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Chapter 10: Problem 64

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

Short Answer

Expert verified
Answer: Within its interval of convergence, the given power series converges to the function: $$f(x) = x\ln(1-x) - x + C_1x + C_2$$ where \(C_1\) and \(C_2\) are constants of integration.

Step by step solution

01

Rewrite as a derivative or integral of a known power series

First, let's notice that we can rewrite the given power series as follows: $$\sum_{k=2}^{\infty} \frac{x^{k}}{k(k-1)} = \sum_{k=2}^{\infty} \frac{1}{k-1} \cdot \frac{x^{k}}{k} = \sum_{k=2}^{\infty} \frac{1}{k-1} \cdot x^{k} \cdot \frac{1}{k}$$ Now, we can rewrite the sum by changing the index and noticing a pattern from k=2 to k=n: $$\sum_{n=1}^{\infty} \frac{1}{n} \cdot x^{n+1} \cdot \frac{1}{n+1}$$ Now we can notice that it can be the second derivative of the geometric series, with respect to x (without first term of the sum): $$\sum_{n=1}^{\infty} \frac{1}{n} \cdot x^{n+1} \cdot \frac{1}{n+1} = \frac{d^2}{dx^2} \sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1}$$
02

Determine the interval of convergence

Now, we will determine the interval of convergence of the rewritten power series. For a geometric series, we know that the series converges if \(|x| < 1\). So, our interval of convergence is \((-1, 1)\).
03

Identify the known power series and find the function represented

We can recognize power series in second derivative as a geometric series without first term. The geometric series can be expressed as $$\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$$ Now, we will integrate twice to find the original series function, making sure to add the constants of integration: First integration: $$\int \frac{1}{1-x} \, dx = -\ln(1-x) + C_1$$ Second integration: $$\int -\ln(1-x) + C_1 \, dx = x\ln(1-x) - x + C_1x + C_2$$ Thus, the function represented by the given power series is: $$f(x) = x\ln(1-x) - x + C_1x + C_2$$ Within the interval of convergence, \((-1, 1)\), the given power series converges to the function \(f(x) = x\ln(1-x) - x + C_1x + C_2\).

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Most popular questions from this chapter

Use properties of power series, substitution, and factoring to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. Give the interval of convergence for the new series. Use the Taylor series. $$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^{2}}{8}+\frac{x^{3}}{16}-\cdots, \text { for }-1 < x \leq 1$$ $$\sqrt{4-16 x^{2}}$$

Determine whether the following statements are true and give an explanation or counterexample. a. The function \(f(x)=\sqrt{x}\) has a Taylor series centered at 0 . b. The function \(f(x)=\csc x\) has a Taylor series centered at \(\pi / 2\) c. If \(f\) has a Taylor series that converges only on \((-2,2),\) then \(f\left(x^{2}\right)\) has a Taylor series that also converges only on (-2,2) d. If \(p(x)\) is the Taylor series for \(f\) centered at \(0,\) then \(p(x-1)\) is the Taylor series for \(f\) centered at 1 e. The Taylor series for an even function about 0 has only even powers of \(x\)

By comparing the first four terms, show that the Maclaurin series for \(\sin ^{2} x\) can be found (a) by squaring the Maclaurin series for \(\sin x,\) (b) by using the identity \(\sin ^{2} x=(1-\cos 2 x) / 2,\) or \((\mathrm{c})\) by computing the coefficients using the definition.

Explain why the Mean Value Theorem is a special case of Taylor's Theorem.

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