# Use the table of Taylor series to find the first four non-zero terms of Taylor series for the...

## Question:

Use the table of Taylor series to find the first four non-zero terms of Taylor series for the following functions, centered at {eq}0 {/eq}.

a) {eq}f(x) = \ln(1 + 2x) {/eq}.

b) {eq}f(x) = \sin(x^{2}) {/eq}.

## Power Series:

A power series which is centered at c with {eq}\displaystyle a_m {/eq} coefficients is of the form {eq}\displaystyle \sum_{m=0}^{\infty} a_m(x-c)^m\\ {/eq}

Here are some of the useful power series which are centered at 0,

{eq}\displaystyle \sin(ax)=ax-\frac{(ax)^3}{3!}+\frac{(ax)^5}{5!}-\frac{(ax)^7}{7!}+\cdot\cdot\cdot\\ \displaystyle \frac{1}{1-ax}=1+ax+(ax)^2+(ax)^3+(ax)^4+\cdot\cdot\cdot\\ \displaystyle \frac{1}{(1-ax)^2}=1+2ax+3(ax)^2+4(ax)^3+5(ax)^4+\cdot\cdot\cdot\\ \displaystyle e^{ax}=1+(ax)+\frac{(ax)^2}{2!}+\frac{(ax)^3}{3!}+\frac{(ax)^4}{4!}+\frac{(ax)^5}{5!}+\cdot\cdot\cdot\\ \displaystyle \cos(ax)=1-\frac{(ax)^2}{2!}+\frac{(ax)^4}{4!}-\frac{(ax)^6}{6!}+\cdot\cdot\cdot\\ \displaystyle \ln(1+ax)=x-\frac{(ax)^2}{2}+\frac{(ax)^3}{3}-\frac{(ax)^4}{4}+\cdot\cdot\cdot\\ {/eq}

## Answer and Explanation:

a) The given function is {eq}\displaystyle f(x) = \ln(1 + 2x) {/eq}

The power series of {eq}\displaystyle...

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