Problem 33 - Calculus Early Transcendentals

Question

Use the geometric series $$f(x)=\frac{1}{1-x}=\sum_{k=0}^{\infty} x^{k}, \quad \text { for }|x|<1$$ to find the power series representation for the following functions (centered at 0 ). Give the interval of convergence of the new series. $$p(x)=\frac{4 x^{12}}{1-x}$$

Step-by-Step Solution

Answer

Answer: The power series representation for the function p(x) is p(x) = 4x^12 + 4x^13 + 4x^14 + 4x^15 + 4x^16 + ..., and its interval of convergence is -1 < x < 1.

Step 1: Identify the power series representation for the geometric series f(x)

Given the geometric series: $$f(x)=\frac{1}{1-x}=\sum_{k=0}^{\infty} x^{k}, \quad \text { for } |x|<1$$ This implies that the power series representation for f(x) is: $$f(x) = 1 + x + x^2 + x^3 + x^4 + ...$$

Step 2: Find the power series representation for p(x)

Now, let's look at the function p(x): $$p(x)=\frac{4 x^{12}}{1-x}$$ To find the power series representation for p(x), we need to modify the power series representation for f(x) from step 1. We can do this by multiplying each term in the series by 4 and x^12: $$p(x) = 4x^{12} (1 + x + x^2 + x^3 + x^4 + ...) = 4x^{12} + 4x^{13} + 4x^{14} + 4x^{15} + 4x^{16} + ...$$

Step 3: Find the interval of convergence for the power series representation of p(x)

Since the given geometric series f(x) converges for |x| < 1, the same will hold true for the power series representation of p(x), as the power series converges under the same conditions. Therefore, the interval of convergence for the power series representation of p(x) is: $$|x|<1 \implies -1 < x < 1$$ Thus, the power series representation for the function p(x) is: $$p(x) = 4x^{12} + 4x^{13} + 4x^{14} + 4x^{15} + 4x^{16} + ...$$ and it converges for -1 < x < 1.