Taylor Series

**PensFan10** · November 19th, 2008, 11:30 AM

Find the Taylor Series for (im not asking you to do it. i just want to make sure that what i want to do will work)
Sin4x and e^x+e^-x

For Sin 4x can I use the sinx Taylor series and multiply each component by 4 and get:
4x- (4x^3)/3! + (4x^5)/5! ...

and for e^x+e^-x can i just find 2 separate series (one for e^x and one for e^-x) and then just add them together?

**Mathstud28** · November 19th, 2008, 12:26 PM

Originally Posted by PensFan10

Find the Taylor Series for (im not asking you to do it. i just want to make sure that what i want to do will work)
Sin4x and e^x+e^-x

For Sin 4x can I use the sinx Taylor series and multiply each component by 4 and get:
4x- (4x^3)/3! + (4x^5)/5! ...

and for e^x+e^-x can i just find 2 separate series (one for e^x and one for e^-x) and then just add them together?

$\begin{aligned}\forall{x}\in\mathbb{R}~\sin(x)&=\s um_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}\\ &\implies\forall{x}\in\mathbb{R}~\sin(4x)=\sum_{n= 0}^{\infty}\frac{(-1)^n(4x)^{2n+1}}{(2n+1)!}\\ &=4\sum_{n=0}^{\infty}\frac{(-1)^n16^nx^{2n+1}}{(2n+1)!} \end{aligned}$

Yes, there are two approaches. The first is to just note that

$\forall{x}\in\mathbb{R}~e^{x}=\sum_{n=0}^{\infty}\ frac{x^n}{n!}$

And

$\begin{aligned}\forall{x}\in\mathbb{R}~e^{-x}&=\sum_{n=0}^{\infty}\frac{(-x)^n}{n!}\\ &=\sum_{n=0}^{\infty}\frac{(-1)^nx^n}{n!} \end{aligned}$

So

$\begin{aligned}\forall{x}\in\mathbb{R}\cap\mathbb{ R}=\mathbb{R}~e^{x}+e^{-x}&=\sum_{n=0}^{\infty}\frac{x^n}{n!}+\sum_{n=0}^{ \infty}\frac{(-1)^nx^n}{n!}\\ &=\sum_{n=0}^{\infty}\frac{\left(1+(-1)^n\right)x^n}{n!} \end{aligned}$

And realize that
$\left(1+(-1)^n\right) = \left\{ \begin{array}{rcl} 2 & \mbox{if} & n\in2\mathbb{N} \\ 0 & \mbox{if} & n\in2\mathbb{N}+1 \end{array}\right.$

So

$\begin{aligned}\sum_{n=0}^{\infty}\frac{\left(1+(-1)^n\right)x^n}{n!}&=\sum_{n=1,3,5\cdots}^{\infty} \frac{\left(1+(-1)^n\right)x^n}{n!}+\sum_{n=0,2,4,6\cdots}^{\infty }\frac{\left(1+(-1)^n\right)x^n}{n!}\\ &=0+2\sum_{n=0,2,4,6\cdots}^{\infty}\frac{x^n}{n!} \\ &=2\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!} \end{aligned}$

So we can conclude that

$\forall{x}\in\mathbb{R}~e^x+e^{-x}=2\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}$

The other easier approach is to see that

$\begin{aligned}e^x+e^{-x}&=2\cosh(x)\\ &=2\cos(ix)\\ &=2\sum_{n=0}^{\infty}\frac{(-1)^n(ix)^{2n}}{(2n)!}\\ &=2\sum_{n=0}^{\infty}\frac{(-1)^n\left(i^2\right)^nx^{2n}}{(2n)!}\\ &=2\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!} \end{aligned}$

**PensFan10** · November 19th, 2008, 06:52 PM

For the Sin4x i got the same thing you got until the last part. I do not know where the 16^n came from.

Then for the e^x+e^-x i got the same thing for the end result. Is your "realize that" part just the place where it converges? If so, did you use the ratio test to find that?

**Mathstud28** · November 19th, 2008, 06:57 PM

Originally Posted by PensFan10

For the Sin4x i got the same thing you got until the last part. I do not know where the 16^n came from.

$4^{2n}=16^n$

Then for the e^x+e^-x i got the same thing for the end result. Is your "realize that" part just the place where it converges? If so, did you use the ratio test to find that?

Do you mean $ \left(1+(-1)^n\right) = \left\{ \begin{array}{rcl} 2 & \mbox{if} & n\in2\mathbb{N} \\ 0 & \mbox{if} & n\in2\mathbb{N}+1 \end{array}\right.$ ? If so I was not talking about where it converges but what its behavior was when n was odd or even.

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