# Taylor series

by arl146
Tags: series, taylor
 P: 343 Find the Taylor series for f(x) centered at a. Assume power series expansion. f(x) = sin(x) at a= pi/2 My answer ended up being the summation from n=0 to infinity of ((x-pi/2)^n ) / n! And the radius of convergence is R=infinity This is right ?
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,569 No, that is not right. (In fact it looks like a series for $e^{x-\pi/2}$. If you show how you got that, we might be able to point out an error.
 P: 343 Oh yea I followed the example in the book for e^x that's probly why So where do I start then? All I know about Taylor series stuff is: F(x)= summation [ (f^n)(a)*(x-a)^n ] / n! Ok well when I followed it in the book you have (f^n)(pi/2) = sin(pi/2) and then you put the a value which is pi/2 into the definition of a Taylor series so you get: Summation [ sin(pi/2)(x-pi/2)^n ] / n!
Mentor
P: 21,321
Taylor series

 Quote by arl146 Oh yea I followed the example in the book for e^x that's probly why So where do I start then? All I know about Taylor series stuff is: F(x)= summation [ (f^n)(a)*(x-a)^n ] / n! Ok well when I followed it in the book you have (f^n)(pi/2) = sin(pi/2)
This is not true.
f(x) = sin(x), f(pi/2) = 1
f'(x) = cos(x), f'(pi/2) = 0
f''(x) = -sin(x), f''(pi/2) = -1
etc.
 Quote by arl146 and then you put the a value which is pi/2 into the definition of a Taylor series so you get: Summation [ sin(pi/2)(x-pi/2)^n ] / n!
 P: 343 Ok so how does it go (I don't have my book with me) It's like f(x) = f'(x)*(x-a) + f''(x)*(x-a)^2 + f'''(x)*(x-a)^3 + .... With the first part has 1! Under it, the second part has 2! Under it, and the third part had 4! Under it?? So you just plug in that stuff? Idk
 Mentor P: 21,321 It goes like this: f(x) = f(a) + f'(a) * (x - a)/1! + f''(a) * (x - a)2/2! + f'''(a) * (x - a)3/3! + ... + f(n)(a) * (x - a)n/n! + ...
 P: 343 Oh yea I think that's what I meant! So when you plug in values you get 1+0-1(((x-pi/2)^3)/2)+0+... right? So it has (-1)^n. I don't really know what else the series would be... Umm I get this: Summation n=0 to infinity: (-1)^n * ((x-pi/2)^(n+2)) / (n+2)! Does that work?
Mentor
P: 21,321
 Quote by arl146 Oh yea I think that's what I meant! So when you plug in values you get 1+0-1(((x-pi/2)^3)/2)+0+... right?
No, your exponent on the third term is wrong, and you should show more terms. With too few terms, you won't be able to get the general term correctly.

Make a table with the values for n, f(n)(x), and f(n)($\pi/2$), similar to the one I did in post #4, but with more entries than I showed.
 Quote by arl146 So it has (-1)^n. I don't really know what else the series would be... Umm I get this: Summation n=0 to infinity: (-1)^n * ((x-pi/2)^(n+2)) / (n+2)! Does that work?
No.
 P: 343 oh, yea i meant this: 1+0-1(((x-pi/2)^2)/2)+0+... i just had a typo. but okay, i did all the way up to f'''''''(x) and when you plug in the a value (pi/2) you get" 1, 0, -1, 0, 1, 0, -1, 0 i ended up getting: f(x) = 1 - (x-pi/2)2 / 2! + (x-pi/2)4 / 4! + (x-pi/2)6 / 6! so i dont know what the series would be. all i know is (-1)n should be there.
Mentor
P: 21,321
 Quote by arl146 oh, yea i meant this: 1+0-1(((x-pi/2)^2)/2)+0+... i just had a typo. but okay, i did all the way up to f'''''''(x) and when you plug in the a value (pi/2) you get" 1, 0, -1, 0, 1, 0, -1, 0 i ended up getting: f(x) = 1 - (x-pi/2)2 / 2! + (x-pi/2)4 / 4! + (x-pi/2)6 / 6! so i dont know what the series would be. all i know is (-1)n should be there.
That is the series (sort of), if you make a couple of corrections.
1) You have a sign error on the last term you show.
2) This is supposed to be an infinite series, so there is no last term. To indicate this, add "+ ..." after the last term you show.

With these corrections, that is your Taylor series for f(x) = sin(x). Does the series have to be shown in closed form (i.e., as a summation)? If so, you need to be able to identify the general term of the series.

Some things to note:
Only even degree terms show up. IOW, the exponents are 0, 2, 4, 6, 8, ... This means that the exponent should be 2n.
The terms alternate in sign. You can show this by having a factor of (-1)n or possibly (-1)n + 1.
 P: 343 Yes I realize those corrections. Typo and the second is just me typing slower than im thinking. But it doesn't say if it had to be in closed form but I think that's how all the answers are so yea. Ok so te answer is just summation (-1)^n * ((x-pi/2)^(2n))/(2n)! Right?
 P: 343 Is my series in the last post at the end the correct answer?
Emeritus