Problem 57 - Calculus Early Transcendentals

Question

Identify the functions represented by the following power series. $$\sum_{k=0}^{\infty}(-1)^{k} \frac{x^{2 k}}{4^{k}}$$

Step-by-Step Solution

Answer

Question: Identify the function represented by the given power series \(\sum_{k=0}^{\infty}(-1)^{k} \frac{x^{2 k}}{4^{k}}\). Answer: The function represented by the given power series is \(\cos\left(\frac{x}{2}\right)\).

Step 1: Identify the form of the power series

The given power series is in the following form: $$\sum_{k=0}^{\infty}(-1)^{k} \frac{x^{2 k}}{4^{k}}$$ Notice that this series has alternating signs, which can be written as \((-1)^k\). The quadratic power of x in the numerator is \(x^{2k}\). The power of 4 in the denominator is \(4^k\).

Step 2: Rewrite the power series

We can rewrite the power series to see if it can be written in terms of a well-known function: $$\sum_{k=0}^{\infty}(-1)^{k} \frac{x^{2 k}}{4^{k}} = \sum_{k=0}^{\infty}(-1)^{k} \frac{x^{2 k}}{(2^2)^{k}} = \sum_{k=0}^{\infty}(-1)^{k} \frac{x^{2 k}}{2^{2k}}$$

Step 3: Compare with known power series

The power series representation of the cosine function is: $$\cos(x) = \sum_{k=0}^{\infty}(-1)^{k} \frac{x^{2 k}}{(2k)!}$$ Comparing this with our rewritten power series, we can see that they have a similar form. They have alternating signs, a quadratic power of x, but the denominator is different. If we can rewrite the denominator in our power series to be in the form of the cosine function, we can identify the function.

Step 4: Rewrite the denominator

Since the denominator in our power series is a power of 2, we can divide the whole power series by a constant factor to make it appear similar to the cosine function: $$\frac{1}{2^0} \sum_{k=0}^{\infty}(-1)^{k} \frac{x^{2 k}}{2^{2k}} = \sum_{k=0}^{\infty}(-1)^{k} \frac{x^{2 k}}{(2k)!}$$ Now the power series resembles the cosine function with an argument of \(x/2\): $$\cos\left(\frac{x}{2}\right) = \sum_{k=0}^{\infty}(-1)^{k} \frac{\left(\frac{x}{2}\right)^{2 k}}{(2k)!}$$

Step 5: Identify the function

By comparing the power series, we can now identify the function represented by the given power series as: $$\sum_{k=0}^{\infty}(-1)^{k} \frac{x^{2 k}}{4^{k}} = \cos\left(\frac{x}{2}\right)$$