Question

Taylor Series Question! f(x) = e^(-x)+sinx -1; I'm trying to propose a reformation using taylor series to clear up cancelation near x = 0. Please help!

Answer 1

I don't understand why my answer to finding the series should be f(x) = [1+x + x^2/2!....]; I get the rest of the summation, but don't get where I should be getting "1+"

Answer 2

B/c I know "x" comes from ((-1)^(1) * (x)(1))/1!

Answer 3

I'm not sure I understand the question... Do you want to find the Taylor series of \(f(x)=e^{-x}+\sin x-1\) ?

Answer 4

Yes, and I know some of the general summation which make solving it easier, like e^(-x) = SUM ((-1)^(n) * (x)^(n))/n!

Answer 5

But I'm not getting my answer right.

Answer 6

First off, I do know that the function is terminal at n =5, and then it starts to repeat. So I know I'm going to have a remainder term at n = 5, so I'm solving the taylor series for only up to that point. And I also know I have to solve e^(-x) separately and add it to the taylor series part of sinx, and then -1, in a string (I hope this makes since). But I don't know what I 'm doing that makes my answer wrong.

Answer 7

\[e^{x}=\sum_{n=0}^\infty\frac{x^n}{n!}\]
So
\[e^{-x}=\sum_{n=0}^\infty\frac{(-x)^n}{n!}\]
Also,
\[\sin x=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)!}\]
So you have
\[f(x)=\left(1-x+\frac{x^2}{2}-\frac{x^3}{3!}+\cdots\right)+\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\right)-1\]
Like you mentioned, all the odd powers past 5 and including 1 and 5 will disappear.
\[f(x)=\left(1+\frac{x^2}{2}-\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^6}{6!}+\cdots\right)+\left(-\frac{x^3}{3!}\right)-1\]
\[f(x)=\frac{x^2}{2}-\frac{2x^3}{3!}+\frac{x^4}{4!}+\frac{x^6}{6!}+\cdots\]

Answer 8

I'm not seeing a compact version of this, except for the even powers...

Answer 9

Thanks so much for your reply! So, at n = 0, I'm doing -x^(0)/(0!), correct? Why am I not getting 1?

Answer 10

But \(\dfrac{(-x)^0}{0!}=\dfrac{1}{1}=1\)

Answer 11

That's my confusion. What do I put into x?

Answer 12

I know e^(-0) = 1. But I thought I had to use the sum? Sorry, this may be a dumb questions, but I haven't looked at calc 2 for about 3 years...

Answer 13

It doesn't matter what \(x\) is. You only have to worry about the index \(n\). Notice how we keep the \(x\)'s in all the other terms?

Answer 14

Correction: \(x\) can't be 0, since \(0^0\) is indeterminate, but like I said, we're only worrying about the \(n\).

Answer 15

Oh, so we treat it as 1 * x? So I can compute -x^(0) as -1^(0)/0!...no, that doesn't give me the same answer

Answer 16

What is the answer you should be getting?

Answer 17

I should get [1-x+x^2/2!....Ox^5]...(I understand the sinx part, I think). The O I think stands for my error/remainder part.

Answer 18

Instead of O(X^5), I wrote (-\xi)^(5)/5!, which I think means the same thing.

Answer 19

Hmm, my understanding of big O notation is rusty... Wolfram gives an answer different from that one.

Answer 20

That's okay. I just need understanding about the first term. For my e^(-x), my first term is (-x)^(0)/0! How do I know how to calculate that? Or for the sinx term:" -x/1!---I'm supposed to get x out of that one, but I have -x. I don't know know what I'm doing wrong.