Definition of Substitution (Equality vs Equivalence relations) I'm confused as to wether substitution is a consequence of the properties about equivalence relations (transitivity, reflexivity, and symetry) or rather it is an independent property of equality. For example, in general $(x=y) \wedge (x<z) \Longrightarrow (y<z)$. But what if I have an equivalence relation , say $"\equiv$", instead of equality. Can I use substitution directly or do I need to prove it?

EDIT: I have to admit that I asked this question without being very clear and also without thinking much about it. So, my appologies for that. I'll do some aditional details and please if you consider that my question doesn't make sense, vote to close it.

Like I said, I'm confused because I don't know if "equality" ($=$) is a term more general than "equivalence relation" (~) or else they are the same. I was asking about the property of substitution because aparently when using "$=$" everything can be substituted whereas it is not the case when using "~".

Let's define, for example, "$\equiv_{n}$" such that $\forall a,b\in \mathbb{Z}(a\equiv_{n} b \iff n|b-a)$. Now, suppose that $a\equiv_{n} b$ and $b=6^{m}$ for some $m\in \mathbb{Z}$. Then $a\equiv_{n}6^{m}$ by substitution. Also if $m=k+7$ we can again make a substitution to get $a\equiv_{n}6^{k+7}$. But, this same process cannot always be done if we consider $b\equiv_{n}6^{m}$ and $m\equiv_{n}k+7$.

Since both "$=$" and "~" are transitive, reflexive and symmetric my guess was then that substitution is a property that belongs to $"="$ but no to "~", but I'm not sure and I'm totally confused.

You would need to prove the substitution property. Suppose we have some equivalence relation $\cong$, an operation $\circ$, and a relation $R$.
A lot of mathematics can be done by painstakingly using $\cong$ every time one normally uses $=$. We could define the following:

$\circ$ is a $\cong$-compatible operation iff for every $x,y,z$:
$$x \cong y \implies z \circ x \cong z \circ y\land x \circ z \cong y \circ z$$
$R$ is a $\cong$-compatible relation iff for every $x,y,z$:
$$x \cong y \implies (z R x \iff z R y \land x R z \iff y R z)$$

Effectively, then, $\circ$ and $R$ do not "distinguish" $\cong$-equivalence classes. The former of these usually is defined the other way around, and $\cong$ is called compatible with $\circ$ (see ProofWiki).
Similar definitions can be made for operations and relations of different arities. It is clear that these definitions amount to the substitution property.
They also provide us an insight into what equality is. For $x = y$ to be meaningful, it is obvious that all operations and relations should be $=$-compatible.
On the other hand, this property intuitively says everything there is to say about equality. For, if we cannot distinguish two things by any operation or relation, are they really different?
There is a philosophical aspect to this, but for the purposes of mathematics, it is perfectly viable to define $=$ by requiring it to universally satisfy the substitution property.

What I believe you call substitution, isn't always possible in ordinary mathematics.  You can easily end up with nonsense if you aren't careful.  For instance, no one around here (well there does exist one exception, but I think he's on holiday at the moment), will deny that 5-3=2.  Nor will anyone deny 4-1=3.  Given that you can substitute arbitrarily, it follows that 5-4-1=2.  Except, 5-4-1=1-1=0 makes sense under our assumptions, and who would ever dare deny one of our assumptions?  No wise person would do that, right?

The fool currently on holiday has told me on more than one occasion that the problem is not hard to resolve at all.  Substitutions for equals can happen in mathematics easily if one starts doing things differently.  The fool on holiday has also told me that no one around here much cares to take up any sort of method which can resolve such a problem.  The fool on holiday has also told me he has personally concluded that either mathematicians don't usually believe in substitutions for equals, or there have to exist a massive number of contradictions derivable from the principles employed by many mathematical texts.  And yes, the fool is a he.

The fool left me with some notes which indicate how to resolve the problem in several different ways, which I can share if you have not happened upon a solution yourself so far.

You can have "substitution for equals" in a fairly mechanical fashion which requires no insight.  There exist at least 4 ways to do this.  You can't use all 4 ways simultaneously.  The 4 ways the fool indicated to me for binary functions are

Write all functions before numerical symbols, variables, and other constants.  Instead writing x+y-x*y write - + x y * x y.
Put brackets or other parenthetical symbols around the expressions.  Instead of writing x+y-x*y, write [[x+y]-[x*y]].
Put brackets or other parentheses around the variables, numerical symbols, and other constants.  Instead of writing x+y-x*y write [[x]+[y]]-[[x]*[y]].
Write all function after the numerical symbols, variables, and other constants.  Instead writing x+y-x*y write x y + x y * - .

The fool kindly left me with notes which encapsulate rules which enable us to determine when an expression can get "substituted for an equal" mechanically, as somehow the fool managed to convince me that things would work out better if more texts did that thoroughly.

Let U stand for any unary function, and B the class of binary functions.  For the first way of writing things, the rules go:

If x is a variable or constant, then Fx equals a variable or constant.
If x and y are variables or constant, then Bxy equals a variable or constant.

For the second way:

If x is a variable or constant, then Fx equals a variable or constant.
If x and y are variables or constants, then [xBy] equals a variable or constant.

For the third way:

If x is a variable or constant, then Fx equals a variable or constant.
If x and y are variables or constants, then [x]B[y] equals a variable or constant.

For the fourth way:

If x is a variable or constant, then Fx equals a variable or constant.
If x and y are variables or constants, then xyB equals a variable or constant.

The fool informed me you can freely space things out in any of the 4 ways as well as adding more parenthetical symbols in the first and fourth way if you like.  And that all four ways allow you to "substitute for equals" just by manipulating symbols, so to speak.


Differential equation Euler substitution I was trying to solve this differential equation but can't figure out the final integral I get by variable separable method

  The equation is
  $$
x^3 \, y' = y^3 + y^2 \, \sqrt{y^2-x^2}
$$

I got the integral
$$
\frac{dv}{v^3 + v^2 \sqrt{(v^2 - 1)} - v}
$$
but can't figure out how to solve it.

The Euler substitution $v= \frac{u^2 + 1}{2u}$ could work but I can't seem to proceed with it.

Any help with this problem is much appreciated.

Simply use a partial fraction decomposition to obtain:

$$ \int \frac{ dx}{x^3 + x^2 \sqrt{ x^2 -1} - x } = \int \frac{ \sqrt{x^2 -1} }{2 (x-1) } - \frac{ \sqrt{ x^2 - 1}}{ 2 (x+1)} - \frac{1}{x} dx$$

Use $u$ substitute with $x-1$ and $x+1$ since $x^2-1 = (x+1)(x-1)$. I think that should do it.


Does $2764976 + 3734045\,\sqrt[3]{7} -2707603\,\sqrt[3]{7^2} = 0$? In the process of my numerical computations I have found a very special identity:

$\;\;1264483 + 1707789 \,\sqrt[3]{7} - 1238313\,\sqrt[3]{7^2} = 9.313225746154785 \times 10^{-10}$
$
-1500493 - 2026256\,\sqrt[3]{7} + 1469290\,\sqrt[3]{7^2}
 = 9.313225746154785 \times 10^{-10}$

Therefore if we subtract these two equations one should find the difference is zero.  On my computer I found:

$$
 2764976 +  3734045\,\sqrt[3]{7} -2707603\,\sqrt[3]{7^2} \stackrel{?}{=} \left\{
\begin{array}{cl} 0 & \text{by hand} \\
-1.862645149230957 \times 10^{-9} & \text{by computer} \end{array}\right.
 $$

Subtracting these two numbers - which might be the same - we have gotten twice the number!

To expand on the comments of Fabio and Jyrki:

You're claiming that $\sqrt[3]7$ is a root of the quadratic equation

$$2764976 + 3734045x -2707603x^2 = 0$$

but that leads to a contradiction.

The solutions of a quadratic with integer coefficients can be written as

$$x = \frac{p \pm \sqrt q}{r} $$

for some integers $p, q, r$.

If $x^3 = 7$ then

$$\begin{align}
\left(\frac{p \pm \sqrt q}{r}\right)^3 = 7\\
\frac{p^3 \pm 3p^2\sqrt q + 3pq +q\sqrt q}{r^3} = 7\\
\frac{(p^3 + 3pq) \pm \sqrt q(3p^2 +q)}{r^3} = 7\\
\end{align}$$

Now the only way the expression on the LHS can be rational (let alone integer) is if $\sqrt q$ is rational. But $q$ is an integer, so its square root can only be rational if $q$ is a perfect square. But that would make $x = \frac{p \pm \sqrt q}{r}$ rational, and we know that $x = \sqrt[3]7$ is not rational.

$$
276497+3734045x−2707603x^2
$$
is a quadratic with two roots:
$$
x = \frac{-3734045\pm \sqrt{3734045^2 + 4 \cdot 2764976 \cdot 2707603}}{2 \cdot (−2707603)}
$$
neither of which is a cube root of $7$, which can be verified by cubing them (or by knowing about the degrees of algebraic numbers, per @JackD'Aurizio).


Prove that $\sqrt{5}$ exists Prove that $\sqrt{5}$ exists; in other words prove that there exists a positive number $x\in \mathbb R$ satisfying $x^2=5$

Here's what I've done:

I let $A= \{x>0:x^2\leq 5\}$

We know that $A$ is not empty because clearly $2$ is in it: $2^2<5$ and we also know that $A$ is bounded by $3$ because

$$x>3\implies x^2>9$$ so by completeness, $\sup A$ exists. I let $α = \sup A$

So now I'm trying to prove that $α^2=5$ (that there exists a positive number $α\in \mathbb R$ satisfying $α^2=5$) and to do this I'm going to try to show that $α^2<5$ and $α^2>5$ are impossible.

Case 1: $α^2<5$ to do this I'm assuming a proof by contradiction will work.

So, suppose $α^2<5$ then ...

Here's where I'm a bit lost, I don't know how to proceed with a proof by contradiction here. If anyone can give me hints or explain what I should do next that would be really appreciated.

If we use your strategy, we need to show that we cannot have $\alpha^2\lt 5$, and we cannot have $\alpha^2\gt 5$. We do this with minimal machinery. That makes the argument much longer than if there is some already developed theory to appeal to.

Suppose that $\alpha^2\lt 5$. Let $\alpha^2+\delta=5$ for some positive $\delta$. Note that $\delta \le 1$ and $\alpha\ge 2$. Then
$$\left(\alpha +\frac{\delta}{4\alpha}\right)^2=\alpha^2+\frac{\delta}{2}+\frac{\delta^2}{16\alpha^2}=\alpha^2+\delta\left(\frac{1}{2}+\frac{1}{64}\right)\lt 5.$$
It follows that $\alpha +\frac{\delta}{4\alpha}\in A$, contradicting the fact that $\alpha$ is an upper bound for $A$.

Suppose now that $\alpha^2\gt 5$.  Let $\alpha^2=5+\delta$ for some positive $\delta$. We leave it to you to find a positive $x$ smaller than $\alpha$ such that $x^2\gt 5$, contradicting the fact that $\alpha$ is the least upper bound of $A$. Hint: Something like what we did will work, except that we must subtract some small number from $\alpha$, instead of adding a small number, as we did earlier.

Hint
Show that $f(x) := x^2-5$ is continuous on $\mathbb R$ and use the intermediate value thm. on
$(2, -1) \in {\rm graph}(f)$ and $(3,4) \in {\rm graph}(f)$


Series that converge to $\pi$ quickly I know the series, $4-{4\over3}+{4\over5}-{4\over7}...$ converges to $\pi$ but I have heard many people say that while this is a classic example, there are series that converge much faster.  Does anyone know of any?

The series $$ \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!} \left(\frac{1}{2}\right)^n = \frac{\pi}{2}$$ converges quickly. Here $!!$ is the double factorial defined by $0!! = 1!! = 1$ and $n!! = n (n-2)!!$

This is series is not too hard to derive. Start by defining $$f(t) = \sum _{n=0}^{\infty } \frac{(-1)^n}{(2n+1)}t^n.$$ Note that $f(1) = \pi/4$ is the series you referenced. Now we take what is called the Euler Transform of the series which gives us $$ \left(\frac{1}{1-t}\right)f\left(\frac{t}{1-t}\right) = \sum _{n=0}^{\infty } \left(\sum _{k=0}^n {n \choose k}\frac{(-1)^k}{(2k+1)}\right)t^n.$$

Now $$\sum _{k=0}^n {n \choose k}\frac{(-1)^k}{(2k+1)} = \frac{(2n)!!}{(2n+1)!!}$$ for hints on how to prove this identity see Proving a binomial sum identity $\sum _{k=0}^n \binom nk \frac{(-1)^k}{2k+1} = \frac{(2n)!!}{(2n+1)!!}$. Now put $t = 1/2$ and the identity follows. Showing the error term for the nth partial sum is less than $(1/2)^n$ is not too difficult.

The convergence can be arbitrary fast unless you don't specify what kind of series you are looking. Let $k$ be a positive integer, $a_n=\pi/k$ if $n\leq k$ and zero elsewhere. Then $\sum_{n=1}^\infty a_n$ converges to $\pi$ after $k$ summands.


calculating coordinates along a clothoid betwen 2 curves I need to write a program that will calculate coordinates along the designed rails of a proposed subway.  The designed polyline representing the rail is composed of lines, arcs (circular curves) and transition spirals which are clothoids (Euler spirals).  My problem concerns the clothoids.  The simplest case is as follows: A tangent line to a clothoid to an arc to another clothoid to another tangent line. See the first image below.
standard clothoid transition spiral

After searching the internet I found a solution for the standard case.  It is possible to calculate the variables y and x which are the distance along the tangent and the perpendicular distance to a point on the spiral, defined by the length along the spiral.  Using y and x it is easy to calculate the coordinate of the point on the spiral in relation to the tangent line.  The given or known values for the solutions are extracted from the Autocad design drawing.  They include the coordinates of the tangent points, start and end of arcs, start and end of clothoids, radiuses of arcs and lengths of clothoids.  The formula used to calculate y and x are shown above.  The calculation is an expansion of Fresnel integrals.  A clothoid is used as a transition spiral because the radius changes linearly along the length of the spiral from infinity at the tangent point to the radius of the arc segment at the end of the spiral.
     My problem is with the more complicated cases the first of which is as follows:
Tangent line to Spiral to Curve to Spiral to CURVE to Spiral to Tangent line.  The difference between this and the first case is, how do I solve a spiral that is between two arcs?  See the spiral labeled S2 below:
a spiral between 2 arcs

Thank you
John

September 15, 2019
Hello
The only way I will be able to understand the solution is by seeing a numerical solution. Here is an example spiral between 2 circular curves.  If someone has the time to do this I would be extremely grateful.  I have the data in other formats but I can't post the files here.
Thanks
John

Basic clothoid knowlege

The clothoid has generic equation $R*L = A^2$, where $R$ is radius of curvature, $L$ is the length along the arc (not straight) and $A$ is a scale factor.

Derived from this, we have some other useful relations, like $t=\frac{L}{2R} = \frac{L^2}{2A^2} = \frac{A^2}{2R^2}$
where $t$ is the angle from the tangent at $R=\infty$ (usually handled as $R=0$) and the tangent at some point with given $L,R$. In the sketch the tangent at $R=0$ is the green line $x'$. It isn't an angle from the origin, but between tangent lines; see $t2$ in the sketch.

There isn't a formula of type $y=f(x)$ for the clothoid. You may find some polynomial approximations.
Instead, $x,y$ are defined by expanded Fresnel integrals:
$$x' = A * \sqrt {2t} * \sum_{n=1}^\infty (-1)^{n+1} * \frac {t^{2n-2}}{(4n-3)*(2n-2)!}  = A * \sqrt {2t} * (1 - \frac{t^2}{10} + \frac {t^4}{216} - ...) $$

$$y' = A * \sqrt {2t} * \sum_{n=1}^\infty (-1)^{n+1} * \frac {t^{2n-1}}{(4n-1)*(2n-1)!}  = A * \sqrt {2t} * (\frac{t}{3} - \frac{t^3}{42} + \frac {t^5}{1320} - ...) $$

Notice these x', y' coordinates are in clothoid space, not in general x,yspace.
To go from "general" to "clothoid" space use the known equations:
$$ x' = (x-a)cos(\delta) - (y-b)sin(\delta) $$
$$ y' = (x-a)sin(\delta) + (y-b)cos(\delta) $$

where a,b are the coordinates in general space of the point at $R=\infty, t=0, L=0$ and $\delta$ is the angle from "general" to "clotoid" space.

It's common in surveying to use a different angle criterium: Start at north, grow clockwise. The space transformation formulas above work with normal criterium: Start at east, grow counter-clockwise.
I don't show here how those transformations change with this "survey criterium", nor the formulas for $x=f(x')$ direction. But all of them are easy to derive.

Clothoid between two circumferences

Given data: Radius and centers of both arcs, clothoid parameter A

Circumferences must be tangent to clothoids, and radius of curvature must be the same at each tangent point.

This means that at point $P1$ we have $R=R1$ and then $t1= \frac{A^2}{2R1^2}$.
Now we know $t$ we can calculate $x_1',y_1'$ by the Fresnel expansions. Repeat the same proccess for the other circumference, calculate $t2$ and $x_2',y_2'$

Case 1: P1 or P2 is given
Then you can calculate the angle $\delta$ by combining angles $C1 to P1$ and $t1$ or $C2 to P2$ and $t2$
Then calculate $a,b$ using this $\delta$ and the $x,y$ and $x',y'$ data by the rotation&translation transformation formulas.

Be aware of the direction on the clothoid: From $R1>R2$ to $R2$ or from $R1<R2$ to $R2$. This affects the point at $R=0$ and all of $\delta, a, b$. Pay attention to what $t$ angles you are calculating.

Case 2: P1 and P2 are unknown
In this case you build the triangle $P1C1P2$ with the goal of calculating the distance $r_p=C1toP2$

You already know $C1toP1=R1$ and $d_{12}= P1toP2 = (x_1',y_1')to(x_2',y_2')$
Working in "clothoid space" you can calculate the angle $P1P2$ with the $x'$ axis. Adding the $t1$ angle you get the angle for the line $P1toC1$. With the distances and this angle you solve the triangle can calculate $r_p$

Now build a circumference of center = $C1$ and radius $r_p$. Intersect it with the given circumference $C2,R2$. That is the $P2$ point.
The $P1$ point comes from another intersection: center=$P2$ radius $d_{12}$ and given circumference $C1,R1$
You may check that you use the proper intersection (circ-circ has two intersections).

You can play with angles $P2toC2$ and $t2$ to obtain angle $\delta$. Anyhow, you are now in case 1, solved above.

I'm sure there are many optimizations to this solution, I don't have them at hand.

Then

Now you have all data $a,b,\delta$ you can get any point in the clothoid by Fresnel expansions and then calculate the "general" coordinates by undoing the rotation&translation transformation.

The interpolation problem is a system of three nonlinear equations with multiple solutions which is difficult to solve also numerically. the solution of this system is reduced to the computation of the zeros of one single function in one variable
A solution exists as a G1 Fitting:
https://github.com/ebertolazzi/G1fitting
Additional information for the maths behind it:
https://www.researchgate.net/publication/237062806_Fast_and_accurate_G1_fitting_of_clothoid_curves


How to calculate $\int_a^b \frac{1}{x^2} dx$ by the definition of Riemann integral. ($0<a<b$) How to calculate $\int_a^b \frac{1}{x^2} dx$ by the definition of Riemann integral. ($0<a<b$)

I have tried to let $x_k=a+\frac{(b-a)}{n}k$. However, I have no idea to deal with the denominator in the following expression:
$$
\sum_{k=1}^n\frac{b-a}{n}\cdot\left(\frac{1}{a+\frac{b-a}{n}k}\right)^2
$$
I expanded the quadratic term, but it seems become more difficult.

Using the end points for the Riemann sum is not a very good idea as you have observed the difficulty.

Instead, let $x_k^*$ be the geometric mean of $x_{k-1}$ and $x_k$, that is $x_k^*=\sqrt{x_{k-1}x_k}$. Then the Riemann sum becomes
$$
R_n=\sum_{k=1}^n\frac{b-a}{n}\frac{1}{(x_k^*)^2}
=\sum_{k=1}^n\frac{b-a}{n}\frac{1}{x_{k-1}x_k}.
$$
Now observe that
$$
\begin{align}
\frac{1}{x_{k-1}x_k}&=\frac{1}{(a+h_nk-h_n)\cdot(a+h_nk)}\\
&=\left(\frac{1}{a+h_nk-h_n}-\frac{1}{a+h_nk}\right)\frac{n}{b-a},\quad h_n:=\frac{b-a}{n}.
\end{align}
$$
So
$$
R_n=\sum_{k=1}^n\left(\frac{1}{a+h_n(k-1)}-\frac{1}{a+h_nk}\right)
=\frac{1}{a}-\frac{1}{a+h_nn}.
$$
Now the limit $\lim_{n\to\infty}R_n$ should be easy.

Divide the interval from $a$ to $b$ into $n$ pieces.  Each piece will have length $(b-a)/n$.  Let $x_k^*$ be a point in the $k$th interval.  We approximate the value all across each interval by its value at $x_k^*$, $\frac{1}{x_k^*}$ so approximate the "area under the curve" by the area $\frac{1}{x_k^{*2}}(b-a)/n$.

The area under the curve from $a$ to $b$ is approximated by the sum
$\sum_{k=1}^n \frac{1}{x_k^{*2}}(b- a)/n= \frac{b-a}{n}\sum_{k=1}
\frac{1}{x_k^{*2}}$.

We can, in particular, take $x_k^*$ to be the right hand endpoint of each interval, $k(b-a)/n$, so the Riemann sum is $\frac{b-a}{n}\sum_{k=1}^n \frac{1}{k^2(b-a)^2/n^2}= \frac{n}{b-a}\sum_{k=1}^n \frac{1}{k^2}$

Now, can you find a "closed form" for $\sum_{k=1}^n \frac{1}{k^2}$?


If $\#(S)<\#(\Bbb N)$, then prove that $S$ is finite, without using the Axiom of Choice. If we have a set $S$ such that $\#(S)<\#(\Bbb N)$, where $\Bbb N$ is the set of the natural numbers, how would one prove that $S$ must be finite?

Here is a proof using the Axiom of Choice:
Proof: From $\#(S)<\#(\mathbb{N})$, we know that there exists an injection $f:S\to \mathbb{N}$. On the other hand, $S$ cannot be an infinite set: If $S$ is infinite, it contains a copy of $\mathbb{N}$ by the Axiom of choice. Therefore, we have an injection $g:\mathbb{N}\to S$. By the Schröder–Bernstein Theorem, $\#(S)=\#(\mathbb{N})$, which is a contradiction.
Can we prove the theorem without the Axiom of Choice?

By definition of $<$ for cardinalities, there exists an  injection $f\colon S\to \Bbb N$. For $n\in\Bbb N$, let $$h(n)=\Big|\big\{\,x\in f(S)\mid x<n\,\big\}\Big|\,.$$ If $h$ is bounded, $h(n)<M$ for all $n$,  this shows that $f(s)<M$ for all $s\in S$, whence $f$ can be viewed as map $\to\{0,\ldots, M\}$ and $S$ is finite; on the other hand, if $h$ is not bounded, for $n\in\Bbb N$, let $$m=\min\big\{\,k\in\Bbb N\mid h(j)\ge n\,\big\}\,,$$ observe that then $m\in f(S)$, and define $g(n)=f^{-1}(m)$. This gives us an injection (in fact, bijection)  $\Bbb N\to S$.

I think you can prove this by assuming that a set is infinite, you can clearly select an element and by definition of infinity you can select another element ad infinitum and you can enumerate the element that you selected at step as n.


Compute $e^5$ to 2 decimal digits precision How to compute $e^5$ without calculator with 2 decimal digits precision?

I was thinking about Taylor series:
$e^x=\Sigma_{k=0}^\infty \dfrac{x^k}{k!}$
, but it doesn't work.

(It's an interview question.)

For starters, $e - s_n = \frac{1}{(n+1)!} + \frac{1}{(n+2)!} + ... < \frac{1}{(n+1)!} * \{1 + \frac{1}{n+1} + \frac{1}{(n+1)^2} + ...\} = \frac{1}{n!n}$, where $s_n$ is the nth partial sum of taylor series for $e$. See if you can use this to approximate $e^5$.

You should trust the Taylor Series expansion more, it is simpler than you probably think. If $e^x$ is approximately $1+x+{x^2}/{2}+{x^3}/{6}+...$, we can approximate your expression as $1+5+12.5+125/6+625/24=18.5+(20+5/6)+(20+145/24)$ which is approximately $59.3+6$. The next term will be $3125/120$, after which is $15625/720$, after which is about $78000/5000$. The terms now become smaller and smaller as the factorial is being multiplied faster than 5 each term. Some rough guessing gives $65.3+27+22+16$. Perhaps we can guess the next terms can continue to get smaller such as $10,5,2$ and $1$. Adding these up yields $142.3$. If you feel lucky and throw in an extra $5$ for the infinite terms, you will be only $1.03$ lower than the actual value of $148.3362...$.

No calculator used until the very last "actual" figure.


$a<b$ and $f(a)=f(b)$ then $f \equiv 0$ in $[a,b]$ when $f´(x)=[f(x)]^2$ Let $f:\mathbb{R} \to \mathbb{R}$ a diff. function satisfying $f´(x)=[f(x)]^2 \;\;\forall x \in \mathbb{R}$. Show that if $a<b$ and $f(a)=f(b)$ then $f$ is the null function in $[a,b]$.

I tried this: $f´(x)=[f(x)]^2\geq 0$, so $f$ is non decreasing on $[a,b]$,  which means $f(x)\geq f(a) \;\; \forall x \in [a,b]$. If there is $c \in [a,b]$ such that $f(c) \neq 0$, then, since $a<c<b$, there never can be the case that $f(a)=f(b)$.

Is this correct? I don´t know if I am being rigorous enough...

$f$ is increasing as you observed, so $f(a)=f(b)$ is possible only when $f$ is a constant. [ $f(b)=f(a) \leq f(x) \leq f(b)$  gives $f(x)=f(b)$]. But then $[f(x)]^{2}=f'(x)=0$ for all $x$ so $f \equiv 0$.

I think there are some arguments missing.
The following is not exactly clear in my opinion.

  If there is $c \in [a,b]$ such that $f(c) \neq 0$, then, since $a<c<b$, there never can be the case that $f(a)=f(b)$.

If $f(x)\neq0$, why is $a<c$? Nowhere in your assumptions does it say that $f(a)=0$.

However, there is a way to fix this argument. You have to use your assumption that $f$ is non decreasing on $[a,b]$. Can you continue?


Can we cover a self similar set using another self similar set with smaller dimension? Suppose we have two self similar sets $A$ and $B$ constructed as follows:

Let $\varphi_i(x):=\frac{x+i}{5},i=0,1,2,3,4$ be similarities, then $A$ and $B$ are self similar sets with:

$$A=\varphi_0(A)\cup\varphi_2(A)\cup\varphi_4(A)\text{ and }B=\varphi_1(A)\cup\varphi_3(A).$$
Can we cover $A$ with countably many translates of the set $B$?

I believe not, but I am stuck on one small implication.

My attempt:

Using the open set condition on $(0,1)$ we can easily calculate the dimension of $A$ and $B$ using Morans Theorem and show that $\dim_H(B)<\dim_H(A).$
Since a translation is a bi-Lipshitz map, it preserves the Hausdorff dimension (informally, moving $B$ doesn't change its Hausdorff dimension). Also, the dimension of a countable union of sets is the supremum of the dimension of sets, hence any countable union of translations of $B$ have dimension $\dim_H(B)$ which is strictly less than $\dim_H(A)$.

So finally, can I conclude that if the dimension of countably many translations of $B$ is still less than the dimension of $A$, we can never cover it this way?

The Hausdorff dimension has the following relevant properties:

Countable stability: Let $\{ E_j \}_{j\in \mathbb{N}}$ be a countable collection of sets.  Then
$$ \dim_H\left( \bigcup_{j\in\mathbb{N}} E_j \right) \le \sup_{j\in\mathbb{N}} \dim_{H}(E_j).$$
Monotonicity: If $E \subseteq F$, then $\dim_H(E) \subseteq \dim_H(F)$.

Suppose that $A$ and $B$ are arbitrary subsets of some Euclidean space $\mathbb{R}^n$ such that
$$ \dim_{H}(B) < \dim_H(A). $$
Assume for contradiction that there exists a countable collection of translations $\{\tau_j\}_{j\in\mathbb{N}}$ such that
$$ A \subseteq \bigcup_{j\in\mathbb{N}} \tau_j B.$$
That is, assume that $A$ may be covered by a countable number of copies of $B$.  Translation is an isometry, hence it preserves the Hausdorff dimension.  Thus $\dim_{H}( \tau_j B) = \dim_H(B)$ for all $j$.  The Haudsorff dimension is countably stable and monotonic, thus
$$ \dim_{H}(A)
\le \dim_{H}\left( \bigcup_{j\in\mathbb{N}} \tau_j B\right)
\le \sup_{j\in\mathbb{N}} \dim_{H}(\tau_j B)
= \dim_{H} (B)
< \dim_{H}(A),$$
which is a contradiction.


Uniform Convergence of Dirichlet Series $\sum_{n=1}^{\infty} (-1)^n/(n^z)$ Show that the series $\sum_{n=1}^{\infty} (-1)^n/(n^z)$  converges uniformly on a region $\{z \in \mathbb{C} | Re(z) \geq \epsilon\}$ for $\epsilon>0$ and defines an analytic function on $\{z \in \mathbb{C} | Re(z)>0\}$..

Surely the $(-1)^n$ factor ensures the uniform convergence since the summands' modulus goes to zero but to actually show this has become tricky for me. I've thought about partial sums but it becomes real messy and the Dirichlet convergence theorem doesn't necessarily give uniform convergence

I'm not looking for a solution, instead a nudge in the right direction would be ideal.

Thanks in advanced for any help :)

Hint: $\sum\limits_{k=1}^{N}\frac {(-1)^{k}} {k^{z}} =\sum\limits_{k=1}^{N} s_k (\frac 1 {k^{z}}-\frac 1 {(k+1)^{z}}) +\frac {s_N} {(N+1)^{z}}$ (summation by parts) where $s_n=\sum\limits_{k=1}^{n} (-1)^{k}$. Use the boundedness of $(s_n)$ to show that $\sum\limits_{k=1}^{\infty} s_k (\frac 1 {k^{z}}-\frac 1 {(k+1)^{z}}) $ converges uniformly and $\frac {s_N} {(N+1)^{z}} \to 0$ uniformly on compact subsets of $\{z \in \mathbb C: Re(z) >0\}$.

It doesn't converge uniformly.
$$\eta(s) = \lim_{N \to \infty} \eta_N(s),\qquad \eta_N(s)=\sum_{n=1}^N (-1)^{n+1} n^{-s}, \Re(s) > 0$$
If the convergence was uniform on $\Re(s) \ge 1/2$, since $n^{-1/2-it}$ is bounded we would have that $\eta(1/2+it),t \in \Bbb{R}$ is bounded and
$$\lim\sup_{T \to\infty} \frac1{2T}\int_{-T}^T |\eta(1/2+it)|^2 dt <\infty$$
But again by uniform convergence it would be
$$ = \lim_{N \to \infty}\lim\sup_{T \to\infty} \frac1{2T}\int_{-T}^T |\eta_N(1/2+it)|^2 dt $$
$$= \lim_{N \to \infty} \lim\sup_{T \to\infty}\sum_{n=1}^N\sum_{m=1}^N (-1)^{m+1} (-1)^{n+1} \frac1{2T}\int_{-T}^T  n^{-1/2-it}m^{-1/2+it} dt $$
$$= \lim_{N \to \infty} \sum_{n=1}^N\sum_{m=1}^N (-1)^{m+n} (nm)^{-1/2}\lim\sup_{T \to\infty}\frac1{2T}\int_{-T}^T  (n/m)^{-it} dt $$
$$= \lim_{N \to \infty} \sum_{n=1}^N\sum_{m=1}^N (-1)^{m+n} (nm)^{-1/2} 1_{n/m=1}$$
$$ =  \lim_{N \to \infty}\sum_{n=1}^N (-1)^{n+n} (nn)^{-1/2}= \infty$$
A contradiction.

Thus it doesn't converge uniformly.

A stronger result is that $\eta(1+\epsilon+it)$ is unbounded, which can be shown from the Euler product valid for $\Re(s) > 1$ $$\eta(s) = (1-2^{1-s})\exp(-\sum_{p \ prime} \log(1-p^{-s}))$$
and the fact the $\log p$ are $\Bbb{Q}$-linearly independent.


Find $f'(0)$ if $f(x)+f(2x)=x\space\space\forall x$ Find $f'(0)$ if $f(x)+f(2x)=x\space\space\forall x$

If we assume $f'(0)\in\mathbb R$, then obviously, $f'(0)=\frac{1}{3}$.

But what if we don't assume the derivative exists?

I get this question when I am taking an exam, and I then asked the professor about whether there exist $f$ such that $f'(0)$ doesn't exist but satisfies the condition, but he says he think both exist or not is likely...

What I have tried:

If $f(x)+f(2x)=x\space\space\forall x\in\mathbb R$, $f(x)=\frac{x}{3}$ is the only nice function I have ever thought up$.

If we restrict $dom(f)\in\mathbb R$, I think that any polynomial besides $\frac{x}{3}$ doesn't satisfy the condition, and somethings like $\vert x\vert$ doesn't help neither.

On the other hand, I have tried to think about the equivalence statements with $f(x)+f(2x)=x\space\space\forall x$ in order to prove $f$ must be specific kind of problem s.t.$f'(0)$ exists.

$f(2x)+f(4x)=2x$, so $f(4x)-f(x)=x$.

In general, $$f(2^{2^n}x)-f(x)=x\prod_{k=1}^{n-1}(2^{2^k}+1)\forall n\in\mathbb N, x\in\mathbb R$$

But I don't think this helps.

Can we find $f(2x)-f(x)$ by given condition? I havn't get an idea. But while I am thinking of it, I observe that $g(x)=f(2x)-f(x)$ is on its own satisfying $g(2x)+g(x)=x$.

Will thinking about $f\circ f\circ f\circ f\circ f\circ\dots$ be useful? $f\biggl(f(x)+f(2x)\biggr)+f\biggl(2\bigl(f(x)+f(2x)\bigr)\biggr)=x\space\space\forall x$

Any help will be appreciate. Thank you!

Consider the function
$$
f(x)=\left\{\begin{array}{}\sin(\pi\log_2(|x|))+\frac x3&\text{if }x\ne0\\ 0&\text{if }x=0\end{array}\right.
$$
That is, the conditions given do not imply that $f'(0)$ exists.

$f(x)$ is separated into independent subsets defined by equivalence relationship $a = b$ given by $\exists k : 2^k \cdot a = b$.

This is due to two reorderings of the original equation, which can be iterated:

$$f(2x) = x - f(x)$$
$$f(x/2) = x/2 - f(x)$$

Let $f(1) = 1$ by definition. Then we have iteration:

0.5 -0.5
0.25 0.75
0.125 -0.625
0.0625 0.6875
0.03125 -0.65625
0.015625 0.671875
0.0078125 -0.6640625
0.00390625 0.66796875
0.001953125 -0.666015625

It is clear that there is no derivative at $f(0)$.


Convergence of Cauchy sample mean If $X_i$'s are iid Cauchy random variables, does the sample mean $\frac{\sum^{n}_{i=1}X_i}{n}$ converge in distribution to any limit? If yes, can someone explain to me how the limit can be computed step by step? I don't think it would, since the variance of a Cauchy random variable is infinity, but I am not so sure about this.

For every $n$ $\frac 1 n \sum\limits_{i=1}^{n} X_i$ also has the Cauchy distribution (!) so the Cauchy means certainly converge in distribution to the Cauchy distribution. The easiest way to prove this is to use characteristic functions: $Ee^{it \frac 1 n \sum\limits_{i=1}^{n} X_i}=(Ee^{itX_1/n})^{n}=(e^{-|t/n|})^{n}=e^{-|t|}$.
For the additional question in the comment below I will give a hint: If $S_n =\frac 1 n \sum\limits_{i=1}^{n} X_i$ converges in probability then $S_{2n}-S_n \to 0$ in probability, hence in distribution. So its characteristic function must tend to $1$ at every point. But the characteristic function converges to $e^{-|t|}$.


How to show that the limit $\lim_{n \in\mathbb N}\sqrt[n]n$ exists using epsilon-delta method. Finding the limit using L'Hopital's method might seem fine for me, but how can you prove that the limit exists :

$\lim_{n \in\mathbb N}\sqrt[n]n$

It is clear that $(\forall n\in\mathbb{N}):\sqrt[n]n>1$. Let, for each $n\in\mathbb N$, $\varepsilon_n=\sqrt[n]n-1$. Then, if $n>1$,\begin{align*}n&=\bigl(\sqrt[n]n\bigr)^n\\&=(1+\varepsilon_n)^n\\&=1+n\varepsilon_n+\frac{n(n-1)}2{\varepsilon_n}^2+\cdots\\&>\frac{n(n-1)}2{\varepsilon_n}^2.\end{align*}Therefore,$$(\forall n\in\mathbb{N}\setminus\{1\}):0<\varepsilon_n<\sqrt{\frac2{n-1}}$$and this proves that$$\lim_{n\in\mathbb N}\sqrt[n]n=\lim_{n\in\mathbb N}(1+\varepsilon_n)=1.$$

$$ \sqrt[n+1]{n+1}< \sqrt[n]{n}$$

$$ (n+1)^{n} < n^{n+1}$$

$$ n > \left(\frac{n+1}{n}\right)^{n}= \left(1 +\frac{1}{n}\right)^{n},\ \  n > n_{0}.$$

Sequence  $ \left(\left(1 + \frac{1}{n}\right)^{n}\right)$ is bounded.

Sequence $(\sqrt[n]{n})$ is decresing and lower bounded  number 1.

From Bolzano-Weierstrass theorem - sequence  $\sqrt[2n]{2n}$ is convergence to the same limit g.

$ g^2 = g\cdot g =\lim_{n\to \infty}\sqrt[2n]{2n}\cdot \lim_{n\to \infty}\sqrt[2n]{2n}= \lim_{n\to \infty}(\sqrt[2n]{2n})^{2}= \lim_{n\to \infty}(\sqrt[n]{2}\cdot \sqrt[n]{n}) = \lim_{n\to\infty}\sqrt[n]{2}\cdot \sqrt[n]{n}=1\cdot g $

$$ g = 1. $$


Localization of regular local ring by prime ideal Let $\quad R := k[X_1,\ldots,X_n]/(I)\quad$ be quotient of polynomial ring by some prime ideal, $\mathfrak{p} \subset \mathfrak{m}$ two ideals of $R$, where $\mathfrak{p}$ - prime ideal and $\mathfrak{m}$ - maximal ideal containing it. What is the easiest proof that if $R_{\mathfrak{m}}$ (localization of $R$ by $\mathfrak{m}$) is regular local ring then $R_{\mathfrak{p}}$ is also regular local ring.

There is well-known result that if $A$ is regular local ring then so is any of its localizations by prime ideals. There are proof of this in many references, but it is very nonelementary. So can someone indicate an easier proofs or give some references for the special case I given above.

Thank you!

I don't know an elementary proof. I know that $R$ regular $\Rightarrow R_P$ regular is a consequence of Auslander-Buchsbaum-Serre's theorem, which involves indeed homological algebra. See the book "Cohen-Macaulay rings", J.Herzog and W.Bruns, a.k.a. the bible of commutative algebra.


Hitting times in random walks I'm having a little trouble obtaining an intuition wrt hitting times in random walks.
Let $X_n$ be the position of a simple $1$-dimensional random walk on $\mathbb{Z}$ and $T_k$ the first hitting time of $k\in\mathbb{Z}$.

  Why do we have $\mathbb{P}(X_n\geq k)\leq \mathbb{P}(T_k\leq n)?$

I understand that $\mathbb{P}(X_n\geq k)$ is the probability that our RW is at a position greater or equal to $k$ after $n$ steps and $\mathbb{P}(T_k\leq n)$ is the probability that $k$ will first be reached before $n$ steps, but why is the latter more probable than the former?

I guess that k>0 and that you are starting from zero, right?

For to be $X_n\ge k$ it is needed that $T_k\le n$ and not the opposite.


How to prove $x^3-y^3 = (x-y)(x^2+xy+y^2)$ without expand the right side? I can prove that $x^3-y^3 = (x-y)(x^2+xy+y^2)$ by expanding the right side.

$x^3-y^3 = (x-y)x^2 + (x-y)(xy) + (x-y)y^2$
$\implies x^3 - x^2y + x^2y -xy^2 + xy^2 - y^3$
$\implies x^3 - y^3$

I was wondering what are other ways to prove that $x^3-y^3 = (x-y)(x^2+xy+y^2)$

Divide $x^3-y^3$  by  $x-y$ as polynomials

You can use "Long Division of Polynomials"

$\frac {1-(\frac y x)^3}{1-(\frac y x)}=1+\frac y x+(\frac y x)^2$

Multiply both sides by $x^3$

$\frac {x^3-y^3}{1-(\frac y x)}=x(x^2+xy+y^2)$

Then you have

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


What is exactly the inverse of the function $f(x)=\frac{x}{1-x^2}$? How do we find the inverse of the function $f:(-1,1)\to \Bbb R$ by $f(x)=\frac{x}{1-x^2}$?

The problem has been posted here and the answer below says "To show that $f$ has a continuous inverse, you just need to show that $f$ is an open map..."

Is it impossible to find the inverse analytically?

For there to be an inverse, the original function has to pass the horizontal line test-there must not be two points on the curve with the same $y$ value.  Your function satisfies that.  A graph (below) is not a proof, but very suggestive.  In this case you can prove it by showing $y' \gt 0$, which shows the function is monotonic.  If you solve $y=\frac x{1-x^2}$ for $x$ you get $$y=\frac x{1-x^2}\\y-x^2y=x\\x^2y+x-y=0\\x=\frac {-1}{2y} \pm \frac {\sqrt {1+4y^2}}{2y}$$  You need to recognize that you can't divide by zero when $y=0$.  We resolve the sign ambiguity by noting that the signs of $x$ and $y$ are always the same.  This will be true with the plus sign so the inverse is the unique $$x=\begin {cases} \frac {-1}{2y} + \frac {\sqrt {1+4y^2}}{2y} & y \neq 0\\0&y=0\end {cases}$$ The resulting curve is a nice reflection in $x=y$

Inverse function is
$$ x = y/(1-y^2)$$
solving for y(x) we can choose any one of the two valid inverse functions
$$ y(x)= -\frac{1}{2x}\pm \sqrt { 1+\frac{1}{(2x)^2} }$$


Power Mean Inequality based question For $a,b,c \in R^+$ prove that  $8(a^3 + b^3 + c^3)^2 \geq 9(a^2 + bc)(b^2 + ca)(c^2 + ab)$.
Well this one is supposed to be done using power-mean inequality, Weighted means and regular means.
Soo power mean tells us that $(\frac{a^3 + b^3 + c^3}{3})^2 \geq (\frac{a^2+b^2+c^2}{3})^3$.
now we have 3 factors of $a^2+b^2+c^2$ so it would be very very nice if we could show that each of these   $\geq$ the three factors on the RHS. however we get it as $a^2+b^2+c^2 \geq a^2 +2bc$ and not $a^2+bc$ making my life very much harder. Now i think taking that extra factor from each mean (namely $bc,ca,ab$) and multiplying them out and separating them from the desired inequality and expressing them as squares which is $\geq 0$ does seem like a way. However even if it is correct its too much brute force and m sure there exists something smarter about which i would love to know. It would be fantastic to see a solution of your own as well as continuation of my work (which will help me learn to clean my mess ;). Thanks a lot.


expected value of brownian motion How can you find this expected value?
$$
\mathbb{E}[|W_{t}^2 - t|]
$$

where $W_{t}$ is a brownian motion.

$W_t$ is a normal random variable with mean $0$ and variance $t$.  If $f(x)$ is the density of a standard normal distribution, you're looking at
$t \int_{-\infty}^\infty |x^2 - 1| f(x)\ dx$, which according to Maple is
$ 2 t e^{-1/2} \sqrt{2/\pi}$

The above expression is a martingale, just use Ito calulus to produce a formula that does not include an integral which is integrating w.r.t time, hence it does not change with a change in time. Therefore the expected value of any martingale is 0.


Which of these is larger by combinatorial reasoning Determine which of the following expression's is larger using a combinatorial reasoning $$\binom{n}{k}-\binom{n-3}{k}$$ and $$\binom{n-1}{k-1}+\binom{n-2}{k-1}+\binom{n-3}{k-1}$$I can solve this algebraicly however I need to determine which is larger using combinatorial reasoning which is something I'm not very good at, If someone could provide some advice on how to approach these kind of problems, it would be greatly appreciated. Thanks in advance :)

As usual let $[n]=\{1,2,\ldots,n\}$.
First, $\binom{n}k-\binom{n-3}k$ is the number of $k$-element subsets of $[n]$ that are not subsets of $[n-3]$, so it’s the number of $k$-element subsets of $[n]$ that contain at least one of the numbers $n,n-1$, and $n-2$. Note that these are precisely the $k$-element subsets of $[n]$ that have one of $n,n-1$, and $n-2$ as largest element.
$\binom{n-1}{k-1}$ is the number of $(k-1)$-element subsets of $[n-1]$, so it’s the number of $k$ element subsets of $[n]$ that have $n$ as largest element. $\binom{n-2}{k-1}$ is the number of $(k-1)$-element subsets of $[n-2]$, so it’s the number of $k$-element subsets of $[n]$ that have $n-1$ as largest element. And $\binom{n-3}{k-1}$ is the number of $(k-1)$-element subsets of $[n-3]$, so it’s the number of $k$-element subsets of $[n]$ that have $n-2$ as largest element. The sum of these three binomial coefficients is therefore the number of $k$-element subsets of $[n]$ that have one of $n,n-1$, and $n-2$ as largest element, and the two expressions are equal.

First of all, the 2 quantities that you have written in the question are equal.
$$\begin{align}\binom nk-\binom{n-3}k&=\binom{n-1}{k-1}+\binom{n-1}k-\binom{n-3}k\\&=\binom{n-1}{k-1}+\binom{n-2}{k-1}+\binom{n-2}{k}-\binom{n-3}{k}\\&=\binom{n-1}{k-1}+\binom{n-2}{k-1}+\binom{n-3}{k-1}+\binom{n-3}{k}-\binom{n-3}{k}\\&=\binom{n-1}{k-1}+\binom{n-2}{k-1}+\binom{n-3}{k-1}\end{align}$$
So algebraically we have proved that they are the same quantities.

Now to look at it in the view of combinatorics, $\binom nk$ is the number of ways of selecting a group of $k$ people out of $n$. So your first expression shows the number of ways of selecting a group of $k$ people with at least one of $3$ special people, say person $A, B, C$.

But this can be done in another way. Say you already select the person $A$ and put him into the group of $k$ people. So now you have to choose $k-1$ people out of $n-1$ people (which includes $B$ and $C$).

Then you select $B$ and put him into a group and don't put $A$ into that group. This ensures that you don't overcount the cases in which $A$ and $B$ are present in a group. So now you have to choose $k-1$ people to put in the group from $n-2$ people (which contains the person $C$ also).

Then you select $C$ and put him into the group of $k$ people and remove $A$ and $B$ from that group. This ensures that the cases of all $3$ being present in the group are not overcounted. So to form the group you need to choose $k-1$ people from the remaining $n-3$ people.

Hence if you sum up these $3$ cases, then you get the second expression in your question which is another way of forming a group of $k$ people with at least one of $3$ special people in a gathering of $n$ people.


How does this resolve? Supposedly the conclusion follows but I don't see how, since if we imagine $S$ as false but $P$ as true, then things seem to check out. (Classical predicate calculus, use elementary inference rules in a discrete math course, for example).

HYP 1: $P\vee \neg Q$

HYP 2: $Q \vee R$

HYP 3: $\neg R \vee S$

HYP 4: $P$

Thus $S$

Thanks, sorry for missing proper inference symbols. My professor is insistent that this works but I used contradiction and there was no contradiction.

I hope this can help:

p ∧(p ∨ ⊣q)⋀(q⋁r)⋀(⊣r⋁s)

≡p∧(q∨r)∧(⊣r⋁s)

Since p is always true, according to hypothesis we have:

((q∨r)∧⊣r)∨((q∨r)∧s)

≡((q∧⊣r)⋀F)⋁((q⋀s)⋁(r⋀s))

≡(q∧⊣r)⋁(q⋀s)⋁(r⋀s)

≡(q∧(⊣r⋁s))⋁(r⋀s)

As mentioned in the problem, ⊣r⋁s is always true. Now we have:

q∨(r∧s)≡T

(q∨r)∧(q∨s)≡T

And we know q∨r is always true, so:

q∨s≡T

This shows that we can conclude (q or s is true) and we can't conclude that s is true.

This is what I think about this problem, I hope I have solved it truly!!!


Please explain this step in proving the square root of 3 is irrational Assume that

$$3 = \frac{p^2}{q^2}$$

So,

$$ 3 q^2 = p^2$$

So $p^2$ is divisible by $3$. How we can conclude this?

Since $p$ is divisible by $3$, it is $3$ times some number, let's say $m$. Then we substitute, and we get
$$
3q^2 = (3m)^2 = 9m^2
$$
$$
q^2 = 3m^2
$$
Now $q$ has to be divisible by 3, so by the same argument, for some $n$,
$$
(3n)^2 = 3m^2
$$
$$
3n^2 = m^2
$$
And then you can keep dividing by $3$ indefinetely, and still end up with integers, but this is a contradiction, so there can be no such $p$ and $q$ in the first place.

Every number can be written uniquely as the product of primes, since $3q^2 = p^2$ their representation as the product of the primes must be the same, and since 3 is in the representation of the left hand side, it must also be in the representation  of the right hand side, $p^2$ and so it divides $p^2$


How to solve or transform this 2nd-order linear ODE? Is it possible to solve the eigenvalue problem of $$y''(x) - 2\gamma\, y'(x) + [\lambda^2 + \gamma^2 - (\frac{x^2}{2}+\alpha)^2 + x]\, y(x)=0$$ where $\lambda$ is the eigenvalue and $\alpha,\gamma$ are parameters. The boundary condition is $y(\pm\infty)=0$.
Or instead of the eigenvalue problem, can I just solve it with freely running $\lambda$? Then probably I can tackle the eigenproblem by imposing the boundary condition. Hopefully, it is related to or can be transformed into some classical form with special function?

$y''(x)-2\gamma y'(x)+\left(\lambda^2+\gamma^2-\left(\dfrac{x^2}{2}+\alpha\right)^2+x\right)y(x)=0$
$y''(x)-2\gamma y'(x)-\left(\dfrac{x^4}{4}+\alpha x^2-x+\alpha^2-\lambda^2-\gamma^2\right)y(x)=0$
Let $y(x)=e^{nx^3}u(x)$ ,
Then $y'(x)=e^{nx^3}u'(x)+3nx^2e^{nx^3}u(x)$
$y''(x)=e^{nx^3}u''(x)+3nx^2e^{nx^3}u'(x)+3nx^2e^{nx^3}u'(x)+(9n^2x^4+6nx)e^{nx^3}u(x)=e^{nx^3}u''(x)+6nx^2e^{nx^3}u'(x)+(9n^2x^4+6nx)e^{nx^3}u(x)$
$\therefore e^{nx^3}u''(x)+6nx^2e^{nx^3}u'(x)+(9n^2x^4+6nx)e^{nx^3}u(x)-2\gamma(e^{nx^3}u'(x)+3nx^2e^{nx^3}u(x))-\left(\dfrac{x^4}{4}+\alpha x^2-x+\alpha^2-\lambda^2-\gamma^2\right)e^{nx^3}u(x)=0$
$u''(x)+(6nx^2-2\gamma)u'(x)+\left(\dfrac{(36n^2-1)x^4}{4}-(6\gamma n+\alpha)x^2+(6n+1)x-\alpha^2+\lambda^2+\gamma^2\right)u(x)=0$
Choose $36n^2-1=0$ , i.e. $n=-\dfrac{1}{6}$ , the ODE becomes
$u''(x)-(x^2+2\gamma)u'(x)-((\alpha-\gamma)x^2+\alpha^2-\lambda^2-\gamma^2)u(x)=0$
Let $u(x)=e^{kx}v(x)$ ,
Then $u'(x)=e^{kx}v'(x)+ke^{kx}v(x)$
$u''(x)=e^{kx}v''(x)+ke^{kx}v'(x)+ke^{kx}v'(x)+k^2e^{kx}v(x)=e^{kx}v''(x)+2ke^{kx}v'(x)+k^2e^{kx}v(x)$
$\therefore e^{kx}v''(x)+2ke^{kx}v'(x)+k^2e^{kx}v(x)-(x^2+2\gamma)(e^{kx}v'(x)+ke^{kx}v(x))-((\alpha-\gamma)x^2+\alpha^2-\lambda^2-\gamma^2)e^{kx}v(x)=0$
$v''(x)-(x^2+2\gamma-2k)v'(x)-((\alpha-\gamma+k)x^2+\alpha^2-\lambda^2-\gamma^2-k^2+2\gamma k)v(x)=0$
Choose $k=\gamma-\alpha$ , the ODE becomes
$v''(x)-(x^2+2\alpha)v'(x)+\lambda^2v(x)=0$
Which relates to Heun's Triconfluent Equation.
Alternatively, choose $n=\dfrac{1}{6}$ , the ODE becomes
$u''(x)+(x^2-2\gamma)u'(x)-((\alpha+\gamma)x^2-2x+\alpha^2-\lambda^2-\gamma^2)u(x)=0$
Choose another $k=\alpha+\gamma$ , the ODE simplify to
$v''(x)+(x^2+2\alpha)v'(x)+(2x+\lambda^2)v(x)=0$
Which relates to Heun's Triconfluent Equation.

Starting with
$$
   y''(x) - 2\gamma\, y'(x) + [\lambda^2 + \gamma^2 - (\frac{x^2}{2}+\alpha)^2 + x]\, y(x)=0,
$$
let $y = e^{\gamma t} f$.  Then the above is reduced to potential form
$$
           (e^{\gamma t}f''+2\gamma e^{\gamma t}f'+\gamma^2e^{\gamma t}f)
      -2\gamma(e^{\gamma t}f'+\gamma e^{\gamma t}f)
      +[\lambda^2 + \gamma^2 - (\frac{x^2}{2}+\alpha)^2 + x]e^{\gamma t}f=0 \\
      f''+(\lambda^2-(\frac{x^2}{2}+\alpha)^2+x)f=0.
$$
This may be written as an eigenfunction problem in standard potential form:
$$
                 -f''+\left((\frac{x^2}{2}+\alpha)^2-x\right)f=\lambda^2 f.
$$


How to find percentage for a given set of numbers? If I have a set of numbers known in advance, how do I know which percentage is each number. What is the mathematical equation to do that?
For example, if I know I have the numbers 1,2,3,4,5. I know that 1 is 20%, 2 is 40%, 3 is 60%, 4 is 80% and 5 is 100%, but I don't know how to write the formula to get that percentage.
I'm in an excel file, where in cells I have 1,2,3,4,5 and I need to write down a percentage based on these numbers, but I don't know how to do that, and I don't want to do IF(1) then 20%.

I’m going to assume that you have a list $\{1, 2,3, ..., n\}$
Then the percentage corresponding to a number $k$ is given by $\frac{k \times 100}{n}$


Prove that if $x_1 + x_2 + ... + x_n = n$, then $x_1^k + x_2^k + ... + x_n^k \ge n$ $x_1, x_2, ..., x_n \in \mathbb R$ are nonegative and $k \in \mathbb R$, $k \ge 1$. Prove that if $x_1 + x_2 + ... + x_n = n$, then $x_1^k + x_2^k + ... + x_n^k \ge n$. I tried to find the smallest value of $x_1^k + ... + x_n^k$ and it turned out to be $n$ for $x_1 = ... = x_n = 1$. I did it by proving that if we assume that this expression assumes smallest value for sequence $x_1', x_2', ..., x_n'$ in which not every elements are equal, then we can find a sequence which gives smaller value, by just replacing greatest and smallest $x'$ with their arithmetic mean, which contradicts the "minimality" of this sequence.

Is it a proper proof or does it have some logical faults? I'm not really sure if it is fully correct.

Because by Holder
$$(x_1^k+x_2^k+...+x_n^k)(1+1+...+1)^{k-1}\geq(x_1+x_2+...+x_n)^k,$$
which gives
$$(x_1^k+x_2^k+...+x_n^k)n^{k-1}\geq n^k$$ or
$$x_1^k+x_2^k+...+x_n^k\geq n.$$
Also, we can use Jensen for $f(x)=x^k.$

I think your reasoning is not correct, at lest it's not full for the proof.

Fill in details:

$$n\le \left(x_1+\ldots+x_n\right)^k=x_1^k+\ldots+x_n^k+\left\{\text{non-negative number}\right\}$$

Where did we use that $\;1\le k\in\Bbb R\;$, that the numbers are non-negative, etc ?


Recursion $T(n) = T(\log (n)) + O(\log(\log (n)))$ $T(n) = T(\log (n)) + O(\log(\log (n)))$

assuming $n =2 \ ^ {m}$ for $m \in N $

I need to prove by induction an upper bound.

I thought of doing the following:

$$T(2\ ^{m}  )= T(m) + O(log(m)) $$

define $S(m) := T(2 \ ^ {m})  $ we get :
$$S(m) = S(log(m)) + O(log(m)) $$

I'm not sure how to continue.

You don't have upper bound cos $T(2^m)-T(m)\ge t>0$  since you can continue doing infinite times $T(2^{2^{.^{.^{.^m}}}})$ and each next is for a non infinitely small number bigger then previous.

Hint:

Assume $T(n)=O(\log(\log(n)))$. Then the recurrence says

$$T(n)=O(\log(\log(n)))=O(\log(\log(\log(n))))+O(\log(\log(n)))=O(\log(\log(n))),$$ which is true.


Given $f$ is strictly increasing, is $f$ bounded? Let $f \in C^1(\Bbb{R})$ such that $f'(x)>0$ always with $\lim_{x\to \infty} f'(x)=0$, then is $f$ bounded above?

Tried many functions but couldn't find any counter, all seems to be satisfying, not sure if the statement is true or not.

No, consider,

$$f(x) = \begin{cases}x-1 & x\leq 1,\\ \log x & x>1 \end{cases}$$

Try $\tanh(x)$, it fullfills all of your assumptions.

\edit
This is a copunterexample:
$f(x)=\sqrt{x}$ for $x\ge1$ and $f(x)=\frac{x}{2}+\frac{1}{2}$ for $x\le 1$


Find two numbers whose $AM+...$ Find two numbers whose $AM + GM =25$ and $AM:GM=5:3$.

My Attempt;

Given,
$\frac {AM}{GM}=\frac {5}{3} = k (let) $
$\AM=5k,
GM=3k$.

Also,
$AM+GM=25$
$5k+3k=25$
$8k=25$
$k=\frac {25}{8}$.

Am I going right? Or, is there any other simple alternative.?

Your process of working has so far been correct, but there is still work to do.

Substituting we have AM=$\frac {a+b}{2}=\frac {125}{8} $ and GM=$\sqrt {ab}=\frac {75}{8} $ where $a $ and $b $ are the two numbers. Thus, find $(a-b) $ from the relation $$(a-b)=\sqrt {(a+b)^2-4ab} $$ and solve for the numbers. Hope it helps.

we have $$AM=\frac{5}{3}GM$$ from here we get with the first equation:
$$\frac{5}{3}GM+GM=25$$ thus we have $$\frac{8}{3}GM=25$$ and $$GM=\frac{75}{8}$$
from here we get $$ab=\left(\frac{75}{8}\right)^2$$ and $$a+b=\frac{375}{12}$$ you can solve one equation e.g. for $a$ and plug these equation in the other one
ok with $$b=\frac{375}{12}-a$$ we get
$$a\left(\frac{375}{12}-a\right)=\left(\frac{75}{8}\right)^2$$
this is equivalent to $$0=a^2-\frac{375}{12}a+\left(\frac{75}{8}\right)^2$$
solveing this we get $$a=\frac{225}{8}$$ or $$a=\frac{25}{8}$$


Prove $\sin(kx) \rightharpoonup 0$ as $k \to \infty$ in $L^2(0,1)$ I want to show that $u_k(x)= \sin(kx) \rightharpoonup 0$ as $k \to \infty$ in $L^2(0,1)$.

We know trivially that $0 \in L^2(0,1)$.

I need to show that $\langle u^*,\sin(kx) \rangle \to \langle u^*, 0 \rangle$ for each bounded linear functional $u^* \in L^2(U)$, where $L^2(U)$ is a dual space of itself (since $L^2$ is a Hilbert space).

I think I need to show that, as $k \to \infty$,

$$\int_0^1 u^* \sin(kx) \, dx \to 0.$$

Let $f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \sin (2\pi i \,n x) + b_n \cos (2\pi i \,n x) \in L^2[0,1]$ then we have Bessel inequality:

$$ \frac{a_0}{2} + \sum_{n=1}^\infty (a_n^2 + b_n^2) < \bigg|\bigg|\int_0^1 f(x)^2 \, dx\bigg|\bigg|^2 < \infty$$

Then the Fourier coefficients tend to zero $|a_n| =|\langle f ,\sin (2\pi i \,n x)\rangle | \to 0$

Any orthogonormal basis tends the intuition here is that $\{ \sin kx \}$ or $\{e^{2\pi i \, kx}\}$ oscillate so violently we may as well just substitute $0$ there.  Weak convergence formalizes this asymptotic behavior in many cases.


An isomorphism is a mono and epi This must be very easy, but how to prove that an isomorphism is an epi and also a mono?

Suppose $f:X\to Y$ is an isomorphism. This means there is an arrow $g:Y\to X$ such that $fg=1_Y,\ gf=1_X$.

To show $f$ is an epi, suppose $h,h':Y\to Z$ are arrows and $hf=h'f$. We must show that $h=h'$. There must be some kind of trick with composing something with identities, but I don't know what the trick is. For example, we can write $hf=h1_Yf=hfgf$ or something like that, but I don't see how this can be helpful.

And similarly, I don't see what to use to prove that $f$ is an mono. To this end we need to show that if $h,h':Z\to X$ are arrows, then $fh=fh'\implies h=h'$.

To show that it is epi, simply precompose with $g$. I.e.,
$$hf=h'f \qquad \Rightarrow \qquad hfg=h'fg \qquad \Rightarrow \qquad h=h'$$
since $fg=1$. Similarly, to show that it is mono, postcompose with $g$
$$fh=fh' \qquad \Rightarrow \qquad gfh=gfh' \qquad \Rightarrow \qquad h=h'$$
since $gf=1$.

We will shot that $f$ is mono.

Lets suppose that there are $a,b\in X$ such that $f(a)=f(b)$. Then, we will apply $g$:

$$g(f(a))=g(f(b))~\Rightarrow~ 1_X(a) = 1_X(b) ~ \Rightarrow~a=b.$$

We will shot now that $f$ is epi:

Let be $c\in Y$, since $c=c$, we may write $1_Y(c)=c$.
but: $$1_Y(c)=f(g(c))~\Rightarrow~ \exists d\in X ~ (d=g(c)),~ \mbox{ s.t } f(d)=c $$

which means that $f$ is also epi.


Parabolic sine approximation Problem Find a parabola ($f(x)=ax^2+bx+c$) that approximate the function sine the best on interval [0,$\pi$].

The distance between two solutions is calculated this way (in relation to scalar product): $\langle u,v \rangle=\int_0^\pi fg$.

My (wrong) solution I thought that I would get the solution by calculating the orthogonal projection $w=a+bx+cx^2$ of $v=\sin x$ on subspace $W=\langle u_1,u_2,u_3 \rangle=\langle 1,x,x^2\rangle$ using Gramm matrix. Then I have
$$\begin{pmatrix}\langle u_1, u_1\rangle&\langle u_1, u_2\rangle&\langle u_1, u_3\rangle\\
\langle u_2, u_1\rangle&\langle u_2, u_2\rangle&\langle u_2, u_3\rangle\\
\langle u_3, u_1\rangle&\langle u_3, u_2\rangle& \langle u_3, u_3\rangle\end{pmatrix}
\begin{pmatrix} a\\b\\c\end{pmatrix}=
\begin{pmatrix} \langle u_1, v\rangle\\
\langle u_2,v\rangle\\
\langle u_3, v\rangle \end{pmatrix}$$
So then
$$\begin{pmatrix} \int_0^\pi 1&\int_0^\pi x&\int_0^\pi x^2\\
\int_0^\pi x& \int_0^\pi x^2& \int_0^\pi x^3 \\
\int_0^\pi x^2&\int_0^\pi x^3&\int_0^\pi x^4\end{pmatrix}\begin{pmatrix} a\\b\\c\end{pmatrix}=\begin{pmatrix} \int_0^\pi \sin x\\\int_0^\pi x\sin x\\\int_0^\pi x^2 \sin x \end{pmatrix}$$
But solving these equations didn't give me any good answer. So my question is - is my way of solving it completely wrong (if so, can you give me hints how to do it otherwise)?

Thank you

Edit
Then$$
\begin{pmatrix} \pi&\frac{\pi^2}{2}&\frac{\pi^3}{3}\\
\frac{\pi^2}{2}&\frac{\pi^3}{3}&\frac{\pi^4}{4}\\
\frac{\pi^3}{3}&\frac{\pi^4}{4}&\frac{\pi^5}{5}\end{pmatrix}\begin{pmatrix} a\\b\\c\end{pmatrix}=\begin{pmatrix} 2\\\pi\\\pi^2-4 \end{pmatrix}$$
$$
\begin{pmatrix} \pi&\frac{\pi^2}{2}&\frac{\pi^3}{3}\\
0&\frac{\pi^3}{12}&\frac{\pi^4}{12}\\
0&\frac{\pi^4}{12}&\frac{4\pi^5}{45}\end{pmatrix}\begin{pmatrix} a\\b\\c\end{pmatrix}=\begin{pmatrix} 2\\0\\\frac{1}{3}\pi^2-4 \end{pmatrix}$$
$$
\begin{pmatrix} \pi&\frac{\pi^2}{2}&\frac{\pi^3}{3}\\
0&\frac{\pi^3}{12}&\frac{\pi^4}{12}\\
0&0&\frac{\pi^5}{180}\end{pmatrix}\begin{pmatrix} a\\b\\c\end{pmatrix}=\begin{pmatrix} 2\\0\\\frac{1}{3}\pi^2-4 \end{pmatrix}$$
Edit II My way (I have found the mistake I did) will get the result
$$f(x)=\dfrac{60(\pi^2-12)}{\pi^5}x^2-\dfrac{60(\pi^2-12)}{\pi^4}x+\dfrac{12(\pi^2-10)}{\pi^3}$$
which I hope is the right answer with error approx. $0,000936$

Hint

By similarity with least square fit problems, I think that you need to minimize $$F=\int_0^{\pi}\Big(\sin(x)-(a x^2+bx+c)\Big)^2~dx$$ Compute the expression and say that $$\frac{dF}{da}=\frac{dF}{db}=\frac{dF}{dc}=0$$ This will provide three linear equations for the three unknowns $a,b,c$.

This seems to be very similar to what you posted (and the result is quite good).

Edited later

The answer you obtained is perfectly right but, if I may, I would like to add a comment about the problem itself. Fist, you notices that the function does not reproduce the values of $\sin(x)$ at the bounds.

As mentioned by Lucian, almost 1400 years ago, Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician and astronomer, proposed for the sine developements in terms of $x(\pi-x)$. So, let me consider the approximation $$\sin(x)\approx \sum_{i=1}^j a_i \big(x(\pi-x)\big)^i$$ (I excluded constant terms in order the values of  $\sin(x)$ at the bounds be respected) and repeat the calculations as you made them (you could notice from your final result that the expression corresponds to $j=1$ plus a constant term).

For $j=1$, $F=2.27 \times 10^{-3}$ which is effectively  worse than the $9.36 \times 10^{-4}$ you obtained (because I omitted the constant term).

For $j=2$, $F=7.89 \times 10^{-7}$ which is much better.

For $j=3$, $F=6.77 \times 10^{-10}$ which is much much better.

This was just to show that, depending on the base function we select, for the same number of adjustable parameters (or even less), we can make approximations of very different quality.

Just for your curiosity, the marvelous approximation given 1400 years ago $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}$$ leads to $F=2.98 \times 10^{-6}$.

Is not the direct modern equivalent of Bhaskara's parabola approximation for Sine curve good enough as a start?

$$ y(x) = \dfrac{x(\pi-x)}{2},$$
$$ y(\pi/2)= \pi^2/8;y^{'}(\pi/2) =0; y^{''}(\pi/2) =-1. $$

And cannot the amplitude alone be improved/adjusted from here further to minimize error by least squares?


Expected number of unpecked chicks - NYT article In this article, the winner of the math competition answered this question correctly:

  In a barn, 100 chicks sit peacefully in a circle. Suddenly, each chick randomly pecks the chick immediately to its left or right. What is the expected number of unpecked chicks?

The answer was given as 25. I am interested to know the correct method for solving this problem as well as a general way to find this out for N chicks. I was thinking of trying to solve by figuring out the number of occurrences of (something?) inside a binary string of length 100 (or N), but I don't know if that is the right way to approach it.

For any individual chick, there is a $0.5$ chance that the one on its right won't peck it, and a $0.5$ chance that the one on its left won't peck it. So overall, it has a $0.25$ chance of not being pecked.

Since this is true for every chick, we can add up $0.25(100)=25$ to get the number of unpecked chicks.

"But wait," you might say, "the probabilities that chicks are unpecked aren't independent! If chick $n$ is unpecked, then chicks $n-2$ and $n+2$ will be pecked. Surely we have to take this into account!"

Fortunately, there's a very useful theorem called linearity of expectation. This tells us that for any random variables $X$ and $Y$, independent or not, $E[X+Y]=E[X]+E[Y]$, where $E$ is the expected value. What we're formally doing above is assigning a random variable $X_i$ to the $i$th chick that is equal to $0$ if the chick is pecked and $1$ if it's unpecked. Then the total number of unpecked chickens is $\sum X_i$, so the expected number of unpecked chickens is
$$E\left[\sum X_i\right]=\sum E[X_i]$$
by linearity of expectation. But $E[X_i]$ is just the probability that the $i$th chicken is unpecked, thus justifying our reasoning above.

Represent chickens as variables ($a_{1}, a_{2}, \dots, a_{100})$. Assign it a $1$ if unpecked, $0$ otherwise. We want to find the expected value of the sum $a_{1} + \dots + a_{100}$. Notice that the expected value of each of these is $\frac{1}{4}$, so by linearity of expectation, the expected value of the sum is $100 \cdot \frac{1}{4} = 25$


Show the following is a sigma algebra Let $\mathcal{A}$ be an algebra containing a pi system $\pi$, and let $\mathcal{F}=\sigma(\pi)$. Suppose
$\mathcal{A}\subset \mathcal{F}$.

Define $\mathcal{A}'=\{A\in \mathcal{F} \mid$ there is $A_n\in\mathcal{A}$ with $\mathbb{P}[A_n\triangle A]\to 0\}$.

I am struggling to show that $\mathcal{A}'$ is a sigma algebra. Namely, the final property that it is closed under countable unions.
Any help would be appreciated.

Induction:

Let A,B $\subset$ A'. Then for any x $\in$ A or B, x $\in$ A $\bigcup$ B.  Therefore A $\bigcup$ B $\subset$ A'.

For a collection of sets $$\{A_i \}_{i \in \Lambda}$$s.t. $\Lambda$ is countable and $A_i$ $\subset$ A' $\forall$  $i\in \Lambda$ assume:
$$\bigcup_{i \in \Lambda} A_i \in A'$$.

Then for an additional set $A_j \in \Lambda$ then
$$A_j \bigcup (\bigcup_{i \in \Lambda} A_i) \rightarrow A_j \bigcup B \in A'$$
$$B = \bigcup_{i \in \Lambda} A_i$$
We know this because both A,B $\subset$ A'.


Does there exist $n\in\mathbb{N}$ such that $5^n-2^n$ is a perfect square? I checked up to $n\leq 100000$ then found no example. So I suspect that there doesn't exist $n\in\mathbb{N}$ such that $5^n-2^n$ is a perfect square.
Then I tried to prove by modular arithmetic, but there seems no $k$ such that $\{5^n-2^n \bmod k \mid n\in\mathbb{N}\}\cap\{n^2\bmod k\mid n\in\mathbb{N}\}=\emptyset$.
Thank you for your help.

Checking things modulo $5$ you can see that $n$ has to be even, as $2$ is not a square (and hence so are all odd powers). So say $n = 2m$. Then we can write
$$5^n - 2^n = 5^{2m} - 2^{2m} = (5^m - 2^m)(5^m + 2^m).$$

Assume there exist $n$ such that we get perfect squares here. Then there is a minimal such $n$, that (due to laziness) we will in the following simply call $n$.
Any prime that divides $5^n - 2^n$ must divide it to an even power of at least two (quickly exclude the case $5^n - 2^n = 1$ for completeness sake please). But if all these primes also divide $5^m - 2^m$ to an even power, then $5^m - 2^m$ is also a perfect square, a contradiction to the assumption that $n$ is minimal.
That means that there must exist a prime divisor $p$ of $5^n - 2^n$ that divides both $5^m - 2^m$ and $5^m + 2^m$.
But then $p$ also divides the difference of the two, which is $2^{m+1}$, so $p = 2$.

But $5^n - 2^n$ is always odd, so $p = 2$ is not possible. Hence, we have found a contradiction to the assumption that any such $n$ exists.

One might note that $5^n-2^n$ is a base, and every new value of $n$ brings a new prime, previously not before seen.  If some prime $p$ divides $5^n-2^n$ then it divides $5^{pn}-2^{pn}$ an additional time.

So $5-2=3$ which is not square.  So we replace $n$ with $3n$, to get $5^3-2^3 = 3^2 13$.  But the 13 is not square, so we have to replace $n$ with $13n$.  Here we now introduce two very large coprime numbers, of which the larger is a multiple of 13.

It runs away as a power of exponentials.


$H_k$ summability of a sequence implies its Abel summability to the same sum. Let $\sigma_n^{(k)}=\frac{1}{n+1}\sum_{j=0}^{n}\sigma_j^{(k-1)}$ and $\sigma_n^{(1)}=\frac{1}{n+1}\sum_{j=0}^{n}s_j.$

If $\lim_{n\to \infty}\sigma_n^{(k)}=L$ we call the sequence $(s_n)$ is summable $H_k$ to $L.$

Also the sequence $(s_n)$ is called Abel summable to $L$ if $\lim_{x\to1^{-}}(1-x)\sum_{j=0}^{n}s_jx^j=L.$

Does $H_k$ summability of $(s_n)$ to $L$ imply its Abel summability to $L?$

Also, are there any sequence which is Abel summable but not $H_k$ summable?

The answer to your first question is yes, because Hölder summability (your method $H_k$) is equivalent to Cesàro summability, and Abel summability is stronger than all the Cesàro methods. These are theorems 49 and 55 in Hardy's Divergent Series (1949). So, if a sequence is $H_k$ summable, then it is $(C, k)$ summable and therefore Abel summable.

The answer to your second question is also yes. Hardy's example (p.109) defines $a_n$ by $$f(x) = e^{1/(1+x)} = \sum_{n \ge 0} a_n x^n$$
so that $f(x)$ exists for $|x| < 1$ and $f(x) \to e^{1/2}$ as $x \to 1$. But the sequence $a_n$ is not $O(n^k)$ for any $k$ since this would imply $f(x) = O((1 - |x|)^{-k-1})$ uniformly in the disk $|x| < 1$, whereas $f(x)$ tends to infinity like $e^{1/(1-|x|)}$ (which is much faster) when $x\to-1$ through real values. So, $a_n$ is not $(C, k)$ summable for any $k$, by the limitation theorem (46 op. cit.).


How can I solve $x^{5}\equiv 13 \pmod{18}$. I need to solve the congruence of $ x^{5}\equiv 13 \pmod{18}$ and find whether it is solvable.

Actually I can solve $x^{k}\equiv \pmod{q}$ when $q$ is odd. But now it is composite, and I do not know how to apply the theorem. Can you help me?

Hint: By Euler–Fermat, $x^{6}\equiv 1 \bmod{18}$.


Find all Integral solutions to $x+y+z=3$, $x^3+y^3+z^3=3$. Suppose that $x^3+y^3+z^3=3$ and $x+y+z=3$.

What are all integral solutions of this equation?

I can only find $x=y=z=1$.

Using identity: $(x+y+z)^3 = x^3+y^3+z^3 + 3(x+y)(y+z)(z+x)$, we have:

$3^3 = 3 + 3(x+y)(y+z)(z+x) \Rightarrow (x+y)(y+z)(z+x) = 8$. From this you should be able to deduce the answer.

Using the Power Mean Inequality we have that:

$$ 1=\frac{1}{3}(x^3 + y^3 + z^3)  \leq \left[\frac{1}{3}(x+y+z)\right]^3 =1$$

Equality only holds when $x=y=z=1$.

Let's replace $z \mapsto -z$ and try to solve the system of equations for $x,y,z \geq 0$:

$$ x^3 + y^3 = 3 + z^3 \text{ and } x + y = z + 3 $$

We can still use the power mean inequality (in only two of the varaibles) and use the linear relation

$$ \frac{1}{3}z^3 < \frac{1}{3}(3 + z^3 ) = \frac{1}{3}(x^3 + y^3 )  \leq \left[\frac{1}{3}(x+y)\right]^3 = \left[\frac{1}{3}(z+3)\right]^3
< \frac{1}{24}(z+3)^3 $$

Maybe if I take cube roots the left and right sides are easier to compare.  I also rounded $3^3 \approx 27$

$$ z <  \frac{1}{2}(z + 3)$$

This means $0 \leq z < 3$.  Can it be that $0^3 + 3 = 3, 1^3 + 3 = 4, 2^3 + 3 = 11$ are the sums of two cubes?

If two of the 3 numbers are negative,  the starting point should be:

$$ x^3 + y^3 = -3 + z^3 \text{ and } x + y = z - 3 $$

Then using the power mean inequality we can attempt to deduce a range on $z$

$$  \frac{1}{3}(-3 + z^3 ) < \frac{1}{24}(z-3)^3 $$

This one is harder to estimate.  After fiddling around, we see that $z \leq 0$:

$$   z <  \frac{1}{2}\sqrt[3]{(z-3)^3 + 24}
< \frac{1}{2}z $$

So there are definitely no solutiuons of this kind.


A Question About A Polynomial Problem I'm trying to understand the solution to this problem:https://imgur.com/E2Jm63r. The solution states that the answer is $C$ and $D$, because if $-p(x)=p(y)$, then the equation of the values that satisfy the $-p(x)=p(y)$ must by symmetric with respect to$ y=-x$. I have 2 questions about this.

$1$. Why does the fact that $-p(x)=p(y)$ mean that the equation must be symmetrical over y=-x? Is this only true for polynomials, or is it true for any function?

$2$. Does the fact that $p(x)=p(y)$ mean that the equation must be symmetrical over y=x? Is this only true for polynomials, or is it true for any function?

Thanks in advance

Let $S$ be the set of points $(x, y)$ such that $p(x) + p(y) = 0$.

Suppose $(x, y) \in S$. Then $p(x) + p(y) = 0$. But then also $p(y) + p(x) = 0$, and so $(y, x) \in S$. This means that $S$ is symmetric with respect to the line $y = x$. Thus, the answer (A) is wrong.

In general, $S$ might not be symmetric with respect to the line $y = -x$. Indeed, if $p(x) = x + 1$, then the equation $p(x) + p(y) = 0$ amounts to $x + y + 2 = 0$, which represents a line parallel to the line $y = -x$.

$C$ is true because you can define $p(x)=x^2-\frac{1}{2}$ and you have that

$p(x)+p(y)=x^2+y^2-1=0$

That is the equation of the circumference;

$D$ is true because you can define

$p(x)=x$

and you have that

$p(x)+p(y)=x+y=0$

that is the equation of the second and fourth quadrant;

$A$ is not true because it is the locus of $y=ax^2$ where $a>0$


Convergence of the infinite product $\prod_{n = 1}^{\infty} \frac{z - \alpha_n}{z - \beta_n}$ I've been trying to solve this homework problem for a while but I can't seem to get any significant ideas about how to approach it, so I would really appreciate any hints that could help me solve it.

The problem is exercise 8.14 from Steven Krantz and Robert Greene's book Function Theory of One Complex Variable. It goes as follows:

  Suppose that
  $$\sum |\alpha_n - \beta_n| < \infty$$
  Then determine the largest open set of $z$ for which
  $$\prod_{n = 1}^{\infty} \frac{z - \alpha_n}{z - \beta_n}$$
  converges normally.

What I've tried so far is writing the factors as

$$\frac{z - \alpha_n}{z - \beta_n} = 1 + \frac{z - \alpha_n}{z - \beta_n} - 1 = 1 + \frac{\beta_n - \alpha_n}{z - \beta_n}$$

so as to put the infinite product in the form $\displaystyle{\prod (1 + f_n(z))}$ to try to apply the basic convergence criteria I have available  which says that this product would converge normally if the series

$$\sum |f_n(z)|$$
converges normally. Now I'm kind of stuck here because I think that maybe I would have to bound this sum with the sum $\sum |\alpha_n - \beta_n|$ but I'm not sure about how to proceed (assuming that this is the right way to follow).

So I would really appreciate some hints that would get me in the right track to solve this problem. Thank you very much.

Hint:

Suppose that $z\in\mathbb{C}\setminus\overline{\{b_n\}}$, then there is an $\epsilon>0$ so that $|z-b_n|\ge\epsilon$ for all $n$. Then,
$$
\sum_n\left|\frac{b_n-a_n}{z-b_n}\right|\le\frac{1}{\epsilon}\sum_n|b_n-a_n|<\infty\tag{1}
$$
Inequality $(1)$ implies that
$$
\prod_n\frac{z-a_n}{z-b_n}=\prod_n\left(1+\frac{b_n-a_n}{z-b_n}\right)\tag{2}
$$
converges.

I think you are very close.

The convergence criteria you are using is correct - consider this:

We observe that

$$|\beta_n - \alpha_n| = |\alpha_n - \beta_n| $$

so
$$\left|\frac{\beta_n - \alpha_n}{z - \beta_n}\right| < |\alpha_n - \beta_n| $$
As you know that the absolute series converges as per the question's definition so $$\sum \left|\frac{\beta_n - \alpha_n}{z - \beta_n}\right| \leqslant \sum |\alpha_n - \beta_n| < \infty$$ so it converges by limit comparison theorem.

As per the largest open set, the condition needed for convergence is $\left|\frac{\beta_n - \alpha_n}{z - \beta_n} \right| \leqslant |\alpha_n - \beta_n| $ Solve for $z$.


Why does $(-1) \times (-1)$ give +1? Why is $(-1) \times  (-1)=+1$ ?  What is the intuitive concept ?
 My second question :

How can I show that no triangular number can be of the form $3n-1$ ?

First, let's establish a pattern that can be used to intuitively understand why $1 \times (-1) = (-1)$

$1 \times 3 = 3$

$1 \times 2 = 2$

$1 \times 1 = 1$

$1 \times 0 = 0$

Every time the factor goes down by one, the product goes down by 1 (which is the first factor here).

Extending the pattern:

$1 \times (-1) = -1$

$1 \times (-2) = -2$

and so on.  So $a \times (-b) = - (a \times b)$

Well, we can apply the pattern to the other factor as well:

$(-1) \times 3 = -3$

$(-1) \times 2 = -2$

$(-1) \times 1 = -1$

$(-1) \times 0 = 0$

Every time the factor goes down by one, the product goes down by -1 (which is the first factor here), or in other words, goes up by 1.

Extending the pattern:

$(-1) \times (-1) = 1$

$(-1) \times (-2) = 2$

and so on.  So $(-a) \times (-b) = a \times b$

Now for the second question:

A triangular number is of the form $n(n + 1)/2$.  And of the form $3n - 1$ really means $-1 \pmod 3 \equiv 2 \pmod 3$.

If $n \equiv 0 \pmod 3$, $n(n + 1)/2 \equiv 0 \pmod 3$.

If $n \equiv 1 \pmod 3$, $n(n + 1)/2 \equiv 2/2 \equiv 1 \pmod 3$.

If $n \equiv 2 \pmod 3$, $n(n + 1)/2 \equiv 0 \pmod 3$.

None are equivalent to $2 \pmod 3$, so it can't be expressed as $3n - 1$.

$$(-1) \times (2) = -2$$
$$(-1) \times (1) = -1$$
$$(-1) \times (0) = 0$$
$$(-1) \times (-1) = \cdots$$

Of course, maybe some of these aren't intuitive either!


Does the series $1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \dots$ converge? Does the following variant of the harmonic series converge? If it diverges (which I think it does), can I know if it diverges to $\infty$ or has no limit?

Note that the series is not alternating in the classical sense of the word.
$$1 + \frac{1}{2} -\frac{1}{3} + \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \dots$$

The generic term of the series would have to be something like,

$$a_n = \left\{\begin{array}{ll} -\frac{1}{n}, & \text{if } 3 \mid n  \\ \;\,
\,\, \frac{1}{n}, & \text{otherwise}\end{array}\right.$$
I'm not sure if it's very helpful. The terms divisible by 3 are negative, and the others are positive.

Is there a way to decide and prove whether an alternating series of this sort (e.g. with a period other than 2) converges? Or one where terms are positive or negative according to some other rule?

Almost all convergence tests I've come across are generally limited to either simple alternating series or where all the terms are positive.

No, the series diverges.  Note that the partial sum of length $3N$ can be written as

$$\sum_{n=0}^{N-1} \left(\frac{1}{3n+1} + \frac{1}{3n+2} - \frac{1}{3(n+1)}\right).$$

Combining the terms, we get

$$\sum_{n=0}^{N-1} \frac{9 n^2+18 n+7}{3 (n+1) (3n+1) (3 n+2)}$$

and it's easy to see that this is a divergent series by limit comparison with $\sum 1/n$.  Now, if the original series converged, then this sub-sequence of the sequence of partial sums would also have to converge.

It's quite easy to see it diverges.  It is the difference between two series, 1/n and 2/(n(n+1)).  The first diverges, the second converges.


A way to find this shaded area without calculus? This is a popular problem spreading around. Solve for the shaded reddish/orange area. (more precisely: the area in hex color #FF5600)

$ABCD$ is a square with a side of $10$, $APD$ and $CPD$ are semicircles, and $ADQB$ is a quarter circle. The problem is to find the shaded area $DPQ$.

I was able to solve it with coordinate geometry and calculus, and I verified the exact answer against a numerical calculation on Desmos.

Ultimately the result is 4 terms and not very complicated. So I was wondering: Is there was a way to solve this using trigonometry? Perhaps there is a way to decompose the shapes I am not seeing.

A couple of years ago there was a similar "Find the shaded area" problem for Chinese students. I was able to solve that without calculus, even though it was quite an involved calculation.

Disclosure: I run the YouTube channel MindYourDecisions. I plan to post a video on this topic. I'm okay posting only the calculus solution, but it would be nice to post one using only trigonometry as many have not taken calculus. I will give proper credit to anyone that helps, thanks!

Update: Thanks for everyone's help! I prepared a video for this and presented 3 methods of solving it (the short way like Achille Hui's answer, a slightly longer way like David K and Seyed's answer, and a third way using calculus). I thanked those people in the video on screen, see around 1:30 in this link: https://youtu.be/cPNdvdYn05c.

The area is equal to difference between the area of two lenses.

It is easy to find the area of lenses like the one I did in this question before: How to find the shaded area

Let's assume we have 4 semi circles inside ABCD. There will be 4 intersections and all will be equal to DP.So we can calculate DP by
4 times area of DP + the area of square = 2 times the area of circle (r=5)
so DP= 50/4 (pi=3)
İF we assume one more semi circle in ABCD (APB)
and one more quarter circle in ABCD (CDHB)
so area of 2 quarter circles - the area of the square will give us the area of intersection
DB
10*10*3/2= 150  10*10=100 150-100= 50
In half the area of intersection there are 2 red areas and DP so
50/2= 25 and (25-(25/2))/2=7.5 will be our answer


Characterization of Parseval Frame Let $H$ be a Hilbert space, and let $e_j\in H$ for every $j\in\mathbb N$.

Is it possible to show that

$$\left\|f\right\|^2=\sum_{j=1}^\infty\left|\langle f,e_j\rangle\right|^2\text{ for every }f\in H\text{ if and only if }f=\sum_{j=1}^\infty\langle f,e_j\rangle e_j\text{ for every }f\in H\tag*{?}$$

I get the feeling that we need $e_j\perp e_k$ if $j\neq k$.

For the reverse implication see this answer.
Note that orthogonality is not required for a Parseval Frame.
Assume that $\|f\|^2=\sum_j|\langle f,e_j\rangle|^2$ for all $f$. Given $m>n$,
\begin{align}
\Big\|\sum_{j=1}^m\langle f,e_j\rangle\,e_j-\sum_{j=1}^n\langle f,e_j\rangle e_j\Big\|^2
&=\Big\|\sum_{j=n+1}^m\langle f,e_j\rangle e_j\Big\|^2\\[0.3cm]
&=\sup\Big\{\Big|\Big\langle\sum_{j=n+1}^m\langle f,e_j\rangle e_j,h\Big\rangle\Big|:\ \|h\|=1\Big\}\\[0.3cm]
&=\sup\Big\{\Big|\sum_{j=n+1}^m\Big\langle\langle f,e_j\rangle e_j,h\Big\rangle\Big|:\ \|h\|=1\Big\}\\[0.3cm]
&\leq\sup\Big\{\sum_{j=n+1}^m|\langle f,e_j\rangle\,\langle e_j,h\rangle|:\ \|h\|=1\Big\}\\[0.3cm]
&\leq\bigg(\sum_{j=n+1}^m|\langle f,e_j\rangle|^2\bigg)^{1/2}
\sup\Big\{\bigg(\sum_{j=n+1}^m|\langle e_j,h\rangle|^2\bigg)^{1/2}:\ \|h\|=1\Big\}\\[0.3cm]
&\leq\bigg(\sum_{j=n+1}^m|\langle f,e_j\rangle|^2\bigg)^{1/2}
\sup\Big\{\bigg(\sum_{j=1}^\infty|\langle e_j,h\rangle|^2\bigg)^{1/2}:\ \|h\|=1\Big\}\\[0.3cm]
&=\bigg(\sum_{j=n+1}^m|\langle f,e_j\rangle|^2\bigg)^{1/2}.
\end{align}
The convergence of $\sum_j|\langle f,e_j\rangle|^2$ guarantees that the last series can be made arbitrarily small if $m$ and $n$ are large enough. So
$$
\sum_j\langle f,e_j\rangle e_j
$$
exists. Now we need to show that it equals $f$. We will need the Polarization Identity
$$
\langle f,g\rangle=\tfrac14\,\sum_{k=0}^3i^k\|f+i^kg\|^2,
$$
and also the particular case to complex numbers,
$$
z\overline w=\tfrac14\,\sum_{k=0}^3i^k|z+i^kw|^2.
$$
We have
\begin{align}
\langle f,g\rangle&=\tfrac14\,\sum_{k=0}^3i^k\|f+i^kg\|^2
=\tfrac14\,\sum_{k=0}^3i^k\sum_j|\langle f+i^kg,e_j\rangle|^2\\[0.3cm]
&=\sum_j\tfrac14\,\sum_{k=0}^3i^k|\langle f,e_j\rangle+i^k\langle g,e_j\rangle|^2\\[0.3cm]
&=\sum_j\langle f,e_j\rangle\langle e_j,g\rangle.
\end{align}
Then
\begin{align}
\Big\langle f-\sum_j\langle f,e_j\rangle e_j,g\Big\rangle
&=\langle f,g\rangle-\sum_j\langle f,e_j\rangle\langle e_j,g\rangle=0.
\end{align}
As this can be done for any $g$, it follows that
$$
f=\sum_j\langle f,e_j\rangle e_j.
$$

If $(e_j)$ is a sequence in $H$ with $<e_j,e_k>=\delta_{jk}$, then we have:

$(e_j)$ is an orthonormalbasis of $H$ $ \iff$

$\left\|f\right\|^2=\sum_{j=1}^\infty\left|\langle f,e_j\rangle\right|^2\text{ for every }f\in H \iff\sum_{j=1}^\infty\langle f,e_j\rangle e_j\text{ for every }f\in H\tag*{}$


Show that $\cap _{M>0} \{z: -M<f(z)<M \}$ is connected. Suppose that $f$ is an entire function such that $f(0)=0$ and for every $M>0$, the set $\{z\in \mathbb{C}: -M<f(z)<M\}$ is connected. Show that $\cap_{M>0} \{z\in \mathbb{C}: -M<f(z)<M\}$ is connected.

I know that the intersection of nested, connected, compact sets is connected, but we do not have compactness here. A hint would be most welcome. Thank you.

Assume $f$ is not identically zero. Then the intersection (i.e. $f^{-1}(0)$) is discrete.
Then for sufficiently small $r>0$, $f$ has no zero on $r\cdot S^1$. Let $h_r>0$ be the minimum of the continuous function $|f(z)|$ on this compact set. We conclude that the single connected component of $\{\,z\in\mathbb C:|f|<h_r\,\}$ is fully inside $rS 1$, i.e in $r\mathbb D$. Then
$$\bigcap_{M>0}\{\,z\in\mathbb C:|f|<M\,\} \subseteq \bigcap _{r>0}r\mathbb D=\{0\}.$$

If $f$ is constant, then we must have $f=0$ and the result follows, so assume $f$ is not constant.

Let $L_M = \{ z | |f(z)| < M \}$. Then $z \in L_M$ for all $M>0$ iff $|f(z)| <M$ for all $M>0$ iff $f(z) = 0$.

Let $L^* = \cap_{M>0} L_M = f^{-1}( \{0\})$.
We are given that $f(0) = 0$, hence $0 \in L^*$

Suppose there is another zero $z \neq 0$. Since $L_M$ is open and connected, it is path connected, hence for $M={1 \over n}$ there is a path $\gamma_n$ from $0$ to $z$.

Pick any $\delta>0$ such that $\delta < |z|$, and let $z_n$ be a point on $\gamma_n$ such that $|z_n| = \delta$. The sequence $z_n$ has an accumulation point $z_\delta$ and we must have $f(z_\delta) = 0$. Since $\delta>0$ was arbitrary, it follows that $f=0$, a contradiction.

Hence we see that $L^* = \{0\}$.


Clarification on Rules Differentiation and First Principles Derivatives My grade 11 class has just started differential calculus, the one area seemingly glazed over in our book. We have covered some simple rules of differentiation, like
f(x) = n x^(n-1), and have applied these with subsequent rules like the constant rule, constant multiple rule, and sum or difference rule, child's play to you, I am sure. Anyway, we also covered the derivative, defined as:

The limit as h approaches zero = (f(x+h)-f(x))/h...

We were told by our teacher that tossing any equation into that formula will result in the spitting out of our derivative function...
We were also told that using the rules of differentiation will break things down into a derivative function; therefore, should using these to methods of derivation not produce the exact same results?

Apparently not:
Take x^2 + 1

From the rules of differentiation we get 2 x right?

But from the above formula we get 2 x+h

Now, clearly just by the presence of the h, which cannot magically come into the formula via the rules of differentiation, the rules and the formula do/produce different things...

Does one produce the function of the tangent line?
Does one produce the function of the gradients of possible tangents?

Which one of the above is the derivative? The specific tangent? Or the function to produce the slopes of said tangents? Which is yielded by which?

Sorry, I understand how stupid and trivial this all seems but I want to set a proper calculus foundation for the future and cannot risk misunderstanding something, and if I do, then I need rectification!

Thank you to any and all who may answer I will certainly appreciate the help.

The problem is with your calculation of the limit:

$$\begin{align*}
\lim_{h\to 0}\frac{f(x+h)-f(x)}h&=\lim_{h\to 0}\frac{\big((x+h)^2+1\big)-\big(x^2+1\big)}h\\\\
&=\lim_{h\to 0}\frac{(x^2+2xh+h^2+1)-(x^2+1)}h\\\\
&=\lim_{h\to 0}\frac{2xh+h^2}h\\\\
&=\lim_{h\to 0}(2x+h)\\\\
&=2x\;,
\end{align*}$$

not $2x+h$. In fact the formula comes from applying the definition to the general case, so it cannot give a different result.

The formula does not give 2x + h.
It is the limit as h goes to 0 of (f(x+h))-f(x))/h yes....
so f(x) = x^2 + 1 and the limit is

limit as h goes to 0 of ((x+h)^2 +1) - (x^2 + 1))/h

If you actually do that limit you will get 2x...


Why is empty set an open set? I thought about it for a long time, but I can't come up some good ideas. I think that empty set has no elements,how to use the definition of an open set to prove the proposition.
The definition of an open set: a set S in n-dimensional space is called open if all its points are interior points.

It's vacuously true. All points $\in\emptyset$ are interior points, have a blue-eyed pet unicorn, and live in Surrey.

Consider the following definition of an open set: For every element in A, there exists a ball of the element contained in A. What is the negation of that statement? Well, it is: There exists an element in A such that every ball is not contained in A. Clearly, the empty set does not satisfy the negation of an open set, therefore it must be vacuously true that the empty set satisfies the definition of an open set.


What is distribution of normal plus minus a constant? If $Y\sim N(\mu, \sigma^2)$,

then $Y+\mu_0=N(\mu+\mu_0,\sigma^2)$

and $nY\sim N(n\mu, (n\sigma)^2)$.

What about $-Y+\mu_0$?

Will it give $N(\mu+\mu_0,(-\sigma)^2)$?

If $Y∼N(\mu,\sigma^2)$ then $-Y∼N(-\mu,\sigma^2)$. Thus $−Y+\mu_0~N(-\mu,\sigma^2)$


Find the derivative of y^2=x as a function of y. Find the derivative of $$y^2=x$$ as a function of y.

i have found for the function of x, it will be $$\pm\frac{1}{2\sqrt x}$$

however for the function of y will be $$\frac{dy}{dx}=2y$$ ?
  it looks too simple, and I'm sure it's wrong.

As a function fo $y$, you should have $\frac{dx}{dy}=\frac{1}{2y}$ which is the same as $\frac{1}{2\sqrt{x}}$. Note however that square is a function only if we choose the positive square root.   So there is no need for the plus/or minus.


Identity map with change of basis Let a map with the field $K$, from $f:K^3 \rightarrow K^2$ be defined as $f\left (\begin{bmatrix}
x_1\\
x_2\\
x_3
\end{bmatrix}  \right )=\begin{pmatrix}
x_1+2x_2+3x_3\\
3x_1+2x_2+x_3
\end{pmatrix}$
and the basis $\begin{pmatrix}
-1\\
0\\
2
\end{pmatrix}, \begin{pmatrix}
3\\
1\\
-1
\end{pmatrix}, \begin{pmatrix}
1\\
-1\\
1
\end{pmatrix}$ for $K^3$ and the standard basis $(e_1,e_2,e_3)$ for the codomain
Question
Find the matrix that represents $Id:K^3 \rightarrow K^3$ with respect to the given basis for the domain and the standard basis $(e_1,e_2,e_3)$ codomain.
I have found the matrix that represents $f:K^3 \rightarrow K^2$ with respect to the given standardbasis for the domain and the codomain but this identity map is really difficult for me to get.

Hint: Given any linear transformation, the columns of the matrix that represents that map are exactly the images of the basis vectors of the domain, expressed in the basis of the codomain.

Sorry, but the question as you asked it doesn't make any sense. I'm pretty sure you mixed together two unrelated questions. I suggest you check the source (the textbook or the instructor's original assignment), and I bet these are two separate parts being two separate questions, not a single question.
For starters, if the given function is $f:K^3\to K^2$, then we can't have "the standard basis $(e_1,e_2,e_3)$ for the codomain", because the dimension of the codomain is $2$, not $3$.
My guess is that there's a separate question about finding the matrix representing the given transformation $f$ with respect to some bases in the domain $K^3$ and in the codomain $K^2$, and there's a separate question about finding the matrix representing the identity map $\operatorname{Id}:K^3\to K^3$ with respect to the given basis in the domain $K^3$ and the standard basis in the codomain $K^3$ — but the map $f$ has nothing to do with this question.
To address this latter question, the answer is already in the question. :-) Technically speaking, you need to do what other responses have already said. To find the matrix of a linear transformation $T$, such as e.g. $T=\operatorname{Id}$, you do the following: for each given basis vector $\mathbf{b}_i$ (of the "domain" basis), calculate $T(\mathbf{b}_i)=\operatorname{Id}(\mathbf{b}_i)$, and the components of the resulting vectors with respect to the codomain basis will be the corresponding columns of the desired matrix. And since in the codomain we're using the standard basis, what are we going to see as these columns?


How to simplify the summation kk! without using induction? $$\sum_{k=1}^nk(k!)$$

I know the answer is (n+1)!-1..I can solve this question using principle of mathematical induction...but I would like to know if there is any other alternative approach

Set a permutation $\pi = \pi_1 \ldots \pi_{n+1}$ in $S_{n+1}$.

Now let $m = m(\pi)$ be the maximal index such that $\pi_1 = 1, \pi_2 = 2, \ldots, \pi_m = m$.

The number of permutations such that $m(\pi) = m$ for $m < n$ is $(n-m) (n-m)!$: now $n-m$ is the number of choices for $\pi_{m+1} \neq m+1$, and $(n-m)!$ is the number of permutations of the remaining $n-m$ numbers.

No permutation can satisfy $m(\pi) = n$, and there is only one permutation such that $m(\pi) = n+1$.

Since there are $(n+1)!$ permutations in $S_{n+1}$, one has

$$ (n+1)! = \sum_{m=0}^{n-1} (n-m)(n-m)! + 1 = \sum_{k=1}^n k \cdot k! + 1. $$


Proving $\frac{1-\exp(-2st)}{s}-\frac{t}{0.5+st}\geq(\sqrt{\frac{1- \exp(-2st)}{s}} -\sqrt{ \frac{t}{1 + st}})^2$ for positive $s$ and $t$ I am trying to prove the following

$$\frac{1- \exp(-2st)}{s} - \frac{t}{0.5 + st} \geq \left(\sqrt{\frac{1- \exp(-2st)}{s}} -\sqrt{ \frac{t}{1 + st}} \right)^2 $$
for $s>0$ and $t>0$.

I am able to simplify it, but not able to go beyond it. Following is the simplification.
$$\frac{1- \exp(-2st)}{s} - \frac{t}{0.5 + st} \geq \frac{1- \exp(-2st)}{s} + \frac{t}{1 + st} - 2\left(\sqrt{\frac{1- \exp(-2st)}{s}}\sqrt{ \frac{t}{1 + st}} \right) \\
\Rightarrow 2\left(\sqrt{\frac{1- \exp(-2st)}{s}}\sqrt{ \frac{t}{1 + st}} \right) \geq \frac{t}{0.5 + st} + \frac{t}{1 + st} \\
\Rightarrow 4\left(\frac{1- \exp(-2st)}{st} \frac{1}{1 + st} \right) \geq \left(\frac{1}{0.5 + st} + \frac{1}{1 + st}\right)^2
 $$
The last line comes after squaring.

Okay here is my improved answer
$$\frac{1-e^{-2st}}{s} - \frac{t}{0.5 + st}\geq \left(\sqrt{\frac{1-e^{-2st}}{s}} - \sqrt{\frac{t}{1 + st}}\right)^2$$
for $s,t>0$\
Simplifying the given expression we have:
$$2\left(\sqrt{\frac{1-e^{-2st}}{s}}\sqrt{\frac{t}{1 + st}}\right)\geq \frac{t}{0.5 + st} + \frac{t}{1 + st} $$
Since  $\frac{t}{0.5 + st} \geq \frac{t}{1 + st}$
\Then,
$$2\left(\sqrt{\frac{1-e^{-2st}}{s}}\sqrt{\frac{t}{1 + st}}\right)\geq\frac{t}{1 + st} + \frac{t}{1 + st}$$
$$2\left(\sqrt{\frac{1-e^{-2st}}{s}}\sqrt{\frac{t}{1 + st}}\right)\geq \frac{2t}{1 + st}$$
squaring both sides we have
$$4\left(\sqrt{\frac{1-e^{-2st}}{s}}\sqrt{\frac{t}{1 + st}}\right)^2\geq 4\left(\frac{t}{1 + st}\right)^2$$
$$\frac{1-e^{-2st}}{s}\frac{t}{1 + st}\geq \frac{t^2}{(1 + st)^2}$$
Simplifying this we have
$$1-e^{-2st} \geq \frac{st}{1+st}$$
But if $0 < s < t < 1,$  $1- e^{-2st}<0$
Which implies that the inequality is no true for $s,t>0$


How to find solutions of linear Diophantine ax + by = c? I want to find a set of integer solutions of Diophantine equation: $ax + by = c$, and apparently $\gcd(a,b)|c$. Then by what formula can I use to find $x$ and $y$ ?

I tried to play around with it:
$x = (c - by)/a$, hence $a|(c - by)$.

$a$, $c$ and $b$ are known. So to obtain integer solution for $a$, then $c - by = ak$, and I lost from here, because $y = (c - ak)/b$. I kept repeating this routine and could not find a way to get rid of it? Any hint?

Thanks,
Chan

The diophantine equation $ax+by = c$ has solutions if and only if $\gcd(a,b)|c$. If so, it has infinitely many solutions, and any one solution can be used to generate all the other ones.

To see this, note that the greatest common divisor of $a$ and $b$ divides both $ax$ and $by$, hence divides $c$ if there is a solution. This gives the necessity of the condition (which you have backwards). (fixed in edits)

The converse is actually a constructive proof, that you can find in pretty much every elementary number theory course or book, and which is essentially the same as yunone's answer above (but without dividing through first).

From the Extended Euclidean Algorithm, given any integers $a$ and $b$ you can find integers $s$ and $t$ such that $as+bt = \gcd(a,b)$; the numbers $s$ and $t$ are not unique, but you only need one pair. Once you find $s$ and $t$, since we are assuming that $\gcd(a,b)$ divides $c$, there exists an integer $k$ such that $\gcd(a,b)k = c$. Multiplying $as+bt=\gcd(a,b)$ through by $k$ you get
$$a(sk) + b(tk) = \gcd(a,b)k = c.$$

So this gives one solution, with $x=sk$ and $y=tk$.

Now suppose that $ax_1 + by_1 = c$ is a solution, and $ax+by=c$ is some other solution. Taking the difference between the two, we get
$$a(x_1-x) + b(y_1-y) = 0.$$
Therefore, $a(x_1-x) = b(y-y_1)$. That means that $a$ divides $b(y-y_1)$, and therefore $\frac{a}{\gcd(a,b)}$ divides $y-y_1$. Therefore, $y = y_1 + r\frac{a}{\gcd(a,b)}$ for some integer $r$. Substituting into the equation $a(x_1-x) = b(y-y_1)$ gives
$$a(x_1 - x) = rb\left(\frac{a}{\gcd(a,b)}\right)$$
which yields
$$\gcd(a,b)a(x_1-x) = rba$$
or $x = x_1 - r\frac{b}{\gcd(a,b)}$.

Thus, if $ax_1+by_1 = c$ is any solution, then all solutions are of the form
$$x = x_1 - r\frac{b}{\gcd(a,b)},\qquad y = y_1 + r\frac{a}{\gcd(a,b)}$$
exactly as yunone said.

To give you an example of this in action, suppose we want to find all integer solutions to $$258x + 147y = 369.$$

First, we use the Euclidean Algorithm to find $\gcd(147,258)$; the parenthetical equation on the far right is how we will use this equality after we are done with the computation.
\begin{align*}
258 &= 147(1) + 111 &\quad&\mbox{(equivalently, $111=258 - 147$)}\\
147 &= 111(1) + 36&&\mbox{(equivalently, $36 = 147 - 111$)}\\
111 &= 36(3) + 3&&\mbox{(equivalently, $3 = 111-3(36)$)}\\
36 &= 3(12).
\end{align*}
So $\gcd(147,258)=3$. Since $3|369$, the equation has integral solutions.

Then we find a way of writing $3$ as a linear combination of $147$ and $258$, using the Euclidean algorithm computation above, and the equalities on the far right. We have:
\begin{align*}
3 &= 111 - 3(36)\\
&= 111 - 3(147 - 111) = 4(111) - 3(147)\\
&= 4(258 - 147) - 3(147)\\
&= 4(258) -7(147).
\end{align*}
Then, we take $258(4) + 147(-7)=3$, and multiply through by $123$; why $123$? Because $3\times 123 = 369$. We get:
$$258(492) + 147(-861) = 369.$$
So one solution is $x=492$ and $y=-861$. All other solutions will have the form
\begin{align*}
x &= 492 - \frac{147r}{3} = 492 - 49r,\\
y &= -861 + \frac{258r}{3} =86r - 861, &\qquad&r\in\mathbb{Z}.
\end{align*}
You can reduce those constants by making a simple change of variable. For example, if we let $r=t+10$, then
\begin{align*}
x &= 492 - 49(t+10) = 2 - 49t,\\
y &= 86(t+10) - 861 = 86t - 1,&\qquad&t\in\mathbb{Z}.
\end{align*}

Look can be deceiving. The integer solution to the equation $ax + by = c$ is anything but easy. Please endure a rather long derivation.
To make it more comprehensible let's first solve the equation for y:
\begin{align*}
ax + by = c\\
by = c - ax\\
y = \frac{c - ax}{b}
\end{align*}
To have an integer solution, $y$ must be an integer, and that is if $c - ax$ is a multiple of $b$, or $c - ax = -nb \iff ax = c + nb$. This has the same meaning as $ax \equiv c \: (mod \: n)$.
To continue, we need this Theorem 1:

The congruence $ax \equiv c \: (mod \: n)$ has a solution iff $gcd(a, n) \: | \: c$.

And this Lemma 2:

If $gcd(p, q) = 1$, then $px \equiv r \: (mod \: q)$ has a solution modulo $q$.

To keep this answer manageable, I would like to skip the proof of Theorem 1 and Lemma 2 (which can be found by googling). Just post a question and comment me if you encounter some trouble with them.
Let's define $d = gcd(a, n)$, and continue the derivation:
\begin{align*}
ax \equiv c \: (mod \: n)\\
ax = c + bn\\
\frac{a}{d} x = \frac{c}{d} + \frac{b}{d} n
\end{align*}
Now we want to switch $b$ and $n$ so $\frac{n}{d}$ could be seen more clearly as the modulo and continue it as following:
\begin{align*}
\frac{a}{d} x = \frac{c}{d} + b \frac{n}{d}\\
\frac{a}{d} x \equiv \frac{c}{d} \: (mod \: \frac{n}{d})
\end{align*}
Note that $\frac{a}{d}$ and $\frac{n}{d}$ from our derivation above is the $p$ and $q$ in the Lemma 2 respectively. Also note that as $d$ is $gcd(a, n)$, so $gcd(\frac{a}{d}, \frac{n}{d}) = 1$. Hence by Lemma 2:
\begin{align*}
\frac{a}{d} x \equiv \frac{c}{d} \: (mod \: \frac{n}{d})
\end{align*}
is our solution to equation $ax + by = c$.

As an example, let us solve $6x - 10y = 4 \iff 6x = 4 + 10y \iff 6x ≡ 4 \: (mod \: 10)$. $Gcd(6, 10) = 2$, and $2 \: | \: 4$, so by Theorem 1 that equation has a solution.
From our derivation, the solution is $\frac{6}{2} x = \frac{4}{2} \: (mod \: \frac{10}{2}) \iff 3x = 2 \: (mod \: 5)$.
By Lemma 2, we have a solution $modulo \: 5$. What it means is if we write the solution in $Z_5$, we would have:
\begin{align*}
\bar{3} \bar{x} = \bar{2}\\
\bar{x} = \bar{3}^{-1} \: \bar{2}
\end{align*}
As in $Z_5$, $\bar{3} \: \bar{2} = \bar{1} = \bar{3} \: \bar{3}^{-1}$, so $\bar{3}^{-1} = \bar{2}$ and we have:
\begin{align*}
\bar{x} = \bar{3}^{-1} \: \bar{2}\\
\bar{x} = \bar{2} \: \bar{2} = \bar{4}
\end{align*}
In $Z_5$, $\bar{x} = \bar{4} \iff x \equiv 4 \: (mod \: 5) \iff x = 4 + 5s \iff x = 4, 9, 13, \cdots$. You can check that indeed $x \equiv 4 \: (mod \:5)$ is the solution.


show that $11\nmid n^6-2n^5+4n^4-8n^3+16n^2-32n+64 $ Let $n$ be postive integer,show that
  $$11\nmid n^6-2n^5+4n^4-8n^3+16n^2-32n+64 $$

I have use ugly metods to solve consider this case,$n\equiv 0,\pm 1,\pm 2,\pm 3,\pm 4,\pm 5\pmod {11}$,have other more simple methods it?

Starting from @Crostul's comment, we have
$$
f(n)=n^6-2n^5+4n^4-8n^3+16n^2-32n+64 = \frac{n^7+2^7}{n+2}
$$
and so $f(n)(n+2)=n^7+2^7$.
Therefore, if $11$ divides $f(n)$, then $11$ divides $n^7+2^7$.

Now $n \mapsto n^7$ is an injective map mod $11$ because $\gcd(7,\phi(11))=1$.

Therefore, if $11$ divides $f(n)$, then $n^7 \equiv -2^7 = (-2)^7 \bmod 11$ and so $n \equiv -2 \bmod 11$.

So you only have to check that $11$ does not divide $f(n)$ for $n=-2$.


Proof that the set of irrational numbers is dense in reals I'm being asked to prove that the set of irrational number is dense in the real numbers.  While I do understand the general idea of the proof:

Given an interval $(x,y)$, choose a positive rational number (say) $z=\sqrt{2}$.  By density of rationals, there exists a rational number $p$ in the interval $(x/z, y/z)$, which essentially means that $$\frac{x}{\sqrt{2}} < p < \frac{y}{\sqrt{2}}.$$  I find that $pz$ is irrational, since it is the product of a rational and irrational number. However, my instructions besides writing that proof down is to specifically verify that $y=xz$ is irrational.  What does this have to do with anything and how does it prove denseness of irrationals?  Regardless, assume that $x$ is a nonzero rational number and that $z$ is irrational. For the sake of contradiction, assume that $y=xz$ is rational. This should mean that $y/x$ rational as well, and therefore that $z$ is rational, a contradiction.

Another argument:

$\mathbb{Q}$ is dense in $\mathbb{R}$, so $\mathbb{Q} + \sqrt{2}$ is dense in $\mathbb{R} + \sqrt{2} = \mathbb{R}$. Since $\mathbb{Q} + \sqrt{2}$ is a subset of the irrationals, we conclude that the irrationals are also dense in $\mathbb{R}$.

A set $A\subset \mathbb R$ is dense in $\mathbb R$ if $Cl(A)=\mathbb R$, where $Cl(A)=A\cup der(A)$ is closure of $A$.