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1

Here's a way of thinking about it. To be clear however, this does not prove the Pythagorean theorem. Given two 2-dimensional vectors $\mathbb{x},\mathbb{y} \in \Bbb R^2,$ with entries $\mathbb{x} = (x_1,x_2)$ and $\mathbb{y} = (y_1,y_2),$ we define a norm as, $$ |\mathbb{x}| = \sqrt{x_1^2 + x_2^2} $$ We say two vectors are orthogonal if, $$ x_1y_1 + x_2y_2 ...


3

The question: Prove that there are infinitely many solutions in positive integers to $x^2+y^2=z^2$ is not Pythagoras' theorem, because Pythagoras' theorem is a theorem about side lengths of right-angled triangles. Consequently, it doesn't make sense to talk about a non-geometrical proof: it's a statement about geometry. If you want to prove your ...


1

For your new question: Take triplets $(t^2-1)^2+(2t)^2=(t^2+1)^2$, in which $t\ge 2$ in order that $x,y,z\in \mathbb Z^+$


1

You could try to use something like this, but as you'll see, it would be a circular argument, because in the derivation of these results the Pythagorean theorem is already used. You could just Define arc-length to be the linked expression, but then I think you'll loose your high school students.


2

An open problem I find surprising, the PAC (Perimeter to Area Conjecture) due to Keleti (1998): Conjecture: The perimeter to area ratio of the union of finitely many unit squares in the plane does not exceed 4. See for example Bounded - Yes, but 4? and references therein.


1

This one is probably already included here. But just for the image...


0

I have a specific application in mind, but Gunnar Carlsson gave a mini-course at this year's YTM, and in this he used many statistical methods along with topological methods to answer problems in computing. I think the broader title of this field would be somethign like Topological Data Analysis! Along this flavour, a paper by Gunnar Carlsson entitled ...


3

Let $$ f(x) = \sum_{n = 1}^{\infty} nx^{n-1}, \quad |x| < 1. $$ Then $$f'(x) = \sum_{n = 2}^{\infty} n(n-1)x^{n-2} = 2 \left( \sum_{n=2}^{\infty} \frac{n(n-1)}2x^{n-2} \right ).$$ Note $\frac{n(n-1)}2 = \dbinom{n}2$ so $$f'(x) = 2 \sum_{n=2}^{\infty} \dbinom{n}2x^{n-2} .$$ Now consider the coefficient of $x^k$ in $\frac{1}{(1-x)^3} = (1+x+x^2+ \cdots ...


4

Here is another variation. Assuming the geometric series $\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$ is known, we consider functions $f,g:(-1,1) \rightarrow \mathbb{R}$ \begin{align*} f(x)&=\sum_{n=0}^{\infty}(n+1)x^n\qquad\qquad g(x)=\frac{1}{(1-x)^2} \end{align*} We obtain \begin{align*} \int f(x) dx&=\int\left(\sum_{n=0}^{\infty}(n+1)x^n\right) ...


3

As I'd suggested like Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $, Using Generalized Binomial Expansion, $$(1+y)^n=1+ny+\frac{n(n-1)}{2!}y^2+\frac{n(n-1)(n-2)}{3!}y^3+\cdots$$ given the converge holds Comparing with given Series $ny=2x\ \ \ \ (1)$ ...


0

There is a series of math children books in Russian by Владимир Артурович Лёвшин. To list some: Магистр рассеянных наук (translates roughly as Master (as in M.Sci) of the absent-minded sciences, though google translates it as Master scattered Sciences), Новые рассказы Рассеянного Магистра (New stories of the absent-minded Master), Путешествие по Карликании и ...


1

In my opinion, one of the most elegant is the "calculus proof" by John Molokach: Given a right triangle with legs $a$ and $b$ and hypotenuse $c$, construct a semicircle in a rectangular coordinate system like the picture below. The function $y$, whose graph is the semicircle, satisfies the IVP $$\left\{\begin{align} &y'=-x/y\\ ...


2

The sequence $y=(1,2,3,4,\ldots)$ is an output of the linear system $$ y_{k+1}=y_k+u_k,\qquad y_0=1 $$ for the input $u=(1,1,1,1,\ldots)$. Perform the $Z$-transform (multiply by $z^k$ and add up for all $k$) $$ \sum_{k=0}^\infty y_{k+1}z^k=\underbrace{\sum_{k=0}^\infty y_kz^k}_{y(z)}+\underbrace{\sum_{k=0}^\infty u_kz^k}_{u(z)}\quad\Rightarrow\quad ...


1

Use differences of a sum. $$ \begin{array}{r} S &=& +1 & +2 x & +3 x^2 & +4 x^3 & +5 x^4 & +6 x^5 & +7 x^6 & \cdots\\ -2 x S &=& & -2x & -4 x & -6 x^2 & -8 x^3 & -10 x^4 & -12 x^5 & \cdots\\ x^2 S &=& & & +x^2 & +2 x^2 & +3 x^4 & +4 x^5 & +5 x^6 & ...


2

The following proof is far to complicated, but it is a new one and I think it is somewhat funny too. For $x\in\left[0;1\right)$ we have: $$ \frac{1}{\left(1-x\right)^2}=\frac{1}{1-\left(2x-x^2\right)}=\sum_{k=0}^{\infty}\left(2x-x^2\right)^k=\sum_{k=0}^{\infty}{\sum_{r=0}^{k}\binom{k}{r}(-1)^r2^{k-r}x^{k+r}}=\sum_{k=0}^{\infty}c_kx^k $$ We have ...


3

Problem: For a given integer $N$, how many integers $n_1$ and $n_2$ larger than or equal to zero are there that satisfy the equation: $$n_1 + n_2 = N$$ We note that it is the coefficient of $x^N$ in the expansion of $$\left(\sum_{k=0}^{\infty}x^k\right)^2 = \frac{1}{(1-x)^2}$$ We can also solve the problem by noting that it is the number of ways you can ...


4

Let $X \sim G(1-x)$, a geometric random variable with success probability $1-x$. We have $$ \mathbb{E}[X] = \sum_{n=1}^\infty nx^{n-1}(1-x). $$ On the other hand, we know that $\mathbb{E}[X] = 1/(1-x)$, and we deduce the formula. We can argue that $\mathbb{E}[X] = 1/(1-x)$ in many ways. One way is to consider $N$ different trials with success probability ...


6

The effect of multiplication by $1/(1-x)$ to the sequence of coefficients is to calculate partial sums: if the original sequence is $c_0,c_1,\ldots$ then the new one is $$ d_i = c_0 + \cdots + c_i. $$ The starting point is the sequence $1,0,0,\ldots$. Applying this operator twice, we get $$ 1,0,0,0,0,\ldots \\ 1,1,1,1,1,\ldots \\ 1,2,3,4,5,\ldots $$ In this ...


3

ie. $(1 - x)^2(1 + 2x + 3x^2 + ...) = 1$


23

$${1\over(1-x)^2}={1\over 1-x}\cdot{1\over 1-x}=\sum_{j\geq0} x^j\cdot\sum_{k\geq0}x^k =\sum_{r\geq0} x^r\left(\sum_{j+k=r}1\right)=\sum_{r\geq0}(r+1)x^r\ .$$


6

Let $S=1+2x+3x^2+4x^3+\dotsb$. \begin{align} \phantom{-x^2}S&=1+2x+3x^2+4x^3+\dotsb\\ \phantom{^2}-xS&=\phantom1-\phantom2x-2x^2-3x^3-\dotsb\\ \phantom{^2}-xS&=\phantom1-\phantom2x-2x^2-3x^3-\dotsb\\ \phantom{-}x^2S&=\phantom{1+2x+2}x^2+2x^3+\dotsb \end{align} Adding them together: \begin{align} (1-2x&+x^2)S\\ ...


-1

If I have $n$ cakes (and no cake can be partitioned) and $m$ people, with $n\lt m$, then someone won't get a cake.


0

Ramsey's theorem For any given number of colors c, and any given integers n1, . . . ,nc, there is a number, R(n1, ..., nc), such that if the edges of a complete graph of order R(n1, . . . , nc) are colored with c different colors, then for some i between 1 and c, it must contain a complete subgraph of order ni whose edges are all color i. The two ...


0

The best book as far as i know are these two: Discrete Mathematics By Norman L.Biggs or Discrete Mathematics and Its Applications by Kenneth H. Rosen


0

$$\lim \limits_{n \to \infty} {D_2(n) \over D_2'(n)}=\lim \limits_{n \to \infty} {F_{n+1} \over F_n} = {1+\sqrt 5 \over 2} = \phi$$ where $D_2(n) $ and $D_2'(n)$ are the partition functions occuring in the Rogers-Ramanujan indentities and $F_n$ is the $n$th Fibonacci number.


2

For $n\geq k\in\mathbb{N}$, you can prove that $\frac{n!}{k!\cdot(n-k)!}$ is an integer from a combinatorial/counting argument. Establish that the formula gives the number of ways to choose a subset of $k$ from a set of $n$, and that automatically makes it an integer. Or you can prove that $\frac{n!}{k!\cdot(n-k)!}$ is an integer by showing that the ...


2

One interesting generalization follows Fourier's original analysis for Ordinary Differential Equations on $\mathbb{R}$. It's easiest to break first into ODEs on a half line $[0,\infty)$ for Sturm-Liouville problems $$ Lf=-\frac{d^{2}f}{dx^{2}}+q(x)f(x) = g(x),\\ \cos\alpha f(0)+\sin\alpha f'(0) = 0. $$ where ...


1

Perhaps not the most squarely on-the-mark example, but an exercise requesting a proof of the irrationality of $e-\sqrt{2}$ or similar had me stymied (I was young and had lived a sheltered, analysis-centric life). I shelved it, looked at it the next day, misread 'irrational' as 'transcendental', and went 'Uh... Oh yeah!'


9

Not a visual proof, but by the Binomial Theorem, $$(1-x)^{-2}=\sum_0^{\infty}{-2\choose n}(-1)^nx^n$$ Now $${-2\choose n}={-2\cdot-3\cdots(-1-n)\over n!}=(-1)^n(n+1)$$ so $(1-x)^{-2}=\sum_0^{\infty}(n+1)x^n$, as desired.


16

For the special case $x=\dfrac12$: If you accept that $1+x+x^2+\dotsb=\dfrac1{1-x}$, the same picture works — just move the horizontal and vertical lines. Instead of them being at $1,1\frac12,1\frac34,\dotsb,2$, you should put them at $1,1+x,1+x+x^2,\dotsb,\dfrac1{1-x}$. The area of the square is then $\left(\dfrac1{1-x}\right){}^2$.


0

The work done by Donaldson on 4-manifolds used techniques borrowed from gauge theory. Witten also used techniques from Physics to solve mathematical problem, including if I remember correctly, a simpler proof of the index theorem using supersymmetry.


0

Count the number of ways of placing them in line, in a circle; how many ways to form a line of, say, 5 of the students. How many ways to choose, say, 5 of them gives binomial coefficients. How many ways are there to make up a 5-person group out of 2 boys and 3 girls, this generalizes to Vandermonde's identity. Fool around with a 3-way split of the class, and ...


1

One I really like is this one : If $f$ is a continuous function for which $f(a+b-t)=f(t)$ then $$\int_a^b t\cdot f(t) \mathrm{d}t=\frac{a+b}{2}\int_a^bf(t) \mathrm{d}t$$ Example : $$\begin{align} \int_0^{\pi} \frac{x\sin(x)}{1+\cos^2 (x)}\mathrm{d}x &=\frac{\pi}{2}\int_0^{\pi} \frac{\sin(x)}{1+\cos^2 (x)}\mathrm{d}x\\ &=\frac{\pi}{2} ...


0

This proof will look much better if someone can add a diagram to it. Suppose $\sqrt2=a/b$ with $a$ and $b$ positive integers chosen as small as possible. Draw a square of side $a$. In the upper left corner, place a square of side $b$, and another one in the lower right corner. The two $b$-squares have total area $2b^2=a^2$, the same as the area of the ...


1

To calculate those facts, you would need some statistics. For example, to calculate the death by vending machine problem, you would need statistics on how many people each year use vending machines, and how many die from vending machines. For high school students, I think it would be cool to arrange the seats in a circle and have them experiment on the ...


1

If in your "etc" you are willing to include some set theory, the two volumes by Just & Weese, Discovering Modern Set Theory, are a must-read in this vein. I think it is relevant to add that, apart from being very well written, the first volume ("The Basics") is the only math textbook that made me laugh out loud!


1

Although not being one of the subjects you asked for right now, John Oprea's Differential Geometry and Its Applications might fit the bill: there's lot of exercises in the middle of the text, eventually some details of proofs turn into exercises too, he doesn't have any section labeled "exercises", working through most of the examples given are exercises ...


0

I've always been a fan of Rudin's books on analysis; Principles of Mathematica Analysis, Real and Complex Analysis. These books have numerous exercises, although the latter can be quite tough. Also worth looking at is Mathematical Analysis by Apostol. This book covers all the basics of mathematical analysis and has a lot of exercises. For Topology, I ...


2

You might be talking about "problem books" which contain very little exposition, and ask the reader to prove most of the propositions. For topology, you might try Elementary Topology, by Oleg Viro. Basically the structure is that of a guided inquiry from definitions through the major theorems. There are hints and answers in the book if you get stuck in ...


0

Claim: Given a set of $n$ points. Then these points lay on one line. Proof: Inductive Basis: Clearly, one point lays on one line. Inductive Hypothesis: Given a set of $k$ points. Then these points lay on one line. Inductive Step: Consider a set of $k+1$ points. Consider a subset of $k$ points. Then these lay on a line. Consider another subset of $k$ ...


2

The cross-ratio is the unique rational function of four complex numbers up to multiplication by a constant such that the cross-ratio of four distinct points is real if and only if the points are concyclic.


12

I'm hardly qualified to answer this question, but you might find the following references useful. Let's start with the two classical examples of the Fourier transform: the Fourier transforms for $L^2$ functions on the line and the circle. Generalizing the notion of space The line and the circle are both topological abelian groups, and the Fourier ...


0

Pick's Theorem says that the area of a lattice polygon is the number of interior lattice points, plus half the number of lattice points on the boundary, minus 1. It can be proved by induction on the number of sides of the polygon. It's not hard to prove that if you divide a lattice polygon up into smaller lattice polygons, and if the formula holds for the ...


2

Robin Chapman gives 14 proofs of $$\sum_{n=1}^{\infty}{1\over n^2}={\pi^2\over6}$$ here.


0

The Product Rule, $$(f_1f_2\cdots f_n)'=f_1'f_2\cdots f_n+\cdots+f_1f_2\cdots f_n'$$ To prove the base case, $n=2$, $(fg)'=f'g+fg'$, you need to apply the definition of the derivative, and properties of limits. But then you can deduce the $n+1$ case very simply from the base case and the induction hypothesis: $$(f_1f_2\cdots f_{n+1})'=\bigl((f_1f_2\cdots ...


7

Let $z_1, z_2, z_3, z_4 \in \Bbb P ^1 (\Bbb C) = \Bbb C \cup \{ \infty \}$, all of them distinct. The cross-ratio is essentially the only projective invariant of this system of points. More rigorously, let $$X = \{ (z_1, z_2, z_3, z_4) \in \big( \Bbb P ^1 (\Bbb C) \big) ^4 \mid z_i \ne z_j \; \forall i \ne j \} ;$$ then $F : X \to \Bbb C$ is invariant ...


1

Suppose $\sqrt2$ is arational, then there exist $p,q$ two natural numbers such that $$\color{Red}{p^2=2q^2.}$$ Then by the parametric solution of Pythagoras Equation there exist two natural numbers $a,b$ such that $a\gt b\ge 1$ and $p=a^2+b^2,\,\,\color{Green}{q=a^2-b^2=2ab.}$ Now, if $r=a+b$ and $s=a,$ then $$\color{Red}{r^2=2s^2}$$ with $r\lt p,\ s\lt q.$ ...


2

An example would be Tim Gowers' beautiful proof of the following problem: Let $n$ be an even integer. How many subsets of the set $\{1,2,\dots,n\}$ can you pick if they all have to have odd size but the intersection of any two of them has to have even size? using basic linear algebra over finite fields. Concisely, he considers the characteristic ...


1

This application is maybe not for undergrads, but if you take Hensel's lemma for granted, it's a nice application, in my opinion. Hensel's lemma: Let $f(t)$ be a polynomial with coefficients in $\mathbb{Z}_p$ (the $p$-adic integers, $p$ prime). If there exists $a\in\mathbb{Z}_p$ such that $f(a)\equiv0\pmod{p}$ and $f'(a)\not\equiv0\pmod{p}$, then there ...


1

I can think of one possible example. As $\phi(m)$ is the number of natural numbers less than $m$ which are coprime to $m$, Euler's theorem provides a good upper bound to a lot of problems which involve coprime numbers. For example, if we consider the problem of Quadratic Congruence, solving the equation $x^{2} \equiv a\ (mod\ p)$, where p is a prime. ...



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