Evaluating the Limit of an Infinite Product

In summary, an infinite product is a mathematical expression that involves multiplying an infinite number of terms together. To evaluate its limit, the limit of the individual terms must be taken as the number of terms approaches infinity. If the limit is a non-zero number, the product converges, while a limit of zero or non-existence results in divergence. Infinite products have various applications in science, such as modeling growth or decay in physics, biology, and finance, as well as solving problems in calculus.
  • #1
anemone
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Evaluate $\displaystyle \lim_{{n}\to{\infty}} \prod_{k=3}^{n}\left(1-\tan^4\dfrac{\pi}{2^k}\right)$.
 
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  • #2
Hint:

Think along the line of telescoping product...:)
 
  • #3
My solution:
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,
$$ \frac{\sqrt{2}}{2} \left ( \frac{4}{\pi \sqrt{2}} \right )^{-3} = \frac{\pi^3}{32}.$$
 
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  • #4
jacobi said:
My solution:
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! :)
 
  • #5


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.
 

FAQ: Evaluating the Limit of an Infinite Product

1. What is an infinite product?

An infinite product is a mathematical expression that involves multiplying an infinite number of terms together. It is similar to an infinite series, but instead of adding the terms, you are multiplying them.

2. How do you evaluate the limit of an infinite product?

To evaluate the limit of an infinite product, you need to take the limit of the individual terms as the number of terms approaches infinity. If the limit of the terms is a non-zero number, then the infinite product will converge to that number. If the limit is zero or does not exist, then the infinite product will diverge.

3. What is convergence and divergence of an infinite product?

If the limit of an infinite product is a non-zero number, then it is said to converge. This means that the product is approaching a finite value as the number of terms increases. If the limit is zero or does not exist, then the product is said to diverge. This means that the product is either approaching infinity or oscillating between positive and negative infinity as the number of terms increases.

4. Can an infinite product converge to zero?

Yes, an infinite product can converge to zero if the limit of the terms is zero. This means that the product is approaching zero as the number of terms increases. However, it is also possible for an infinite product to diverge to zero, in which case it is approaching zero but never quite reaches it.

5. What are some applications of infinite products in science?

Infinite products have many applications in science, particularly in fields that involve growth or decay. For example, they can be used in physics to model radioactive decay, in biology to model population growth, and in finance to model compound interest. They are also used in calculus to solve problems involving infinite series and improper integrals.

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