r = a$ if $r$ is a sequence ending by $0$, $B_r = B^{\prod B_s}$
where the product is on the $s$, the sequences which has $r$ as initial segment and one term more than r,  and $x_n = B_{(0)} .. B_{(n)}$

It is quite technical, but it enables to understand how the towers of $B$ and $a$ are distributed, and can be helpful if you want to have some asymptotics.


Sequence of functions uniformly integrable but not bounded Find a sequence $\{ f_n : E \to \mathbb{R}, m(E) < \infty\}$ of pointwise convergent functions that is uniformly integrable and not bounded by a single integrable function (i.e. there is no integrable function $g$ such that $|f_n| \leq g$ for all $n$). Show your Claim. Show that, for each example that one can find in the above problem, the limit function is integrable.

I was thinking $f_n =\frac{1}{n} \chi_{[0,n]}$ but I am not sure, anyone can help? Thank.

Check $E = (0, 1]$, $f_n(x) = \frac{n}{\log n} I_{(0, n^{-1})}(x), n = 1, 2, \ldots$. And $m$ is the Lebesgue measure on $E$.

You are on the right track, but an even simpler, more degenerate case can happen, an example that is a nice counterexample to a lot of things in analysis: set $f_n=\chi_{[n,n+1]}$. Then $\{f_n\}$ is obviously uniformly integrable, converges pointwise a.e. to $0$, and the "best" function that dominates is $g\equiv1$, which is of course not integrable.


Pooling Log Inequality I am trying to find a proof for this log inequality I found in an Information Theory. The inequality is $$-(a+b)log(a+b)\leq-aloga-blogb\leq-(a+b)log\frac{a+b}{2}$$ for $a,b\geq0$ and $a+b>0$.

The first inequality seems quite clear, by the fact that $log$ is an increasing function, but I am struggling to see how to prove the second inequality. Any hints or guidance would be appreciated. Thank you!

Recall the weighted AM-GM-HM inequality: for any finite list of non-negative reals $x_i$ and $w_i,$ with $\sum w_i >0,$   $$ \frac{ \sum w_i x_i}{\sum w_i } \ge \left( \prod  {x_i}^{w_i}\right)^{1/\sum w_i} \ge \frac{\sum w_i}{ \sum \frac{w_i}{x_i}}.$$

We'll apply this with $w_1 = x_1 = a, w_2 = x_2 = b$. This gives $$ \frac{a^2 + b^2}{a+b} \ge \left( a^a b^b \right)^{1/a+b} \ge \frac{a+b}{ a/a + b/b} = \frac{a+b}{2}.$$

Since $-\log$ is a decreasing function, this gives the inequality (after raising by $a+b$, taking $-\log$, and rearranging, and using the convention $0\log 0 = 0$,)

$$ -(a+b)\log \frac{a^2 + b^2}{a+b} \le - a\log a - b\log b \le -(a+b) \log \frac{a+b}{2}.$$

Finally, note that $ (a+ b)^2 \ge a^2+b^2$ for $a , b \ge 0,$ which means that $-\log(a+b) \le -\log \frac{a^2 + b^2}{a+b}$, finishing the argument.

Rewriting your inequality in the form
$$\log((a+b)^{a+b})\geq \log (a^a b^b)$$
let$$a=bt$$ then we get after a rfew steps of algebra:
$$(t+1)^{t+1}\geq t^t$$
this is $$(t+1)\ln(t+1)\geq t\ln(t)$$
$$h(t)=(1+t)\ln(1+t)-t\ln(t)$$
$$h'(t)=ln(1+t)-\ln(t)=\ln(\frac{1+t}{t})$$
$$h'(t)=\ln(1+\frac{1}{t})>0$$ for $t>0$
and $$\lim_{t\to 0^+}(1+t)\ln(1+t)-t\ln(t))=0$$


Distance between point and a spiral I'm trying to work out an algorithm where, given the equation for a spiral in polar coordinates, $r(\theta)$, and a point rectilinear coordinates, $P(x,y)$, I can work out the minimum distance between that point and the curve traced by that spiral. I was hoping to have a general solution for any spiral form, but if that's not possible, then a solution for a Fermat spiral $$r(\theta)=\pm a\sqrt{\theta}$$ would be most preferable.

I've tried to find a global minima using the second and third differential of the distance equation, but I keep getting stumped. Any one have any ideas?
Thanks very much in advance.

Suppose the spiral is $r = f(\theta)$. So, any point on the spiral can be represented as $(f(\theta),\theta)$. The reference point $P(x,y)$ is fixed. For easier computation, convert P to polar coordinates as $P(r_0,\theta_0)$. Let $D(\theta)$ denote the squared distance between the point $(f(\theta),\theta)$ on the spiral and the reference point P.

$$D(\theta) = f^2(\theta)+r_0^2 - 2f(\theta)r_0\cos(\theta-\theta_0)$$
For the spiral that you have mentioned, consider the positive branch $f(\theta) = a\sqrt{\theta}$ .

$$D(\theta) = a^2\theta + r_0^2 -2 a r_0\sqrt{\theta}\cos(\theta-\theta_0)$$
$$D'(\theta) = a^2 +2ar_0\sqrt{\theta}\sin(\theta-\theta_0)-\frac{ar_0\cos(\theta-\theta_0)}{\sqrt{\theta}}$$

If $D'(\theta)=0$, $$g(\theta)=\frac{\cos(\theta-\theta_0)-2\theta\sin(\theta-\theta_0)}{\sqrt{\theta}} = \frac{a}{r_0}$$
Though no easy closed form solution presents itself, the problem is very easy to solve numerically as the LHS is just a function of $\theta$ while the RHS is a constant. You just need to keep evaluating $g(\theta)$ in a suitable interval until $g(\theta)\approx\frac{a}{r_0}$ up to the desired tolerance.

In order to find out the required interval, note that $g(\theta) \le \frac{1+2\theta}{\sqrt{\theta}}$. So, we get an upper bound on $\theta$, say $\theta^*$, for the given $a$. Search in $(0,\theta^*)$ for a value of $\theta$ that satisfies $g(\theta) =\frac{a}{r_0}$.

Similarly, you can deal with the negative branch $f(\theta) = -a\sqrt{\theta}$.

Use a golden logarithmic spiral. The distance from any point to the origin (in either polar or rectangular coordinates) is equal to a(phi)^(theta/(pi/2)) where a is the distance from an intersection of the spiral an the x axis, and theta is the angle of rotation (of the spiral) when said point lies on its curve.

To find the minimum distance of the point to an arm of the spiral when the spiral arm is between the point and origin, simply subtract the radius value of the closest arm to the point from the distance of the point to the origin.

When the point is between the spiral arm and the origin, its minimum distance from the spiral arm is the absolute value of the radius of that part of the spiral minus the distance from the point to the origin.

Since all spirals have rotational symmetry, this method works for all spirals (Fermat's included). Using the golden spiral is simply easiest because of how "intuitive" it is and the properties of phi (makes calculations far simpler than any other spiral).

Feel free to contact me at andrewbarton15@yahoo.com if you have any questions or need clarification.


Prove that $HK$ is a subgroup iff $HK=KH$. Let $H$ and $K$ be subgroups of a group $G$, and let $HK=\{hk: h
\in H, k \in K\}$, $KH=\{kh: k \in K, h \in H\}$. How can we prove that $HK$ is a subgroup iff $HK=KH$?

It is usually helpful to recall that $ L$ is a subgroup if and only if $ L^ 2=L $, $ L^{- 1}=  L $ and $  L\neq \varnothing$.

If $HK$ is a subgroup, then $HK=(HK)^{-1}=K^{ -1}H^{-1}=KH$. Hence the condition is met. Conversely,
$$ \begin{align}(HK)^{ -1} &= K^{ -1}H^{-1}\\&=KH =HK\text{, }\\(HK)(HK)=H(KH)K&=\\H( HK)K&=  H^2K^ 2=   HK\end{align} $$ and of course $ HK\neq \varnothing$. We made repeated use of the fact $ H $ and $ K $ are subgroups and $ H K = KH $.

HK is a group iff each hk has an inverse in HK iff  each k^-1 h^-1 is in HK  iff each kh is in HK (and similarly each hk is in KH) iff HK and KH are identical.


Explicit solution for a particular linear second-order elliptic PDE with boundary conditions? I have come across an elliptic PDE of the form:

$$C\dfrac{\partial^2f}{\partial x^2}+D\dfrac{\partial^2f}{\partial y^2}-E(y-F)\dfrac{\partial f}{\partial y}+G\left(\dfrac{\partial f}{\partial x}+y\right)-H=0$$

where all of the constants $C,D,E,F,G,H\geq0$. We seek a function

$$f(x,y):[0,B]\times\mathbb{R}\rightarrow\mathbb{R}$$ for $B>0$ satisfying the PDE on $(0,B)\times\mathbb{R}$ subject to the boundary conditions

$$\dfrac{\partial f}{\partial x}(0,y)=-y, \quad  \dfrac{\partial f}{\partial x}(B,y)=0$$ for all $y\in\mathbb{R}$ .

Anyone know if there is any hope for obtaining a solution in any kind of explicit form? What would be the best approach to attack this?

Thanks.

Since you specify this PDE is elliptic, this also implies that $C,D\neq0$ .

$C\dfrac{\partial^2f}{\partial x^2}+D\dfrac{\partial^2f}{\partial y^2}-E(y-F)\dfrac{\partial f}{\partial y}+G\left(\dfrac{\partial f}{\partial x}+y\right)-H=0$

$C\dfrac{\partial^2f}{\partial x^2}+D\dfrac{\partial^2f}{\partial y^2}-E(y-F)\dfrac{\partial f}{\partial y}+G\dfrac{\partial f}{\partial x}+Gy-H=0$

Let $f=g+\dfrac{x^2y}{2B}-xy$ ,

Then $\dfrac{\partial f}{\partial x}=\dfrac{\partial g}{\partial x}+\dfrac{xy}{B}-y$

$\dfrac{\partial^2f}{\partial x^2}=\dfrac{\partial^2g}{\partial x^2}+\dfrac{y}{B}$

$\dfrac{\partial f}{\partial y}=\dfrac{\partial g}{\partial y}+\dfrac{x^2}{2B}-x$

$\dfrac{\partial^2f}{\partial y^2}=\dfrac{\partial^2g}{\partial y^2}$

$\therefore C\biggl(\dfrac{\partial^2g}{\partial x^2}+\dfrac{y}{B}\biggr)+D\dfrac{\partial^2g}{\partial y^2}-E(y-F)\biggl(\dfrac{\partial g}{\partial y}+\dfrac{x^2}{2B}-x\biggr)+G\left(\dfrac{\partial g}{\partial x}+\dfrac{xy}{B}-y\right)+Gy-H=0$

$C\dfrac{\partial^2g}{\partial x^2}+\dfrac{Cy}{B}+D\dfrac{\partial^2g}{\partial y^2}-E(y-F)\dfrac{\partial g}{\partial y}-\dfrac{Ex^2(y-F)}{2B}+Ex(y-F)+G\dfrac{\partial g}{\partial x}+\dfrac{Gxy}{B}-Gy+Gy-H=0$

$C\dfrac{\partial^2g}{\partial x^2}+D\dfrac{\partial^2g}{\partial y^2}-E(y-F)\dfrac{\partial g}{\partial y}+G\dfrac{\partial g}{\partial x}=\dfrac{Ex^2(y-F)}{2B}-Ex(y-F)-\dfrac{Gxy}{B}-\dfrac{Cy}{B}+H$ with $\dfrac{\partial g}{\partial x}(0,y)=0$ and $\dfrac{\partial g}{\partial x}(B,y)=0$

When solving $C\dfrac{\partial^2g}{\partial x^2}+D\dfrac{\partial^2g}{\partial y^2}-E(y-F)\dfrac{\partial g}{\partial y}+G\dfrac{\partial g}{\partial x}=0$ by separation of variables:

Let $g(x,y)=X(x)Y(y)$ ,

Then $CX''(x)Y(y)+DX(x)Y''(y)-E(y-F)X(x)Y'(y)+GX'(x)Y(y)=0$

$(CX''(x)+GX'(x))Y(y)=-(DY''(y)-E(y-F)Y'(y))X(x)$

$\dfrac{CX''(x)+GX'(x)}{X(x)}=-\dfrac{DY''(y)-E(y-F)Y'(y)}{Y(y)}=-\dfrac{4C^2t^2+G^2}{4C}$

$\begin{cases}CX''(x)+GX'(x)+\dfrac{4C^2t^2+G^2}{4C}X(x)=0\\DY''(y)-E(y-F)Y'(y)-\dfrac{4C^2t^2+G^2}{4C}Y(y)=0\end{cases}$

$\therefore$ Let $g(x,y)=\sum\limits_{n=0}^\infty C(n,y)e^{-\frac{Gx}{2C}}\cos\dfrac{n\pi x}{B}$ so that it automatically satisfies $\dfrac{\partial g}{\partial x}(0,y)=0$ and $\dfrac{\partial g}{\partial x}(B,y)=0$ ,

For starting point
$$df(x,y)=g(x,y)dx+h(x,y)dy$$
where
$$g(0,y)=-y\qquad g(B,y)=0\Rightarrow g(x,y)=y\frac{x-B}{B}$$
which leads to $$\frac{\partial^2 f}{\partial x^2}=\frac{\partial g}{\partial x}=\frac{y}{B}$$


Can't understand why Master Theorem fits or not a given recurrence relation I'm posting here $2$ little parts of "Introduction to Algorithms" by Thomas H. Cormen (2009):

First example:

Second example:

As you can see, in the first picture we can apply the Master Theorem (MT) and we are in the third MT case.
Whereas in the second picture we can't apply the Master Theorem because The ratio $f(n) / n^{\log_b a}$ that is equal to $n \lg n$ is asymptotically less than $n^\epsilon$ for any positive constant $\epsilon$.

I can't understand why MT fits the first example, because if we try to solve the ratio (like we did in the second example) $f(n) / n^{\log_b a}$ we get $(n^{0.207}) \lg n$ that it doesn't seem to be greater than $n^\epsilon$.

------

I think that I forgot a little detail: $ε$ should be an arbitrarily small quantity.

Here is the Master Theorem definition in the textbook:

The textbook just indicates that $ε>0$, so I have considered all positive numbers greater than $0$.

But

if we assume that $ε$ is a small quantity near the zero, like $ε=0.1$ or $ε=0.2$,
and we take a sufficiently large $n$, like $n=2^{60}$,

then $(n^{0.207}) \lg n$ is greater than $n^ε$.

This is an assumption I've made, help me understand if it's correct.

Thank you.

$(n^{0.207})\lg n$ must be asymptotically greater than $n^{\epsilon}$ for some $\epsilon \gt 0$ . And that is true, for example $\epsilon = 0.207$ will do the job.

I think that I forgot a little detail:
$\epsilon$ should be an arbitrarily small quantity.

Here is the Master Theorem definition in "Introduction to Algorithms" by Thomas H. Cormen(2009):

The textbook just indicates that $\epsilon\gt 0$, so I have considered all positive numbers greater than $0$.

But if we assume that $\epsilon$ is a small quantity near the zero, like $\epsilon = 0.1$ or $\epsilon = 0.2$, then $(n^{0.207}) \lg n$ is greater than $n^\epsilon$.

For example, with $n=1024$ and $\epsilon \le 0.5$ we have that $(n^{0.207}) \lg n$ is greater than $n^\epsilon$.


Exercise measure theory: properties of a function Let $\Omega \subset R^n$ ($n\geq 2$) a bounded domain. Let $u \in L^{1}(\Omega)$.
For each number $c >0$ define $A_c :=\{ x \in \Omega; u(x) > c\}$. Define

$$ f(k) := \int_{A_k} (u(x)-k) dx$$.

I am reading  a book and the author says that

$$f(k) = \int_{k}^{+\infty} |A_t| dt  \ \ \ (*)$$

and

$$ - f^{'}(k) = |A_k|\ \ \ \ (**)$$.

I have no idea of how to prove these two identities. Someone could help me ?

Thanks in advance.

That depends a bit on which definition of integrals you're using. I take the liberty to use the simplest for the purpose - sloppily expressed the area (or measure) of the region under the graph (and taking it negative if below the x-axis).

The (*) should be rather straight forward using this definition as it's basically the same set we're measuring. You have to do the outer-inner measurement construct first to see this though.

The (**) follows from that we can use the estimate $f(k+h)-f(k)$ to be between $|h||A_k|$ and $|h||A_{k+h}|$.

If you're not using this definition it will be required to prove that $\int f(x)dx = |P|-|N|$ where $P = \{x, y: 0\le y\le f(x)\}$ and $N=\{x, y: f(x)\le y \le 0\}$.


Property about algebraic integers Recently I am reading about transcendental number theory, there is a statement that all algebraic integers are norm at least 1 except 0, I don't know how to prove it.

The norm of an algebraic integer is the last coefficient of the minimal polynomial. So it is integer. If it is zero then the number is zero because minimal polynomial is irreducible

In number theory, an algebraic integer, is a complex number that is the root of some monic polynomial (leading coefficient is 1) with integer coefficients. From WIKI.


Isomorphism and $\mathrm{id}$ In a category I have two objects $a$ and $b$ and a morphism $m$ from $a$ to $b$ and one $n$ from $b$ to $a$. Is this always an isomorphism? Why is it emphasized that this has to be true, too: $m \circ n = \mathrm{id}_b$ and $n \circ m = \mathrm{id}_a$?

I am looking for an example in which the id-part is not true and therefore $m$ and $n$ are not isomorphic.

If you have no restrictions on m and n, then clearly they cannot be isomorphisms in general.  For instance, take any two groups G and H and let m: G --> H, n: H --> G be the zero homomorphisms.

Even if you say that n and m are monomorphisms, then it is still not true in general that they are isomorphisms.  I believe it is true however if your category is one whose objects are sets with additional structure. See this question: If $A$ is a subobject of $B$, and $B$ a subobject of $A$, are they isomorphic?.

To make this perhaps even clearer, say we have the objects $a=\{x, y\}$ and $b=\{z\}$.  Define $m\colon a\to b$ by $x\mapsto z$, $y\mapsto z$ (the only possible morphism here).  Define $n\colon b\to a$ by $z\mapsto x$.  But, of course, $n\circ m$ doesn't equal $\mathrm{id}_a$, since $(n\circ m)(y)=n(m(y))=n(z)=x$, $(n\circ m)(x)=n(m(x))=n(z)=x$, so we have $n\circ m\colon a\to a$ is $x\mapsto x$, $y\mapsto x$.

Perhaps if the cardinality of objects $a$ and $b$ equal each other, then a morphism $m$ from $a$ to $b$ and a morphism $n$ from $b$ to $a$, then both $m$ qualifies as an isomorphism and $n$ qualifies as an isomorphism also.


An extra condition for composition of relations which are not necessarily functions Given two binary relations $\mathcal R$ and $\mathcal S$ over $A \times B$ and $B \times C$,then their composition relation is denoted by $\mathcal S \circ \mathcal R$ and is defined as:
$$\mathcal S \circ \mathcal R:=\left\{(a,c) \mid \exists \;b \in B: a\mathcal R b \wedge b\mathcal S c\right\}$$
The converse (inverse) of composition of relations is given by:
$$(\mathcal S \circ \mathcal R)^T=\mathcal  R^T \circ \mathcal S^T$$
The proof is as follows:
$$
\begin{align}
\\
&(c,a) \in (\mathcal S \circ \mathcal R)^T\\
&\iff (a,c) \in (\mathcal S \circ \mathcal R)\tag{1}\\
&\iff\exists \;b \in B: a\mathcal R b \wedge b\mathcal S c \tag{2}\\
&\iff\exists \;b \in B:   c\mathcal S^T b \wedge b\mathcal R^T a\tag{3}\\
&\iff(c,a) \in   \mathcal R^T \circ  \mathcal S^T \tag{4}\\
&\iff (\mathcal S \circ \mathcal R)^T=\mathcal R^T \circ  \mathcal S^T\tag{5}
\end{align}
$$
Explanation:
$(1)\;,(3)$: Definition of converse relations
$(2)\;,(4)$: Definition of composition of relations
$(5)$: Definition of equality of sets

Proofwiki gives another condition for $(\mathcal S \circ \mathcal R)^T$ to exist:

From the definition of an inverse relation, we have:

$\mathsf{Domain}(\mathcal R_2)=\mathsf{Codomain}(\mathcal R^T_2)$

$\mathsf{Codomain}(\mathcal R_1)=\mathsf{Domain}(\mathcal R^T_1)$

So we confirm that $\mathcal R^T_1∘\mathcal R^T_2$ is defined.

I don't understand why do we need to check the existence of $(\mathcal S \circ \mathcal R)^T$,since we are not talking about functions and their left-totality property,so it's not the case and we don't need such a restriction,unless the relations are functions.
Am I right?If my argument is not true then please explain.

I don't think this is an extra condition. We known that a necessary condition for $\mathcal S \circ \mathcal R$ to be well-defined is that $\text{Codomain } \mathcal R = \text{Domain } \mathcal S$ - this is the set that you refer to as $B$ in you definition. The author is simply pointing out that this means that $\text{Codomain } \mathcal S^T = \text{Domain } \mathcal R^T$ and so $\mathcal R^T \circ \mathcal S^T$ is also well-defined.
"We confirm" is short for "we can confirm" or "it is clear that", not an instruction that "we must confirm".


Linear differential system of Bessel equations I have the following system:
\begin{cases}
g_{1}^{\prime\prime}+\rho^{-1}g_{1}^{\prime}-(1+\rho^{-2})g_{1}-f_{3} & =0\\
f_{3}^{\prime\prime}+\rho^{-1}f_{3}^{\prime}-(1+9\rho^{-2})f_{3}-g_{1} & =0
\end{cases}

which I rewrote as

$$\left(\begin{array}{cc}
\mathcal{L}_{1}^{(\tilde{\kappa})} & -1\\
-1 & \mathcal{L}_{3}^{(\tilde{\kappa})}
\end{array}\right)\left(\begin{array}{c}g_{1}\\
f_{3}\end{array}\right)=\left(\begin{array}{c}0\\0
\end{array}\right)$$

where $\mathcal L^{(\kappa)}_n$ is the Bessel operator with solutions $K_n(\kappa r)$, $I_n(\kappa r)$.

But I am not sure how to solve it.

I assume $g_3$ and $f_1$ depend on $\rho$, since you write the system in terms of Bessel operators. Combining the two equations, it can be concisely written as
\begin{equation}
\left( \mathcal{L}_3 \mathcal{L}_1 -1 \right) g_1 = 0, \tag{1}
\end{equation}
where $\mathcal{L}_\nu$ is the operator giving the modified (!) Bessel equation
\begin{equation}
 \mathcal{L}_\nu \phi = \rho^2 \phi_{\rho\rho} + \rho \phi_\rho - (\rho^2 + \nu^2)\phi = 0.
\end{equation}
I tried to find two second order Sturm-Liouville operators $D_i$ such that $(1)$ can be written in the form
\begin{equation}
 D_1 D_2 \phi = 0,
\end{equation}
such that you could infer that $D_2\phi \in \text{ker} D_1$, which would allow you to solve the inhomogeneous second order Sturm-Liouville ODE
\begin{equation}
 D_2 \phi = \psi,\quad \psi \in \text{ker }D_1
\end{equation}
by variation of parameters. Sadly, I didn't succeed in finding such Sturm-Liouville operators, but maybe you (or someone else) will have more luck.

A general idea is to use the Fourier-Bessel series to express a general solution as a sum of Bessel functions (actually, modified Bessel functions in your case, but the idea is the same). Because these functions are eigenfunctions of the differential operators $\mathcal{L}_\nu$, equation $(1)$ might give you a particularly simple relation between the series coefficients.

Solving by large computer algebra system (Mathematica 10.3.1), does not turn up Bessel functions.  The form is similar to that of liniear fourht order equations, suggesting that the solution technique is to solve, for example, your second equation for $g_1$, then substitute that into the first equation, yielding a fourth order linear equation.

[The original solution without cleanup is preserved far below.]

Solution from Mathematica is below.  The four constants of integration are $c_1$, $c_2$, $c_3$, and $c_4$.  The code to generate this output is

However, somewhat more readable is

FullSimplify[FullSimplify[
 {f3[x], g1[x]} /. DSolve[{
    g1''[x] + ρ^-1 g1'[x] - (1 + ρ^-2) g1[x] - f3[x] == 0,
    f3''[x] + ρ^-1 f3'[x] - (1 + 9 ρ^-2) f3[x] - g1[x] == 0
    },
   {f3, g1},
   x]
 ] //. {Sqrt[ρ^4 (16 + ρ^4)] -> R1, 1/Sqrt[ρ^4 (16 + ρ^4)] -> 1/R1,
Sqrt[-4 R1 + 21 ρ^2 + 4 ρ^4] -> R2, 1/Sqrt[-4 R1 + 21 ρ^2 + 4 ρ^4] -> 1/R2,
Sqrt[4 R1 + 21 ρ^2 + 4 ρ^4] -> R3,  1/Sqrt[4 R1 + 21 ρ^2 + 4 ρ^4] -> 1/R3}
] // TraditionalForm // TeXForm

giving (where I have made one manual change, moving the denominator to a fraction in front) $\{f_3(x), g_1(x)\} = {}$

$\left\{\frac{1}{4 R_1 R_2 R_3} \left( R_3 \left(R_1 \left(\rho  \left(2 c_2 \rho +c_1\right)+c_1
   R_2\right)-\rho ^2 \left(\rho  \left(\rho  \left(\rho  \left(2 c_4 \rho
   +c_3\right)+8 c_2\right)+4 c_1\right)+R_2 \left(c_3 \rho ^2+4
   c_1\right)\right)\right) e^{\frac{x \left(R_2-\rho \right)}{2 \rho
   ^2}}+R_3 \left(\rho ^3 \left(\rho  \left(\rho  \left(2 c_4 \rho
   +c_3\right)+8 c_2\right)+4 c_1\right)+R_2 \left(-c_3 \rho ^4-4 c_1 \rho
   ^2+c_1 R_1\right)-\rho  R_1 \left(2 c_2 \rho +c_1\right)\right)
   e^{-\frac{x \left(\rho +R_2\right)}{2 \rho ^2}}+R_2 \left(\rho ^3
   \left(-\left(\rho  \left(\rho  \left(2 c_4 \rho +c_3\right)+8
   c_2\right)+4 c_1\right)\right)+R_3 \left(c_3 \rho ^4+4 c_1 \rho ^2+c_1
   R_1\right)-\rho  R_1 \left(2 c_2 \rho +c_1\right)\right) e^{-\frac{x
   \left(\rho +R_3\right)}{2 \rho ^2}}+R_2 \left(\rho ^3 \left(\rho
   \left(\rho  \left(2 c_4 \rho +c_3\right)+8 c_2\right)+4 c_1\right)+R_3
   \left(c_3 \rho ^4+4 c_1 \rho ^2+c_1 R_1\right)+\rho  R_1 \left(2 c_2 \rho
   +c_1\right)\right) e^{\frac{x \left(R_3-\rho \right)}{2 \rho ^2}}\right),\frac{1}{4 R_1 R_2 R_3} \left( R_3 \left(\rho ^3 \left(\rho  \left(\rho  \left(2 c_2 \rho
   +c_1\right)-8 c_4\right)-4 c_3\right)+R_2 \left(-c_1 \rho ^4+4 c_3 \rho
   ^2+c_3 R_1\right)-\rho  R_1 \left(2 c_4 \rho +c_3\right)\right)
   e^{-\frac{x \left(\rho +R_2\right)}{2 \rho ^2}}+R_3 \left(\rho ^3
   \left(\rho  \left(8 c_4-\rho  \left(2 c_2 \rho +c_1\right)\right)+4
   c_3\right)+R_2 \left(-c_1 \rho ^4+4 c_3 \rho ^2+c_3 R_1\right)+\rho  R_1
   \left(2 c_4 \rho +c_3\right)\right) e^{\frac{x \left(R_2-\rho \right)}{2
   \rho ^2}}+R_2 \left(\rho ^3 \left(\rho  \left(8 c_4-\rho  \left(2 c_2
   \rho +c_1\right)\right)+4 c_3\right)+R_3 \left(c_1 \rho ^4-4 c_3 \rho
   ^2+c_3 R_1\right)-\rho  R_1 \left(2 c_4 \rho +c_3\right)\right)
   e^{-\frac{x \left(\rho +R_3\right)}{2 \rho ^2}}+R_2 \left(\rho ^3
   \left(\rho  \left(\rho  \left(2 c_2 \rho +c_1\right)-8 c_4\right)-4
   c_3\right)+R_3 \left(c_1 \rho ^4-4 c_3 \rho ^2+c_3 R_1\right)+\rho  R_1
   \left(2 c_4 \rho +c_3\right)\right) e^{\frac{x \left(R_3-\rho \right)}{2
   \rho ^2}} \right)\right\}$

where

$$\begin{align}
    R_1 &= \sqrt{\rho^4(16+\rho^4)} \\
    R_2 &= \sqrt{4 \rho^4 + 21 \rho^2 - 4 R_1} \\
    R_3 &= \sqrt{4 \rho^4 + 21 \rho^2 + 4 R_1}
\end{align}$$

Note that $x$ appears only in the exponents.  The method of undetermined coefficients applied to these exponentials (and "$1$") should clean this up substantially.  However, my original observation is unchanged: these exponents are from the characteristic equation arising from the fourth order equations you get by solving your first equation for $f_3$ and substituting into your second equation and vice versa for $g_1$.

Original mess below here.  Note that there is some room for compaction by extracting frequently repeated (constant) subexpressions.  The four constants of integration are $c_1$, $c_2$, $c_3$, and $c_4$.  The code to generate this output is

FullSimplify[
 {f3[x], g1[x]} /. DSolve[{
    g1''[x] + ρ^-1 g1'[x] - (1 + ρ^-2) g1[x] - f3[x] == 0,
    f3''[x] + ρ^-1 f3'[x] - (1 + 9 ρ^-2) f3[x] - g1[x] == 0
    },
   {f3, g1},
   x]
 ]
% // TeXForm

$\{ f_3(x), g_1(x)\} = \\
\left\{\frac{e^{\frac{x \left(\sqrt{4 \rho ^4+21 \rho ^2-4
   \sqrt{\rho ^4 \left(\rho ^4+16\right)}}-\rho \right)}{2 \rho ^2}} \sqrt{4
   \rho ^4+21 \rho ^2+4 \sqrt{\rho ^4 \left(\rho ^4+16\right)}} \left(-2 c_4
   \rho ^6-c_3 \rho ^5-\left(8 c_2+\sqrt{4 \rho ^4+21 \rho ^2-4 \sqrt{\rho
   ^4 \left(\rho ^4+16\right)}} c_3\right) \rho ^4-4 c_1 \rho ^3+\left(2
   \sqrt{\rho ^4 \left(\rho ^4+16\right)} c_2-4 \sqrt{4 \rho ^4+21 \rho ^2-4
   \sqrt{\rho ^4 \left(\rho ^4+16\right)}} c_1\right) \rho ^2+\sqrt{\rho ^4
   \left(\rho ^4+16\right)} c_1 \rho +\sqrt{\rho ^4 \left(\rho ^4+16\right)}
   \sqrt{4 \rho ^4+21 \rho ^2-4 \sqrt{\rho ^4 \left(\rho ^4+16\right)}}
   c_1\right)+e^{-\frac{x \left(\rho +\sqrt{4 \rho ^4+21 \rho ^2-4
   \sqrt{\rho ^4 \left(\rho ^4+16\right)}}\right)}{2 \rho ^2}} \sqrt{4 \rho
   ^4+21 \rho ^2+4 \sqrt{\rho ^4 \left(\rho ^4+16\right)}} \left(2 c_4 \rho
   ^6+c_3 \rho ^5+\left(8 c_2-\sqrt{4 \rho ^4+21 \rho ^2-4 \sqrt{\rho ^4
   \left(\rho ^4+16\right)}} c_3\right) \rho ^4+4 c_1 \rho ^3-2 \left(2
   \sqrt{4 \rho ^4+21 \rho ^2-4 \sqrt{\rho ^4 \left(\rho ^4+16\right)}}
   c_1+\sqrt{\rho ^4 \left(\rho ^4+16\right)} c_2\right) \rho ^2-\sqrt{\rho
   ^4 \left(\rho ^4+16\right)} c_1 \rho +\sqrt{\rho ^4 \left(\rho
   ^4+16\right)} \sqrt{4 \rho ^4+21 \rho ^2-4 \sqrt{\rho ^4 \left(\rho
   ^4+16\right)}} c_1\right)-e^{-\frac{x \left(\rho +\sqrt{4 \rho ^4+21 \rho
   ^2+4 \sqrt{\rho ^4 \left(\rho ^4+16\right)}}\right)}{2 \rho ^2}} \sqrt{4
   \rho ^4+21 \rho ^2-4 \sqrt{\rho ^4 \left(\rho ^4+16\right)}} \left(2 c_4
   \rho ^6+c_3 \rho ^5+\left(8 c_2-\sqrt{4 \rho ^4+21 \rho ^2+4 \sqrt{\rho
   ^4 \left(\rho ^4+16\right)}} c_3\right) \rho ^4+4 c_1 \rho ^3+\left(2
   \sqrt{\rho ^4 \left(\rho ^4+16\right)} c_2-4 \sqrt{4 \rho ^4+21 \rho ^2+4
   \sqrt{\rho ^4 \left(\rho ^4+16\right)}} c_1\right) \rho ^2+\sqrt{\rho ^4
   \left(\rho ^4+16\right)} c_1 \rho -\sqrt{\rho ^4 \left(\rho ^4+16\right)}
   \sqrt{4 \rho ^4+21 \rho ^2+4 \sqrt{\rho ^4 \left(\rho ^4+16\right)}}
   c_1\right)+e^{\frac{x \left(\sqrt{4 \rho ^4+21 \rho ^2+4 \sqrt{\rho ^4
   \left(\rho ^4+16\right)}}-\rho \right)}{2 \rho ^2}} \sqrt{4 \rho ^4+21
   \rho ^2-4 \sqrt{\rho ^4 \left(\rho ^4+16\right)}} \left(2 c_4 \rho ^6+c_3
   \rho ^5+\left(8 c_2+\sqrt{4 \rho ^4+21 \rho ^2+4 \sqrt{\rho ^4 \left(\rho
   ^4+16\right)}} c_3\right) \rho ^4+4 c_1 \rho ^3+2 \left(2 \sqrt{4 \rho
   ^4+21 \rho ^2+4 \sqrt{\rho ^4 \left(\rho ^4+16\right)}} c_1+\sqrt{\rho ^4
   \left(\rho ^4+16\right)} c_2\right) \rho ^2+\sqrt{\rho ^4 \left(\rho
   ^4+16\right)} c_1 \rho +\sqrt{\rho ^4 \left(\rho ^4+16\right)} \sqrt{4
   \rho ^4+21 \rho ^2+4 \sqrt{\rho ^4 \left(\rho ^4+16\right)}}
   c_1\right)}{4 \sqrt{\rho ^4 \left(\rho ^4+16\right)} \sqrt{4 \rho ^4+21
   \rho ^2-4 \sqrt{\rho ^4 \left(\rho ^4+16\right)}} \sqrt{4 \rho ^4+21 \rho
   ^2+4 \sqrt{\rho ^4 \left(\rho ^4+16\right)}}},\frac{e^{-\frac{x
   \left(\rho +\sqrt{4 \rho ^4+21 \rho ^2-4 \sqrt{\rho ^4 \left(\rho
   ^4+16\right)}}\right)}{2 \rho ^2}} \sqrt{4 \rho ^4+21 \rho ^2+4
   \sqrt{\rho ^4 \left(\rho ^4+16\right)}} \left(2 c_2 \rho ^6+c_1 \rho
   ^5-\left(\sqrt{4 \rho ^4+21 \rho ^2-4 \sqrt{\rho ^4 \left(\rho
   ^4+16\right)}} c_1+8 c_4\right) \rho ^4-4 c_3 \rho ^3+\left(4 \sqrt{4
   \rho ^4+21 \rho ^2-4 \sqrt{\rho ^4 \left(\rho ^4+16\right)}} c_3-2
   \sqrt{\rho ^4 \left(\rho ^4+16\right)} c_4\right) \rho ^2-\sqrt{\rho ^4
   \left(\rho ^4+16\right)} c_3 \rho +\sqrt{\rho ^4 \left(\rho ^4+16\right)}
   \sqrt{4 \rho ^4+21 \rho ^2-4 \sqrt{\rho ^4 \left(\rho ^4+16\right)}}
   c_3\right)+e^{\frac{x \left(\sqrt{4 \rho ^4+21 \rho ^2-4 \sqrt{\rho ^4
   \left(\rho ^4+16\right)}}-\rho \right)}{2 \rho ^2}} \sqrt{4 \rho ^4+21
   \rho ^2+4 \sqrt{\rho ^4 \left(\rho ^4+16\right)}} \left(-2 c_2 \rho
   ^6-c_1 \rho ^5+\left(8 c_4-\sqrt{4 \rho ^4+21 \rho ^2-4 \sqrt{\rho ^4
   \left(\rho ^4+16\right)}} c_1\right) \rho ^4+4 c_3 \rho ^3+2 \left(2
   \sqrt{4 \rho ^4+21 \rho ^2-4 \sqrt{\rho ^4 \left(\rho ^4+16\right)}}
   c_3+\sqrt{\rho ^4 \left(\rho ^4+16\right)} c_4\right) \rho ^2+\sqrt{\rho
   ^4 \left(\rho ^4+16\right)} c_3 \rho +\sqrt{\rho ^4 \left(\rho
   ^4+16\right)} \sqrt{4 \rho ^4+21 \rho ^2-4 \sqrt{\rho ^4 \left(\rho
   ^4+16\right)}} c_3\right)+e^{-\frac{x \left(\rho +\sqrt{4 \rho ^4+21 \rho
   ^2+4 \sqrt{\rho ^4 \left(\rho ^4+16\right)}}\right)}{2 \rho ^2}} \sqrt{4
   \rho ^4+21 \rho ^2-4 \sqrt{\rho ^4 \left(\rho ^4+16\right)}} \left(-2 c_2
   \rho ^6-c_1 \rho ^5+\left(\sqrt{4 \rho ^4+21 \rho ^2+4 \sqrt{\rho ^4
   \left(\rho ^4+16\right)}} c_1+8 c_4\right) \rho ^4+4 c_3 \rho ^3-2
   \left(2 \sqrt{4 \rho ^4+21 \rho ^2+4 \sqrt{\rho ^4 \left(\rho
   ^4+16\right)}} c_3+\sqrt{\rho ^4 \left(\rho ^4+16\right)} c_4\right) \rho
   ^2-\sqrt{\rho ^4 \left(\rho ^4+16\right)} c_3 \rho +\sqrt{\rho ^4
   \left(\rho ^4+16\right)} \sqrt{4 \rho ^4+21 \rho ^2+4 \sqrt{\rho ^4
   \left(\rho ^4+16\right)}} c_3\right)+e^{\frac{x \left(\sqrt{4 \rho ^4+21
   \rho ^2+4 \sqrt{\rho ^4 \left(\rho ^4+16\right)}}-\rho \right)}{2 \rho
   ^2}} \sqrt{4 \rho ^4+21 \rho ^2-4 \sqrt{\rho ^4 \left(\rho ^4+16\right)}}
   \left(2 c_2 \rho ^6+c_1 \rho ^5+\left(\sqrt{4 \rho ^4+21 \rho ^2+4
   \sqrt{\rho ^4 \left(\rho ^4+16\right)}} c_1-8 c_4\right) \rho ^4-4 c_3
   \rho ^3+\left(2 \sqrt{\rho ^4 \left(\rho ^4+16\right)} c_4-4 \sqrt{4 \rho
   ^4+21 \rho ^2+4 \sqrt{\rho ^4 \left(\rho ^4+16\right)}} c_3\right) \rho
   ^2+\sqrt{\rho ^4 \left(\rho ^4+16\right)} c_3 \rho +\sqrt{\rho ^4
   \left(\rho ^4+16\right)} \sqrt{4 \rho ^4+21 \rho ^2+4 \sqrt{\rho ^4
   \left(\rho ^4+16\right)}} c_3\right)}{4 \sqrt{\rho ^4 \left(\rho
   ^4+16\right)} \sqrt{4 \rho ^4+21 \rho ^2-4 \sqrt{\rho ^4 \left(\rho
   ^4+16\right)}} \sqrt{4 \rho ^4+21 \rho ^2+4 \sqrt{\rho ^4 \left(\rho
   ^4+16\right)}}}\right\}$


Showing that $\frac{x!}{x^{x}}$ tends to zero as x tends to infinity The question is pretty much in the title, I'm having difficulty formally showing that $\lim\limits _{x\to\infty}\frac{x!}{x^{x}}=0$ (despite intuitively it's fairly obvious).

Thanks in advance for helping ;)

We have that

$$n! \leq \frac{n^n}{2^n} \text{ for } n \geq 6$$

so

$$\lim_{n \to \infty} \frac{n!}{n^n} \leq \lim_{n\to \infty} \frac{1}{2^n} = 0.$$

For $x \in \mathbb{R}$, we know that $\Gamma$ function is increasing for inputs $\geq 2$, so the limit still holds (assume $x > n \geq 6$):

\begin{align}
\Gamma(n) &= (n-1)! \\
\Gamma(x) &\leq \Gamma(\lfloor x+1 \rfloor) = \lfloor x \rfloor!
\end{align}

so

$$\lim_{x \to \infty} \frac{\Gamma(x+1)}{x^x} \leq
\lim_{x \to \infty} \frac{\lfloor x+1 \rfloor!}{\lfloor x \rfloor^{\lfloor x \rfloor}} =
\lim_{n \to \infty} \frac{(n+1)!}{n^n} \leq
\lim_{n\to \infty} \frac{n+1}{2^n} = 0.$$

I hope this helps ;-)

Edit: Fixed some minor issues, thanks to @julien for noticing.

Hint: What is $\log(n!)$? Then replace a sum with integrals


Is it possible to split a triangle in two, making a two right triangles, and how? I have a problem, where I want to figure out how to split a triangle in two making two right triangles. The line has to be 90 degrees from line c(red font in picture), and has to meet with point C(black font)
We know that The triangle is 73degrees, 75 degrees, and 32 degrees, we also know that line c is 8cm. Is it possible to split it through mathematics?

Need help with this

The triangles are well defined.  For each you know all the angles (75,90,15) and (72,90,18). You can use the law of sines to get the side lengths when you know one side.

I saw nobody answered this and the few "starter" answers were wrong.

I began explaining a solution where the first step is to divide the triangle into a small right triangle using the left side (T1 in the diagram) to define the height of the triangle using angles of 73, 90 and 17 (all must equal 180 and thus we know the acute angle at the top) and solving for the base of triangle T1 using sin(17) = x/8.  (sine t = opposite/base)

From here, using the hypotenuse and base, the leg, or height is calculated as sqrt(8^2 - 2.34^) = 7.65

Now that we also have the base leg of T2 we can solve for the hypotenuse using the acute angle 32 degrees as sin(32) = 7.65/x - 14.44 and solving for the missing side of T2.

Next add the missing side of T2 to the base of T1 and the result is 2 known hypotenuse lengths of 2 new triangles NT1 (73, 90, 17) and NT2 (75, 90, 15) which are right triangles meeting on line C.

Use each acute angle to find the base as above and you end up with base of NT1 of 4.26 and a base leg of NT2 of 3.74 mathematically bisecting the 8cm side at two right triangles.

*** NOTE - I saw on the image of the solution there was a line which states
   "further by dividing both sides by cos(32)"
which should state
   "further by dividing both sides by sin(32)"

I corrected that, labelled and color coded the labels for the 2 subsets of triangles and side lengths and added a Pythagorean theorem for sub triangle T2 as proof that law of sines provided the correct answer.

I saw a down vote and took the optimistic approach that the person couldn't follow the solution due to the mentioned typo and lack of labels rather than could not follow geometry and trigonometry.

one solution


$4y^2+y=3x^2+x$ implies that $x-y$ is a perfect square. I need a help to prove the following statement. (Sorry for my bad english).

  If $x,y\in\mathbb{Z}$ are solutions of $4y^2+y=3x^2+x$, then $x-y$ is a perfect square.

I've tried to rewrite the equation as $\cfrac{y^2}{x-y}=3(x+y)+1 $ and conclude that $x-y\mid y^2$, but I do not think that's useful. I found two solutions: $(x,y)=(2,-2)$  and $(x,y)=(30,26)$. If I could find a recursive way to build more solutions, I would probably show that $x-y\in\mathcal{Q}:=\big\{ n^2\mid n\in \mathbb{N} \big\}$.

Thanks to everyone which will help me to solve (prove) it.

You are on the right track: $4y^2+y=3x^2+x$ implies
$$(x-y)(3(x+y)+1)=y^2\tag{1}$$
which means that $(x-y)$ divides $y^2$.

Now assume that $(x-y)$ is not a square. Then there exists a prime $p$ and an odd positive integer $k$ such that the largest power of $p$  which divides $(x-y)$ is $p^k$. Then $p$ divides $y^2$ and therefore it divides also $y$ and $x+y=2y+(x-y)$.
Since $k$ is odd and the RHS of (1) is a square, it follows that $p$ divides also $(3(x+y)+1)$. Hence
$$p\mid\gcd(x+y,3(x+y)+1)=1$$
Contradiction!

The equation
$$
4Y^2 + Y = 3X^2 + X
$$.
Let me proof why $X - Y = c$.
where $c$ is a perfect square, note: the values $X$ and $Y$ here are the points $( X, Y )$ on the graph of the equation.
$4Y^2+Y = 3X^2+X$.
make $Y$ the subject of the formula there.
$Y = \frac{-1 \pm \sqrt{48X^2+16X+1}}{8}$.
Now say $X-Y = c$.
So that.
$
\begin{align}
X - \frac{-1 \pm \sqrt{48X^2+16X+1}}{8} = c\\
8X - \frac{-1 \pm \sqrt{48X^2+16X+1}}{1} = 8c\\
8X + 1 \mp \sqrt{48X^2+16X+1} = 8c\\
8X + 1 - 8c = \pm \sqrt{48X^2+16X+1}\\
(8X+1-8c)^2 = 48X^2+16X+1\\
1-16c+64c^2+16X-128cX+64X^2 = 48X^2+16X+1\\
16X^2-128cX+64c^2-16c = 0
\end{align}
$.
This equation would give us the value of the $X$ coordinate for which $X-Y = c$.
$X^2-8cX+4c^2-c = 0$.
solving for $X$ shows that.
$X = \frac{8c \pm \sqrt{64c^2-4(4c^2-c)}}{2}$.
which can be further depressed to.
$X = 4c \pm \sqrt{12c^2+c}$.
if we had put $c = 4$, we would get coordinate point $X = 2 \text{and} 30$ as solution.
the determinant here is $12c^2+c$, were $c$ was defined to be a perfect square, let's assume we are looking for integers values of $X$.
Then $12c^2+c = c(12c+1)$ must also be a perfect square, then the number $(12c+1)$ must be a perfect square also, but as far I can say, this worked only when $c = 4$ where $(12×4+1) = 49$, and $49$ is also a perfect number.
Let's recall that $X^2-8cX+4c^2-c = 0$.
now we want to see the value of $X$ that can make $c$ a perfect square.
put as $4c^2-(8X+1)c+X^2 = 0$.
$c = \frac{8X+1 \pm \sqrt{48X^2+16X+1}}{8}$.
one of these condition is that $(48X^2+16X+1)$ must be a perfect square.
we are still limited to only two values that can fit this $X = 2 \text{and} 30$


$0<x<y<z<1, P(z-x<1/3)=$? I'm trying to solve a question about 'order statistics'.
$X, Y, Z$ follow uniform distributions between 0 to 1.
$(X , Y, Z \sim U(0,1))$

An unequality is given that $0< X < Y < Z < 1$.
What is the probability of $(Z - X < 1/3)$?

I can easily calculate the density function of $X, Y, Z$ that $f(x, y, z) = 6.$

I think I should solve a triple integral problem.
A cube is given that $0 < X , Y, Z < 1$.

And two conditions are given

$$0 < X < Y < Z < 1,$$

$$Z-X <1/3.$$

Is it a right way to solve this question?
I'm a bit of a dummy so a thorough explanation would be much appreciated. Thanks in advance!

I think it is easier to find $P(z-x>1/3)$.
Then $1/3 < z < 1$, $0 < x < z-1/3$, and $x<y<z$.


Proof attempt for collection of all open intervals being a basis of $\Bbb R$ with the standard topology Show that the collection of all open intervals $\{(a,b)\}$ is a basis of $\Bbb R$ with the standard topology:

My attempt:

I believe we want to show two things:

1) All elements, $x\in\Bbb R$ are contained in some basis element:

$\forall x\in\Bbb R$ $x\in(x-1,x+1)$ $\square$

2) If $x\in B_1\cap B_2$ then there is some $B_3$ such that $x\in B_3$ and $B_3\subset B_1\cap B_2$

If $x\in B_1\cap B_2$ and $B_1=(a_1,b_1)$ and $B_2=(a_2,b_2)$ then $B_1\cap B_2 = (\max(a_1,a_2),(\min(b_1,b_2))$, if we let $B_3=(\max(a_1,a_2),(\min(b_1,b_2))$ then $B_3\subset B_1\cap B_2$ and clearly $x\in B_1\cap B_2=B_3 \iff x\in B_3$

Is it acceptable that I simply let $B_3=B_1\cap B_2$ by construction? Or should my $B_3$ be a proper subset, i.e. not equal?

Also is my proof acceptable in general?

What you have shown is that the collection of open intervals is a basis for some topology, but not that it is a basis for the standard topology on $\mathbb R$.  If $X$ is a topological space, then we say that a collection $\mathcal B$ of open sets of $X$ is a basis (of open sets) for the topology if every open set in $X$ can be written as a union of open sets drawn from $\mathcal B$.

Equivalently, $\mathcal B$ is a basis if the open subsets of $X$ are precisely those sets $U$ such that for every $x\in U$ there exists $V\in\mathcal B$ such that $x\in V\subset U$.

Using the second version, it should be clear that the collection of open intervals in $\mathbb R$ is a basis for the standard topology.

Where do the two conditions you talked about come in?  Well, if $X$ is a set, then it can be shown that a collection $\mathcal B\subset\mathcal P(X)$ of subsets of $X$ is a basis of open sets for some topology on $X$ if and only if it satisfies the two conditions you mentioned:

For every $p\in X$, there is $V\in\mathcal B$ such that $x\in V$.
For every $U,V\in\mathcal B$, and any $q\in U\cap V$, there is $W\in\mathcal B$ such that $q\in W\subset U\cap V$.  Equivalently, $U\cap V$ can be written as the union of sets drawn from $\mathcal B$.

But the topology might not be the one you are interested in.  For instance, the singleton collection $\{\mathbb R\}$ satisfies conditions (1) and (2) (for $X=\mathbb R$), and it is indeed a basis of open sets for a topology on $\mathbb R$ (the indiscrete topology), but it is not a basis for the standard topology.  Similarly, $\mathbb P(\mathbb R)$ satisfies the conditions for a basis, but it is a basis for the discrete topology on $\mathbb R$, not the standard topology.  Finally, the collection $\mathcal B$ of all half-open intervals $[x, y)$ satisfies (1) and (2), but it is a basis for the half-open interval topology, which is not the same as the standard topology (indeed, $[0,1)$ is not open in the standard topology).

It's a good idea to work through the proofs of the claims I have made.  Namely, that:

Show that the two versions of a definition of a basis for a topology given above are equivalent.
If $X$ is a topological space, and $\mathcal B$ is a basis of open sets for the topology on $X$, show that $\mathcal B$ satisfies (1) and (2).
If $X$ is a set, and $\mathcal B\subset\mathcal P(X)$ satisfies (1) and (2), show that the collection of all unions of sets contained in $\mathcal B$ is a topology on $X$.

Let us know if you have any trouble with those.

I think you have taken a long winded approach in part 2. Suppose $x \in B_1 \cap B_2$. That means $B_1 \cap B_2 \neq \phi$. If you say $B_1=(a_1,b_1)$, $B_2=(a_2,b_2)$ then it must hold that $a_2 < b_1$. So $B_1 \cap B_2 = (a_2,b_1)$.


Logical Equivalence Question - basic $(p \wedge \neg q) \vee(p \vee\neg q) \Leftrightarrow q\rightarrow p$
I am learning discrete math and I was stuck in this problem for hours.
I tried using De Morgan's law as well as other equivalence laws but I ended up with the same hypothesis.
Is there any advice that I could obtain? Thank you.

DeMorgan's Laws are not required.

Begin with: $~~~(p\wedge\neg q)\vee(p\vee\neg q)$
Associate to: $~((p\wedge \neg q)\vee p)\vee \neg q$
Commute to: $~\neg q\vee(p\vee(p\wedge\neg q))$
...
End with: $q\to p$

All steps are equivalences therefore $((p\wedge\neg q)\vee(p\vee\neg q))\iff (q\to p)$

You can try to use a truth table to find if there is a logical equivalency. However, you should probably try to simplify the left side as (p∧¬q)∨(p∨¬q) is equivalent to (p∨¬q) since any case where (p∧¬q) is true, (p∨¬q) also holds true.


How do you know one approximation to $\pi$ is more accurate than another? I have always been confused  regarding the accuracy of $\pi$.

In the books which are written on this subject $\pi$ , there are references of people and their methods for finding the value of $\pi$. Person A found the value of $\pi$ and then it says that person B found more accurate value of $\pi$ (may be up to $100$ decimal places) and then person C found much more accurate value of $\pi$ and this goes on.

My question is: How does the mathematician know that the value of $\pi$ which he has calculated based on whatever algorithm he applied is more accurate then the value calculated by the previous mathematician?

Imagine a scenario, where the world has just started and there are only $2$ mathematicians (A & B) in it. Mathematician A has calculated the value of of $\pi$ as $3.1547$ (this value is in ancient Chinese text), now the other mathematician says he has calculated a more accurate value, i.e. $3.1416$.

My question is How is the Mathematician B so sure that $3.1416$ is a more accurate value?

I mean there is no standard with which to compare.

Wikipedia says:
The Indian astronomer Aryabhata used a value of $3.1416$.
Fibonacci in c. 1220 computed $3.1418$.
Italian author Dante apparently employed the value $3.14142$.

When there is no standard to compare, how will I know which is the correct value of $\pi$?

We can prove that $$\pi=4\arctan 1 =4\sum_{i=1}^\infty \frac{-(-1)^i}{2i-1}=4\left(1-\frac 13+\frac 15-\frac 17+\dots\right)$$  There are series that converge much more quickly, but let's pretend that this is the only series so far known.  The alternating series theorem says that the error truncating a series like this is of the sign of and smaller than the first neglected term.  A lazy mathematician might compute the first hundred terms, getting $3.13159\frac {+0.02}{-0}$.  Another might sum the first thousand terms, which would give a maximum error of $0.002$ and would clearly be better.  Other formulas also come with a bound on the error, so we know how bad it can be.

This claims π is rational, which is known to be false.

Proof No.1

( π-3)×3^3= 4
A circle of diameter 1 is inscribed within a square where the side of the square and the diameter of the circle is one and the same.

Now the circumference of the square 4 is divided by 3 diameters and the quotient of 1.33333333…is further divided by 3^2 (3 diameters squared) the result is 0.148148148… Accordingly, the rational pi is 3.148148148…

Applying this quantum to the equation ( π-3)×3^3= 4

  (3.148148148…- 3)×3^3= 4

We can arrive back to 4 which is the perimeter of the square. But if we substitute the irrational pi, then the result is

(3.141592653…- 3)×3^3= 3.823001631…

which is lesser than 4. But as the perimeter of the square is 4, therefore, the side of the square and the diameter of the circle inscribed within that square ought to be 1 each but it is only(3.823001631/4)0.9557504078.

Proof No.2. (Pythagorean)
( π-3)×3^3= 4 = 2^2

We can carve out an equilateral triangle from the Archimedean method of hemming a circle between two regular hexagons to approximate the circumference of the circle. Now we can divide the equilateral triangle into two right angles. It becomes a right angle of 30, 60 and 90 degrees.

It is in the proportion of hypotenuse being 2, adjacent being √3 and the opposite being 1 unit wise, the quantum rational pi could be arrived at.

( π-3)={(hypotenuse^2)/(adjacent^2 )}÷(hypotenuse +opposite)^2

( π-3)= {2^2/√(3^2 )}÷(2+1)^2 = 0.148148148........

Applying this quantum to the equation ( π-3)×3^3=4 = 2^2

(3.148148148…-3)×3^3= 2^2

We can return back to the unit of the hypotenuse. But it is impossible with the pi of 3.141592653…..which is irrational.


5 sets of 5 options, how many possible combinations? Apologies in advance if this was already answered somewhere, I tried searching but didn't really know what to search for which is the same issue I'm having with Google. I really appreciate your help with this!!

I need to figure out how many combinations are possible of the following sets. I think that I just multiply everything together can anyone confirm that for me?

Example 1: 5 sets of 5 options. So there's (a1, a2, a3, a4, and a5) in one set, (b1, b2, b3, b4, and b5) in the next and there's c, d, and e all with 5 in the set. To figure out all the possible combinations would that just be 5*5*5*5*5= 3125?

Do I have that right?

If you need exactly one element of each set, and in the specific order of
$\\$
a -> b -> c -> d -> e, $\\$
then yes.
If not, it depends on how you pick the elements to make the new randomized set.


What does it mean denumerable sequence of sets? I've been searching for the definition of $\textbf{denumerable sequence of sets}$, but I have not found it yet.

Can someone tell me what doest it mean a denumerable sequence of sets? Thank you!

If for each $n \in \mathbb N$ a set $M_n$ is given, then you have the sequence

$$(M_1,M_2,M_3,....).$$

Such a sequence is a denumerable sequence of sets.

If you can label each element in the set with a real number, it is denumerable. If a set is denumerable, it is countable. This is a way, in which we may have an infinite countable set


How to detect when continued fractions period terminates I'm doing continued fractions arithmetic. Is there a method to detect when a continued fractions period terminates?
Let me give you an example:
$\sqrt{2} = [1; \overline{2}]$, $\sqrt{7} = [2; \overline{1, 1, 1, 4}]$ and $\sqrt{14} = [3; \overline{1, 2, 1, 6}]$.
Now clearly $\sqrt{2} \times \sqrt{7} = \sqrt{14}$, but if we do continued fractions arithmetic we get: $[1; \overline{2}] \times [2; \overline{1, 1, 1, 4}] = 3, 1, 2, 1, 6, 1, 2, 1, 6, 1, 2, 1, 6, \dots$. Obviously this sequence never ends, since the continued fraction can always input a term from either $\sqrt{2}$ or $\sqrt{7}$ (as they are infinite).

So my question is: is there a mathematical method to detect when the period ends (during addition, subtraction, multiplication and division), rather than guessing and checking?
Maybe using the $z$ object (this "thing")? For example, after it emits the first $1$ it is:
$\left \langle \begin{matrix}13 & 31 & 28 & 67\\
 11 & 27 2& 8 & 20
\end{matrix}\right \rangle$

and when the period starts again:
$\left \langle \begin{matrix} 549 & 1317 & 2226 & 5335\\ 95 & 239 & 814 & 2010 \end{matrix} \right \rangle$
and again: $\left \langle \begin{matrix} 92029 & 222405 & 122610 &  296283\\ 41317 & 99487 & 37841 & 91039 \end{matrix} \right \rangle$.

Is there some kind of a pattern? I've noted none.

Thank you,
rubik

This is more for J.M... it is not necessary to use cycle-detection algorithms if you stick with reduced quadratic forms, http://www.numbertheory.org/php/reduce.html for one description.

As I wrote in https://mathoverflow.net/questions/22811/upper-bound-of-period-length-of-continued-fraction-representation-of-very-composi/23014#23014
  if you are looking at only reduced forms $$ \langle a,b,c \rangle \approx a x^2 + b x y + c y^2,  $$ with discriminant $\Delta = b^2 - 4 a c$ positive but not a square, with
$$  0 < b < \sqrt \Delta   $$
and
$$  \sqrt \Delta - b < 2 |a| <  \sqrt \Delta + b.   $$

Edit, April 2015: THEOREM: a form $ \langle a,b,c \rangle$ with discriminant $\Delta = b^2 - 4 a c$ positive but not a square is REDUCED if and only if
$$ ac < 0 \; \; \mbox{AND} \; \; \; b > |a+c|.  $$

Then, at any given step, find $\delta$ such that
$$ \delta c > 0 $$
and
$$  |\delta| = \left\lfloor \left| \frac{b +\sqrt \Delta}{2 c} \right|\right\rfloor $$
Now, the right adjacent form, or right neighbor, is given by
$$ \langle a,b,c \rangle \rightarrow  \langle c, -b + 2 c \delta, a - b \delta + c \delta^2 \rangle.  $$
Lather, rinse, repeat.

If you begin with a reduced form $\langle a_0,b_0,c_0 \rangle$ then, after a finite number of steps, call it $n$ steps, we get an exact match of all three coefficients,  $$\langle a_n,b_n,c_n \rangle = \langle a_0,b_0,c_0 \rangle$$

An indefinite binary quadratic form, with integer coefficients, corresponds to one of two quadratic irrationals given by the quadratic formula, where we are solving for either $x/y$ or $y/x$ according to taste. As the discriminant is positive and not a square, the roots are real numbers. It turns out that the number defined by the continued fraction (take the absolute values of the $\delta$'s) is not just positive but is larger than 1. Just one of those things.

I noticed in your code for $\sqrt N$ you had it terminate when the first coefficient returned to $\pm 1.$ Thts is not going to work if the reduced form you start with is $\langle 7,3,-7 \rangle$ where the exact method I describe is giving the continued fraction for
$$  \frac{3 + \sqrt{205}}{14}  $$

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle
Input three coefficients a b c for indef f(x,y)= a x^2 + b x y + c y^2
7  3  -7

0  form   7 3 -7   delta  -1
1  form   -7 11 3   delta  4
2  form   3 13 -3   delta  -4
3  form   -3 11 7   delta  1
4  form   7 3 -7
minimum was   3rep -1 -1 disc   205 dSqrt 14.317821063  M_Ratio  4.183673
Automorph, written on right of Gram matrix:
17  21
21  26
 Trace:  43   gcd(a21, a22 - a11, a12) : 3
=========================================
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$

To repeat, reduced indefinite integral quadratic form correspond to quadratic irrationals with purely periodic continued fractions. The cycle is complete if and only if the initial reduced form has been repeated. No other cycle detection is required.

The one caution is that, because we are taking the absolute values of the $\delta$'s, it is possible for the continued fraction period to be exactly half the period of the quadratic form. For example,

=========================================
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./Pell
Input n for Pell
61

0  form   1 14 -12   delta  -1
1  form   -12 10 3   delta  4
2  form   3 14 -4   delta  -3
3  form   -4 10 9   delta  1
4  form   9 8 -5   delta  -2
5  form   -5 12 5   delta  2
6  form   5 8 -9   delta  -1
7  form   -9 10 4   delta  3
8  form   4 14 -3   delta  -4
9  form   -3 10 12   delta  1
10  form   12 14 -1   delta  -14
11  form   -1 14 12   delta  1
12  form   12 10 -3   delta  -4
13  form   -3 14 4   delta  3
14  form   4 10 -9   delta  -1
15  form   -9 8 5   delta  2
16  form   5 12 -5   delta  -2
17  form   -5 8 9   delta  1
18  form   9 10 -4   delta  -3
19  form   -4 14 3   delta  4
20  form   3 10 -12   delta  -1
21  form   -12 14 1   delta  14
22  form   1 14 -12

 disc   244
Automorph, written on right of Gram matrix:
-183241189  1581119536
-226153980  945570387

 Pell automorph
-1766319049  -910490892
-226153980  -1766319049

Pell unit
1766319049^2 - 61 * 226153980^2 = 1

=========================================
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$

The above is giving the continued fraction for
$$  \frac{7 + \sqrt{61}}{12},$$
So I think we are always getting the largest value of $x/y$ rather than $y/x.$   For your purpose, this is just right, you are putting a 7 in front, which uses the reciprocal of the above, to get
$$  7 + \; \;  \frac{12}{ \sqrt{61} + 7}    $$
which works out to $\sqrt 61$ when you rationalize the denominator.

Please, read chapter 3, pages 21-48, in Binary Quadratic Forms by Duncan A. Buell,  AMAZON

It may sound a bit trivial, but actually, the important qustion would be: When has a fraction a period, i.e., when is there repetitive bahvior in the digits. It is known that irrational numbers are non-terminating, non-periodic decimal fractions. Using simple set-theory, i.e., taking the complement of the set of irrational numbers in the set of real numbers, we find, that either the rational numbers are terminating decimal fractions or periodic fractions. In either case, by the construction of the rational numbers, it is possible to find an expression like this: $q=\dfrac{n}{m}, n\in\mathbb{Z}, m\in\mathbb{N}$. Note that this expression needn't be unique!. Hence, after having added $m$ times the number $q$ we're done. So far about the theory.
My experience has told me that these questions are the practical ones, i.e., those of little mathematical interest (b.t.w.: I am a pure mathematician working in global analysis, differntial topology and differential geometry <---- Purest mathematics apart from logic). I think that numerically one could find the answer to your problem, but in general, you will have to stick to guessing and checking. But perhaps when you learn abstract algebra and number theory combined with logic, you come across some solution (I have heard abstract algebra in the shortened version for geometers :D)


Which shapes would I use to figure out the total surface area of this composite shape? I am currently working on a problem where I have to figure out the formula(s) to find the total surface area of this composite figure. I have to find the individual areas of the shapes in the figure but I am having a hard time figuring out the shapes. I posted the composite figure and the answers that I already picked for the shapes. Can someone please tell me what the shapes are?

Here is the composite figure:

Here is the shapes question that I tried answering:

I hate these kinds of quizzes, because they teach you to anticipate others' thoughts, and not to solve problems.  We are humans, solvers of problems, investigators of science, manipulators of our environs; not betazeds, the readers of others minds!

My point is, there are many different roads to the same truths.  Every one of them is as valid as the others.  Thus, this quiz is incorrect; stupid, really.

We have an object which resembles a halved log, and we are asked to find its surface area.

The two ends are simple: they are halves of circular discs.  We know the area of a circular disc,
$$A_{○} = \pi r^2 \tag{1a}\label{AC1a}$$
where $r$ is the radius.  We know the diameter $d$, which is twice the radius,
$$d = 2 r \quad \iff \quad r = \frac{d}{2}$$
which means that the area of a full circular disc is
$$A_{○} = \pi \left( \frac{d}{2} \right )^2 = \frac{\pi}{4} d^2 \tag{1b}\label{AC1b}$$
Two halves is one whole: $2 \frac{1}{2} = 1$, so $A_{○}$ describes the sum surface area of the two ends.

The flat side is also simple, as it is just a rectangle.  We know its length/height $h$ and its width ($w = d$, the same as the diameter of the circular disc).  The area of a rectangle is
$$A_{▭} = w h \tag{2}\label{AC2}$$

That leaves the curved side, and here we get to the crux of the matter.

We can either calculate the area of the curved side as half the lateral surface area of the sides of the original "log" (cylinder) this is a half of, or we can think outside the box and consider it just a bent rectangle.

(Think of a sheet of paper.  It is easy to bend into a half cylinder.)

Mathematically, the two end up being the same thing.  The perimeter of a full circle is
$$p_{○} = \pi d = 2 \pi r$$
so the perimeter of a half circle is half that, $\pi r$.  If we take a sheet of paper and bend an edge of length $\pi r$ to a half cylinder, it will have radius $r$.

In other words, the area of the curved side is
$$A_{◡} = \pi h r = \pi h \frac{d}{2} \tag{3}\label{AC3}$$

Combining all these we get the total surface area of the object, given its cross section is exactly half a circular disc, length $h$, and width (diameter of those circular discs) $d$:
$$A = 2 \frac{1}{2} A_{○} ~ + ~ A_{▭} ~ + ~ A_{◡} \tag{4a}\label{AC4a}$$
which is equivalent to
$$A = \frac{\pi}{4} d^2 ~ + ~ d h ~ + ~ \pi h \frac{d}{2} \tag{4b}\label{AC4b}$$
In this particular case, $d = 2 ft$ and $h = 15 ft$, so we can calculate $A \approx 80 {ft}^2$, or about $80$ square foot.

(I like to build the formula like this first, for several reasons.  The main one being that this way I avoid re-rounding already rounded values, which increases the numerical error.  In real life you often get asked a similar question but with slightly different values, so instead of redoing all the work, you just reuse the formula.  It ends up often being less total work, really.)

What did we use to get to this answer?  The area of a rectangle, the area of a circular disc, and the perimeter of a circle.  But the last one wasn't even an option!  Another way to think of the area of the curved part is as the half of the lateral area of a cylinder; that would make the set the area of a rectangle, the area of a circular disc, and half the lateral area of a cylinder.

See?  The hard part of this question is not the math, but trying to discover what the person who set the question was thinking.  And I hate that.  I might be a bit autistic, but I am functional; penalizing people like me just because we find it difficult to discern what the person who stated the quiz question was thinking is evil.


The sum of three consecutive terms of an Arithmetic progression is $18$ and their product is $120$. Find the terms. I'm currently going through some questions on series and stumbled on this.

The sum of three consecutive terms of an Arithmetic progression is $18$ and their product is $120$. Find the terms.

I attempted it like this but I think I'm wrong cause the products is giving me $210$.

$x + (x+1) +(x+2) = 18$;

$3x +3 = 18;$

$3x = 15;$

$x=5,$ The three consecutive numbers are

$5, 6, 7$;

Appreciate the assistance.

In general, the consecutive terms can differ by any number, say $d$, instead of just $1$. So, the three numbers are better represented as $x-d,x,$ and $x+d$.

Then,
$$3x=18 \Rightarrow x=6$$
and
$$x(x^2-d^2)=120$$
$$6(36-d^2)=120$$
$$36-d^2=20$$
$$d^2=16$$
$$d=\pm4$$

The numbers thus are $\{2,6,10\}$ or $\{10,6,2\}$.

Let $a_0=0$ and $a_1=1$. Then let the arithmetic progression for your sequence be

$a_n = a_(n-1) + a_(n-2)$

You will get a sequence {3,5,8} which 3*5*8=120

This is the Fibonacci sequence.


Find the number of ordered triplets $(a, b, c)$ of positive integers such that $30a + 50b + 70c < 343$ Find the number of ordered triplets $(a, b, c)$ of positive integers such that $30a + 50b + 70c < 343$

My Attempt:

$c$ cannot be $5$ since $70 \times 5 > 343$.

It can't be $4$ either since if we put $c = 4$, we get $30a + 50b < 63$, which would mean there are no positive integer solutions for at least one of $a$ or $b$.

So, there are 3 options for $c$.

Proceeding similarly, we will get that $b = 1,2,3,4$ or there are $4$ options for b.

Similarly, there are $7$ options for $c$.

Since any value of $a$ can be paired with any value of $b$ and $c$, we get  $3 \times 4 \times 7=84$ total triplets which is not the answer

The problem is that if $c=3$ you need $30a+50b \lt 133$.  In that case $b$ can only be $1$ or $2$.  If $b=2$ you must have $a=1$, while if $b=1$ you can have $a=1$ or $2$.  The choices are not independent, so you cannot multiply.

You can just continue the casework.  There are not too many cases for $b,c$.

$(x^3+x^6+x^9+x^{12}+...x^{33})\cdot (x^5+x^{10}+x^{15}+x^{20}+x^{30}) \cdot (x^7+x^{14}+x^{21}+x^{28})$

Expand the above using geometric progression sum, and then find the sum of coefficients of $x^k$ where $k\leq 34$


For $z \in \mathbb{C}\setminus \{0\}$, is $|1/z| = 1/|z|$? I have been trying to prove that for $z \in \mathbb{C}$ and $z \neq 0$ that $|1/z| = 1/|z|$ but with no success.  Not all the properties of real numbers can be applied to complex numbers, of course.

Yes.
Note that $1/|z|$ is, by definition, the real multiplicative inverse of $|z|$, so by uniqueness of such an inverse, since
$$|1/z|\times |z|=|z/z|=1,$$
we have $|1/z|=1/|z|$.

The main problem that you have here is that you have been trying to re-invent the wheel, rather than actually consulting a Complex Analysis textbook.  The following results are based on Chapter 1 of "An Introduction To Complex Function Theory" [Bruce Palka].

For $z,w \in \mathbb{C}, |z| \times |w| = |z \times w|.$

For $z = (x + iy) \in \mathbb{C}$, let $\overline{z}$ denote $(x - iy).$

Then $\frac{1}{z} = \frac{\overline{z}}{z \times \overline{z}}.$

$z \times \overline{z} = |z|^2.$

$|\overline{z}| = |z|$.

The result that you are trying to prove is a direct consequence of the above intermediate results.


$(a+b)^2\le(1+c)a^2+(1+\frac{1}{c})b^2$ Proof Having $c\gt0$
$a, c \in \Bbb{R}$
$$(a+b)^2\le(1+c)a^2+(1+\frac{1}{c})b^2$$
I don't know how to solve this inequality, I tried with the AM-GM inequality and I reached this point so far.
$$(a+b)^2\le a^2(c+\frac{1}{c})+b^2(c+\frac{1}{c})$$
I don't know how to continue, and I don't even know if this is the right path.
If you have any hints, those would be much appreciated.
Thanks

Hi others have given you solutions. I only want to comment on your partial results but since I cannot comment now I'll type it in the answer box.
If you have already proved $(a+b)^2 \leqslant (c+\frac{1}{c}) (a^2+b^2)$ then you are done because $RHS\geqslant 2(a^2+b^2) \geqslant (a+b)^2=LHS$.
However I believe you made a mistake because the following inequality cannot always hold:
$(1+c)a^2 + (1+\frac{1}{c})b^2 \geqslant (c+\frac{1}{c})(a^2+b^2)$.
For example if $c=2$, it reduces to $3a^2+\frac{3}{2}b^2 \geqslant \frac{5}{2} (a^2+b^2) \iff a^2\geqslant 2b^2$ so if $a=1, b=1$ it's reversed.


Does $R \subsetneq \mathbb{N}$ contradict '$f:\mathbb{N}\rightarrow R$ is one-to-one and onto' There is a paradox I can't resolve. If we take the set of natural numbers $\mathbb{N}^0$. We could take the element $0$ as our starting position and reference to all other elements by defining a successor function $S(n)=n+1$ (i.e. $1$ is $S(0)$, 2 is $S(S(0))$ and so on).

If we take the set of all references it would look like $R=\{S, SS, SSS, SSSS, ...\}$. Now it seems obvious that for every element in $\mathbb{N}$ (without 0) there is a corresponding element in $R$ ($\forall x \in \mathbb{N}:x \in R$). Or in other terms, there is a function $f:\mathbb{N}\rightarrow R$ that is one-to-one and onto.

However, one can rewrite the set $R$ by (notationally) replacing $S$ with $1$. $R$ would then look like $R=\{1, 11, 111, 1111, ...\}$. In this case $R$ is the set of all natural numbers only composed of a single digit. Of course this is only a subset of all natural numbers ($R \subsetneq \mathbb{N}$).

Does $R \subsetneq \mathbb{N}$ contradict '$f:\mathbb{N}\rightarrow R$ is one-to-one and onto'? Where is the mistake in the above argument?

I'm not entirely sure if this answers your question, but you can indeed form a bijection $f: \mathbb{N} \to \{1,\ 11,\ 111, \ \dots\}$ via:
$$
f(n) = 10^{n-1} + 10^{n-2} + \ \cdots \ + 1 = \sum_{i= 0}^{n-1}10^i
$$

This is an injective mapping, because $f$ is monotone, and surjective because for each number $l =\sum_{i= 0}^M10^i \in R$, $f(M+1) = l$.

Edit: also, as someone else has said, in general $A \subsetneq B$ does not imply $\#A < \#B$. For example, $\mathbb{Z}$ is in a bijective correspondence with $\mathbb{Q}$.

There is no contradiction because $\Bbb N$ is infinite and infinite sets can biject with its proper subset (assuming axiom of choice).


Limit of composite function: $\lim_{t \to 0} [\frac{\sin(\tan(t))}{\tan(t)}]$ Limit of composite function: $\lim_{t \to 0} [\frac{\sin(\tan(t))}{\tan(t)}]$

This is an exercise from Apostol's chapter on composite functions and continuity, in which the following theorem is stated:

  Let $v$ be continuous on $p$; let $u$ be continuous on $v(p)$, then $u∘v$ is continuous on $p$

So, I'm assuming this type of limits are meant to be evaluated using this theorem.

I define $v(t):=\tan(t)$, and $u(t):=\frac{\sin(t)}{t}$.

It's clear $v$ is continuous on $0$ since $\lim_{t \to 0} \tan(t) = \tan(0) = 0 $
Which means that, to apply the theorem, $\frac{\sin(t)}{t}$ must be continuous on $v(0)$, this is, on $0$. That is not the case, since $u(0)$ isn't even defined.

However, in this solutions manual they still get to use the theorem, it seems: http://www.stumblingrobot.com/2015/09/02/evaluate-the-given-limit-4/

How should I follow up from here? am I missing something that still allows me to use the theorem? Are they using something else in the solutions manual, and I'm just confused? In case I can't use that theorem here, how should I compute this limit, then?

Thanks in advance.

$u(0)$ is not involved in finding the limit. The required limit is $\lim_{u \to 0} u(s)=1$ because $s =\tan\, t \to 0$ as $t \to 0$ and $u(s) \to 1$ as $ s\to 0$.

Just apply definition of limit. There is no need to apply any theorem except the one which says  $u(s) \to 1$ as $ s\to 0$.

$\sin(x)$ has the same limit as $x$ when $x$ approaches $0$. This means that $\sin(\tan(x))$ will approach $\tan(x)$ at $x=0$. This means that we are looking at limit as $x$ approaches $0$ of $\tan(x)/\tan(x)=1$.


Probability of Bride entering the Church? A Bride is standing at the entrance of a church with her father (one step forward will take her into the church). Her father has a basket containing $10$ White roses and $10$ Red roses. He takes $1$ rose at a time from the basket and gives it to the Bride. If the rose is Red, the Bride takes $1$ step towards the church and if it is White, she takes $1$ step away from the church. What is the probability that the Bride enters the church? Assume that Bride's father can not see the rose until he takes the rose out of the basket.

I am stuck at this point: If the first rose is Red then the Bride enters the Church and in that case the probability is $\frac{10}{20}$. But then come the cases when the first rose is White: WRR, WWRRR, WRWRR and so on. No matter what, if the first rose is White, the last two roses must be Red. And the total number of roses required (⩽20) to enter the church is Odd, where the number of Red roses will never exceed that of the White roses but once when the Bride finally enters the church. Leaves me in doldrums, though

This problem can be stated in terms of Dyck Paths. Let $n$ be the number of roses of each colour, so there are a total of $2n$ roses. A rose sequence will take the bride into the church if at any point in the sequence the number of red roses exceeds the number of white roses. So the sequences that keep the bride out of the church are those where the number of red roses never exceeds the number of white roses. Those sequences that keep her out of the church correspond to Dyck paths of length $2n$.

Dyck paths / Dyck words are often represented using parentheses, with a Dyck path corresponding to a correctly-nested sequence of parentheses.

To illustrate, here are the sequences for $n = 3$, using parentheses, w & r for the roses, and - and + for the bride's steps.

1 ()()() wrwrwr -+-+-+
2 ()(()) wrwwrr -+--++
3 (())() wwrrwr --++-+
4 (()()) wwrwrr --+-++
5 ((())) wwwrrr ---+++

From the '-+' strings we can easily see that the number of '+'s never exceeds the number of '-'s.

The Wikipedia article on Catalan numbers has some good information on this topic. In particular, see the second and third proofs, which have helpful diagrams.

The total number of Dyck paths of length $2n$ is

$$\frac{1}{n+1} \binom{2n}{n}$$

where $\binom{m}{r}$ is the binomial coefficient. $\binom{m}{r} = \frac{m!}{r!(m-r)!}$

The total number of all the paths is

$$\binom{2n}{n}$$

So the number of non-Dyck paths is

$$\binom{2n}{n} - \frac{1}{n+1} \binom{2n}{n} = \frac{n}{n+1} \binom{2n}{n}$$

Hence the probability we seek is

$$\frac{\frac{n}{n+1} \binom{2n}{n}}{\binom{2n}{n}} = \frac{n}{n+1}$$

For the case of $n = 10$, the probability is 10/11 = 0.909090...

In the comments, Ant asks

  What's the expected number of steps that the bride has to take to enter the church, given that she enters it?

The answer is given in OEIS A008549,

Number of ways of choosing at most n-1 items from a set of size 2n+1.

  Area under Dyck excursions (paths ending in 0): a(n) is the sum of the
  areas under all Dyck excursions of length 2*n (nonnegative walks
  beginning and ending in 0 with jumps -1,+1).

The relevant formula says that the total number of steps is given by

$$4^n - \binom{2n+1}{n}$$

So to get the expected number of steps we need to divide that by the number of successful paths, i.e., non-Dyck paths.

$$\left(\frac{4^n - \binom{2n+1}{n}}{\binom{2n}{n}}\right)\frac{n+1}{n}$$
$$= \left(\frac{4^n}{\binom{2n}{n}} - \frac{2n+1}{n+1}\right)\frac{n+1}{n}$$
$$= \frac{4^n}{\binom{2n}{n}}\frac{n+1}{n} - \frac{2n+1}{n}$$
$$= \frac{4^n}{\binom{2n}{n-1}} - \frac{2n+1}{n}$$

Here's a table showing the relevant numbers for $n$ = 1..10

 n: paths success     prob :  steps expected
 1:      2      1 0.500000 :      1 1.000000
 2:      6      4 0.666667 :      6 1.500000
 3:     20     15 0.750000 :     29 1.933333
 4:     70     56 0.800000 :    130 2.321429
 5:    252    210 0.833333 :    562 2.676190
 6:    924    792 0.857143 :   2380 3.005051
 7:   3432   3003 0.875000 :   9949 3.313020
 8:  12870  11440 0.888889 :  41226 3.603671
 9:  48620  43758 0.900000 : 169766 3.879656
10: 184756 167960 0.909091 : 695860 4.143010

That table was created using this Python 3 code:

from itertools import product

def lexico_permute(a):
    a = list(a)
    yield a
    n = len(a) - 1
    while True:
        for j in range(n-1, -1, -1):
            if a[j] < a[j + 1]:
                break
        else:
            return

        v = a[j]
        for k in range(n, j, -1):
            if v < a[k]:
                break

        a[j], a[k] = a[k], a[j]
        a[j+1:] = a[j+1:][::-1]
        yield a

def bride(num):
    success = 0
    steps = 0
    base = [-1] * num + [1] * num
    for i, seq in enumerate(lexico_permute(base), 1):
        pos = 0
        for j, u in enumerate(seq, 1):
            pos += u
            if pos > 0:
                success += 1
                steps += j
                break
    return i, success, steps

print(' n: paths success     prob :  steps expected')
fmt = '{:2}: {:6} {:6} {:0.6f} : {:6} {:0.6f}'
for n in range(1, 11):
    total, success, steps = bride(n)
    prob = success / total
    expected = steps / success
    print(fmt.format(n, total, success, prob, steps, expected))

I guess it's worth mentioning (especially in relation to the expected number of steps) that

$$\binom{2n}{n} \approx \frac{4^n}{\sqrt{\pi n}}$$

as mentioned in Central binomial coefficient.

A better approximation is

$$\frac{4^n}{\sqrt{\pi n}} \frac{16n-1}{16n+1}$$

Thus the expected number of steps is approximately

$$\sqrt{\pi n} \left(\frac{16n+1}{16n-1}\right) \left(\frac{n+1}{n}\right) - \frac{2n+1}{n}$$

The obvious answer should be $1\over2$, but it is not. If you get a red rose at first, then she enters. Since the Probability of this is $1\over2$, therefore $P(x)>{1\over2}$. But if she tales a white one, then you need two consecutive red ones to get into the church. This continues.

I can't figure out the final answer but it should help.

Edit

The only outcomes in which the bride enters the church is where $r>w$, and the order does not matter. But, of all possible outcomes the only ones in which $r=w+1$ will do, since if it was more than one, it would not make sense, as the father would not choose roses after the bride had entered the Church.

There are only 20 possiblities in which $r=w+1$, namely:

$(1,0),(2,1),(3,2),(4,3),(5,4),\dots,(17,16),(18,17),(19,18),(20,19)$.

Out of the $20\cdot20=400$ possiblites. Therefore, the probablity is ${20\over400}={1\over20}$.


How can I control the 'range' of weighted sum of inputs from a neural network? Typically we use activation functions to get e.g. probability distribution from softmax. However, is there a way to control the values of the weighted sum of inputs even before applying the activation functions? such as clever ways to formulate loss functions etc. Thank you!

Yes! There are several strategies you can use to attempt to normalize the input before being sent of the a softmax function. This is often a desired behavior due to softmax's asymptotes that quickly rise with even a relatively small input. However, there are no ways where you can truly specify a range short of normalizing the inputs along each layer. What you can do is try to keep them small in proportion to keeping them accurate. This is on of the main battles of correctly fitting a neural network.
There are two main approaches you can take to do this. I believe the prior is much more of what you're looking for.
1. Input Control
Controlling input to a network involves two things. Controlling the values of the inputs themselves, and controlling the weights that mutate those inputs.
a) Normalize Your Input
You are going to want to use some sort of normalization on your initial input to scale your feature vectors appropriately. A standard approach is to scale the inputs to have mean 0 and a variance of 1. Also linear decorrelation/whitening/pca helps a lot.
b) Use a regularization term in your loss function.
When adding a regularization term in your loss function, you can allow back prop to train for keeping the weights of your network small as to control the value going into the activation function. There are two main regularization terms for the kernel (weights).
$l1$ regularization:
$new\_loss\_f = old\_loss\_f + kernel$
and $l2$ regularization:
$new\_loss\_f = old\_loss\_f + \frac{1}{2}kernel^2$
For example, cross entropy loss with l2 regularization becomes:
$$\mathcal{L}(X, Y) = -\frac{1}{n} \sum_{i=1}^n y^{(i)} \ln a(x^{(i)}) + \left(1 - y^{(i)}\right) \ln \left(1 - a(x^{(i)})\right) + \frac{1}{2}K^2$$
Thus, your loss function accounts for the size of the kernel, or weights, and punishes them being larger. In optimization (back-prop), the weights will be adjusted accordingly to of course mitigate this additional term.
2. Batch Normalization
This is a less common technique, but some choose to use  batch normalization to control the inputs to the hidden layers. This is mostly used to speed up training, but does have theoretical control over the inputs to different hidden layers.


$f$ is an entire function such that $f(0)=0$ Let $f$ be a non-constant entire function satisfying the following conditions:
$f(0)=0$ and for each $N \gt 0$ the set $\{z \mid \left| f(z)\right| < N\}$ is connected.

Prove that $f(z)=cz^n$ for some constant $c$ and positive integer $n$.

$f(0)=0$ implies there is $r>0$ such that $f(z)\neq 0$ for any $z\in \{z: 0<|z|\leq r\}$

All I can see it $0$ is the only root of $f(z)$. Suppose not, for small $M$,
the set $\{z: |f(z)|<M\}$ contains disjoint open sets which contradicts connectedness of the set. So, there is only one root.

So, $0$ is the only root of $f(z)$.

Can we now say that we are forced to have $f(z)=cz^n$ for some $n$?

I could not see more than this.. please give some hints.

Picard is a big gun, maybe even a WMD (weapon of math destruction), for this problem. We know the only zero of $f$ is $0$ from the connectivity condition. Thus $f(z) = z^ng(z)$ for some $n \in \mathbb N$ and some entire $g$ that never vanishes.

We also know there is $r>0$ such that $\min_{|z|=r} |f| = m > 0.$ It then follows, again from the connectivity condition, that $\{|f|<m/2\} \subset D(0,r).$ Thus $|f(z)| \ge m/2$ for $|z| \ge r.$

Putting this together shows

$$\frac{1}{|f(z)|} =  \frac{1}{|z|^n|g(z)|}\le \frac{2}{m}, \,\,|z|\ge r.$$

Now $1/g$ is entire, and the above gives $|1/g(z)| \le (2/m)|z|^n$ for $|z|\ge r.$ So $1/g$ is an entire function that grows no faster than a polynomial. A well known result then implies $1/g$ is a polynomial. But $1/g$ never vanishes. The only polynomials that never vanish are the nonzero constants. This gives $f(z) = cz^n$ as desired.

Let's start by assuming that $f(z)$ is a polynomial. Since we know it is an entire function, we know that  it can at least be written as a power series, so we'll begin with this simplification.

Since it is a polynomial, we can factor it over $\mathbb{C}$ as $f(z) = z(z - a_1) \cdots (z - a_k)$ for some $a_i \in \mathbb{C}$. Suppose without loss of generality that $a_k \neq 0$. Locally around $a_k$ (and 0) the function looks like $a(z- a_k)^{b_k}$. In particular, we can choose $N > 0$ small enough so that $|f|^{-1}([0, N])$ is (at least) two disjoint neigbourhoods of $a_k$ and 0, respectively. So if it's a polynomial, you're fine.

So what if it is not a polynomial? Well, then it is an entire function. Since entire functions attain every single value (except perhaps one) infinitely often, it follows that there are infinitely many $\lambda \in \mathbb{C}$ such that $f(\lambda) = 0$. Using a similar argument to the polynomial case, you can then show that the set is disconnected for $N > 0$ sufficiently small.

Edit As pointed out in the comments, this last part doesn't quite work; for example, our function could be $f(z) = z e^z$ which attains the value 0 exactly once. However, one should be able to note that in such a case, it must attain the value $\varepsilon$ infinitely often, so choosing $N > \varepsilon > 0$ one should be able to make a similar argument (since the function attains the value $\varepsilon$ close to zero, and also infinitely often close to $\infty$).


Does every uncountable Borel subset of $\mathbb R$ contains a perfect subset？ This question came from (London Mathematical Society Student Texts) Krzysztof Ciesielski-Set Theory for the Working Mathematician-Cambridge University Press. Chapter 6.2 Exercise 5.

I have thought about it for a few weeks and asked some friends.

The extension question is whether every Borel subset of $\mathbb R$ is either countable, or contains a perfect subset？

Yes - this is called the perfect set property. The analytic (= continuous image of Borel) sets also have this property; for more complicated sets (e.g. complements of analytic sets), the perfect set question is undecidable from the usual axioms of set theory.

While it's trivially true for open sets, already for closed sets it takes some work (the easiest approach, given a closed set $C$, is to consider the set of elements of $C$ around which $C$ is "locally uncountable" - that is, those $c\in C$ such that every open $U$ containing $c$ has uncountable intersection with $C$).

The full result follows from Borel determinacy, which is a very hard theorem; off the top of my head I don't know a proof that doesn't use this. Kechris' book Descriptive set theory is a very good source (as is Moschovakis' book).

I have an idea mainly from Noah Schweber's advice. I don't know is it exactly right or not?

The simple idea is a subset of a Polish space (like the real numbers) has the perfect set property if it is either countable or has a nonempty perfect subset (Kechris 1995, p. 150).

A proof from Noah Schweber' idea is as follow:

Firstly, it is trivially true for open sets, (since the definition of borel set is from the open set). If a borel subset of R is/or contains an open set, we may let it be $B(p, \epsilon)$，then in $R$ there is a $\delta < \epsilon$, such that $[p-\delta, p+\delta]$ is a close set ，and this closed set has no isolated point，and this set is perfect set. From the definition of perfect set, that is,  in the field of topology, a subset of a topological space is perfect if it is closed and has no isolated points.

But for closed sets it takes some difficult work. The easiest approach, given a closed set $C$, if it doesn't contain a perfect set ,that is every point/element of $C$,let it be  $c \in C$ is isolated points, but this is impossible. Because every family U of pairwise-disjoint open subsets of R is at most countable. For every isolated point of $C$, we could find an enough "small" open set $B(c, \epsilon)$, to cover the isolated point, and we could ensure all the open sets are pairwise-disjoint. So we have only at most countable isolated points. But this closed set of Borel subset is uncountable. So this closed set contains a part of uncountable closed set.

Then from (Cantor–Bendixson theorem) ,every uncountable, closed subset F of R can be represented as a disjoint union of a perfect set P and an at most countable set C.So we have a perfect set.


What is the following expression simplified to? $$x \mod 1000 \mod 5$$

I would have thought that it was $x \mod 5000$  except that it doesn't hold true for $x = 5005$ since you'll get zero, but $5005 \mod 5000 = 5$.

Since 5 is a divisor of 1000, I believe the expression would simplify to "x mod 5" as adding multiples of 1000 wouldn't change the remainder when divided by 5.

The Chinese Remainder Theorem provides solution to the following
$$\begin{cases}
a_1 & \mod n_1\\
a_2 & \mod n_2\\
\dots
\end{cases}$$
if $n_1, n_2, \dots$ are relative primes. If not, you can simply factorize $n_x = p_1 p_2 \dots$ and express $a_x$ with $\mod p_1, p_2, \dots$ and in your question, $1000 = 5 \times 200$, so $x \mod 5$ is just tautology.


Why should $|2^\mathbb{N}|>|\mathbb{N}^2|$ be true? I've been thinking a bit about infinite things lately, and this question I had wondered about came back to me.

One of the classic expository demonstrations of Cantor's work is the two equally surprising facts that there are as many rationals as natural numbers, but there are more reals than natural numbers. This can be reduced to a statement in cardinal arithmetic, namely that
$$|2^\mathbb{N}|>|\mathbb{N}^2|$$

Now, we can think of these sets as two collections of functions, $2^\mathbb{N}=\{f:\mathbb{N}\to \{0,1\}\}$ and $\mathbb{N}^2=\{f:\{0,1\}\to\mathbb{N}\}$. So it seems that what this statement (and others like it) is saying, is that if you want more functions, you're better off having a large domain than a large codomain.

Is there an intuitive explanation for why this should be true?

The inequality is made somewhat plausible by the observation that $|\mathbb N^2|$ is almost counting the number of $1$-element and $2$-element sets of natural numbers, while $|2^{\mathbb N}|$ counts all the sets of natural numbers.  ("Almost" refers to the facts that each $2$-element set $\{a,b\}$ is counted twice, as $(a,b)$ and as $(b,a)$.)

There is an interesting paper about this theme that claims that the inequality   $|2^\mathbb{N}|>|\mathbb{N}^2|$ need not be strict! So in stead of $|2^\mathbb{N}|>|\mathbb{N}^2|$ we have $|2^\mathbb{N}|\geq|\mathbb{N}^2|$.

Here is the link: http://www.gauge-institute.org/cantorset/CantorSetP.pdf.


Does a number get destroyed when it is multiplied by zero? We consider following subtraction-
   $$3-1=2$$
If we subtract 1 from 3 than result is 2.
in This operation 1 is not destroyed but  only replaced from 3
now we consider following equation-
  $$1\times 0=0$$
in this multiplication,  what happened with 1?
where 1 is replaced?  does it mean 1 is destroyed?

If we say that 1 is changed into 0 after multiplication by zero than
what is the meaning of changed into zero in mathematics?
Is the number   destroyed or replaced anywhere?

Doing arithmetic operations does not change numbers, but just produces new numbers as a result. If you compute $1+1=2$ then you did not change $1$ into $2$, you just produced the value$~2$ by adding $1$ and $1$ together. If one did assume that one of those $1$s were changed into$~2$, all kinds of strange questions would arise. Which one was changed, or did they both change? And what would $(1+1)+1$ be; if the final $1$ has in fact been changed into $2$, then the result should be $4$ rather than $3$. But this is not how mathematics works; whenever we write a the same number we mean the same value, so numbers themselves never change, and always represent the same value.

As for multiplication, it just happens that the value of $a\times b$ is always $0$ when $a=0$, regardless of what $b$ was, just like it always is $0$ if $b=0$, regardless of what $a$ was. The result being independent of one of the values, provide one knows the other to have a specific value, does not happen with addition or subtraction. However it really is not a big deal that it does sometimes happens with multiplication. Nothing gets destroyed, just ignored in some cases.

This is a philosophical question. It is related to the fact that $1/0$ if undefined in mathematics. There are several approaches to define divition by zero... One of them is to define every number as set with one element the number itself. And define zero as the set of all numbers. Multiplication then defined as the set of all products $ab$ where $a$ and $b$ are element of the multiplied sets... Therefore multiplication of any number by zero which group of all numbers will be group of all number which is a zero...
So is the number 1 is destroyed? We can say it is not because it is embedded inside the zero... It's always been there you just didn't know....


Probability - XOR of subset There are 15 numbers (546,861,69,868,751,562,755,43,989,120,35,947,601,651,935) and you have to create a subset of those.
0.5,0.5,0.1,0.4,0.6,0.5,0.2,0.1,0.1,0.8,0.8,0.2,0.3,0.5,0.7 are the respective probabilities to include them in the subset.
Find the expected XOR of the subset.
I am using this algorithm to solve: https://discuss.codechef.com/t/expxor-editorial/25883 as I find both the questions very similar, but I am not getting the correct answer.
511.5 is what I got, but it's incorrect. Edit: This turned out to be correct, the answer checker had a glitch.
Any help will be appreciated.

If you think each of the ten bits has a probability of about $0.5$ of surviving the XORs then the expectation should be about $\frac{2^{10}-1}2=511.5$
If you think each of the ten bits has a probability of exactly $0.5$ of surviving the XORs then the expectation should be exactly $\frac{2^{10}-1}2=511.5$
Four of the values have a probability of $0.5$ of going into the $XOR$ calculation.  Written in binary they are
546   1000100010
861   1101011101
562   1000110010
651   1010001011

and by observation each of the ten bits appears in at least one of these.
So we can conclude that $511.5$ is the correct answer.  To confirm, here is some R code which considers all the of the $2^{15}$ possibilities and their associated probabilities, confirming $511.5$
values <- c(546,861,69,868,751,562,755,43,989,120,35,947,601,651,935)
probs <- c(0.5,0.5,0.1,0.4,0.6,0.5,0.2,0.1,0.1,0.8,0.8,0.2,0.3,0.5,0.7)
combovalues <- 0
comboprobs <- 1

for (n in 1:length(values)){
  combovalues <- c(combovalues, bitwXor(combovalues, values[n]))
  comboprobs <- c(comboprobs * (1 - probs[n]), comboprobs * probs[n])
  }

sum(combovalues * comboprobs)
# 511.5

this was a question in the code decode event of haywire dude. I am from mcs and this is cheating. Thanks for participating!


Knowing that $a_m = 2a_{m - 1}(a_{m - 1} + 1)$, find $n$ such that $a_1^2 + a_2^2 + \cdots + a_{m - 1}^2 + a_m^2 + n^2$ is a square number. Knowing that $$\large a_0 \in \mathbb Z \mid a_m = 2a_{m - 1}(a_{m - 1} + 1), \forall m \in \mathbb Z^+$$, find $n$ such that $$\large a_1^2 + a_2^2 + \cdots + a_{m - 1}^2 + a_m^2 + n^2$$ is a square number for $\forall m \in \mathbb Z^+$.

I'm not sure how to solve this problem. The solution might include mathematical induction and I don't know about it.

In general, this is not possible:
Assume there are $b_m\ge0$ with
$$a_1^2+\cdots +a_m^2+n^2=b_m^2.$$
For example, if $a_0=1$, we find $a_1=4$, so
$$b_1^2=16+n^2.$$
Then $16=b_1^2-n^2=(b_1+n)(b_1-n)$, where both factirs have same, hence even, parity. This allows only $b_1+n=8$, $b_1-n=2$, so necessarily $n=3$, $b_1=5$.
But $a_2=2a_1(a_1+1)=40$, and we'd need
$$b_2^2=a_1^2+a_2^2+n^2=1625,$$
which is not a square.


If $a^3 \equiv b^3\pmod{11}$ then $a \equiv b \pmod{11}$. If $a^3 \equiv b^3\pmod{11}$ then $a \equiv b \pmod{11}$.

I verified the above statement for smaller values of $a$ and $b$ but I am unable to prove it in general. Please prove or disprove it. Can we replace $11$ by any other integer?

Here is a general statement:

  If $p$ is a prime and $m$ is coprime with $p-1$, then $a^m \equiv b^m \bmod p$ iff $a \equiv b\bmod p$.

This follows from Fermat's theorem by writing $(p-1)(-u)+mv=1$ for $u,v \in \mathbb N$.

In your example, $p=11$ and $m=3$. We can take $u=2$ and $v=7$ to get $10\cdot (-2) + 3\cdot(7)=1$. Write this as $1 + 10 \cdot 2 = 3\cdot 7 $. Then
$$
a \equiv a \cdot a^{10 \cdot 2} = a^{1 + 10 \cdot 2} = a^{3\cdot7}
\equiv
b^{3\cdot7} = b^{1 + 10 \cdot 2} = b \cdot b^{10 \cdot 2} \equiv b
\bmod 11
$$

Note $$1^{3} \equiv 2^{3}\pmod{7}$$ doesnt imply $1\equiv 2\pmod{7}$


Introduction to Analysis: Bounded and Defined This question has been rattling my brain for a while. If $f(x)$ is bounded and defined on the interval $[a,b]$ does this imply $(f(x))^2$ is bounded and defined on $[a,b]$? I would say so. Just not sure if it is a given fact or not.

Yes, $f(x)^2$ is defined and is also bounded.

To see that it is defined, note that $f(x)^2 = f(x) \cdot f(x)$. Since $f(x)$ is defined for all $x \in [a,b]$ and since multiplication of two real numbers is defined, we have $f(x)^2$ is defined.

To see that it is bounded, since $\vert f(x) \vert \leq M$ for some $M \in \mathbb{R}^+$, we have $f(x)^2 = \vert f(x) \vert^2 \leq M^2$.

Remember that the product of two functions is continuous. Now use the extreme value theorem. This is another way to think about it.


matrix exponential property Take $\alpha\in\mathrm{C}$ and $A$ a complex matrix.

Is is true that :

$$\exp(\alpha A) = (\exp{A})^\alpha$$

My intuition tells me that this is true. But I can't prove it and I can't find
this property anywhere. If this property isn't true, could you give me a
counter-example.

Assume that $\exp(A)$ has no eigenvalues in $(-\infty,0]$ and let $\log(.)$ be the principal logarithm. Then we can define $(\exp(A))^{\alpha}=\exp(\alpha \log(e^A))$. For the sake of simplicity, assume that $A=diag(\lambda_j)$; then $\log(e^A)=diag(\log(e^{\lambda_j}))=diag(\lambda_j+2k_ji\pi)$ where $k_j\in\mathbb{Z}$. Finally $\exp(\alpha\log(e^A))=diag(\exp(\alpha(\lambda_j+2k_ji\pi))=diag(\exp(\alpha\lambda_j)\exp(2k_j\alpha i\pi))=$
$\exp(\alpha A)diag(\exp(2k_j\alpha i\pi))$, that is not $\exp(\alpha A)$ when some $k_j$ is not $0$.

EDIT. Answer to PinkFloyd. Let $\theta\in \mathbb{R}\setminus \pi(1+2\mathbb{Z})$. If $A=\begin{pmatrix}0&-\theta\\\theta&0\end{pmatrix}$ then  $e^A=\begin{pmatrix}\cos(\theta)&-\sin(\theta)\\\sin(\theta)&\cos(\theta)\end{pmatrix}$ and $\log(e^A)=\begin{pmatrix}0&-\theta-2k\pi \\ \theta+2k\pi &0\end{pmatrix}$ where $k\in \mathbb{Z}$ is s.t. $\theta+2k\pi\in(-\pi,\pi)$.

Hint: write the matrix exponential in the form
$$
e^A=\sum_{k=0}^{\infty}\frac{1}{k!}A^k
$$