Is there for all $N$ an integer that is the sum of, say, twenty $N$-th powers in two different ways? I believe the following to be true, but I have been unable to find any reference concerning the question.

Statement : for every $k\geq 1$, there exists $N = N(k)$ such that there is no non-trivial solution in non-negative integers to the equation $$x_1^N + \cdots +x_k^N = y_1^N + \cdots + y_k^N.$$

That is, there is at most one (up to order) representation of any natural number as a sum of $k$ $N$-th powers, for some sufficiently large $N$ with respect to $k$. (I also believe that it should be true for all sufficiently large $N$).

I know (from the Wikipedia page on generalized taxicab numbers) that, for $k=2$, it is unknown whether $N=5$ works, but I don't know if it is unknown whether any $N$ works.

Is the statement true, and if yes, how to prove it/where can I find a proof ?

Out of curiosity, where does this question come from? I remember hearing about it very recently...

Anyways, for sufficiently large $M$, we can always represent $M$ as a sum of $k$ $N$th powers.

Assume all the $x_i$ lie between $1$ and $m$. Then there are $\binom{m}{k}$ different sums of $\sum x_i^N$. But there at most $km^N$ sums, meaning that $$\binom{m}{k}< km^N$$ which doesn't hold for large $m$ - so there are at least two sums which are the same.

"OP" inquired about equal sums of around $20$ (nth) power's

in two ways. Well for degree two, Albert Beiler in his book

"Recreations in the theory of numbers" page $159$ has given

a solution for $28$ squares in two ways. The solution is

mentioned below.

[2,18,22,37,38,39,41,43,49,67,72,80,85,103,116,154,164,

175,178,192,200,207,215,222,230,247,422,593] equal to

[13,16,17,23,30,43,47,84,92,93,119,120,142,163,165,167,

177,183,188,199,215,219,261,270,280,363,372,382]

The sum of above is $1030225$ which is a square of $(1015)^2$

From above solution, many more similar solution's can also be

arrived at.


Is this a circular proof of Pythagorean Theorem? If not, are there better ones? I found this proof of Pythagorean Theorem from  what 3Blue1Brown shows in his Lockdown math lecture  "Trigonometry Fundamentals":
In the following right triangle, project $\cos\alpha$ and $\sin\alpha$ back to the hypotenuse of length $1$, they become $\cos^2\alpha$ and $\sin^2\alpha$:
$\to\cos^2\alpha+\sin^2\alpha =1 $.

I thought this is what a simple, visual, and elegant proof!  My questions:

Is it a circular proof, why?
If not, are there better ones?

I want clarity about whether or not this proof is circular.  The argument only depends on the definition of the sine and cosine functions and nothing more (please correct me if I am wrong).  This is why I love it.  Many other geometry solutions involve more than three shapes.

This is a great question. The original proof-as-I-read-it is valid, but some proofs can be "better" or worse in other respects. (By "as I read it", I mean with the implication that it calculates projections with similar triangles as I describe below, rather than stating their values without proof.) There are many criteria for saying one proof is "better" than another, so I'll focus on one germane to this question: making it obvious to all readers that there's no circularity (spoiler alert: there isn't).
Let the legs adjacent to and opposite $\alpha$ have respective lengths $a,\,b$,and let $c$ denote the hypotenuse's length. By similar triangles, your projections are $c(a/c)^2,\,c(b/c)^2$. Since they sum to the full hypotenuse $c$, $a^2+b^2=c^2$.
Since we define $\cos\alpha,\,\sin\alpha$ respectively as $a/c,\,b/c$, the above calculation can be restated as$$c\cos^2\alpha+c\sin^2\alpha=1\implies c(a/c)^2+c(b/c)^2=1\implies a^2+b^2=c^2,$$which is basically what your proof does. (It takes $c=1$ throughout, but that just changes the units to tidy the algebra; it's not an extra assumption as such, so it covers the general case.) But as I've shown above, there's no need to work with anything other than rational functions of $a,\,b,\,c$. The trigonometric approach switches from such rational functions to something more conceptually advanced, then "uninstalls" the trigonometry later.
One could argue this is a worse proof than the one I've presented, because in taking unnecessary detours it not only asks more of us, it also obscures the underlying reasoning. Well, whether it does is a psychological question, but people mistakenly thinking it's circular suggests as much.
Well, I want to be fair with an edit: as best I can tell from recent comments, those people either didn't realize which proof strategy is implicit in the OP's discussion of projections, or are making the actually salient point that the drawing itself doesn't necessarily communicate that as clearly as one would like in a proof without words. But the facts are these:

the proof can easily be spelled out in the aforementioned non-circular way,
it takes a bit more work to do that than is obtained in at least some readings of the picture and the text above it, and
none of these subtleties matter if we skip the trigonometry altogether, as aforesaid.

I hope the author wasn't regarding that picture as a proof of Pythagoras. Since trigonometry depends on the Pythagorean theorem, the above "demonstration" employs circular reasoning and wouldn't count as a valid proof. So yes, there are better proofs.


Show that every solution $x(t)$ of $x'(t)= A(t)x(t)$ converges to some limit (Long-time asymptotics). Suppose

$$\int_0^∞\|A(t)\|\,dt < ∞.$$

Show that every solution $x(t)$ of $x'(t)= A(t)x(t)$ converges to some limit:
$\lim_{t→∞} x(t) = x_∞.$

(Hint: First show that all solutions are bounded and then use the corresponding integral equation.)

I was able to solve a first part:

I used that the solutions are limited by $\|\phi(t,t_0)\| \leq e^ {\int_0^∞\|A(t)\|\,dt}$. As $\int_0^∞\|A(t)\|\,dt < ∞$, then $\|\phi(t,t_0)\| \leq K $. The problem now is to show that the above limit always exists. Someone can help

Hint

Once you know that $x$ is bounded by some constant $K$ you can use the integral equation to show that $x(t)$ is Cauchy, for $t_1 \leq t_2$ you have:
$$\|x(t_1)-x(t_2)\| =\left\|\int_{t_1}^{t_2} A(s) x(s) ds \right\| \leq K \int_{t_1} ^{t_2} \|A(s)\| ds$$
And $\int_{t_1} ^{t_2} \|A(s)\|ds$ goes to $0$ as $t_1$ goes to $+\infty$.

So using the sequential characterization of convergence and Cauchy sequences you can show that, as for Cauchy sequences, as long as your space is complete the function converges.


Proving the Exterior (Outer) Measure of Rectangle is Equal to Volume I'm having trouble understanding one step of Stein and Shakarchi's proof that the exterior measure of a rectangle is equal to its volume. The proof I reference is part of Example 4 in section 1.2 of the Real Analysis book.

To summarize the proof, first they show $|R| \leq m_*(E)$. To show the reverse inequality, the book divides $\mathbb{R}^d$ into cubes of side length $1/k$. Then, it considers the sets $\mathcal{Q}$ and $\mathcal{Q}'$.

The claim is that if $\mathcal{Q}$ consists of the finite collection of all cubes entirely contained in rectangle $R$ and $\mathcal{Q}'$ the finite collection of all cubes that intersect the complement of $R$, then there are $O(k^{d-1})$ cubes in $\mathcal{Q}'$, where $O$ denotes standard Big-Oh notation.

From here, using the volumes of the cubes, the text uses:

$$
\sum_{Q \in (\mathcal{Q} \cup \mathcal{Q}')} |Q| \leq |R| + O(1/k)
$$

to complete the proof.

Question: Why are there $O(k^{d-1})$ cubes in $\mathcal{Q}'$? I interpret $\mathcal{Q}'$ to be the set of all cubes in the complement, i.e. $\mathcal{Q} \cup \mathcal{Q}' = \mathbb{R}^d$. If this is the case, then I believe $\mathcal{Q}'$ is not a finite collection of cubes and a Big-Oh approximation of its size doesn't make sense.

I'll try to answer why the number of cubes in $\mathcal Q'$ is $O(k^{d-1})$ as $k\to\infty$. The set $\mathcal Q'$, by the way, is the set of cubes that intersect the original rectangle and its complement. In $3$-D, this would be the cubes that cover the faces of the rectangular box.

First let's consider the case where $R$ is actually a cube, so all of its edges are the same common length, which we will call $a$.

If we consider just one of these edges, then the number of cubes we need to cover the edge is almost $a/k^{-1} = ak$. This is what it would be if the cubes fit perfectly into the edge, but there may be a little space leftover, so the number of cubes we need is surely no more than $ak + 2$, adding one for each endpoint of the edge.

Thus the number of cubes in a $(d-1)$-dimensional face is no more than the product of the lengths of all the $d-1$ edges:
$$
\#\{\text{cubes in a face}\} \le (ak + 2)^{d-1} \le (a+1)^{d-1}k^{d-1}, \qquad {(2\le k).}
$$
Now, there are $2d$ faces, so the number of cubes in $\mathcal Q'$ is no more than $2d(a+1)^{d-1}k^{d-1}$. Thus, if $2\le k$ and if $k$ is so large that $1/k < a$, then we have the desired result.

To apply this argument to a general rectangle, let $a$ be the length of the longest edge of the rectangle. Then it is clear that the rectangle $R$ is contained in a cube of edge length $a$. The number of cubes in $\mathcal Q'$ for the rectangle can be no more than that of the larger cube, which we know to be $O(k^{d-1})$. Hence there are $O(k^{d-1})$ cubes in $\mathcal Q'$ for the rectangle, as desired.

One side is quite trivial, following from Example two. Another side (to prove $m_*(R) \leq |R|$): First observe that if each side of rectangle is rational then it is trivial. However, suppose not all sides are rational. Then $\forall \epsilon>0$ we know exists $r \in \mathbf{Q}$ such that $r- |R| < \epsilon$. Then try to construct a rectangle with each side rational and its volume equal to $r$. We know such rectangle exists (Since $\mathbf{Q}$ is dense in $\mathbf{R}$).


Converting to logarithm What logarithm rule can convert:

$$\left(\frac n4\right)^i = 1$$

to:

$$i = log_4(n)$$

When I view cheat-sheets for logarithm rules, I only see conversions where both sides of the equation have log in it. Thank you.

The basic rule

$$\log_ax^n=n\log_ax$$

so

$$\left(\frac n4\right)^i=1\iff i\log\frac n4=\log 1=0$$

What you have seems to be incorrect, thus.

No rule can get you from the first equation to the second. What is true is that
$$\begin{align}
0=\log_41&=\log_4\left(\left(\frac{n}{4}\right)^i\right)\\
  &=i\log_4\left(\frac{n}{4}\right)\\
  &=i(\log_4n-\log_44)\\
  &=i(\log_4n-1)
\end{align}$$
and this is only satisfied when $i=0$ or $n=4$ (which makes perfect sense by looking at your first equation).


Series Solution of an ODE The ODE below is required to help compute the coefficients of function.
There isnt any information about this topic in my textbook so i am just wondering how i would go about this question?

In this question you will construct the series solution of the following ODE $dy/dx = 2y$

Of course we are making a big assumption here – that the solution $y(x)$ can be written as a Taylor series.

(a) Assume that $y(0) = A$ for some given number $A$. Use the ODE to compute the
successive derivatives of $y(x)$ at $x = 0$.

(b) Use the result of part (a) to deduce the coefficients $a_0, a_1, a_2, a_3$ and an in the infinite
series
$y(x)=a_0 +a_1x+a_2x^2 +a_3x^3 +···+a_nx^n$

(c) Use the result of part (b) to write down $y(x)$ as an infinite series. Your answer should contain only one constant $A$ and should include the general term in the infinite series.

(d) Solve the ODE using any other method. Does this solution agree with what you found in part (c)? Justify your answer.

you can differentiate the equation repeatedly to get the following: $$y' = 2y, y'' = 2y' = 2^2y, y^{(3)}= 2^3y, \cdots, y^{n} = 2^ny, \cdots \tag 1$$

suppose $y(0) = a.$ then substituting $x = 0,$ in $(1),$ we get $$y'(0) = 2a, y''(0)=2^2 a,\cdots,y^{(n)} = 2^n a, \cdots \tag2 $$

now using the  maclaurin series $$y =y(0)+xy'(0)+\frac{x^2}{2!}y''(0)+ \cdots+ \frac{x^n}{n!}y^{(n)}(0)+\cdots$$ which gives us
$$\begin{align}y &=a+2ax+2^2\frac{x^2}{2!}a+ \cdots+ 2^n\frac{x^n}{n!}a+\cdots
\\&=a\left( 1+2x+\frac{(2x)^2}{2!}+ \cdots+ \frac{(2x)^n}{n!}+\cdots \right)\\
&= ae^{2x}\end{align}$$

Series solution of a first order differential equation? Also you have written $dy/dy = 2y$:

Shurely shome mishtake mish Moneypenny?

EDIT Why downvote? I know it's fine to use series solution for first order; wlog that is; but it's not really needed. Also the typo is important.


Prove that if $R$ is a principal ideal ring and $S$ is a multiplicatively closed subset of $R$ then $S^{-1}R$ is also a principal ideal ring. Prove that if $R$ is a principal ideal ring and $S$ is a multiplicatively closed
  subset of $R$ then $S^{-1}R$ is also a principal ideal ring.

Thanks for any insight.

More generally, prove the following: Let $\phi : R \to R'$ be a ring homomorphism such that every element of $R'$ is associated to an element in $\phi(R)$. Note that localizations satisfy this property by construction, but also quotients. If $R$ is a principal ideal ring, then the same is true for $R'$. The proof is straight forward, since there is only one thing you can do: Use the assumption. Every ideal of $R'$ has some generating set, and by the assumption it may be choosen from $\phi(R)$ (since multiplication of units doesn't change an ideal). Take preimages in $R$, use that $R$ is a principal ideal ring, etc.

$S^{-1}R$ is a subring of a field (the field of fractions of $R$), and thus is an integral domain.

Let $I \subseteq S^{-1}R$ be an ideal. Let $D = \{a \in R \ |\ \frac{a}{b} \in I \}$.

claim that $D$ is an ideal of $R$.

note that $\frac{0}{s} \in I$, so that $0 \in D$. Now let $\frac{a}{d}$, $\frac{b}{e} \in I$; then
$\frac{a}{d} - \frac{e}{d}\frac{b}{e} = \frac{a-b}{d}$, so that $a-b \in D$.

let $r \in R$ and $a \in D$, with $d \in S$. Then $\frac{r}{d} \frac{a}{d} = \frac{ra}{d^2}$, so that $ra \in D$. Thus $D$ is an ideal of $R$, and thus we have $D = (\alpha)$. Let $d \in S$ such that $\frac{\alpha}{d} \in I$; then $(\frac{\alpha}{d}) \subseteq I$. Now let $\frac{a}{b} \in I $with $a = c\alpha$; we have $\frac{a}{b} = \frac{cd}{b} \frac{\alpha}{d}$, so that $I = (\frac{\alpha}{d})$ is principal.

Thus,$S^{-1}R$ is a principal ideal domain.


Proof of inner product Let $V$ be a vector space with inner product $\langle\,,\,\rangle$ and $\{v_1,v_2,\dots,v_n\}$ an orthonormal basis of $V$. Prove that $\sum_{i=1}^n \langle x,v_i\rangle^2=\|x\|^2$, for $x\in V$.

I´m not very fond of proofs but this is what I have:

\begin{align}
\sum_{i=1}^n \langle x,v_i\rangle^2
&=\sum_{i=1}^n (\|x\|\,\|v_i\| \cos(\theta))^2\\
&=\sum_{i=1}^n \|x\|^2 \|v_i\|^2 \cos^2(\theta)\\
&=\sum_{i=1}^n (\|x\|^2\cos(\theta)) (\|v_i\|^2 \cos(\theta))\\
&=\sum_{i=1}^n \langle x,x\rangle \langle v_i,v_i\rangle\\
&=\langle x,x\rangle\sum_{i=1}^n \langle v_i,v_i\rangle\\
&=\|x\|^2 n
\end{align}

What am I wrong? Any suggestions to improve my proofs would be helpful. Thanks.

Your mistake is this:
$$\|x\|^2\cos\theta \ \ne\ \langle x,x\rangle$$
Instead, we have $\langle x,x\rangle = \|x\|^2$ without the $\cos$ term.

And, hint for the solution: write $x=\xi_1 v_1+\xi_2 v_2+\dots$, then calculate $\|x\|^2=\langle x,x\rangle$ by substituting this into the right argument.

$$x=\sum_{i=1}^n v_ix_i,$$ where $x_i=\langle x,v_i\rangle$ are scalars.

Indeed, by linearity and orthonormality, $$\langle x,v_j\rangle=\sum_{i=1}^n \langle v_j,v_i\rangle x_i=\sum_{i=1}^n \delta_{ij} x_i=x_j.$$

Then

$$\|x\|^2=x^2=\sum_{i=1}^n\sum_{j=1}^n \langle v_i,v_j\rangle x_i x_j=\sum_{i=1}^n\sum_{j=1}^n \delta_{ij} x_i x_j=\sum_{i=1}^n x_i^2.$$


bound of integrable function I want to prove the following conjecture:

if an integrable function $f(x)$ is continuous on (0,T] and unbounded at $x=0$, then there exists positive $M$ and $\alpha\in(0,1]$ such that
$$
|f(x)|\leq M|x|^{\alpha-1} \quad\mbox{for}\quad x\in[0,T].
$$

First of all, is this statement valid??
If so, how to prove it?

The following isn't really rigorous, but it can be made rigorous with more work.

Let $T(x) = \begin{cases}1-|x| & \text{if} \ |x| \le 1 \\ 0 & \text{if} \ |x| > 1\end{cases}$, and define $f(x) = \displaystyle\sum_{n = 1}^{\infty}e^nT(e^{2n}(x-\tfrac{1}{n}))$.

Graphically, $T(x)$ is a spike centered at $x = 0$ with width $2$ and height $1$.

Hence, $e^nT(e^{2n}(x-\tfrac{1}{n}))$ is a spike centered at $x = \tfrac{1}{n}$ with width $2e^{-2n}$ and height $e^n$.

It is easy to see that the functions $\{e^nT(e^{2n}(x-\tfrac{1}{n}))\}_{n = 1}^{\infty}$ have pairwise disjoint supports (i.e. the spikes don't overlap). Then since $e^nT(e^{2n}(x-\tfrac{1}{n}))$ is continuous for each $n$, $f(x)$ is continuous.

Also, $\displaystyle\int_{0}^{2}f(x)\,dx = \sum_{n = 1}^{\infty}\dfrac{1}{2} \cdot e^n \cdot 2e^{-2n} = \dfrac{1}{e-1} < \infty$, i.e. $f(x)$ is integrable.

However, $f(\tfrac{1}{n}) = e^n$ for all positive integers $n$.

You can then show that for any $M$ and any $\alpha \in (0,1]$, there exists a large enough integer $n$ such that $f(\tfrac{1}{n}) = e^{n} > M\left(\tfrac{1}{n}\right)^{\alpha-1}$.


Use the Cauchy-Schwarz Inequality to prove that $a^2+b^2+c^2 \ge ab+ac+bc $ for all positive $a,b,c$. Use the Cauchy-Schwarz Inequality to prove that $a^2+b^2+c^2 \ge ab+ac+bc $ for all positive $a,b,c$.

That's what I've tried:

Let a Cauchy-Schwarz Inequality be :

\begin{array}
(((\sqrt{a} )^2+(\sqrt{b})^2+(\sqrt{c})^2 )\left(\left(\cfrac{ab}{\sqrt{a}}\right)^2 +\left(\cfrac{ac} {\sqrt{b}}\right)^2+\left(\cfrac{bc}{\sqrt{c}}\right)^2\right)& \ge (ab+ac+bc)^2 \\
(a+b+c)(ab^2+\cfrac{a^2 c^2}{b}+b^2 c) &\ge (ab+ac+bc)^2 \\
\end{array}

However how should I now factorize the left hand side of the inequality as $(a^2+b^2+c^2) \cdot Something $ ?

I think I've made the problem harder than it needs to be .

If you insist on C-S, let
$$
u=(a,b,c),\quad v=(b,c,a).
$$
Then,
$$
\text{RHS}=u\cdot v\underbrace{\leq}_{\text{C-S}}||u||\times||v||=\text{LHS}.
$$

I have given a detailed proof using an easier method. Please click on this link to see the proof


Distance between surfaces I'm trying to find the minimal distance between the surfaces described by $z=x^2+y^2$ and $x+y-2z=8$. I would imagine there are several approaches including the use of Lagrange multipliers. I attempted to find the spot when their normal vectors are parallel ( since i believe if these vectors are not parallel there is a direction that will bring the distance lower). So if $U(x,y,z)=x^2+y^2$ and $V(x,y,z)=x+y-2z$ are functions describing these surfaces then their normal vectors are $(2x,2y,-1)$ and $(1,1-2)$ respectively. Then I want their cross product to be zero; this is at $(1/4,1/4,z)$. But when i solve for $z$ and I cannot simultaneously satisfy both equations. What gives?! Am I along the right line of thinking?

You won't be able to solve for $z $ on both equations  if the surfaces dont intersect. You have to find the $z $ coordinate of the point of each surface with $x = y = \frac14$ and then find the distance between the two points.

You have correctly identified that the normal to the bottom surface is $(1, 1, -2)$. You have correctly identified that the only point on the top surface with the same normal (up to scaling) is $(\frac14, \frac14, \frac18)$, which I will call $P$. It remains to find the point on the bottom surface that is closest to $P$.

We can rewrite the formula for the bottom surface as $z = \frac12(x+y-8)$. As we vary $x$ and $y$, the point $Q = (x, y, \frac12(x+y-8))$ explores the whole surface. The distance we need to minimise is

$$|PQ| =
\sqrt{(x-\frac14)^2 + (y-\frac14)^2 + (\frac12(x+y-8)-\frac18)^2}$$

For the sake of an easy life, let's minimise $|PQ|^2$ instead.

$$\begin{eqnarray}
\text{d}|PQ|^2
  &=& 2(x-\frac14)\text{d}x + 2(y-\frac14)\text{d}y + 2(\frac12(x+y-8)-\frac18)(\text{d}x + \text{d}y) \\
  &=& (2x-\frac12+x+y-8+\frac14)\text{d}x + (2x-\frac12+x+y-8+\frac14)\text{d}y \\
  &=& (3x+y-\frac{33}4)\text{d}x + (x+3y-\frac{33}4)\text{d}y
\end{eqnarray}$$

This is zero when $3x+y = x+3y = \frac{33}4$, i.e. when $x = y = \frac{33}{16}$. Plugging this into the formula above gives:

$$\begin{eqnarray}
|PQ| &=& \sqrt{(\frac{33}{16}-\frac14)^2 + (\frac{33}{16}-\frac14)^2 + (\frac12(\frac{33}{16}+\frac{33}{16}-8)-\frac18)^2} \\
  &=& \sqrt{\frac{29^2}{16^2} + \frac{29^2}{16^2} + \frac{33^2}{16^2}} \\
  &=& \frac{\sqrt{2771}}{16} \approx 3.290018047062964
\end{eqnarray}$$


How many ways can $8$ rings be put on $3$ fingers? I've been trying to understand an example question in my textbook but after looking around the answer seems unclear to me.

The question:

  "How many combinations/permutations of $8$ different rings can be put on $3$ different fingers? Both if the order matters and also if it does not."

My attempt, if order does not matter would be :

$8^3$

If the order does matter:

$8P3 = 8 \cdot 7 \cdot 6$

Is this the correct way to interpret such a question?

Some online resources seem to have varying and sometimes odd solutions such as this: $C(n+r -1, r-1)$

Any assistance in understanding if I'm heading to the correct interpretation would be helpful.

Edit

Note : More than $1$ ring can be put on each finger

For order doesn't matter, each ring can be placed on one of $3$ fingers. This results in a unique string, for example $12323113$, which results in finger $1$ having rings $1,6,7$, finger $2$ has rings $2,4$ and finger $3$ has rings $3,5,8$.

This clearly groups each finger's rings as distinct, and gives $3^8$ strings.

$8^3$ is three rings onto $8$ fingers.

If order does matter, using $3^8\cdot8!$ doesn't work, for example in the above case there are only $3!2!3!=72$ permutations.

Instead, use stars and bars to give the number of available patterns as $\dbinom{10}{2}=45$.

To feed the fingers their rings, first arrange the rings into one of the $8!$ permutations, and feed from finger $1$ through to finger $3$. This ensures uniqueness, and gives the result as:

$$\binom{10}{2}\cdot8!=45\cdot40320=1814400$$

If you're like me the problem is finding a strategy.

And the strategy that figuring each ring has $3$ choices of fingers so there are $3^8$ ways to choose fingers for each ring, is a dead end as order matters and we can't straightforwardly multiply by any choice or permutation value for each finger as we don't know how many rings are on each finger.

We can choose to say each finger has $aPb = \frac {a!}{(a-b)!}$ where $a$ are the number of rings left, and $b$ are the number of rings on the finger:

Then the answer is $\sum_{k= 8,-1}^0 (8 P k)\sum_{j=0}^k(k P j)(k-j P k-j)$.  Which seems intimidating but: $(k P j)(k-j P k-j) = \frac {k!}{(k-j)!}(k-j)! = k!$ and so $sum_{j=0}^k (k P j)(k-j P k-j) = \sum k! = (k+1)k! = (k+1)!$ and $\sum_{k= 8,-1}^0 (8 P k)\sum_{j=0}^k(k P j)(k-j P k-j) = \sum_{k = 0}^8 \frac{8!}{k!}*(k+1)! = 8! \sum_0^8 (k+1) = 8!\sum_{i=1}^9 i = 8!\frac{9*10}2 = 8!*45$.

But simpler way of thinking of it is that if all the rings were identical and there were $N$ ways to place $8$ things on $3$ fingers.  And there are $8!$ ways to arrange the 8 rings.  There would be $8!N$ ways to arrange 8 different rings.

There are two ways to solve $N$.  You may put $a$= zero to $8$ rings on finger $1$ and you may put $b$ = 0 to $8-a$ on finger 2 and all the rest on finger 3.  That is $\sum_{a=0}^8\sum_{b=0}^{8-a}1 = \sum_{a=0}^8 9-a = \sum{a=8,-1}^0 a+1 = \sum_{i=1}^9 i = \frac {9*10}2 = 45$.

Or: you need to divide $8$ onto three fingers. Put the on one after the other.  There are nine points in time between $0$ and $8$ that you can put in a "go to next finger marker".  Among the $0$ to $8$ ring and the marker there are 10 places to put a second marker.  There are ${10 \choose 2} =\frac {10!}{2!8!} = \frac {10*9}n = 45$.

Either way there are $45*8!$ ways to do this.

==== old answer which was stream of thought with mistakes along the way below === it could be illuminating for thought process... or it could be frustrating as it takes a while to get the right answer =====

If order doesn't matter then each ring can go on any one finger is $3^8$ (not $8^3$) is correct.

But if order on the fingers matters (the emerald on the middle finger under the gold is different than the emerald on the middle finger over the gold) it's a different question.

One way:  Arrange the rings in order that you will put them on.  There are $8*7*6= \frac {8!}{(8-3)!}$ (is that what $8 P 3$ defined to be?  I think so.) ways to arrange  .  Then each ring (in order) has a choice of three fingers to be placed on.  So the answer is $8P3*3^8$.

==== Oh, F###; that's over counting ====

If diamond ring is 1 and we choose finger one and emerald is 2 and we choose finger two is the same as emerald is 1 and we choose finger two and diamond is 2 and we choose finger one.

Oh, well, I'm going to leave this up as foood for thought. but... it is a wrong answer.

=====

My first thought was the hardest. You can choose $a$ rings for the first finger and $b$ rings for the second and $8 - a -b$ for the third and so

$\sum_{a= 0}^8 (8Pa)\sum_{b=0}^{8-b}([8-b]Pb)*([8-b-a]!)$

Logic tells me that sum must add up to $8P3*3^8$ (EDIT: It doesn't) but ... I'd hate to actually work it out (EDIT: but I may have to... it seems like the correct answer still.).

(Actually my very first thought was order didn't matter and rings were all the same, so it'd be $\sum_{a=0}^8 \sum_{b=0}^{8-a} 1 = \sum_{a=0}^8 9-a = 9^2 - \frac{8*9}4= 45$).

====

Here's a second way.  We are going to put all the rings on in order all the rings on the first finger first, then the rings on the second, and then the third.

There are $8!$ ways to arrange there rings and $\sum_{a=0}^8 \sum_{b=0}^{8-a} 1 = 45$ ways to place the two "breaks" on when we stop putting rings on one finger and start putting them on the other.

So $8!*45$.  Is that right?

Anyone with time on his/her hands want to see if $\sum_{a= 0}^8 (8Pa)\sum_{b=0}^{8-b}([8-b]Pb)*([8-b-a]!) = 8!*45=8! {10 \choose 2}$?


Prove $\frac{2ab}{a+b}\leq\sqrt {ab}$ $a$ and $b$ are both positive real numbers. I'm supposed to work backwards (i.e. start with what I'm trying to prove and change it until something is absolutely true, then start from what is absolutely true in my proof).

Here's my attempt:

$\frac{2ab}{a+b}\leq\sqrt {ab}$

$\frac{2a^2b^2}{(a+b)^2}\leq{ab}$

$\frac{2a^2b^2}{a^2+2ab+b^2} - ab \leq 0$

$\frac{2a^2b^2-ab(a^2+2ab+b^2)}{a^2+2ab+b^2}\leq 0$

$\frac{2a^2b^2-a^3b-2a^2b^2-ab^3}{a^2+2ab+b^2}\leq 0$

$\frac{-a^3b-ab^3}{a^2+2ab+b^2}\leq0$

Because $a$ and $b$ are positive, this guarantees the left side is negative, making the inequality true.

This is just a rearrangement of the famous AM-GM inequality!

It states that $\frac{a+b}{2} \geq \sqrt{ab}$

for $a,b \geq 0$.

The proof is clear from the picture!

Now multiply by $\sqrt{ab}$ on both sides and rearrange with ease to get the desired: $\frac{2ab}{a+b}\leq\sqrt {ab}$.

A much simpler way is to use the relation between the harmonic and geometric means:
$$ \frac{2}{\frac{1}{a}+\frac{1}{b}} \leq \sqrt{ab}$$
adding the fraction in the denominator of the left hand side: we have:
$$ \frac{2}{\frac{b}{ab}+\frac{a}{ab}} \leq \sqrt{ab}$$
And the relation you listed follows.

See also https://en.wikipedia.org/wiki/Mean#Relationship_between_AM.2C_GM.2C_and_HM


Having Trouble Understanding Meaning Of A Finite-Dimensional Vector Space The definition I have is the following:

  A vector space V is said to be finite-dimensional if there is a finite set of vectors in V that spans V and is said to be infinite-dimensional if no such set exists.

However, with this definition I can't determine whether the vector space $\mathbb{R}^3$ is finite-dimensional or infinite-dimensional (I am assuming that it is finite since the dimension of $\mathbb{R}^3$ is $3$)

Going with my thought process, though, I know that $(1,0,0),(0,1,0),(0,0,1)$ spans $\mathbb{R}^3$. However we can also check that $(2,0,0),(0,2,0),(0,0,2)$ spans $\mathbb{R}^3$. Also note that $(3,0,0),(0,3,0),(0,0,3)$ spans $\mathbb{R}^3$. This process could be continued over and over to show that there are infinitely many vectors that span $\mathbb{R}^3$.

Wouldn't this mean that $\mathbb{R}^3$ is infinite-dimensional? Because there isn't a finite number of vectors that span $\mathbb{R}^3$. (Again I want to say this isn't the case and that there is something I am overlooking.)

The question asks whether there exists a finite basis of the vector space. If there exists a finite basis, then this vector space is said to be finite dimensional. If not the vector space is infinite dimensional. An example of an infinite dimensional vector space is the vector space of all power series.

Contrast this with the vector space of all polynomials of degree less than or equal to 3, $\mathbb{P}_3 [t]$ which has finite dimension 4 since one basis consists of $\{1,t,t^2,t^3\}$

Definition: A city is said to be  food-friendly if  one can get three different cuisines in a single restaurant.

In London, there is Restaurant A which serves Indian, Chinese and Continental. ANd also Restaurant B that serves Thai, Mexican and Greek food. And restaurant C serving,  Indian, Russian and Japanese.

Now does London meet the above definition of food-friendly city? or do you say there three restaurants A,B, C instead of a single restaurant? So  London is  not food-friendly??


ZFC Axioms to be extended? Sorry if this is going to be a really loaded question.

I was told several times that for virtually all theorems/corollaries/propositions of mathematics (except those cases not compatible with ZFC or not expressed in FOL) we can turn it into a first order language, and obtain a formal proof of it from the logical axioms of first order logic (propositional logic tautologies, first order tautologies, axiom of equality) along with the axioms of ZFC, by using a proof calculi of our choice.
So for example, that means that if I have a theorem that states $A → C$ (and is compatible with ZFC), then its corresponding formula in first order logic is provable (from logical axioms + ZFC). At the same time, this would also make $C$ be deduced from $A$ (by deduction theorem).

On the semantical side, by soundness of first order logic, we would have that $A → C$  is true in every interpretation where all our initial axioms (logical ones and ZFC ones) are true, that is, for all models of our initial axioms. And along with that we could infer that $C$ is true in every interpretation where $A$ is true, that is, $C$ is a semantical consequence of $A$.

If i'm wrong in something until now, it would be good to know.

My question is about the role of the additional (or extended) axioms like  Peano axioms ,Group axioms, etc  in theorems of the kind $A → C$ that use concepts defined by them.
Do they really act like axioms, or more like definitions?

I know that its shown that in any model of ZFC, there exists exactly one set that satisfy the Peano axioms, and a bunch of sets that satisfy the group axioms.

For example, suppose a theorem $A → C$ involves a fact about a natural number (or a group). ie : some basic theorem in elementary number theory or the theorem that states that every cyclic group is abelian.
Would the theorem be saying something about the set(s) ( inside a model of the ZFC ) which was "defined" by the Peano or Group axioms ?

What would mean to be saying that $A → C$ is provable , in light of its reference to some specific sets of ZFC ( defined by the Peano or Group axioms ) ?
Would it still be true that  $A → C$ is provable from our initial axioms (logical ones) ?

Would we need to consider the validity of the Peano axioms (or group axioms, respectively) to  say that $A → C$ is provable?
Would it still be provable from ZFC  + logical axioms only?
 Or would we have to augment ZFC + logical axioms with peano axioms (or group axioms)?
 Or would it be provable directly from Peano (or group axioms), maybe with logical axioms, withouth reference to ZFC?

I hope i could deliver the essence of my doubt. Thanks in advance.

Okay, first off the bat, let me correct something here. In every model of $\sf ZFC$ there is only one model (up to isomorphism) of the second-order version of the Peano axioms. This is important, because the first-order theory is subjected to things like compactness and Lowenheim-Skolem which ensure that there are models of arbitrary cardinality.

But now that we have relaxed about this point, let's get back to the actual question at hand. What is the role of the additional axioms.

I want to answer a whole other question, which in turn is actually the more important question that you've asked (without asking). What is the role of $\sf ZFC$ in the additional axioms and mathematics?

We can develop first-order logic without $\sf ZFC$, perhaps we won't have the model theory and the semantics of logic, but we can talk about languages, terms, formulas, proofs and so on and so forth. We can even do all that in a fairly finitary theory like $\sf PRA$.

Why is that important? Because we want some sort of one "processor" for our mathematics, that will be able to tell us whether or not we are doing okay, and will allow some sort of compatibility between our theories (for example, we want our meta-theory to be able and translate statements about rings to statements about modules, even if those are completely different objects).

So without ever using any axiom of $\sf ZFC$ we can start with $\sf PRA$ and talk about the theory of groups, or abelian groups, and prove all sort of things that we would prove usually about groups. But this will be harder because we are used to taking an "actual" group and arguing the existence, or non-existence of certain objects which would satisfy certain properties. Sometimes we would take ultrapowers, or want to use advanced model theoretic arguments. And these require semantics which we might not have in $\sf PRA$.

The solution is simple. If we want a "processing unit" for mathematics, which can encode and decode internally "pretty much everything we want" -- which include interpretation of objects like the real numbers or very large objects, and if we want model theoretic tools, then we have a good incentive to use set theory instead of arithmetic. And $\sf ZFC$ makes a relatively natural theory to use (for myriad of reasons, for example it doesn't require us to keep track of the syntax of its own meta-theory like in $\sf NF$).

If so, the role of $\sf ZFC$ is to allow us to build mathematical objects. But building them requires logic, and since mathematical logic is also a mathematical object, we can begin by building a "copy" of first-order logic within $\sf ZFC$ and then use that to talk about proofs from $\sf PA$ or group theory, or whatever. And later on we can formalize $\sf ZFC$ within something like $\sf PRA$ and then have something like "$\sf ZFC$ proves that group theory proves that there is exactly one elements such that $\forall x(x\cdot e=x)$". But that's not important right now.

So, to your question. The working mathematician (read: not a set theorist) is not working within $\sf ZFC$. It is working within her own settings, whatever they might be: group theory, measure theory, low-dimensional topology, etc. Each of these have their own "usual" axioms and their own "commonplace logic" (sometimes it's first-order logic, sometimes it's another thing). And they all do what they want to do: prove things using the rules of that logic and the axioms of these theories. And the proofs are proofs from that theory, with that specific logic.

The role of $\sf ZFC$ is to take all that, and translate it back to a unifying mechanism which allows us to worry (from a foundational point of view) about just one theory, rather than many theories.

So when you prove a statement in number theory about the prime numbers, you don't prove it from $\sf ZFC$. You prove it in, say, second-order logic and the axioms of $\sf PA_2$ (second-order Peano). But somewhere in the back of your mind, if you care enough about these things at all, you know that all that is in fact a theorem that $\sf ZFC$ proves that in second-order arithmetic it is true that this and that theory is true.

Let's take the "every cyclic group is abelian" proposition as an example.  It's easy to express abelianess in a first-order language of groups, that $gh=hg$ for all $g,h$ (in your group).

It's less easy to express that a group is cyclic.  Set theory, or at least some inductive scheme, would be a natural way to express that every group element can be generated by some fixed element.

Thus much of what is considered elementary group theory needs more infrastructure than just logical axioms.  To get a handle on "normal subgroups" or "exact sequences" takes us beyond the operations on group elements themselves to a discussion of sets and functions, etc.


Column and row space basis $Let \ A = \left[\begin{array}{cccc}1&2&-1&1\\2&4&-3&0\\1&2&1&5\end{array}\right]$

Using Gauss elimination, lead matrix A to row reduced echelon form:

$\left[\begin{array}{cccc}1&2&0&3\\0&0&1&2\\0&0&0&0\end{array}\right]$

And now using that form I am supposed to say what are column and row space basis.
I think that Column space basis is:
$\left[\begin{array}{cc}1&-1\\2&-3\\1&1\end{array}\right]$
But what is rowspace basis???

I dont know why I am asking all these question, when wikipedia exist but whatever http://en.wikipedia.org/wiki/Row_space says that row space are all non zero row vectors in (reduced row) echelon form.
So i Guess its:
$\left[\begin{array}{cccc}1&2&0&3\\0&0&1&2\end{array}\right]$
Good?

Those two rows do provide a basis of the row space, yes - it's not the only one, but it answers the question so it will do.

The point is that the rows have the same span after Gaussian elimination as they do before, but in row reduced echelon form the non-zero rows are linearly independent, so they are a basis, rather than just a spanning set.

If you know the sifting algorithm you could also just apply that to the rows of the original matrix, to get a basis for the row space whose members are actually rows of $A$.

Please note that elementary row operations as well as column operations don't alter the row space, column space and rank of a matrix.

The objective of performing echelon transformation on a matrix, row wise (column wise), is to identify the linearly dependent rows (columns). The rows (columns) which are linearly dependent on other rows (columns) ultimately become null under such operations.

Again the definition of rank of a matrix is the maximum number of linearly independent rows or columns i.e. the non-null rows or column in the echelon matrix. That's why row rank of a matrix is the number of non-null rows in row echelon form or row reduced echelon form.

Same concept goes for column rank, but here you need to perform elementary column operations.


Why Is a Function Defined As Having Only One Y-Value Output? I know that it is defined such that there is no more than one y-value output for any given x-value input, but I"m wondering WHY it is defined that way? Why can't we apply everything we know about functions to equations that have more than one y-value output for any given x-value input?

In short, what is the benefit or reason to define a function this way?  At the same time, a little historical context would also be appreciated (i.e., when the concept originated and why).

Thanks,

Moshe

Strictly speaking, a "well-defined" function associates one, and only one, output to any particular input. A function would not be well-defined if say $y = f(x)$ such that $y$ can take on any number of values at any particular input. Suppose the input $\;x = a\;$ outputs more than one distinct value, so that $y = f(a) \in\{y_1, y_2, ..., y_k, ...\} $. We'd never be able to say what, precisely, $y$ is when $x = a$.
If $y = f(a) = b$, and $y = f(a) = c$, and $\,b\ne c$, then the value of the function  $\,\,f(x) = y\,$ at $\,a\,$ is ambiguous: $y$ would not specify any particular value at $a$. That is, its value is not well-defined at $a$, and perhaps not well-defined at other inputs, as well. Nor can we say much about the behavior of a function at a particular value, if it can take on many values at a given point.
E.g. How would we define continuity of, say, a real-valued multifunction?
The strict definition of a function, in terms of "outputing" exactly one value for any given input is really no more than an attempt to keeping functions well-defined, and thus properties of functions well-defined.
But you'd might like to explore the following:
See this entry on multi-valued "functions":

A multivalued function (shortly: multifunction, other names: many-valued function, set-valued function, set-valued map, multi-valued map, multimap, correspondence, carrier) is a left-total relation; that is, every input is associated with at least one output.
The term "multivalued function" is, therefore, a misnomer because functions are single-valued. Multivalued functions often arise from functions which are not injective. Such functions do not have an inverse function, but they do have an inverse relation. The multivalued function corresponds to this inverse relation. [bold-face mine]

Obviously this limitation of defining a function f(x) as giving only ONE value of y is specific to the English (or American?) Maths.
In French, we consider y^2 = x as a function, we call it horizontal parabole...
See http://www.alloprof.qc.ca/BV/pages/m1330.aspx
I would be tempted to use pejorative language toward English (or AMerican?) Maths!


Square Inches vs Inches Squared I was just asked the question, is there a difference between 12 square inches and 12 inches squared. At first I assumed that 12 inches squared mean a square with sides of 12 inches. In this case the answer would be no. 12 inches squared would be 144 square inches. I've been searching the web for quite a long time for what is meant by 12 inches squared. I can't find anything. Is there some formal use for the phrase "inches squared"?

In a context in which you actually expect a square -- such as manufacturing -- one can naturally interpret "12 inches square[d]" as "a square 12 inches on a side". This would be the  area of $144$ square inches.

In other context, "inches squared" may be used synonymously with "square inches". This may also depend on the speaker's native language.

Based on comments by Arkamis

I always understood the “squared” part of __inches squared was pertaining to the inch itself, not the quantity of inches. An inch squared is a 1” x 1” square, not a linear one dimensional inch.
That’s why 12 inches squared is 12 square inches, whether 6x2, 3x4, or 12x1.

Taken literally, 12 inches squared would simply be a linear 144” and in no way refer to area. Then it would be written 12^2 inches, not 12”^2.


Using Intermediate Value Theorem for continuous functions Let $f: [0,1] \rightarrow \mathbb{R}$ be continuous with $f(1)=f(0)$. Prove that if $h\in \left(0,\frac{1}{2}\right)$ is not of the form $\frac{1}{n}$, then there does not necessarily exist $|x-y|=h$ satisfying $f(x)=f(y)$. Provide an example that illustrates this using $h=\frac{2}{5}$.

So I was given the hint that I can use a modified $sin$ function, however I'm not really sure how I would go about that. Preferably, an example not using that would be great.

My thoughts so far are that I use a proof by contradiction saying that $\exists$ $x,y\in[0,1]$ such that $f(x)=f(0)$ and $|x-y|=h$. And I need to get to the point where $h=\frac{1}{n}$, which would create the contradiction (I assume that's the endpoint?). However, how would I use $h=\frac{2}{5}$ to prove that? Is it trivial in that $h=\frac{1}{n}$ is false? Also, it isn't given that $n\in \mathbb{N}$, so maybe I shouldn't assume that?

What would it take to modify $\sin x$ to make it satisfy $f(0) = f(1)$? Well, $\sin 0 = \sin \pi$, so how about
$$f(x) = \sin \left(\pi x\right)\,\text{?}$$

Now you just need to show that $$\sin \left(\pi x\right) \ne \sin \left(\pi x + \frac {2\pi}5\right)$$ for any $x \in [0, \frac 35]$, or equivalently, $$\sin x \ne \sin \left(x + \frac {2\pi}5\right)$$ for any $x \in [0, \frac{3\pi}5]$.

A little trig should give you a formula in the tangent satisfied by any $x$ for which the above is true. Show that the only solutions fall outside of $[0, \frac{3\pi}5]$ and you are done.


Prove $\sin^2\theta + \cos^2\theta = 1$ How do you prove the following trigonometric identity: $$ \sin^2\theta+\cos^2\theta=1$$

I'm curious to know of the different ways of proving this depending on different characterizations of sine and cosine.

Let me contribute by this so let
$$f(\theta)=\cos^2\theta+\sin^2\theta$$
then it's simple to see that
$$f'(\theta)=0$$
then
$$f(\theta)=f(0)=1$$

Here are two proofs using only the angle sum identities, the fact the trig functions are periodic, and their values at $0$. It is inspired on the connection to rotations and the fact that rotations don't change the sizes of things, but do not actually assume that connection.

Define the matrix

$$ A(\theta) = \left( \begin{matrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{matrix} \right) $$

By the angle addition formula, we see that $A(\theta + \varphi) = A(\theta) A(\varphi)$; in particular, $A(n \theta) = A(\theta)^n$.

There are arbitrarily large integer multiples of $\theta$ that are arbitrarily close to integer multiples of $2 \pi$, which gives

$$ A(\theta)^n = A(n \theta) \approx I $$

where $I$ is the identity matrix. By taking determinants, we get

$$ (\cos(\theta)^2 + \sin(\theta)^2)^n \approx 1$$

for arbitrarily large $n$; since the determinant is a real nonnegative number, the only possibility is that

$$ \cos(\theta)^2 + \sin(\theta)^2 = 1$$

To get the details right, we select sequences of integers $a_n, b_n$ with

$$ \lim_{n \to \infty} a_n \theta - 2 \pi b_n = 0 $$
$$ \lim_{n \to \infty} a_n = +\infty $$

e.g. this can be done by continued fractions. Then,

$$ \lim_{n \to \infty} A(a_n \theta) =
\lim_{n \to \infty} A(a_n \theta - 2 \pi b_n)
= A(0) = I$$

and consequently

$$ \lim_{n \to \infty}  (\det A(\theta))^{a_n} = 1 $$

Another method in the same vein is to also define vectors

$$ v(\theta) = \left( \begin{matrix} \sin(\theta) \\ \cos(\theta) \end{matrix} \right) $$

so that $A(\theta) v(\varphi) = v(\theta + \varphi)$

However, we know that for every angle $\theta$:

$$\frac{1}{2} \leq \max(\sin(\theta)^2, \cos(\theta)^2) \leq 1 $$

In particular, we have

$$ \frac{1}{2} \leq \| v(\theta) \| \leq 2 $$

and furthermore, the $v(\theta)$ span $\mathbb{R}^2$.

By comparing the lengths of $v(\varphi)$ and $A(\theta) v(\varphi)$, we know that  all of the eigenvalues of $A(\theta)$ must have magnitude lying between $1/2$ and $2$.

However, this remains true for $A(\theta)^n$, and consequently, all of the eigenvalues of $A(\theta)$ must be roots of unity, and thus the determinant is either $1$ or $-1$, and it can't be $-1$.


how to calculate integrate about Heaviside everyone,here I have a question about how to calculate
 $$\int e^t H(t) dt$$
where $H(t)$ is Heaviside step function
thank you for your answering!!

Hint: The Heaviside function is defined as
\begin{aligned}
\text{H}(t) & = 0, \quad t<0,\\
\text{H}(t) & = 1, \quad t\geq0.
\end{aligned}
That means your integrand will be zero for all $t<0$, and $e^{t}$ for $t\geq1$. Can you figure out the rest?

Note: If you use the half-maximum convention though, you will find it to be:
\begin{aligned}
\text{H}(t) & = 0, \quad t<0,\\
\text{H}(t) & = \frac{1}{2}, \quad t=0,\\
\text{H}(t) & = 1, \quad t>0.
\end{aligned}
If so, you will have the integrand is $\frac{1}{2}e^{t}$ in $t=0$, and the rest is similar to the above.

In fact the result is just trivially

$\int e^tH(t)~dt=e^tH(t)+C$


How to prove that $\lim\limits_{n \to \infty} \frac{k^n}{n!} = 0$ It recently came to my mind, how to prove that the factorial grows faster than the exponential, or that the linear grows faster than the logarithmic, etc...

I thought about writing:
$$
a(n) = \frac{k^n}{n!} = \frac{ k \times k \times \dots \times k}{1\times 2\times\dots\times n} = \frac k1 \times \frac k2 \times \dots \times \frac kn = \frac k1 \times \frac k2 \times \dots \times \frac kk \times \frac k{k+1} \times \dots \times \frac kn
$$
It's obvious that after k/k, every factor is smaller than 1, and by increasing n, k/n gets closer to 0, like if we had $\lim_{n \to \infty} (k/n) = 0$, for any constant $k$.

But, I think this is not a clear proof... so any hint is accepted.
Thank you for consideration.

The series for $e^k$
$$
\sum_{n=0}^\infty\frac{k^n}{n!}
$$
converges by the ratio test. The terms of a convergent series must tend to $0$.

For $n\ge2k$, the ratio of terms is $\frac{k^{n+1}/(n+1)!}{k^n/n!}=\frac{k}{n+1}<\frac{1}{2}$. We can remove the reference to series (which seems to have bothered someone) with the following sandwich, valid for $n\ge2k$:
$$
0\le\frac{k^n}{n!}\le\frac{k^{2k}}{(2k)!}\left(\frac{1}{2}\right)^{n-2k}
$$
which shows that $\displaystyle\lim_{n\to\infty}\frac{k^n}{n!}=0$.

Let's observe an example :

$$\frac{9^n}{n!}$$ so we have that :

$$\frac{9^1}{1!}=9 , \frac{9^2}{2!}=40.5 , \frac{9^3}{3!}=121.5$$

but...

$$\frac{9^9}{9!}=\frac{9^8\cdot 9}{8! \cdot 9}=\frac{9^8}{8!}$$

and further...

$$\frac{9^{10}}{10!}=\frac{9^9\cdot 9}{9! \cdot 10} < \frac{9^9}{9!}$$

So we may conclude: after we throw out the first $k-1$ elements that sequence starts to decrease  , therefore sequence :

$$\frac{k^n}{n!}$$

is decreasing if we set $ n\geq k$ , so we may conclude that :

$$n \rightarrow \infty \Rightarrow \frac{k^n}{n!} \rightarrow 0$$


Proof for subtheories using compactness theorem I'm trying to understand the use of the Compactness Theorem to proof certain properties for theories in languages. I've tried to prove the following:

If $\phi$ holds in every model of the theory $T$, then there is a finite subtheory $T'$ such that $\phi$ holds in every model of $T'$.

My proof is:

Because $\phi$ holds in $T$, $T \cup \{\neg \phi\}$ is an inconsistent theory. Compactness theorem then says that there is a finite subtheory of this theory that is inconsistent. This theory will be of the from $T' \cup \{\neg \phi\}$ with $T'$ a finite subtheory of $T$. Because $T' \cup \{\neg \phi\}$  is inconsistent, $\phi$ holds in every model of $T'$.

I'm not too sure if this proof is true, especially the part where I take the inconsistent finite subtheory.

Try using Godel's completeness theorem. If $\phi$ is a statement true in every structure, then $\phi$ is a theorem of $T$ and so $\phi$ has a finite proof in $T$.


How does one get better at real analysis proofs? How does one proceed through a math proof in real analysis? My instructor always says make a diagram, but I am not a visual learner. It seems that whenever I write out the definition of an assumption, then I cannot make the next logical step. Also, when I go to try to verify that my proof is correct, I ask myself questions like, "why must this be true?" but the proof does not end up not being air tight. For those that have had real analysis, what did you do to master proofs and do the exercises?

Most of the theorems in real-analysis (especially those in introductory chapters) are intuitive and based on the concept of inequalities. If one understands the concept of inequalities (not in the sense of memorizing AM greater than equal to GM or other famous inequalities) in terms of comparison of numbers most of the common proofs are trivial applications of the definitions.

I will provide two examples:

1) If $f$ is continuous at $x = a$ and $f(a)$ is positive then there is a neighborhood of $a$ in which $f$ is positive.

Now one has to know what is meant by continuity to prove this. Informally this means that values of $f(x)$ are arbitrarily near $f(a)$ if $x$ is sufficiently near $a$. The $\epsilon, \delta$ are used to quantify "arbitrarily" and "sufficiently" in a formal manner. Now if we see that $f(a)$ is positive then there is a range of values near $f(a)$ which are positive. Hence if $x$ is sufficiently close to $a$, $f(x)$ will take values in the range near $f(a)$ and these are all positive as mentioned in last sentence.

2) If $f(x) \leq g(x)$ in a neighborhood of $a$ and both $\lim_{x \to a}f(x), \lim_{x \to a}g(x)$ exist then $\lim_{x \to a}f(x) \leq \lim_{x \to a}g(x)$.

Clearly let $A = \lim_{x \to a}f(x), B = \lim_{x \to a}g(x)$. Suppose $A > B$. Now values of $f(x)$ are near $A$ when $x$ is near $a$. Similarly values of $g(x)$ are near $B$. Since $A > B$ we can obviously make values of $f$ much nearer to $A$ compared to $B$ and values of $g$ much nearer to $B$ compared to $A$. We will find that this leads to values of $f$ being greater than some values of $g$ and we get contradiction.

Thinking in terms of inequalities as a way of comparing magnitudes and numbers is the key to these kinds of proofs. However thinking in this fashion is not easy for a beginner as he is trained to think in terms of operations like $+, -, \times, /$ and not $< , >$. As a further example consider the two following facts:

a) There is no rational number whose square is equal to $2$ (i.e. $\sqrt{2}$ is an irrational number).

b) If $a$ is a positive rational such that $a^{2} < 2$ then there exists another rational $b$ such that $a < b$ and $b^{2} < 2$.

The proof of statement a) is mostly algebraical and can be figured out easily if we know simple facts about integers and their factorization. The proof of statement b) is not easy unless we know how to deal with inequalities (reader can convince himself by trying to prove this). I consider this to be the fundamental difference between algebraic and analytic approaches and a beginner in analysis must make a transition from understanding statements like a) to understanding statements like b).

Analysis has evolved over three centuries. We need to take a different path with hindsight.
     One should introduce limit using neighborhoods rather than epsilon deltas.
          Also one can define phrases like sufficiently close , and selection of smallest neighborhood or largest n can be formalized in phrases.
          Further several example,s and finding limits in sequences series can be easily tackled using application of L hospital rule treating the variable as continuous.
     One  needs to struggle if one does not use this rule and practices raw analysis
     We need to prove algebra of limits rigorously and here we need  epsilons but not delta as we can use neighborhoods for independent variables.
      Rather than proving theorems on continuous functions it seems worthwhile to introduce metric space but as far as possible stick to neighborhoods. .

But one proves open ball is open set closed ball is closed set. arbitrary union and finite intersection.
One teaches both pathwise contentedness and contentedness but without stress on exotic examples and just to the point and in one of the basic proof, one models using bisection argument may be the theorem that  interval is connected.
   One introduces compactness using  Bolzano Weirstrass property. Proves it for R. One  uses products and show product of complete spaces is complete.
Then one prove cantors theorem on complete spaces and from it derives Bolzano Weirstrass theorem in Euclidean spaces.
       Then one teaches continuous image of compact space is compact and then uniform convergence.nowhere one needs not so intuitive Heine Borel theorem
    Then comes to  basic idea of derivative of one variable . One proves algebra of derivatives Draboux theeorem .
      But then one introduces Henstock-kurzweil integral and using it one derives both Lagranges mean value theorem and L hospitals rules as well as Taylors formula for one variable but vector mappiungs.
       One introduces notion of Banach space and Hilbert space Cauchy Schwartz inequality.
       Then one studies sequences and series tests for convergence in detail.
Admittedly one loses some sight in this synthesis
        For example though contentedness and compactness are different, in case of real line they stem from  one single notion of  completeness.
    Yes we would need separate course on foundations simultaneously treating cantors theory and construction of real numbers.

 I mean with lot of experience and hindsight one needs to approach real Analysis now in this fashion . In calculus one just avoids epsilon and deltas.
  Calculus is motivated geometrically


Is there a way to count the number of unique possible songs able to be created? As in the number of permutations of musical notes, lyrics (all known languages), and so on?

I am no mathematician or musician so this might (probably) be totally stupid. Any ideas would be interesting to hear though.

Any melody can be decently encoded at $128$ kbps, or $1$ MB per minute. Four minutes equals $4$ MB. So the total number of songs no longer than about $4$ minutes in length is lesser than $2^{~4~\cdot~8~\cdot~10^6}$

At least on a cd its possible.
A digital song (not .mid song) is made out of samples.

A cd has a samplerate of 44100hz, this means 44100 samples per second (a sample has a length of 1/44100 seconds). A cd can't be true mono, so it will have one sample for the left channel and one sample for the right channel.

Each sample is a integer value from 32768 a +32767, or 65536 different values.

The cd can have at max 99 songs and each song must have at min 4 seconds.

The max amount of time a cd can hold is 79 minutes and 50 seconds.

Just assuming you make cd with a single song that fill the entire cd, the amount of cds you can make are 65536^422478000


Linear operators well-defined by matrices in $l^2$ are bounded Let $\{a_{jk}\}$ be  an infinite matrix such that corresponding mapping $$A:(x_i) \mapsto (\sum_{j=1}^\infty a_{ij}x_j)$$ is well defined linear operator $A:l^2\to l^2$.
I need  help with showing that this operator will be bounded. I guess it means that i need to check if a unit sphere maps to something bounded, so i need to manage to get some inequality on coefficients of matrix that will allow to write a straight sequence of inequalities and get desired bound. But I don't understand how to get bound from operator being well defined.

This follows from the fact that the pointwise limit of bounded operators is bounded, which follows from the uniform boundedness principle:
If a sequence of bounded operator converges pointwise, then it is bounded in norm

It is helpful to consider the 'finite rank' case first:
Lemma. Let $(a_n)$ be a sequence of real numbers such that for every $x = (x_n) \in \ell^2$ we have $Ax := \sum_{n = 1}^\infty a_n x_n$ convergent. Then $A : \ell^2 \to \mathbb R$ is a bounded linear map.
Proof: Suppose $A$ is not bounded. Take any sequence of positive real numbers $(M_n)$. We may find $x_1 \in \ell^2$ with $\Vert x_1 \Vert = 1$ and $Ax_1 > M_1$. We may find a finite subsequence $x_1'$ of $x_1$ satisfying $Ax_1' \geq M_1/2$. It is no restriction to assume all $a_n x_{1, n} \geq 0$. Because $x_1'$ is a subsequence, $\Vert x_1' \Vert \leq \Vert x_1 \Vert \leq 1$. Let $A_1$ be the corresponding subsequence of $(a_n)$. Then $A_1$ is a finite matrix, so it defines a bounded operator.
Because $A - A_1$ is also unbounded, we may find $x_2 \in \ell^2$ with $\Vert x_2 \Vert = 1$ and $(A-A_1)x_2 > M_2$. We may find a finite subsequence $x_2'$ of $x_2$ with $(A-A_1) x_2' > M_2/2$ and all $a_n x_{2,n} \geq 0$, and a corresponding subsequence $A_2$ of $A - A_1$. By construction, we may assume that $x_1'$ and $x_2'$ have disjoint supports, so that $(A-A_1) x_2' = A x_2'$.
Continuing this procedure, we find a sequence $(x_n')$ with disjoint supports, $\Vert x_n \Vert \leq 1$ and $A x_n' > M_n/2$ and $a_m x_{n,m} \geq 0$. By taking the $M_n$ to grow sufficiently fast, the element $x := \sum_{n=1}^\infty \frac1{M_n} x_n'$ is in $\ell^2$. But because the $x_n'$ have disjoint supports, and because $Ax$ is a sum of nonnegative terms, it is clear that
$$Ax \geq \sum_{n=1}^\infty \frac1{M_n} A x_n > \sum_{n=1}^\infty \frac12 = \infty \,,$$
so that $A$ is not well-defined. $\square$
Corollary. Let $(a_{ij})$ be a matrix of real numbers with $N$ rows, such that for every $x = (x_j) \in \ell^2$ we have $Ax := \left(\sum_{j = 1}^\infty a_j x_j \right)_i \in \mathbb R^N$ convergent. Then $A : \ell^2 \to \mathbb R^N$ is a bounded linear map.
Proof: Each of the projections of $A$ is bounded by the previous lemma, and this suffices because $\mathbb R^N$ is finite-dimensional. $\square$
Now, for the case of $A : \ell^2 \to \ell^2$.
Proof: Suppose $A$ is not bounded. Take any sequence of positive real numbers $(M_n)$. We may find $x_1 \in \ell^2$ with $\Vert x_1 \Vert = 1$ and $\Vert A x_1 \Vert > M_1$. There then exists a submatrix $A_1$ of $A$ with finitely many nonzero rows, and with $\Vert A_1 x_1 \Vert$ 'very close' to $\Vert A x_1 \Vert$.
By the corollary above, $A_1$ is bounded, so that $A - A_1$ is not bounded. Thus we may find $x_2 \in \ell^2$ with $\Vert x_2 \Vert = 1$ and $\Vert (A-A_1) x_2 \Vert > M_2$, and subsequently a submatrix $A_2$ of $A - A_1$ with finitely many rows and with $\Vert A_2 x_2 \Vert$ 'very close' to $\Vert (A-A_1) x_2 \Vert$.
Continuing this process, we find a sequence $(x_n) \in \ell^2$, with $\Vert x_n \Vert = 1$ and $\Vert A_n x_n \Vert$ 'very close' to $\Vert (A - A_1 - A_2 - \ldots - A_{n-1}) x_n \Vert$ and with $\Vert (A - A_1 - A_2 - \ldots - A_{n-1}) x_n \Vert > M_n$. Note how the $A_n x_n$ have disjoint supports, by construction.
Consider the element $x = \sum_{n=1}^\infty \frac1{M_n} x_n$. When the $M_n$ grow sufficiently fast, this lies in $\ell^2$. Using orthogonality of the $A_n x$ and the triangle inequality applied to finite sums,
$$\begin{align*}\Vert Ax \Vert
&\geq \sum_{n=1}^\infty \Vert A_n x \Vert \\
&\geq \sum_{n=1}^\infty \left( \frac1{M_n}\Vert A_n x_n \Vert - \left \Vert A_n \sum_{j \neq n} \frac1{M_j} x_j \right\Vert \right) \\
&\geq \sum_{n=1}^\infty \left( \frac1{M_n}\Vert A_n x_n \Vert - \sum_{j < n} \frac1{M_j} \Vert A_n x_j \Vert - \Vert A_n \Vert \left \Vert \sum_{j> n} \frac1{M_j} x_j \right\Vert \right) \\
&\geq \sum_{n=1}^\infty \left( \frac1{M_n}\Vert A_n x_n \Vert - \sum_{j < n} \frac1{M_j} (\Vert (A - A_1 - A_2 - \ldots - A_{j-1})  x_j \Vert - \Vert A_j x_j \Vert) - \Vert A_n \Vert \left( \sum_{j > n} \frac1{M_j^2}\right)^{1/2} \right) \,.
\end{align*}$$
Now we choose the $M_n$ and $A_n$ so that this sum is infinite: At every step, take $M_{n}$ very large compared to the norms $\Vert A_j \Vert$ for $j < n$. Given this $M_n$, choose $A_n$ so that $\Vert A_n x_n\Vert$ is extremely close to $\Vert (A - A_1 - A_2 - \ldots - A_{n-1}) x_n \Vert$. It is then clear that the negative terms in the above sum can be made arbitrarily small, to get (for example)
$$\Vert Ax \Vert
\geq \sum_{n=1}^\infty \frac1{M_n}\Vert A_n x_n \Vert/2 \geq \sum_{n=1}^\infty \frac12  = \infty \,.$$
More concretely, you could take $M_1 = 1$, $M_n> \max(n, 10n \max_{j < n}(\Vert A_j \Vert))$ and $A_n$ such that $\Vert (A - A_1 - A_2 - \ldots - A_{n-1}) x_n\Vert - \Vert A_n x_n \Vert \leq \frac1{100n^2}$. Then
$$\begin{align*}
&\sum_{n=1}^\infty \left( \frac1{M_n}\Vert A_n x_n \Vert - \sum_{j < n} \frac1{M_j} (\Vert (A - A_1 - A_2 - \ldots - A_{j-1}) x_j \Vert - \Vert A_j x_j \Vert) - \Vert A_n \Vert \left( \sum_{j > n} \frac1{M_j^2}\right)^{1/2} \right) \\
& \geq \sum_{n=1}^\infty \left( \frac9{10} - \sum_{j < n} \frac1{100j^2} - \left( \sum_{j > n} \frac1{100n^2}\right)^{1/2} \right)
\end{align*}$$
and this is fine.


Angle arising from circle rolling along an ellipse In this problem the ellipse and circle are fixed. The ellipse has center $E$ on the origin, its semi-minor axis $r$ is on the $y$-axis, and its semi-major axis $R$ is on the $x$-axis. The circle has center $O$ and radius $c$. The circle and the ellipse intersect at point $P$.

Given $E, O, R, r$, and $c$, is it possible to determine $\angle z=\angle EOP$?

When $P$ lies on the $x$ or $y$-axis, the angle is $0$. But when it lies anywhere else, as in the picture, I'm stumped as to how to easily calculate $z$.

Thank you.

The problem posed is algebraically intractable.

If instead of O you know $P$, then you can find the angle after extending line OP to meet the major axis at $F$.

Label the intersection on the right of the ellipse with its major axis $G$.  Let
$\angle GFO$ be $\alpha$ and $\angle GEO$ be $\beta$; then if we can determine $\alpha$ and $\beta$,  $\angle z = \alpha - \beta$.

But if the projections of $EP$ along the major and minor axes are $(x,y)$ respectively, then
$$
\tan \alpha = \frac{yR^2}{xr^2}
$$
and
$$
\tan \beta = \frac{y+c\sin\alpha}{x+c \cos \alpha}
$$
Take the respective arctangents and subtract.

For the problem posed, you have $O$ so $\beta$ is trivially $\tan^{-1}\frac{y_0}{x_0}$, but to get $\alpha$ you would need to solve for $x, y, \alpha$ in
the equations
$$
\left\{
\begin{array}{c}
\tan \alpha = \frac{yR^2}{xr^2} \\
x + c \cos \alpha = x_0 \\
y + c \sin \alpha = y_0
\end{array}
\right.
$$
You can use trig identities to eliminate $\alpha$ and solve the last equation for $y$ in terms of $x$, but the remaining equation involves square roots of two expressions, one of which has $(x-x_0)^2$ in a denominator, and when you group and square twice you end out with an 8-th degree equation which is much too tough to solve.

Given the data, can you not determine P as the point of intersection of the circle and the ellipse? Then, you can use the method by @MarkFischler.


Argand Diagram - Quadrants help I forgot the rules of adding angles when it comes to argand diagrams.

In the first quadrant, you add 90 degrees to whatever angle you get, what about Q2 Q3 Q4 ?

This picture will explain what i mean :

notice how 18.43 was added with 180 .. can someone list the rules for the other quadrants as well ?

what if i have -3i -3j for example ?

Thanks

When you measure the argument of a complex number you go counterclockwise from the x-axis.

So to find the argument you take $\arctan\left(\frac{b}{a}\right)$. However the range of the $\arctan x$ function is only from $[\frac{-\pi}{2},\frac{\pi}{2}]$. That is to say, you only get values between $-90^{\circ}$ and $90^{\circ}$ for whatever value you put in $\arctan x$. That means you are completely missing the $2$nd and $3$rd quadrant!

To account for the $2$nd and $3$rd quadrant we must first remember that $\tan \theta$ has period $\pi$. With that we realize that while $\theta+\pi$ has a different value from $\theta$; $\tan(\theta+\pi)=\tan(\theta)$. In fact it turns out $\tan(\theta +k\pi)=\tan(\theta),k\in\mathbb{Z}$. That means there are infinitely many solutions to $\tan x =\theta$ of the form $x=\arctan(\theta)+k\pi,k\in\mathbb{Z}$. But since $2\pi$ is one whole round around a circle all a lot of the solutions overlap and we are left with $2$ unique answers: $$x=\arctan(\theta)$$
$$x=\arctan(\theta)+\pi$$
However, your calculator does not know which solution you want and so it always gives you the first one. To get the second solution you simply add $\pi$ (or $180^{\circ}$ if you like degrees).

NLed,  Sorry to be so late with the following answer.  (I found your posting because I am looking for software to construct an Argand diagram just like the one in the gray area in the original posting and need help.  I hope you will contact me at onekevATgmailDOTcom.)

The decision to add 180 degrees to the inverse tangent is based on the sign of the denominator "inside" the inverse tangent. Add 180 degrees only if denominator < 0. Also, for this procedure to work, use angles (in degrees) in (-180, 180] instead of the more typical [0, 360).

The angle interval (-180, 180] is more commonly used in the study of functions of a complex variable and electrical engineering than the (perhaps more intuitive) interval [0, 360) used in trigonometry and pre-calculus.

But if you prefer an angle in [0, 360), use the scheme described above then add 360 degrees only if the angle value found is negative.


Compute $\int^{2 \pi} _0 \frac{1}{a + \sin \theta} d\theta$ I want to compute $\displaystyle \int^{2 \pi} _0 \frac{1}{a + \sin \theta} d\theta$, with $a > 0$, where we may use the Cauchy Integral Formula.

The following hint is given:

Write $sin \theta = (e^{i \theta} - e^{-i \theta})/ 2i$ and interpretate the integral after some manipulation as a complex line integral of a rational function over the unit circle with counterclockwise direction.

What I got so far:

$\displaystyle \int^{2 \pi} _0 \frac{1}{a + \sin \theta} d\theta = \int^{2 \pi} _0 \frac{1}{a + \frac{e^{i \theta} - e^{-i \theta}}{2i}} d\theta = \int^{2 \pi} _0 \frac{1}{\frac{2ia + e^{i \theta} - e^{-i \theta}}{2i}} d\theta = \int^{2 \pi} _0 \frac{2i}{2ia + e^{i \theta} - e^{-i \theta}} d\theta = \int^{2 \pi} _0 \frac{-2}{-2a + i\cdot (e^{i \theta} - e^{-i \theta})} d\theta$

Here I get stuck. It does not look like a complex line integral, where the Cauchy Integral Formula can be used.

We can convert to an integral over the unit circle
$$
\begin{align}
\int_0^{2\pi}\frac{\mathrm{d}\theta}{a+\sin(\theta)}
&=\int_0^{2\pi}\frac{2i\,\mathrm{d}\theta}{2ia+e^{i\theta}-e^{-i\theta}}\\
&=\int_0^{2\pi}\frac{2\,\mathrm{d}e^{i\theta}}{e^{2i\theta}+2iae^{i\theta}-1}\\
&=\oint\frac{2\,\mathrm{d}z}{z^2+2iaz-1}\\
&=\oint\frac{2\,\mathrm{d}z}{(z+ia)^2-(1-a^2)}\\
&=\frac1{\sqrt{1-a^2}}\oint\left(\frac1{z+ia-\sqrt{1-a^2}}-\frac1{z+ia+\sqrt{1-a^2}}\right)\mathrm{d}z\\
\end{align}
$$
If $|a|\lt1$, then $-ia+\sqrt{1-a^2}$ and $-ia-\sqrt{1-a^2}$ are either both inside the unit circle or both outside. In this case the integral is $0$. (If both are outside, then there are no singularities inside the contour. If both are inside, we know the sum of the residues is $0$ since the integral over an arbitrarily large circle vanishes.)

If $|a|\gt1$, then the roots are $-i(a+\sqrt{a^2-1})$ and $-i(a-\sqrt{a^2-1})$. Their product is $-1$, so one must be inside the unit circle and the other is outside.

The residue at the singularities is $\frac{\pm i}{\sqrt{a^2-1}}$. Considering the original integral has the same sign as $a$, we have
$$
\bbox[5px,border:2px solid #C0A000]{
\int_0^{2\pi}\frac{\mathrm{d}\theta}{a+\sin(\theta)}=\left\{\begin{array}{}
\frac{2\pi\,\mathrm{sgn(a)}}{\sqrt{a^2-1}}&\text{if }|a|\gt1\\
0&\text{if }|a|\lt1
\end{array}\right.
}
$$
Of course, when $|a|\le1$, the denominator of the integrand vanishes. The Cauchy Principal Value of the integral is $0$.

Note also that since the integrals are of the form
$$
\frac1{\sqrt{1-a^2}}\oint\frac{\mathrm{d}z}{z-(-ia\pm\sqrt{1-a^2})}
$$
They can also be evaluated using $f(z)=1$ in Cauchy's Integral Formula, arriving at the same answer.

Note that for $z=e^{it}$, $\frac{z-z^{-1}}{2i}=\sin t$. Hence you're integrating $$\frac{1}{a+\dfrac{z-z^{-1}}{2i}}\frac 1{iz}=\frac{2}{z^2+2iaz-1}$$ over the unit circle. Using residues, this is easy.

In general, if $R(x,y)$ is a rational function finite on the unit circle, you can evaluate $$\int_0^{2\pi}R(\cos t,\sin t)dt=\int_{|z|=1}R\left(\frac{z+z^{-1}}2,\frac{z-z^{-1}}{2i}\right)\frac{dz}{zi}$$


How similar are circular and parabolic paths? I'd for a long time thought of parabolas as semi-circles.

However, if you take half of a circle the ends will - look - parallel, where as parabolas continue to extend horizontally and infinitely.

Are the only similarities then the fact that they're curved?

Are sinusoidal graphs connected semi-circles or parabolas?

If they were connected semi-circles, wouldn't the gradient at $x=0$ be $\infty$ or $\frac10$

Notice: I hope the pictures explain themselves and that you enjoy pondering over this post :-)

Circles and parabolas are both second degree equations, meaning that they can be created by cutting open a cone (conic sections): (You could also make these by shining a flashlight)

This is related to eccentricity.

You could imagine eccentricity to be how much these graphs are stretched: ($e=$ eccentricity)

circle: $e=0$

ellipse: $0<e<1$

parabola: $e=1$

hyperbola: $e>1$

Clearly sinusoidal graphs are related to circles (see here for original post):

while probably not what you were thinking of, sinusoidal functions look parabolic locally:

which is due to Taylor's theorem.

Sinusoidal graphs can be approximated by parabolas in some regions ... depending on the type of sinusoid, the graph may be exactly replicated by linking parabolas together.

You are correct in saying that parabolas are linked to circles, but not in a simple way, as circles have tangents that rotate about completely as you rotate around the circle; parabolas do not.

Parabolas also are onto functions, whereas circles cannot be.


Determine if a positive integer is a sum of two positive squares Given a positive integer $N$, what is the best/most efficient way to determine if $N=x^2+y^2$ for some positive integers $x<y$? Only the decision is needed, not explicitly $x,y$.

As shown in this answer, $n$ is the sum of two non-negative squares if and only if for each prime $p\equiv3\pmod4$, $p$ appears with even exponent in the factorization of $n$.

If the factorization of $n$ contains at least one prime $p\equiv1\pmod4$ with positive exponent, or $p=2$ with odd exponent, then $n$ can be written as the sum of two positive squares.

If the factorization of $n$ contains at least one prime $p\equiv1\pmod4$ with positive exponent, then $n$ can be written as the sum of two distinct positive squares.

Fermat's Theorem says,

  If $p$ is an odd prime and $x,y$ are integers numbers. $p=x^2+y^2$ iff $p≡1 \pmod 4$.


For the equation given, evaluate y′ at the point (−2,−1) I must find the derivative for this equation
$$
   (3x−y)^4+4y^3=621
$$
 at the point $(-2,-1)$.

With some differential calculus: differentiating both sides you obtain:
$$4(3x-y)^3(3\,\mathrm dx -\mathrm dy)+12 y^2\,\mathrm dy=0$$
which becomes for $x=-2,\;y=-1$:
$$-500(3\,\mathrm dx -\mathrm dy)+12 \,\mathrm dy=0\iff 512\,\mathrm dy=1500\,\mathrm dx\iff\frac{\mathrm dy}{\mathrm dx}=\frac{1500}{512}=\frac{375}{128}.$$


Good Online Resources to learn Multivariate Statistics? Are there any good online resources to learn multivariate statistics? ( With topics such as Multiple Linear Regression, Principal Component Analysis, Factor Analysis, Cluster Analysis and Discriminant Analysis)

Statsoft has a pretty good comprehensive online guide which I can recommend:

http://www.statsoft.com/textbook/

-Ralph Winters

Hopefully I can help with intuitively understanding PCA with my cartoon:

Visual Intuition (zoom in)

Intuition via utility

I think the main usage for PCA is to be able to categorise different distinct "things" e.g. Shiny cells vs. Dark cells in a way that leads to least error. E.g. Imagine sam was hiding behind me and I pinched a cell off the left side of his body then asked you to guess the color of the cell, by looking at the winning photo, or even the winning line, you can make a very good guess it will be a "dark cell".

Intuition via Analogy

So my understanding is that PCA is like taking a "picture" in a lower dimension, but the various methods used out there attempt to make the picture as informative as possible by deciding which "angle" to take the picture from (notice for 1D the angle of "squishing line" also vary).


What's the shortest length here? Question: An ant starts at one vertex of a solid cube with side of unity length. Calculate
the distance of the shortest route the ant can take to the furthest vertex
from the starting point.

Now, in the answer, the cuboid is unfolded and a rectangle of side 1 and 2 is formed. Calculating the hyptoneuse gives the distance.

That was the proposed solution. I wanted to calculate it via sqrt(1+1+1). Why is my answer wrong?

If you use $\sqrt{1+1+1}$, the answer is drilling straight through the cube.

What a talented ant.

However, we need the ant to go in a path on the surface of a cube, which poses new challenges.

We can quickly visualize the ant going up the cube while going around to the other side.

This, as you mentioned is a straight line along the $1 \times 2$ rectangle formed by the path.

Therefore, the shortest length is $\sqrt{1^2+2^2}=\boxed{\sqrt{5}}$.

The distance from one corner to opposite corner of a 1x2 rectangle is $\sqrt{1^2+2^2}=\sqrt{5}$ and $\sqrt{5} \neq \sqrt{1+1+1} = \sqrt{3}$


$\int_{\gamma}\left(\frac{1}{z}-\frac{1}{z-1}\right)dz=0$ on $\mathbb C\setminus[0,1]$ $\int_{\gamma}\left(\frac{1}{z}-\frac{1}{z-1}\right)dz=0$ on $U:=\mathbb C\setminus[0,1]$ for a closed path with image in $U$

For any analytic function $f$ and a closed path $\gamma$:   $\int_{\gamma}f(w)dw=0$, and I know that $\frac1z$ and $\frac{1}{z-1}$ are analytic, so is their difference ?

Their difference is also anaytic, of course. Perhaps it's easier to see it as follows

$$\frac1z-\frac1{z-1}=-\frac1{z(z-1)}$$

and the above is anaytic on $\;\Bbb C\setminus\{0,1\}\;$, thus its integral is zero on any closed path not containing $\;0,1\;$ . What if it contains both points? Can you take it from here?

If $f$ is holomorphic in a simply connected domain $U,$ then $\int_\gamma f(z)\,dz = 0$ for every closed contour $\gamma$ in $U.$ In this problem, $U =\mathbb {C}\setminus [0,1]$ is simply connected, and $f(z) = 1/z-1/(z-1)$ is holomorphic in $U.$


Variance of a random variable representing the sum of two dice $\newcommand{\Var}{\operatorname{Var}}$

The formula for the variance of the sum of two independent random variables is given $$ \Var (X +X) = \Var(2X) = 2^2\Var(X)$$

How then, does this happen:

Rolling one dice, results in a variance of $\frac{35}{12}$. Rolling two dice, should give a variance of $2^2\Var(\text{one die}) = 4 \times \frac{35}{12} \approx 11.67$. Instead, my Excel spreadsheet sample (and other sources) are giving me 5.83, which can be seen is equal to only $2 \times \Var(X)$.

What am I doing wrong?

$\newcommand{\Var}{\operatorname{Var}}$

The formula you give is not for two independent random variables.  It's for random variables that are as far from independent as you can get.  If $X,Y$ are independent, then you have $\Var(X+Y)=\Var(X)+\Var(Y)$.  If, in addition, $X$ and $Y$ both have the same distribution, then this is equal to $2\Var(X)$.  It is also the case that, as you say, $\Var(X+X)=4\Var(X)$.  But that involves random variables that are nowhere near independent.

X and itself aren't independent. X and itself are correlated with rho = 1.

In general, for two events, X (the first die) and Y (the second die), VAR(X+Y) = VAR(X) + VAR(Y) + 2COV(XY).

Since rolling two dice ARE independent (rho = 0), COV(XY) = 0, so VAR(X+Y) = 35/12 + 35/12 + 0.

VAR(X+X) is 4*VAR(X), but this recognizes that X and itself are perfectly correlated and we could just as well calculate this as VAR(X+X) = 35/12 + 35/12 + 2*35/12 since we know that COV(XX) is the same as VAR(X).


A circle satisfies Power of a Point, but is the converse true? The "Power of a Point" or "Intersecting Chords" theorem states that for any point in a plane, if a line is drawn that intersects a circle, the distance from the point to one of the intersections multiplied by the distance from the point to the other intersection is a constant for that point, no matter what line you draw, as long as it intersects the circle. My question is if the converse is true. If a shape satisfies this, does it have to be a circle? Are there other shapes that satisfy this?

Let $P$ be a point and $\Gamma$ be a curve on the plane. A line through $P$ intersects $S$ at two points $A$ and $B$.
I think your statement on 'Power of a Point' can be interpreted in two ways:
1) $PA\cdot PB$ is constant for a fixed $P$. This is true for some $P$ on the plane.
2) $PA\cdot PB$ is constant for a fixed $P$. This is true for every $P$ on the plane.

If you mean (1), then Aretino provides a neat counter-example. In case you want to know the proof:
Let $X = (0,1)$. Note that $OX = OB = 1$.
Then $OA\cdot OB = \frac{1}{\cos{\angle XOA}}\cdot 2(1)(\cos{\angle XOA}) = 2$

If you mean (2), then the converse is indeed true.
Firstly we have to assume for every line through an arbitrary point $P$ that intersects $\Gamma$, it only intersects $\Gamma$ at 2 points (1 if degenerate).
Let another line through $P$ intersect $\Gamma$ at $C$ and $D$. Then $ABCD$ is cyclic. Let its circumcircle be $\Phi$.
Let $F$ be a point on arc $AB$ in $\Phi$ that does not contain $C$. Let $EC$ intersect $\Gamma$ at $E\ne C$, $AB$ at $Q$.
We have $QA\cdot QB = QC\cdot QE$ from given condition, and $QA\cdot QB = QC\cdot QF$ from Power of a Point in $\Phi$.
So, $QE = QF$.
Note that $E$ cannot be on the side of line $AB$ containing $C$, because that will imply $\Gamma$ is concave, and there exists a line that intersects it 3 times.
So, $E = F$. $F$ is a point in $\Gamma$.
Since $F$ can be any point in that arc, the entire arc is also in $\Gamma$.
Choose any $F$ on the arc and repeat the above process. Then we have the entire arc $AB$ that doesn't contain $F$ (i.e. contains $C$) in $\Gamma$.
Therefore, $\Gamma$ consists of the entire circle $\Phi$.
If there exists some point $G$ in $\Gamma$ that isn't in $\Phi$, then draw a line through $G$ and the centre of $\Phi$. Then the line intersects $\Gamma$ 3 times (once at $G$, twice at $\Phi$).
Thus, $\Gamma$ = $\Phi$ is a circle.

(Sorry for no diagram; couldn't draw it right.)


How to find the following limit? $ \lim_{x \to 0} \frac{x}{\sin(2x)\cos(3x)}$ I am stuck with this limit problem

$$\lim_{x \to 0} \frac{x}{\sin(2x)\cos(3x)} $$

Any hints are appreciated. Also, I can't use L'Hopital's.

We know that: $\cos(3x)\to1$ as $x\to0$, so the only difficulty you're left with is to prove that: $$\lim_{x\to0}\dfrac{x}{\sin(2x)}=\dfrac12$$ and as a hint you can use: $$\lim_{x\to0}\dfrac{{x}}{\sin(2x)}=\lim_{x\to0}\dfrac12\dfrac{2{x}}{\sin(2x)}\quad\color{grey}{\sf and}\quad\lim_{x\to0}\dfrac{x}{\sin(x)}=1.$$


How do I negate the following statement? How do I negate the following statement?

What are the steps which should be taken to negate an argument such as the one given?

$$\exists y \forall x ((y > 0) \land (x < y))$$

Essentially the $\forall$ becomes $\exists$ and vice-versa, and $\land$ becomes $\lor$. Hence, we have:
$$\begin{align}\lnot\left(\exists y\forall x((y>0)\land(x<y))\right)&\iff\forall y\exists x(\lnot(y>0)\lor\lnot(x<y))\\&\iff\forall y\exists x((y\leq 0)\lor(x\geq y))\end{align}$$

Theorem. The negation of $$\exists y \forall x ((y > 0) \land (x < y))$$ is $$\forall y \exists x ((y \le 0) \lor (x \ge y))$$

Proof.
By Lemma 1, 2, 3, 4, and 5.

  Lemma 1. The negation of $\exists y~ P(y)$ is $\forall y,~\lnot P(y)$

Proof.
Calling $Y$ the universe
$$\exists y~ P(y)\equiv\exists y\in Y,~ P(y)\equiv Y\cap P\neq\emptyset$$
where $P=\{z|~P(z)\}$.

So its negation is: $$Y\cap P=\emptyset\equiv Y\subset P^c\equiv\forall y\in Y,~\lnot P(y)$$
where $P^c=\{z|~\lnot P(z)\}$

  Lemma 2. The negation of $\forall y~ P(y)$ is $\exists y,~\lnot P(y)$

Proof.
Calling $Y$ the universe
$$\forall y~ P(y)\equiv\forall y\in Y,~ P(y)\equiv Y\subset P$$

So its negation is: $$Y\not\subset P\equiv Y\cap P^c\neq\emptyset\equiv\exists y\in Y,~\lnot P(y)$$

  Lemma 3. The negation of $a\land b$ is $\bar a\lor\bar b$

Proof.
By de Morgan $$a\land b=\overline{\bar a\lor\bar b}$$

  Lemma 4. The negation of $y>0$ is $y\le 0$

Proof.
Here it is understood that $y\in \mathbb{R}$, though not explicitly stated.
$$y>0\equiv y\in\{z|~z>0\}$$
So its negation is
$$y\not\in\{z|~z>0\}\equiv y\in\{z|~z>0\}^c$$
where $\{z|~z>0\}^c=\mathbb{R}\setminus\{z|~z>0\}=\{z|~z\le0\}$

  Lemma 5. The negation of $x<y$ is $x\ge y$

Proof.
Here it is understood that $x,~ y\in \mathbb{R}$, though not explicitly stated.
$$x<y\equiv (x,y)\in\{(w,z)|~w< z\}$$
So its negation is
$$(x,y)\not\in\{(w,z)|~w< z\}\equiv y\in\{(w,z)|~w< z\}^c$$
where $\{(w,z)|~w< z\}^c=\mathbb{R}^2\setminus \{(w,z)|~w< z\}=\{(w,z)|~w\ge z\}$


Proof that there are infinitely many prime numbers $p$ such that $p-2$ is not prime. Just need verification of this proof by contradiction:

Let $p_k$ be the largest prime number such that $p_k-2$ is not a prime number. Let $p_l$ be some prime number such that $p_l > 3p_k$. We know that $p_l$ exists because there are infinitely many prime numbers as proven elsewhere.

$p_l-2$ is a prime number. $p_l-4$ is a prime number as well. By induction, all numbers $p_k < x < p_l | x \mod 2 = 1 $ are prime numbers. But $p_k<3p_k<p_l$ and $3p_k \mod 2 = 1$ and $3|3p_k$ so there is a contradiction.

Also would love to see more concise proofs if at all possible.

Edit:

Great that I got so many people trying to provide their own proof but the question was primarily asked so that I could have my proof verified. Only skyking addressed my question, though I do not yet understand his criticism.

It is very simple to construct such infinite sequence:

$$35+60n,37+60n$$

$35+60n$ will generate an infinite amount of composite numbers, all of which are divisible by $5$ (in fact, it will generate only composite numbers).

$37+60n$ will generate an infinite amount of prime numbers, since $37$ and $60$ are coprime integers (according to Dirichlet's theorem on arithmetic progressions).

We can have the obvious case where the last digit of p is 7. By drichlet's theorem on arthmetic progressions, we can generate a sequence of 5*n* + 7, where there are infinitely many primes with last digit as 7

Note: I considered the obvious case since (10k+7)-2 is divisible by 5, thus not prime


When the normals to the curve $y = c\cdot sin(x)$are also a tangent $y = c\cdot \sin x$, with $c = 1.7$ then no normal to the graph will also intersect the graph, but $c = 1.8$ allows this (only for a select few of the normals). What is the exact value (I'm not sure if it can be found, but I assume so) of the constant $c$ that will allow normals to also tangent the graph? In other words, what value of $c$ will allow some normals to touch the graph exactly twice(one intersection and one tangent)?

By equating $c\cdot \sin x$ with it's normal(with a parameter of $a$ for the $x$-value of the normal) you get this:

$$\sin x \cos a \cdot c^2 + x = \sin a \cos a \cdot c^2 + a$$

. For some reason I graphed this with $a$ being the $y$ axis and found that the graph looks exactly like $y = x$, until the constant $c$ is raised above $1.7$ again,  then it gets really weird . Not sure if that provides anything useful but it seems important to me.

Did you mean

$$y=c^{\sin x}$$

?

If not, wouldn't the graph be the same shape as a $\sin$ curve but stretched up and down (not sure if that was clear)?


What is the definition of limit superior/inferior of real function? http://en.wikipedia.org/wiki/Upper_limit

Wikipedia has the definition for metric space, but it doesn't have a definition for extended real-field.

It just says, 'Limit inferior and limit superior of a real function are always well-defined in $\overline{\mathbb{R}}$'.

In that sense, I don't think the definition for metric space covers real-valued functions. (Isolated points)

What's the definition it would be?

Say one considers a function $f:(0,+\infty)\to\mathbb R$ and one wants to determine the limsup/liminf of $f$ at $0$. Then one would define $g(t)=\sup\{f(x)\mid0\lt x\lt t\}$, $h(t)=\inf\{f(x)\mid0\lt x \lt t\}$, and
$$
\limsup\limits_{x\to0}f(x)=\inf\{g(t)\mid t\gt0\}=\lim\limits_{t\to0}g(t),
$$
and
$$
\liminf\limits_{x\to0}f(x)=\sup\{h(t)\mid t\gt0\}=\lim\limits_{t\to0}h(t).
$$
Example: $f(x)=\sin(1/x)$ for every $x\gt0$, then $g(t)=+1$ and $h(x)=-1$ for every $t\gt0$ hence $\limsup\limits_{x\to0}f(x)=+1$ and $\liminf\limits_{x\to0}f(x)=-1$.

Similar but more involved: if $f(x)=(2+x^2\cos(1/x))\sin(1/x)$ for every $x\gt0$, then $\limsup\limits_{x\to0}f(x)=+2$ and $\liminf\limits_{x\to0}f(x)=-2$.

Just consider the function

$$ f(x) = \frac{|x|}{x}.$$

Can you determine the limsup and the liminf as $x \to 0$ of this function? See here for the definition.


How can I prove by induction that $n!>n^{\frac n2}$ for $n>2$? First I've checked it by putting $n=3$.

Then, I've assumed it to be true for $n=k$ to get $k!>k^{\frac k2}$.

Then, putting $n=k+1$, I get $(k+1)!=k!(k+1)>k^{\frac k2}(k+1)$. Now, I'm stuck and couldn't proceed further.

It remains to show that $(k^{k/2})(k+1)\geq (k+1)^{(k+1)/2}$, or $k^k\geq (k+1)^{k-1}$. This is equivalent to $$k\geq\left(1+\frac1k\right)^{k-1}.$$
Does the term on the right look familiar?

It's an easy fact that $${k \choose m} \leq k^m.$$  Then $$(k+1)^{k+1} = k^{k+1} + {k \choose 1}k^k + {k \choose 2}k^{k-1}+ \cdots +1 \leq k^{k+1}+\cdots + k^{k+1} = k^{k+1}(k+1)$$ since there are $k+1$ terms in the sum.  If we take the square root of both sides, this gives your last inequality, so apply the induction hypothesis and you're done.

For the "easy fact", note that $${k \choose m} = \frac{k(k-1)\cdots(k-m+1)}{(k-m)!} < \frac{k^m}{(k-m)!} \leq k^m.$$


Showing that an absolute integrable monotone decreasing function $f: [1,\infty[ \rightarrow \mathbb{R}$ is in $L^p([1,\infty[)$ For an exercise in my analysis course, I have to show that: if $f: [1,\infty[ \rightarrow \mathbb{R}$ is monotone decreasing and $f \in L^1([1,\infty[)$, then $f \in L^p([1,\infty[)$ for every $p > 1$.

Define $f_+ = \max(f,0)$ and $f_{-} = \max(-f,0)$

I tried several things: My guess is that I have to use the integral of $|f|$ to provide an upper bound for the integral of $|f|^p$. I tried the following:

$|f| = f_+ + f_-$. Then $f_+$ is monotone decreasing and $f_-$ is monotone increasing. Moreover $|f|^p = f_+^p + f_{-}^p$ since $f^+$ and $f_{-}$ are never nonzero together. Hence
$$\int_{[1,\infty[} |f|^p d\lambda = \int_{[1,\infty[} f_+^p d\lambda + \int_{[1,\infty[} f_+^p d\lambda$$

Now I want to try to get the monotonicity into play. Let $g_k = f_+ \cdot \chi_{[k,k+1[}$, then $f_+ = \sum^\infty_{k=0} g_k$. We can find a similar sequence for $f_{-}$, let's call that one $h_k$.

\begin{align*}\int_{[1,\infty[} f_+^p d\lambda + \int_{[1,\infty[} f_+^p d\lambda &= \int_{[1,\infty[}(\sum^\infty_{k=0} g_k)^p d\lambda+ \int_{[1,\infty[}(\sum^\infty_{k=0} h_k)^p d\lambda \\&= \int_{[1,\infty[} \sum^\infty_{k=0} g_k^p d\lambda+ \int_{[1,\infty[}\sum^\infty_{k=0}h_k^p d\lambda
\\& = \sum^\infty_{k=0} \int_{[1,\infty[} g_k^p d\lambda+ \sum^\infty_{k=0}\int_{[1,\infty[}h_k^p d\lambda\end{align*}
We are allowed to bring in the exponent because for every $x$ only one of them is nonzero.

\begin{align*}
\sum^\infty_{k=0} \int_{[1,\infty[} g_k^p d\lambda+ \sum^\infty_{k=0}\int_{[1,\infty[}h_k^p d\lambda &\leq \sum^\infty_{k=0} \int_{[1,\infty[} f_{+}(k)^p\chi_{[k,k+1[} d\lambda+ \sum^\infty_{k=0}\int_{[1,\infty[}f_{-}(k+1)^p\chi_{[k,k+1[} d\lambda\\
&=  \sum^\infty_{k=0}(f_{+}(k)^p + f_{-}(k+1)^p)\end{align*}

Now I am not sure how to proceed. Am I looking into the wrong direction? Please only provide hints!

First note that $f_- = 0$, as if $f_-(x^*) > 0$, then for each $x \ge x^*$, we have $f(x) \le f(x^*) < 0$, which implies that $f$ isn't integrable. Now choose $x^*$ such that for $x \ge x^*$ we have $f(x) \le 1$. Then
\begin{align*}
  \int_1^\infty f(x)^p\, dx &= \int_1^{x^*} f(x)^p\, dx + \int_{x^*}^\infty f(x)^p\, dx\\
       &\le \int_1^{x^*} f(1)^p\, dx + \int_{x^*}^\infty f(x)\, dx\\
       &\le (x^*-1) \cdot f(1)^p + \int_1^\infty f(x) \, dx\\
       &< \infty.
\end{align*}

If your function is monotone and decreasing on $[1,\infty[$ then your function must be positive (if it's negative at any moment, it will not be integrable).

So I don't know (yet) the way to proceed, but I'm not sure you're on the right direction.

I hope this could help.


How to write an expression in an equivalent form without absolute values? The question I have in front of me is the very first problem in Trench's Introduction to Real Analysis:

Write the following expression in equivalent form not involving absolute values:

$a+b+|a-b|$

Looking at the provided answer in the back of the book its:

$2\max(a,b)$

In the chapter I can't find any examples with solutions involving the min/max functions, so feel at a complete loss as to how to get to the above answer.

All you have to do is write down what the absolute value means.

The definition we have is
$$|x|:=\begin{cases}x&\text{ if }x\ge0\\-x&\text{ if }x<0\end{cases}$$

So now the equation we have means

$$\begin{align}
a+b+|a-b|&=\begin{cases}a+b+a-b&\text{ if }a-b\ge0\\a+b-(a-b)&\text{ if }a-b<0\end{cases}\\
\\&=\begin{cases}2a&\text{ if }a\ge b\\2b&\text{ if }a<b\end{cases}
\end{align}$$

Now all that is required is to recognize
$$\max(a,b)=\begin{cases}a&\text{ if }a\ge b\\b&\text{ if }a<b\end{cases}$$
giving you an answer of
$$a+b+|a-b|=2\max(a,b)$$

There is also a trick. $\sqrt{x}$ means normally the positive square root. Thus, we can use $\sqrt{x^2}=|x|$:

$$a+b+|a-b|=a+b+\sqrt{(a-b)^2}$$


Dividing one equation by another (may one is zero?!) there's something confusing me about how do we divide two equation by each other? but may one is zero .. no?!

lets assume I have like this:
(1) x=y
(2)z=m
how do we do (x/z)=(y/m) without caring about if z is zero or m is zero .. then we can't do division .. so how we do division ?! thanks a lot.

if z or m is zero then we can't do division .. so how we say that we can division two equation one by another?

As you state, if $z = m = 0$ the division makes no sense. You'd have to add that restriction and handle the case $m = 0$ separately.

Without further details, we can't say anything more.

You can write $$\begin{cases}x=y,\\z=m\end{cases}\iff \begin{cases}x=y,\\z=m=0\end{cases}\lor \begin{cases}x=y,\\\dfrac xz=\dfrac ym\end{cases}.$$


Higher dimensional analogues of the argument principle? I know there are higher dimensional analogues of the argument principle.

(See http://en.wikipedia.org/wiki/Variation_of_argument)

But I do not have books about it and I cannot find anything of value on the internet for free.

Plz give me references or explainations.

I would like an example of how one can find the zero's of functions in 3 Dimensions with an integral.

I believe it works as follows :

1) The function needs to be "3D-analytic" , and by that I mean the derivative can be given by the limit

$$ \lim_{h->0} \frac{f(x+h)-f(x)}{h} $$

Where any infinitesimal $h$ will give the same limit as any other infinitesimal $h_2$.
(By "any" I mean in any direction of the dimension.)

This is the natural analogue of complex differentiable.

2) The integral is taken over a closed surface that contains a volume.
The surface must be topologically isomorphic to a sphere. ( no holes , no selfintersections , continu )

3) Just like the argument principle , the direction of the integral matters.
Hence we riemann sum the points with respect to the direction we are going in.
( compare : for contour integration if our (piecewise) path goes from - to + we add if it goes from + to - we substract )

4) The "3D argument principle" then computes $CZ$.
$C$ is some constant and $Z$ is the number of zero's - the number of poles within the volume.

Is this correct ?

The analytic proof of the PNT constructs a counting function for the primes based on the zero's of the Riemann zeta function.

An important part of that proof is the argument principle.

I wonder if there are some similar applications of the "3D argument principle " as described above that have been used in number theory.

I wonder what functions in 3D are associated with (number-theoretical) counting functions*.

(* such as used in the proof of PNT with the argument principle (used on the complex plane).)

Of course, the Argument Principle is in the context of meromorphic functions. However, the higher-dimensional analogue in the context of smooth functions comes from the theory of the degree of maps and winding numbers.

In particular, suppose $W$ is a compact $n$-dimensional oriented manifold with boundary, $f\colon W \to \Bbb R^n$ is smooth, and $f\big|_{\partial W} \ne 0$. If $0$ is a regular value of $f$, then $f^{-1}(0)$ consists of a finite number of points $x_1,\dots,x_k\in W$, each of which appears with a sign $\epsilon_j$ ($+1$ if $df_{x_j}\colon T_{x_j}W\to\Bbb R^n$ is orientation-preserving, and $-1$ if it is orientation-reversing). Then
$$\sum_{j=1}^k \epsilon_j = \text{deg}\left(\frac{f}{\|f\|}\Bigg|_{\partial W}\colon \partial W\to S^{n-1}\right)\,.$$

Comment: In the case $n=2$, $W\subset\Bbb R^2$, and $f$ holomorphic, the $\epsilon_j$ are all always $+1$, and so we merely count the roots in $W$. The degree on the right-hand side is the winding number of $f(\partial W)$, which is exactly what $\dfrac1{2\pi i}\displaystyle\int_{\partial W} \frac{f'(z)}{f(z)}dz$ computes.

The best reference I know on such matters is Guillemin and Pollack's Differential Topology. See pp. 110-111 and 144, in particular.

EDIT: To respond to your additional comments/questions, the one-variable calculus definition of derivative extends to $\Bbb C$ only because $\Bbb C$ is a field. Once you move on to $\Bbb R^n$, the derivative becomes a linear transformation with a limit property but cannot itself be given by the sort of limit you desire. Only directional derivatives have the single-variable calculus definition. (In time, you will learn some rigorously done multivariable calculus.) The degree calculation I have given above can indeed be represented by an integral (but, once again, you'll have a bit to learn to understand this): Choose an $(n-1)$-form $\omega$ on $S^{n-1}$ with $\displaystyle\int_{S^{n-1}}\omega=1$, and compute $\displaystyle\int_{\partial W} \left(\tfrac{f}{\|f\|}\right)^*\omega$. Alternatively, compute
$$\tfrac1{\text{vol}(S^{n-1})}\int_{\partial W} f^*\left(\tfrac{x_1\, dx_2\wedge\dots\wedge dx_n - x_2\, dx_1\wedge dx_3\wedge \dots \wedge dx_n + (-1)^{n-1} x_n\, dx_1\wedge\dots\wedge dx_{n-1}}{(x_1^2+x_2^2+\dots+x_n^2)^{n/2}}\right)\,.$$
 This is in fact an immediate generalization of the integral appearing in the argument principle.

There are generalizations of the Residue Theorem (which is where the Argument Principle comes from) to meromorphic functions in $\Bbb C^n$. This is also a rather rich subject; some comments were made in this post.

Back in the year 1991 I found the higher dimensional complex numbers, they bear many similarities with the ordinary complex plane yet there are all kinds of differences and subtilities that come along with them.

The complex plane is more or less defined via stating $i^2 = -1$, in $\mathbb{R}^3$ you can do the same via stating $j^3 = -1$.
The number one correspondents with a unit vector on the x-axis,
the number $j$ lies on the y-axis while
the number $j^2$ is on the z-axis.

A very important property is the fact that $i$ from the complex plane does not exist in 3-dimensional space.

During the last two years I have made many updates on that, you can find it in the next link:

http://kinkytshirts.nl/rootdirectory/just_some_math/3d_complex_stuff.htm

It takes some time to understand the stuff involved, the best thing is to multiply a few times those 3D numbers and try to understand the Cauchy-Riemann equations for 3 dimensions...

Good luck with it.