a. find the first four nonzero terms of the maclaurin series for the given function. b. write the power…
a. find the first four nonzero terms of the maclaurin series for the given function. b. write the power series using summation notation. c. determine the interval of convergence of the series. f(x)=log₂(1 + 4x)
Answer
Explanation:
Step1: Recall the Maclaurin series formula
The Maclaurin series of $\ln(1 + u)=\sum_{n = 1}^{\infty}\frac{(- 1)^{n+1}}{n}u^{n}=u-\frac{u^{2}}{2}+\frac{u^{3}}{3}-\frac{u^{4}}{4}+\cdots$, for $|u|\lt1$. We know that $\log_{2}(1 + 4x)=\frac{\ln(1 + 4x)}{\ln2}$. Let $u = 4x$.
Step2: Find the first - four non - zero terms
Substitute $u = 4x$ into the Maclaurin series of $\ln(1 + u)$ and then divide by $\ln2$. [ \begin{align*} \log_{2}(1 + 4x)&=\frac{1}{\ln2}\sum_{n = 1}^{\infty}\frac{(-1)^{n + 1}}{n}(4x)^{n}\ &=\frac{1}{\ln2}\left(4x-\frac{(4x)^{2}}{2}+\frac{(4x)^{3}}{3}-\frac{(4x)^{4}}{4}+\cdots\right)\ &=\frac{4x}{\ln2}-\frac{16x^{2}}{2\ln2}+\frac{64x^{3}}{3\ln2}-\frac{256x^{4}}{4\ln2}+\cdots\ &=\frac{4x}{\ln2}-\frac{8x^{2}}{\ln2}+\frac{64x^{3}}{3\ln2}-\frac{64x^{4}}{\ln2}+\cdots \end{align*} ]
Step3: Write the power series using summation notation
The power series of $\log_{2}(1 + 4x)$ is $\sum_{n = 1}^{\infty}\frac{(-1)^{n + 1}4^{n}x^{n}}{n\ln2}$.
Step4: Determine the interval of convergence
We know that the series for $\ln(1 + u)$ converges for $|u|\lt1$. Since $u = 4x$, we set $|4x|\lt1$. Solving for $x$, we get $|x|\lt\frac{1}{4}$. When $x=\frac{1}{4}$, the series becomes $\sum_{n = 1}^{\infty}\frac{(-1)^{n + 1}4^{n}(\frac{1}{4})^{n}}{n\ln2}=\sum_{n = 1}^{\infty}\frac{(-1)^{n + 1}}{n\ln2}$, which is a convergent alternating harmonic - type series. When $x=-\frac{1}{4}$, the series becomes $\sum_{n = 1}^{\infty}\frac{(-1)^{n + 1}4^{n}(-\frac{1}{4})^{n}}{n\ln2}=\sum_{n = 1}^{\infty}\frac{1}{n\ln2}$, which is a divergent harmonic - type series. So the interval of convergence is $(-\frac{1}{4},\frac{1}{4}]$.
Answer:
a. The first four non - zero terms are $\frac{4x}{\ln2}-\frac{8x^{2}}{\ln2}+\frac{64x^{3}}{3\ln2}-\frac{64x^{4}}{\ln2}$. b. The power series in summation notation is $\sum_{n = 1}^{\infty}\frac{(-1)^{n + 1}4^{n}x^{n}}{n\ln2}$. c. The interval of convergence is $(-\frac{1}{4},\frac{1}{4}]$.