Tag Info

New answers tagged

1

I remember watching The fault in our stars which is a great movie, but the wrong math at a very emotional scene, ruined it for me. Shailene Woodley (who has cancer) says this to his boyfriend who also has cancer- “I am not gonna talk about our love story because I can't. So, Instead I am gonna talk about maths. I am not a mathematician but I do know ...


1

Here is an opinionated and detailed 100 page Study Guide to logic textbooks, updated fairly regularly.


0

If you are interested in statistics, then I would heartily recommend the book "Reminiscences of a Statistician: The Company I Kept" by Erich Lehmann. In a sense it is an autobiographical account, but he mainly focuses on his interactions and joint work with other statisticians. It was at this time that the first statistics departments started to show up, ...


1

Mathematics: The Loss of Certainty by Morris Kline fits your requirements. You can view extracts from the book on Amazon.com.


0

You can produce a lot of those with the following tool by Robert Monafo: http://mrob.com/pub/ries/


2

It is more likely to find subtle, nonobvious applications of algebraic graph theory to graph theory. A celebrated early example is the Friendship Theorem https://en.wikipedia.org/wiki/Friendship_graph#Friendship_theorem which has a nice proof using the spectrum of the adjacency matrix. More recently the critical groups of sandpiles on graphs are a major ...


2

Elisha S. Loomis, The Pythagorean Proposition, contains $370$ proofs of the Pythagorean theorem. ERIC has a PDF of NCTM reissue of the $1940$ second edition.


2

The book "The Fundamental Theorem of Algebra" by Fine and Rosenberger (link) contains detailed discussions of at least six proofs of this theorem, all rooted in different areas of mathematics. Links to other papers (not all in English) compiling various proofs of the theorem can be found at this MathOverflow question. H. W. Kuhn gave a combinatorial proof ...


0

An important application of Euler's Theorem is the RSA Cryptosystem. See the paper A Modern Day Application of Euler’s Theorem: The RSA Cryptosystem by M. Maxey for some information about it. You may also find some helpful information in A Concrete Introduction to Higher Algebra by L.N. Childs. The chapter $10$ Applications of Euler’s Theorem is subdivied ...


1

Here is a question on this site with a whole bunch of proofs of $\sum(n+1)x^n=(1-x)^{-2}$.


3

Fourteen Proofs of a Result About Tiling a Rectangle collects $14$ proofs of the fact that a rectangle tiled by rectangles each of which has at least an integer side has an integer side.


0

$\mathbb{R}$ with standard topology is second-countable space. For a second-countable space with a (not necessarily countable) base, any open set can be written as a countable union of basic open set. open set in a second countable space which is not a countable union of basic open sets Clearly, collection of open intervals is a base for the standard ...


2

Consider the integral: $$ I = \int_0^\infty \frac{\arctan(x)}{1+x^2} dx\,. $$ By evaluating this integral through a change of variables: $ u=\arctan x , $ and $ du= {1\over 1+x^2}dx, $ $$ I=\int_0^ {\pi\over 2} u du= {\pi^2\over 8} \ $$ Now,we notice that: $$ \arctan(x)=\int_0^1 \frac{x}{1+x^2y^2} dy\, $$ Utilizing this fact, we can see: $$ I = ...


1

Perhaps not the most elementary, but in Hatcher's Algebraic Topology text (freely available) in the first chapter on the Fundamental Group there is an explanation of two linked circles which lets you intuitively and visually explain a fundamental group which is isomorphic to the additive group of integers (infinite cyclic). Though you might not even get ...


9

Some of my favorite results in topology are: The Borsuk-Ulam Theorem: Given a continuous function $f:S^n\to\mathbb R^n$, there exists a point $x$ such that $f(x)=f(-x)$. My favorite result is that somewhere on the planet, there are antipodal points which have the same temperature and pressure. For this reason, this is sometimes called the Meteorologist's ...


4

I recently came across Prof. Tokieda's series of lectures on Topology at the African Institute for Mathematical Sciences and found the lectures to be very good as well as very entertaining. The first lecture covers a number of nice examples using Möbius strips.


1

The problem is that your criteria are contradictory (and to a lesser extent, subjective): you want "the very basics," and you say you don't want a treatise on elementary number theory or (abstract) algebra, yet your very first criterion is "Show why things are the way they are." These are not mutually compatible requirements. The reason why they are not ...


0

There are many examples related to automata theory. I just mention three of them. (1) A semigroup $S$ with $0$ is nilpotent if there exist $n > 0$ such that, for all $s_1, \dotsm, s_n \in S$, $s_1 \dotsm s_n = 0$. A regular language is finite or cofinite if and only if its syntactic semigroup is nilpotent. This is really a result on semigroup and ...


-2

How about this? Let x be a rational in (0,1). Then x is of the form x = m/M with M>m. If x has a non terminating decimal expansion then x must be of the form x = m/p where p is a prime number. Also, the period of x, say T(x), is less than p. Let A = { x such that x is rational in (0,1) and has non terminating decimal expansion } Let B = { y=T(x) such that x ...


6

There are a couple of applications in PDEs that I am quite fond of. As well as verifying that the Laplace operator $-\Delta$ is positive on $L^2$, I like the application of integration by parts in the energy method to prove uniqueness. Suppose $U$ is an open, bounded and connected subset of $\mathbb{R}^n$. Introduce the BVP \begin{equation*} -\Delta ...


2

In the $n$-dimensional vector space $F_5^n$, every non-zero vector generates a one-dimensional vector space with 5 elements. Each pair of such vector spaces intersect only at the zero vector. Therefore, $F_5^n$ can be divided into one-dimensional sub-spaces (all containing the zero vector). If the zero vector is removed, we have partitioned remaining ...


2

Consider $$(1+x+x^2+x^3+\cdots)(1+x+^2+x^3+\cdots)$$ Coefficient of $x^k$ in this expansion is just $k+1.$ Because $$1.x^k+x.x^{k-1}+x^2.x^{k-2}+\cdots+x^k.1=(k+1)x^k.$$ Therefore $$(1+x+x^2+x^3+\cdots)^2=1+2x+3x^2+4x^3+\cdots.$$ Now evaluate (you can use the same procedure) $$(1-2x+x^2)(1+2x+3x^2+4x^3+\cdots)=?$$


2

"Drawing a figure" is a bit tricky for higher dimensions, but one could consider expanding an "$n-$ cube" with "edge length" of 4 . To make it a bit easier to work with, we would extend each "edge" by $ \ \frac {1}{2} \ $ in both "directions", rather than simply adding 1 . For a line segment , we have $ \ 4 \ = \ 5 \ - \ 2 \ \cdot \ \frac{1}{2} \ $ , ...


4

Look at the following for $n = 2$: $$ \bullet \bullet \bullet \bullet \diamond \\ \bullet \bullet \bullet \bullet \bullet \\ \bullet \bullet \bullet \bullet \bullet \\ \bullet \bullet \bullet \bullet \bullet \\ \bullet \bullet \bullet \bullet \bullet $$ The black dots visualize $5^2 - 1$. Now one easily sees a division of this in sets of $4$ and therefore ...


11

Using base-$5$, this is pretty obvious: $$10^n_5 - 1 = \underbrace{444\ldots4_5}_n = 4\times \underbrace{111\ldots 1_5}_n$$


11

Algebraic proof: Consider the field $K:=\mathbb{F}_{5^n}$. We shall show that the group of units $K^\times =K\setminus\left\{0\right\}$ has a subgroup $H$ of order $4$. By Lagrange's Theorem, $4$ must then divide the order of $K^\times$, which is $5^n-1$. Now, this group $H$ is given by the subset $\{1,2,3,4\}=\mathbb{F}_5\setminus\{0\}$. The result ...


15

Let there be $5$ different characters $\{\_\ , a,b,c,d\}$ to form a string of length $n$. There are $5^n$ of those strings. Apart from the string of all $\_$'s $"\_\ \_\ \_\ldots \_"$, all other strings can be grouped into exactly one of four groups in this way: Take the first character along each string that is not an $\_$, which can be one of $a,b,c$ or ...


1

$$x \to 0 , \cos x \to 1 \\\cos^m(ax) \sim 1-\frac{m*(ax)^2}{2} $$ My proof for this: Based on combining Taylor expansion of $\cos x$ and this $x \to 0$, so, $(1+x)^m \sim 1+mx$. Example for that formula: $$\lim_{x \to 0}\frac{\cos^3(5x)-\sqrt[3]{\cos(3x)}}{1-\sqrt{\cos x}}=\\\lim_{x \to ...


0

hint: $$f(x)=e^{\ln f(x)}\\ \lim_{x \to a}f(x)^{g(x)}=e^{\lim_{x \to a}\ln(f(x)){g(x)}}$$ now not that $f(x) \to 1 \\\ln f(x) \to 0\\$ $$x \to 0 \to \ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-...\\ $$ substitute x by $f(x)-1$ because $f(x) \to 1$ so $f(x)-1 \to 1-1=0$ that mean ...


0

The "Synopsis of Pure and Applied Mathematics" is precisely what it claims to be. It is an abbreviated form of other nineteenth century texts for which it provides reference lists. You are not going to understand the contents of this synopsis properly unless you read these other texts as well, all of which, being 150+years old, are freely available online ...


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...


5

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 ...


7

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) ...


6

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\\ ...


3

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 ...


2

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 & ...


4

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 ...


6

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 ...


6

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 ...


8

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 ...


4

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



Top 50 recent answers are included