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Chapter 10: Problem 59

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

Short Answer

Expert verified
Question: Identify the function represented by the following power series: $$\sum_{k=1}^{\infty} \frac{x^{k}}{k}$$ Answer: The function represented by the given power series is $$f(x) = C - \frac{1}{1-x}$$ where \(C\) is a constant.

Step by step solution

01

Identify and write down the geometric series formula

Recall the geometric series formula: $$\sum_{k=0}^{\infty} x^{k} = \frac{1}{1-x}, \quad |x|<1$$
02

Derive the given power series

We will now derive the geometric series formula with respect to \(x\). Taking the derivative of \(\frac{1}{1-x}\) with respect to \(x\): $$\frac{d}{dx}\left(\frac{1}{1-x}\right) = \frac{1}{(1-x)^{2}}$$ Now, differentiating the power sum term-by-term and incrementing our starting index from \(k=0\) to \(k=1\) (noting that when \(k=0\), our term is simply a constant and will be eliminated when taking the derivative): $$\sum_{k=1}^{\infty} kx^{k-1} =\frac{1}{(1-x)^{2}}$$
03

Integrate both sides with respect to x

Now, in order to identify the given power series, we need to integrate both sides of the equation with respect to \(x\). By doing so, we will get rid off the \(k\) in the numerator on the left side, and obtain the series given in the problem. $$\int\left(\sum_{k=1}^{\infty} kx^{k-1}\right)dx = \int\left(\frac{1}{(1-x)^{2}}\right)dx$$
04

Perform the integration

Now, we will integrate the terms one by one. On the left side, integrate term-by-term: $$\sum_{k=1}^{\infty} \frac{x^{k}}{k} + C_1$$ On the right side: $$-\frac{1}{1-x} + C_2$$ where \(C_1\) and \(C_2\) are constants of integration.
05

Identify the function represented by the power series

Combining both sides and equating the constants of integration, $$\sum_{k=1}^{\infty} \frac{x^{k}}{k} = C - \frac{1}{1-x}$$ where \(C = C_2 - C_1\). Thus, the function represented by the given power series is $$f(x) = C - \frac{1}{1-x}$$

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Most popular questions from this chapter

An essential function in statistics and the study of the normal distribution is the error function $$\operatorname{erf}(x)=\frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-t^{2}} d t$$ a. Compute the derivative of erf \((x)\) b. Expand \(e^{-t^{2}}\) in a Maclaurin series, then integrate to find the first four nonzero terms of the Maclaurin series for erf. c. Use the polynomial in part (b) to approximate erf (0.15) and erf ( -0.09 ). d. Estimate the error in the approximations of part (c).

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 \(x\)

The period of a pendulum is given by $$T=4 \sqrt{\frac{\ell}{g}} \int_{0}^{\pi / 2} \frac{d \theta}{\sqrt{1-k^{2} \sin ^{2} \theta}}=4 \sqrt{\frac{\ell}{g}} F(k)$$ where \(\ell\) is the length of the pendulum, \(g \approx 9.8 \mathrm{m} / \mathrm{s}^{2}\) is the acceleration due to gravity, \(k=\sin \left(\theta_{0} / 2\right),\) and \(\theta_{0}\) is the initial angular displacement of the pendulum (in radians). The integral in this formula \(F(k)\) is called an elliptic integral and it cannot be evaluated analytically. a. Approximate \(F(0.1)\) by expanding the integrand in a Taylor (binomial) series and integrating term by term. b. How many terms of the Taylor series do you suggest using to obtain an approximation to \(F(0.1)\) with an error less than \(10^{-3} ?\) c. Would you expect to use fewer or more terms (than in part (b)) to approximate \(F(0.2)\) to the same accuracy? Explain.

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

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.
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