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Beautiful Proofs of Some Simple Math Propositions
Автор Andycode3759, история, 3 года назад, По-английски

I came up with them all when I was in a daze during classes in school.

This post is a practice translation for helping myself get familiar to English math terms. So if I made any mistakes, it will be grateful if you could correct them in the comments. Original post (Chinese) is here: https://www.luogu.com.cn/blog/andycode3759/math-proofs

Proposition 1

Proposition: Convex n-polygon $$$(n \geq 4)$$$ has $$${{n(n-3)} \over 2}$$$ diagonals.

Proof: Draw a convex n-polygon in a plane randomly and we can get its $$$n$$$ vertices, among which we can connect any two of them to get $$$C_{n}^{2}$$$ line segments in total, in which $$$n$$$ of them are the edges of the polygon, and the rest $$$C_{n}^{2}-n={{n(n-3)} \over 2}$$$ of them are the diagonals. Proved in only one sentence!

Proposition 2

Proposition: There are two independent (thanks nor for correcting it) finite discrete random variables $$$X$$$ and $$$Y$$$ whose expectations are $$$E(X)$$$ and $$$E(Y)$$$ and variances are $$$D(X)$$$ and $$$D(Y)$$$. Then $$$E(X+Y)=E(X)+E(Y), D(X+Y)=D(X)+D(Y)$$$.

Proof:

(i) Assume the distributions of $$$X$$$ and $$$Y$$$ are $$$P(X=x_i)=p_i(1 \leq i \leq n),P(Y=y_i)=q_i(1 \leq i \leq m)$$$, then

$$$ E(X) = \sum_{i=1}^{n}p_ix_i, ~ E(Y) = \sum_{i=1}^{m}q_iy_i $$$

Obviously we have

$$$ \sum_{i=1}^{n}p_i=1, ~ \sum_{i=1}^{m}q_i=1, $$$

So

$$$ \begin{align*} E(X+Y) &= \sum_{i=1}^{n} \sum_{j=1}^{m}p_iq_j(x_i+y_j) \\ &= \sum_{i=1}^{n} p_i(x_i\sum_{j=1}^{m}q_j + \sum_{j=1}^{m}q_jy_j) \\ &= \sum_{i=1}^{n} p_i(x_i + E(Y)) \\ &= \sum_{i=1}^{n} p_ix_i + E(Y) \\ &= E(X) + E(Y) \end{align*} $$$

(ii) For convenience, take $$$dx_i$$$ as $$$x_i-E(X)$$$ and $$$dy_i$$$ as $$$y_i-E(Y)$$$. First we need a lemma:

$$$ \sum_{i=1}^{n}p_idx_i=0 $$$

(Also apply to $$$Y$$$.) The proof is simple:

$$$ \begin{align*} \sum_{i=1}^{n}p_idx_i &= \sum_{i=1}^{n}p_i(x_i-E(X)) \\ &= \sum_{i=1}^{n}p_ix_i - E(X)\sum_{i=1}^{n}p_i \\ &= E(X) - E(X) \\ &= 0 \end{align*} $$$

Since we have

$$$ D(X) = \sum_{i=1}^{n}p_i(x_i-E(X))^2 = \sum_{i=1}^{n}p_idx_i^2 \\ D(Y) = \sum_{i=1}^{n}q_i(y_i-E(Y))^2 = \sum_{i=1}^{m}q_idy_i^2 $$$

Then

$$$ \begin{align*} D(X+Y) &= \sum_{i=1}^{n} \sum_{j=1}^{m}p_iq_j(x_i+y_j-E(X+Y))^2 \\ &= \sum_{i=1}^{n} \sum_{j=1}^{m}p_iq_j(dx_i+dy_j)^2 \\ &= \sum_{i=1}^{n} p_i\sum_{j=1}^{m}(q_jdx_i^2 + 2q_jdx_idy_j + q_jdy_j^2) \\ &= \sum_{i=1}^{n} p_i (dx_i^2\sum_{j=1}^{m}q_j + 2dx_i\sum_{j=1}^{m}q_jdy_j + \sum_{j=1}^{m}q_jdy_j^2) \\ &= \sum_{i=1}^{n} p_i (dx_i^2 + 0 + D(Y)) \\ &= \sum_{i=1}^{n} p_idx_i^2 + D(Y) \\ &= D(X) + D(Y) \end{align*} $$$

Q.E.D.

Proposition 3

Proposition: $$$\forall n \in \mathbb{N}^{+}, 1^2+2^2+3^2+ \dots +n^2 = \sum_{k=1}^{n}k^2 = {1 \over 6}n(n+1)(2n+1)$$$ (The sum of squares of the first $$$n$$$ positive integers).

Proof: It is noticed that:

$$$ \begin{align*} 1^2 &= 1 \\ 2^2 &= 1+3 \\ 3^2 &= 1+3+5 \\ \dots \\ n^2 &= 1+3+5+ \dots +(2n-1) \end{align*} $$$

We can write $$$n$$$ equations like these and add them up, then the left side is $$$1^2+2^2+\dots+n^2$$$, and the right side is the sum of $$$n$$$ times $$$1$$$, $$$(n-1)$$$ times $$$3$$$, ..., $$$(n-k+1)$$$ times $$$(2k-1)$$$, ..., $$$1$$$ times $$$(2n-1)$$$. So we will get

$$$ \sum_{k=1}^{n}k^2 = \sum_{k=1}^{n}(n-k+1)(2k-1) $$$

Reduce the right side of the equation:

$$$ \begin{align*} \sum_{k=1}^{n}(n-k+1)(2k-1) &= \sum_{k=1}^{n}(2nk-2k^2+3k-n-1) \\ &= \sum_{k=1}^{n}(2nk-2k^2+3k)-n(n+1) \\ &= (2n+3)\sum_{k=1}^{n}k - 2\sum_{k=1}^{n}k^2 - n(n+1) \\ &= {n(n+1)(2n+3) \over 2} - n(n+1) - 2\sum_{k=1}^{n}k^2 \\ &= n(n+1)({2n+3 \over 2}-1) - 2\sum_{k=1}^{n}k^2 \\ &= {n(n+1)(2n+1) \over 2} - 2\sum_{k=1}^{n}k^2 \end{align*} $$$

Do the transposition:

$$$ \begin{align*} 3 \sum_{k=1}^{n}k^2 &= {n(n+1)(2n+1) \over 2} \\ \sum_{k=1}^{n}k^2 &= {n(n+1)(2n+1) \over 6} \end{align*} $$$

Q.E.D.

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3 года назад, скрыть # |
 
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I would like to point out that the statement of the second part of proposition 2 is wrong. For example, consider two identical random variables (where they are identical down to their value on each event). The variance of $$$2X$$$ is 4 times the variance of $$$X$$$.

The mistake in the proof is your assumption that the probability that $$$X = x_i$$$ and $$$Y = y_j$$$ at the same time is $$$p_i q_j$$$. Incidentally, this mistake is present in the proof of linearity of expectation as well, but since the conclusion doesn't depend on whether the random variables are independent or not, you get the correct result regardless.

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