Evaluation of limit involving product of cosines

Question

I am looking for alternate approaches to the question:

Evaluate: $$\lim_{x \to 0 } \frac{1 - \cos(a_1 x)\cos(a_2x)\cos(a_3x)...\cos(a_nx)}{x^2}$$

The way I did it:

Apply L Hopital's Rule:

$$=\lim_{x \to 0} \frac{a_1\sin(a_1x)\cos(a_2x)..... + a_2\cos(a_1x)\sin(a_2x)\cos(a_3x).....}{2x}$$

$$=\lim_{x \to 0} \frac{[a_1\tan(a_1x)+a_2\tan(a_2x) +....][\cos(a_1x)\cos(a_2x).....]}{2x}$$

Now the term in numerator with product of cosines will tend to go to 1, so the limit is equivalent to the limt:

$$=\lim_{x \to 0} \sum_{r=1}^n \frac{a_r\tan(a_rx)}{2x} = \sum_{r=1}^{n} \left( \lim_{x \to 0} \frac{a_r\tan(a_rx)}{2x}\right)$$

Now $ \lim_{x \to 0} \frac{\tan x}{x}$ is a standard limit equal to$1$.

$$=\frac{1}{2}\sum_{r=1}^{n} a_r^2$$

Answer 1

Using the Taylor series of $\cos x$, we could say $$\cos ax=1-{a^2x^2\over 2}+o(x^2)$$hence $$\cos a_1x\cos a_2x\cdots \cos a_nx=1-{1\over 2}x^2\sum_{i=1}^{n} a_i^2+o(x^2)$$

Answer 2

We claim that $$L_n(a_1,a_2,\dots,a_n):=\lim_{x \to 0 } \frac{1 - \cos{(a_1 x)}\cos{(a_2x)}\cdots\cos{(a_nx)}}{x^2}=\frac{1}{2}\sum_{i=1}^n a_i^2$$ for all $a_i\in\mathbb{R}$. The proof is by induction on $n$. The base case $n=1$ is clear, since $$\lim_{x\to0}\frac{1-\cos{(a_1x)}}{x^2}=\lim_{x\to0} \frac{2\sin^2{\left(\frac{a_1x}{2}\right)}}{x^2}=\frac{1}{2}a_1^2$$ for each $a_1\in\mathbb{R}$. Suppose that the claim holds for all $n\leq k\in\mathbb{N}$. It suffices to prove the claim for $n=k+1$. Indeed, $$L_{k+1}(a_1,\dots, a_{k+1})=\lim_{x \to 0 } \frac{1 - \cos{(a_{k+1} x)}}{x^2}+\lim_{x \to 0 } \cos{(a_{k+1}x)} \cdot \left(\frac{1 - \cos{(a_1 x)}\cdots\cos{(a_kx)}}{x^2}\right)$$ The first limit on the right-hand side is $\frac{1}{2}a_{k+1}^2$, and the second limit evaluates to $1\cdot\frac{1}{2}\sum_{i=1}^k a_i^2$ by induction hypothesis (note that the second limit is a limit of a product of functions; since the limit of each of the functions exists, we can write this as the product of the limits).

This finishes the proof.

Answer 3

We can go very far beyond the limit composing Taylor series.

Consider $$P_n=\prod_{i=1}^n \cos(a_i x)\implies \log(P_n)=\sum_{i=1}^n \log(\cos(a_i x))$$ $$\log(\cos(a_i x))=-\frac{a_i^2}{2} x^2-\frac{a_i^4}{12} x^4+O\left(x^6\right)$$ $$\log(P_n)=-\frac{\sum_{i=1}^n a_i^2}{2} x^2-\frac{\sum_{i=1}^na_i^4}{12} x^4+O\left(x^6\right)$$ $$P_n=e^{\log(P_n)}=1-\frac{\sum_{i=1}^n a_i^2}{2} x^2+\frac{3 \left(\sum _{i=1}^n a_i^2\right){}^2-2 \sum _{i=1}^n a_i^4}{24}x^4+O\left(x^6\right)$$ $$\frac{1-\prod_{i=1}^n \cos(a_i x)}{x^2}=\frac{\sum_{i=1}^n a_i^2}{2}-\frac{3 \left(\sum _{i=1}^n a_i^2\right){}^2-2 \sum _{i=1}^n a_i^4}{24}x^2+O\left(x^4\right)$$