In Exercises 25 – 30, use the Taylor series given in Key Idea 32 to create the Taylor series of the given func ons.

  1. \(f(x) = cos(x^2)\)

Given that

\(cosx=\sum_{n=0}^{\infty}(-1)^n*\frac{x^{2n}}{(2n)!}\)

substitute \(x^2\) for x in the series, giving

\(cos(x^2)=\sum_{n=1}^{\infty}(-1)^n*\frac{(x^2)^{2n}}{(2n)!}= 1-\frac{x^4}{2!} + \frac{x^8}{4!} - \frac{x^12}{6!}+\frac{x^{16}}{8!}-...\)

  1. \(f(x) = e^{−x}\)

Given that

\(e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!}\)

substitute \(-x\) for \(x\) in the series, giving

\(e^{-x}=\sum_{n=0}^{\infty} \frac{(-x)^n}{n!}=1-\frac{x}{1}+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}-...\)

  1. \(f(x)=sin(2x+3)\)

Given that

\(sinx=\sum_{n=0}^{\infty}(-1)^n*\frac{x^{2n+1}}{(2n+1)!}\)

substitute \(2x+3\) for x in the series, giving

\(sin(2x+3)=\sum_{n=0}^{\infty}(-1)^n*\frac{(2x+3)^{2n+1}}{(2n+1)!} = 2x+3 -\frac{(2x+3)^3}{3!} +\frac{(2x+3)^5}{5!}-\frac{(2x+3)^7}{7!}+...\)

  1. \(f(x) = tan^{−1} (x/2)\)

Given \(tan^{−1}x=\sum_{n=0}^{\infty}(-1)^n*\frac{x^{2n+1}}{2n+1}\)

substitute \(x/2\) for x in the series, giving

\(tan^{−1}x/2=\sum_{n=0}^{\infty}(-1)^n*\frac{(x/2)^{2n+1}}{2n+1}= 1-\frac{(x/2)^3}{3} + \frac{(x/2)^5}{5!} - \frac{(x/2)^7}{7}+\frac{(x/2)^{9}}{9}-...\)

  1. \(f(x) = e^x sin x\) (only find the first 4 terms)

Let’s apply the following theorem

\(f(x)*g(x)=(\sum_{n=0}^{\infty}a_n*x^n)*(\sum_{n=0}^{\infty}b_n*x^n)\)

Given

\(e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!}\)

and

\(sinx=\sum_{n=0}^{\infty}(-1)^n*\frac{x^{2n+1}}{(2n+1)!}\)

we’ve got

\(e^x sin x=(\sum_{n=0}^{\infty} \frac{x^n}{n!})*(\sum_{n=0}^{\infty}(-1)^n*\frac{x^{2n+1}}{(2n+1)!}) = (1+x+\frac{x^2}{2}+\frac{x^3}{3}+...)*(1-\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}-...)= (1-\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}-...) + x*(1-\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}-...)+\frac{x^2}{2}*(1-\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}-...)+\frac{x^3}{3}*(1-\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}-...)+...\)

  1. \(f(x) = (1 + x)^{1/2} cos x\) (only find the first 4 terms)

Let’s apply the following theorem

\(f(x)*g(x)=(\sum_{n=0}^{\infty}a_n*x^n)*(\sum_{n=0}^{\infty}b_n*x^n)\)

Given

\((1+x)^k=\sum_{n=0}^{\infty}\frac{k*(k-1)...(k-(n-1))}{n!}x^n\)

and

\(cosx=\sum_{n=0}^{\infty}(-1)^n*\frac{x^{2n}}{(2n)!}\)

we’ve got

\((1 + x)^{1/2} cos x= (\sum_{n=0}^{\infty}\frac{\frac{1}{2}*(\frac{1}{2}-1)...(\frac{1}{2}-(n-1))}{n!}x^n)*(\sum_{n=0}^{\infty}(-1)^n*\frac{x^{2n}}{(2n)!})\)