# find the infinite series for the following and simplify. (This means to separate (-1) to a power,...

## Question:

find the infinite series for the following and simplify. (This means to separate {eq}(-1){/eq} to a power, put powers of x together, and cancel if possible.)

Use multiplication, division, composition. Simplify by removing parenthesis. {eq}G(x) = x^3 cos(3 \sqrt{x}){/eq}

{eq}f(x) = \frac{e^-4x^2}{x}{/eq}

{eq}\int \frac{sin x}{x} dx{/eq} has no elementary antiderivative. Use a power series to integrate.

## Power Series:

A power series that is centered at c with {eq}\displaystyle p_n {/eq} as coefficients is of the form {eq}\displaystyle \sum_{n=0}^{\infty} p_n(x-c)^n\\ {/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 \arcsin(ax)=ax+\frac{(ax)^3}{6}+\frac{3(ax)^5}{40}+\frac{5(ax)^7}{112}+\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 \tan^{-1} (ax) = ax - \frac{(ax)^3}{3} + \frac{(ax)^5}{5} - \frac{(ax)^7}{7} + \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)=ax-\frac{(ax)^2}{2}+\frac{(ax)^3}{3}-\frac{(ax)^4}{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\\ {/eq}

## Answer and Explanation:

{eq}\displaystyle \eqalign{ & G(x) = {x^3}cos(3\sqrt x ) \cr & {\text{The infinite series for }}\cos x; \cr & \cos x = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{x^{2n}}}}{{2n!}}} \cr & {\text{The infinite series for }}\cos (3\sqrt x ); \cr & \cos (3\sqrt x ) = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{{\left( {3\sqrt x } \right)}^{2n}}}}{{2n!}}} \cr & \cos (3\sqrt x ) = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{{\left( 9 \right)}^n}{{\left( x \right)}^n}}}{{2n!}}} \cr & \cr & G(x) = {x^3}cos(3\sqrt x ) = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{{\left( 9 \right)}^n}{{\left( x \right)}^{n + 3}}}}{{2n!}}} \cr & \cr & f(x) = \frac{{{e^{ - 4{x^2}}}}}{x} \cr & {\text{The infinite series for }}{e^x}; \cr & {e^x} = \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}} \cr & {\text{The infinite series for }}{e^{ - 4{x^2}}}; \cr & {e^{ - 4{x^2}}} = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 4{x^2}} \right)}^n}}}{{n!}}} \cr & {e^{ - 4{x^2}}} = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{{\left( 4 \right)}^n}{x^{2n}}}}{{n!}}} \cr & \cr & f(x) = \frac{{{e^{ - 4{x^2}}}}}{x} = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{{\left( 4 \right)}^n}{x^{2n - 1}}}}{{n!}}} \cr & \cr & \cr & \int {\frac{{sinx}}{x}} dx \cr & {\text{The infinite series for }}\sin x; \cr & \sin x = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{x^{2n + 1}}}}{{\left( {2n + 1} \right)!}}} \cr & \cr & \int {\frac{{sinx}}{x}} dx \cr & \int {\sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{x^{2n}}}}{{\left( {2n + 1} \right)!}}} } dx \cr & \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{\left( {2n + 1} \right)!}}} \int {{x^{2n}}} dx \cr & \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{\left( {2n + 1} \right)!}}} \left[ {\frac{{{x^{2n + 1}}}}{{2n + 1}} + c} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\int {{x^n}} dx = \frac{{{x^{n + 1}}}}{{n + 1}} + c} \right) \cr & \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{\left( {2n + 1} \right)!}}} \frac{{{x^{2n + 1}}}}{{2n + 1}} + c\sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{\left( {2n + 1} \right)!}}} \cr} {/eq}

#### Learn more about this topic:

from GRE Math: Study Guide & Test Prep

Chapter 12 / Lesson 4