Evaluating the Limit of an Infinite Product

[anemone]

Evaluate $\displaystyle \lim_{{n}\to{\infty}} \prod_{k=3}^{n}\left(1-\tan^4\dfrac{\pi}{2^k}\right)$.

 

[anemone]

Hint:

Spoiler

Think along the line of telescoping product...:)

 

[jacobi1]

My solution:

Spoiler

We begin by using the identity $\tan(x) = \frac{\sin(x)}{\cos(x)}$. Substituting and remembering that $1=\frac{\cos^4(x)}{\cos^4(x)}$, the product becomes
$$\prod_{k=3}^{\infty} \frac{\cos^4 \left ( \frac{\pi}{2^k} \right ) -\sin^4 \left ( \frac{\pi}{2^k} \right ) }{\cos^4 \left ( \frac{\pi}{2^k} \right ) }.$$
Factoring, recognizing that $\cos^2(x)+\sin^2(x)= 1$, and recognizing that $\cos^2(x)-\sin^2(x) = \cos(2x)$, we obtain
$$\prod_{k=3}^{\infty} \frac{\cos \left ( \frac{\pi}{2^{k-1}} \right ) }{\cos^4 \left ( \frac{\pi}{2^k} \right ) }.$$ Since $\prod ab = \prod a \prod b$, we can split $\cos^3$ out and leave behind a telescoping product. Multiplying this out, we find that all terms cancel except for $\cos \left (\frac{\pi}{4} \right ) = \frac{\sqrt{2}}{2}$. Now, using the fact that $\prod a^n = \left (\prod a \right )^n$, we obtain
$$\frac{\sqrt{2}}{2} \left (\prod_{k=3}^{\infty} \cos \left (\frac{\pi}{2^k} \right ) \right )^{-3}.$$ Multiplying and dividing by 2 inside the cosine (and forgetting about $\frac{\sqrt{2}}{2}$ and the power of -3 for now), we have
$$\prod_{k=3}^{\infty} \cos \left ( \frac{\frac{\pi}{2}}{2^{k-1}} \right ).$$
Shifting the index by one, we get
$$\prod_{k=2}^{\infty} \cos \left ( \frac{\frac{\pi}{2}}{2^{k}} \right ).$$ The product is now almost in the form of Viete’s formula,
$$ \prod_{k=1}^{\infty} \cos \left (\frac{\theta}{2^k} \right ) = \frac{\sin \theta}{\theta}.$$
However, our index starts from k=2, and not k=1, so we must first divide by $\cos \left ( \frac{\theta}{2} \right )$ to obtain
$$\prod_{k=2}^{\infty} \cos \left ( \frac{\theta}{2^k} \right ) = \frac{\sin \theta}{\theta \cos \left ( \frac{\theta}{2} \right )}.$$ Now, using the formula with $\theta = \frac{\pi}{2},$ we obtain
$$\frac{\sin \left ( \frac{\pi}{2} \right )}{\frac{\pi}{2} \cos \left ( \frac{\pi}{4} \right )} = \frac{4}{\pi \sqrt{2}}.$$
Putting it all together,
$$ \frac{\sqrt{2}}{2} \left ( \frac{4}{\pi \sqrt{2}} \right )^{-3} = \frac{\pi^3}{32}.$$

 

[anemone]

My solution:

Spoiler

We begin by using the identity $\tan(x) = \frac{\sin(x)}{\cos(x)}$. Substituting and remembering that $1=\frac{\cos^4(x)}{\cos^4(x)}$, the product becomes
$$\prod_{k=3}^{\infty} \frac{\cos^4 \left ( \frac{\pi}{2^k} \right ) -\sin^4 \left ( \frac{\pi}{2^k} \right ) }{\cos^4 \left ( \frac{\pi}{2^k} \right ) }.$$
Factoring, recognizing that $\cos^2(x)+\sin^2(x)= 1$, and recognizing that $\cos^2(x)-\sin^2(x) = \cos(2x)$, we obtain
$$\prod_{k=3}^{\infty} \frac{\cos \left ( \frac{\pi}{2^{k-1}} \right ) }{\cos^4 \left ( \frac{\pi}{2^k} \right ) }.$$ Since $\prod ab = \prod a \prod b$, we can split $\cos^3$ out and leave behind a telescoping product. Multiplying this out, we find that all terms cancel except for $\cos \left (\frac{\pi}{4} \right ) = \frac{\sqrt{2}}{2}$. Now, using the fact that $\prod a^n = \left (\prod a \right )^n$, we obtain
$$\frac{\sqrt{2}}{2} \left (\prod_{k=3}^{\infty} \cos \left (\frac{\pi}{2^k} \right ) \right )^{-3}.$$ Multiplying and dividing by 2 inside the cosine (and forgetting about $\frac{\sqrt{2}}{2}$ and the power of -3 for now), we have
$$\prod_{k=3}^{\infty} \cos \left ( \frac{\frac{\pi}{2}}{2^{k-1}} \right ).$$
Shifting the index by one, we get
$$\prod_{k=2}^{\infty} \cos \left ( \frac{\frac{\pi}{2}}{2^{k}} \right ).$$ The product is now almost in the form of Viete’s formula,
$$ \prod_{k=1}^{\infty} \cos \left (\frac{\theta}{2^k} \right ) = \frac{\sin \theta}{\theta}.$$
However, our index starts from k=2, and not k=1, so we must first divide by $\cos \left ( \frac{\theta}{2} \right )$ to obtain
$$\prod_{k=2}^{\infty} \cos \left ( \frac{\theta}{2^k} \right ) = \frac{\sin \theta}{\theta \cos \left ( \frac{\theta}{2} \right )}.$$ Now, using the formula with $\theta = \frac{\pi}{2},$ we obtain
$$\frac{\sin \left ( \frac{\pi}{2} \right )}{\frac{\pi}{2} \cos \left ( \frac{\pi}{4} \right )} = \frac{4}{\pi \sqrt{2}}.$$
Putting it all together,
$$ \frac{\sqrt{2}}{2} \left ( \frac{4}{\pi \sqrt{2}} \right )^{-3} = \frac{\pi^3}{32}.$$

Hi jacobi,

Thanks for participating and also your so neat and great solution! :)

 

[CellCoree]

As an expert in the field of mathematics and science, I would approach this problem by first understanding the concept of an infinite product. An infinite product is a mathematical expression that involves multiplying an infinite number of terms together. In order to evaluate the limit of an infinite product, we must take the limit of each individual term and then multiply them together.

In this specific problem, we have an infinite product that involves the term $\left(1-\tan^4\dfrac{\pi}{2^k}\right)$ where $k$ ranges from 3 to $n$. As $n$ approaches infinity, the term $\dfrac{\pi}{2^k}$ becomes smaller and smaller, approaching 0. This means that the value of $\tan^4\dfrac{\pi}{2^k}$ also approaches 0, since the tangent function has a limit of 0 as its input approaches 0.

Therefore, we can rewrite the original expression as $\displaystyle \lim_{{n}\to{\infty}} \prod_{k=3}^{n}\left(1-\tan^4\dfrac{\pi}{2^k}\right) = \lim_{{n}\to{\infty}} \prod_{k=3}^{n}1 = \lim_{{n}\to{\infty}} 1^n = 1$. This means that as $n$ approaches infinity, the value of the infinite product also approaches 1.

In conclusion, the limit of this infinite product is 1. This result can also be verified by using other mathematical techniques such as logarithms or series expansion. It is important to note that the value of the infinite product may change if the range of $k$ is altered or if the terms in the product are different. Therefore, it is crucial to carefully evaluate the terms and the range of the product in order to accurately determine its limit.