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Proof for fun!

This post will be about several proofs of random theorems or formulas that I enjoy proving and I would like to share them with everyone. I hope that everyone could see the beauty of mathematical proofs from this post.

Formula 1

For a positive integer $n$, it always holds true that:
[\lim_{x \to 0} \frac{1-\prod_{i=1}^n cos(a_i x)}{x^2} = \frac{1}{2} \cdot \sum_{i=1}^n a_i^2]
Note for some who might not be used to the notation, $\prod$ here is just like $\sum$ but instead of summation of terms, it is the product of all the terms. For example $\prod_{i=1}^3 i = 1 \cdot 2 \cdot 3 = 6$


Simple Example:
[\lim_{x \to 0} \frac{1-cos3x \cdot cos4x \cdot cos5x}{x^2} = \frac{3^2+4^2+5^2}{2} = 25]

Proof:
In dealing with problem where there exist $n$ term products, we should try to build it from the most simple case of $n=1$, and then try to prove for $n=k+1$ using the equality from $n=k$ (similar to a domino effect). In mathematics, we call this method as induction.

First step is to proof for $n=1$. Namely, we shall prove that:
[\lim_{x \to 0} \frac{1-cos(a_1 x)}{x^2} = \frac{a_1^2}{2} ]

By using the identity: $1-cosa_1x=2sin^2(\frac{1}{2}a_1x)$, we can obtain the proof as follows.
[\lim_{x \to 0} \frac{1-cos(a_1 x)}{x^2} = \lim_{x \to 0} \frac{2sin^2(\frac{1}{2}a_1x)}{x^2}]
[= \lim_{x \to 0} 2 \cdot \frac{sin(\frac{1}{2}a_1x) \cdot sin(\frac{1}{2}a_1x)}{x \cdot x}]
[= 2 \cdot \frac{1}{2}a_1 \cdot \frac{1}{2}a_1]
[= \frac{a_1^2}{2}]

Thus, for $n=1$, the formula is proven to be true. Now, the next step of induction is to prove the formula for $n=k+1$ assuming that the formula is true for $n=k$.

Since $n=k$ is assumed to be true, we have:
 [\lim_{x \to 0} \frac{1-\prod_{i=1}^k cos(a_i x)}{x^2} = \frac{1}{2} \cdot \sum_{i=1}^k a_i^2]

Now we shall prove for $n=k+1$ using the above equality.
[\lim_{x \to 0} \frac{1-\prod_{i=1}^{k+1} cos(a_i  x)}{x^2}]
[= \lim_{x \to 0} (\frac{cos(a_{k+1} x)-\prod_{i=1}^{k+1} cos(a_i x) + 1-cos(a_{k+1}x)}{x^2})]
[= \lim_{x \to 0} \frac{cos(a_{k+1} x) \cdot (1-\prod_{i=1}^{k} cos(a_i x))}{x^2} + \lim_{x \to 0} \frac{1-cos(a_{k+1}x)}{x^2}]
[= \lim_{x \to 0}cos(a_{k+1}x) \cdot\lim_{x \to 0} \frac{ (1-\prod_{i=1}^{k} cos(a_i x))}{x^2} + \lim_{x \to 0} \frac{1-cos(a_{k+1}x)}{x^2}]
[= 1 \cdot \frac{1}{2} \cdot \sum_{i=1}^k a_i^2 + \frac{1-cos(a_{k+1}x)}{x^2}]
[= \frac{1}{2} \cdot \sum_{i=1}^k a_i^2 + \frac{1}{2} \cdot a_{k+1}^2]
[= \frac{1}{2} \cdot \sum_{i=1}^{k+1} a_i^2]

Hence, the proof is complete and the formula is proven. QED.
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