a. find the first four nonzero terms of the maclaurin series for the given function. b. write the power…

Explanation:

Step1: Recall Maclaurin - series formula

The Maclaurin series of a function (f(x)) is given by (f(x)=\sum_{n = 0}^{\infty}\frac{f^{(n)}(0)}{n!}x^{n}), where (f^{(n)}(x)) is the (n) - th derivative of (f(x)). First, we know that (\log_{a}(u)=\frac{\ln(u)}{\ln(a)}), so (f(x)=\frac{\ln(1 + 5x)}{\ln(5)}). The Maclaurin series of (\ln(1 + t)=\sum_{n=1}^{\infty}\frac{(- 1)^{n + 1}}{n}t^{n}), for (|t|\lt1). Let (t = 5x).

Step2: Find the first four non - zero terms

Substitute (t = 5x) into the series for (\ln(1 + t)): [ \begin{align*} \ln(1+5x)&=\sum_{n = 1}^{\infty}\frac{(-1)^{n + 1}}{n}(5x)^{n}\ &=5x-\frac{(5x)^{2}}{2}+\frac{(5x)^{3}}{3}-\frac{(5x)^{4}}{4}+\cdots \end{align*} ] Then (f(x)=\frac{\ln(1 + 5x)}{\ln(5)}=\frac{1}{\ln(5)}\left(5x-\frac{25x^{2}}{2}+\frac{125x^{3}}{3}-\frac{625x^{4}}{4}+\cdots\right)). The first four non - zero terms are (\frac{5}{\ln(5)}x-\frac{25}{2\ln(5)}x^{2}+\frac{125}{3\ln(5)}x^{3}-\frac{625}{4\ln(5)}x^{4}).

Step3: Write the power series in summation notation

[f(x)=\frac{1}{\ln(5)}\sum_{n = 1}^{\infty}\frac{(-1)^{n + 1}5^{n}}{n}x^{n}]

Step4: Determine the interval of convergence

We know that the series for (\ln(1 + t)) converges for (|t|\lt1). Since (t = 5x), we set (|5x|\lt1). Solving for (x), we get (|x|\lt\frac{1}{5}), or (-\frac{1}{5}\lt x\lt\frac{1}{5}). We need to check the endpoints:

• When (x=\frac{1}{5}), the series becomes (\frac{1}{\ln(5)}\sum_{n = 1}^{\infty}\frac{(-1)^{n + 1}5^{n}}{n}\left(\frac{1}{5}\right)^{n}=\frac{1}{\ln(5)}\sum_{n = 1}^{\infty}\frac{(-1)^{n + 1}}{n}), which is the alternating harmonic series and converges.
• When (x=-\frac{1}{5}), the series becomes (\frac{1}{\ln(5)}\sum_{n = 1}^{\infty}\frac{(-1)^{n + 1}5^{n}}{n}\left(-\frac{1}{5}\right)^{n}=\frac{1}{\ln(5)}\sum_{n = 1}^{\infty}\frac{-1}{n}), which is the negative of the harmonic series and diverges.

So the interval of convergence is (\left(-\frac{1}{5},\frac{1}{5}\right]).

Answer:

a. (\frac{5}{\ln(5)}x-\frac{25}{2\ln(5)}x^{2}+\frac{125}{3\ln(5)}x^{3}-\frac{625}{4\ln(5)}x^{4}) b. (\frac{1}{\ln(5)}\sum_{n = 1}^{\infty}\frac{(-1)^{n + 1}5^{n}}{n}x^{n}) c. (\left(-\frac{1}{5},\frac{1}{5}\right])