Infinite series involving 'x' has a constant value

SatyaDas · Feb 21, 2022

How to prove that \[ \sum_{i=1}^{\infty}\frac{1}{2^{3i}}\left(\csc^{2}\left(\frac{\pi x}{2^{i}}\right)+1\right)\sec^{2}\left(\frac{\pi x}{2^{i}}\right)\sin^{2}\left(\pi x\right)=1 \] for all $ x\in\mathbb{R} $.
Using graph, we can see that the value of this series is 1 for all values of x.
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Opalg · Feb 23, 2022

Use the result $\sin x = 2\sin\frac x2\cos\frac x2$ repeatedly to see that $$\begin{aligned}\sin^2(\pi x) &= 2^2\sin^2\left(\frac{\pi x}2\right)\cos^2\left(\frac{\pi x}2\right) \\ &= 2^4\sin^2\left(\frac{\pi x}4\right)\cos^2\left(\frac{\pi x}4\right)\cos^2\left(\frac{\pi x}2\right) \\ & \vdots \\ &= 2^{2k}\sin^2\left(\frac{\pi x}{2^k}\right) \prod_{i=1}^k \cos^2\left(\frac{\pi x}{2^i}\right) .\end{aligned}$$ You can then write the $k$th term of the series as $$\frac1{2^{3k}}\left(\csc^{2}\left(\frac{\pi x}{2^{k}}\right)+1\right)\sec^{2}\left(\frac{\pi x}{2^{k}}\right)\sin^{2}\left(\pi x\right) = \frac1{2^k} \left(1 + \sin^{2}\left(\frac{\pi x}{2^{k}}\right) \right)\prod_{i=1}^{k-1} \cos^2\left(\frac{\pi x}{2^i}\right) .$$ Using that, you should be able to show by induction that the sum of the first $n$ terms of the series is $$S_n(x) = \sum_{k=1}^{n}\frac{1}{2^{3k}}\left(\csc^{2}\left(\frac{\pi x}{2^{k}}\right)+1\right)\sec^{2}\left(\frac{\pi x}{2^{k}}\right)\sin^{2}\left(\pi x\right) = 1 - \frac1{2^n}\prod_{k=1}^n\cos^2\left(\frac{\pi x}{2^k}\right) .$$ That product of cosines lies between $0$ and $1$. It follows that $S_n(x)$ lies between $1 - \dfrac1{2^n}$ and $1$ for all $x$. Thus $S_n(x)$ converges (uniformly) to the constant function $1$.

SatyaDas · Feb 23, 2022

Great. Thanks.
How did you find the value of $ S_{n} $?

Opalg · Feb 24, 2022

Satya said:

How did you find the value of $ S_{n} $?

It's obvious from the graph at #1 that $S_n(x)$ increases very rapidly to $1$ as $n$ increases. So it seemed helpful to write $S_n(x)$ in the form $1$ - ?. I found that $$S_1(x) = 1 - \frac12\cos^2\left(\frac{\pi x}2\right),$$ $$S_2(x) = 1 - \frac14\cos^2\left(\frac{\pi x}4\right)\cos^2\left(\frac{\pi x}2\right),$$ $$S_3(x) = 1 - \frac18\cos^2\left(\frac{\pi x}8\right)\cos^2\left(\frac{\pi x}4\right)\cos^2\left(\frac{\pi x}2\right).$$ From there, it was easy enough to guess the general formula.

Infinite series involving 'x' has a constant value

1. What is an infinite series involving 'x'?

2. How is the constant value of an infinite series involving 'x' determined?

3. Can an infinite series involving 'x' have multiple constant values?

4. How is the convergence of an infinite series involving 'x' determined?

5. What are some real-world applications of infinite series involving 'x'?

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