# Taylor Series : How to determine coefficient

1. Sep 4, 2011

### cyt91

1. The problem statement, all variables and given/known data
https://skydrive.live.com/?cid=6b041751c72e14ad#!/?cid=6b041751c72e14ad&sc=photos&uc=3&id=6B041751C72E14AD%21149!cid=6B041751C72E14AD&id=6B041751C72E14AD%21154&sc=photos

3. The attempt at a solution

https://skydrive.live.com/?cid=6b041751c72e14ad#!/?cid=6b041751c72e14ad&sc=photos&uc=3&id=6B041751C72E14AD%21149!cid=6B041751C72E14AD&id=6B041751C72E14AD%21155&sc=photos

https://skydrive.live.com/?cid=6b041751c72e14ad#!/?cid=6b041751c72e14ad&sc=photos&uc=3&id=6B041751C72E14AD%21149!cid=6B041751C72E14AD&id=6B041751C72E14AD%21156&sc=photos

Please view the images posted(right-click,view image if not loaded properly) for the question and my attempts at the solution.
How do we determine the coefficient of c18,c19 and c20 respectively?
Thank you for your help.

2. Sep 4, 2011

### HallsofIvy

Staff Emeritus
It is impossible to see your images, even by "right clicking".

3. Sep 5, 2011

### cyt91

I've tried loading the pictures(by right-clicking and selecting open in new tab). They loaded just fine.

Sorry,I'm not that good in Latex.

4. Sep 5, 2011

### EsponV

I think the issue is that we can't find the link to click on in the first place. Nothing in your original statement contains a hyperlink - it's all just plaintext. Unless I'm starting to go blind.

5. Sep 5, 2011

### lanedance

how about using tex?

right click on code below to see how its written
$$f(x) = \Sum_n\frac{1}{n!}\frac{d^{(n)}f(0)}}{dx}x^n$$
[/tex]

you'll generally get a quicker answer, and its easier for people to cut/paste&edit your original post

6. Sep 5, 2011

### Tomer

Hi cyt91,

well, I believe you did most of the job. Try to write the actual sum you just found, element by element... You can easily see what the coefficients of x18, x19 and x20 are...

7. Sep 5, 2011

### vela

Staff Emeritus
Fixed your post.

8. Sep 5, 2011

### HallsofIvy

Staff Emeritus
Okay, so you have written "$c_nx^n$" as $x^{3m}- x^{3k- 1}$. To find the coefficient of $x^{18}$, say, you need to find m so that 3m= 18 and k such that 3k- 1= 18. If there is no such integer, the coefficient is 0. 3m= 18 gives m= 6 so there is a term "$x^{3(6)}= 18$" and it has coefficient 1. There is no integer such that 3m- 1= 18 so that adds nothing. The coeffiicent of $x^{18}$ is 1.

Now do 19 and 20.

9. Sep 5, 2011

### cyt91

Hi.

Thanks a lot for your help.
I've found c_19 = -1 and c_20 = 0.

Just one more clarification: when we are finding c_19 / c_18 /c _20 ,
we are in fact finding the 19th / 18th/ 20th order of Taylor's polynomial right?

So,for instance,if we want to look for c_18, we'll be looking at 3n = 18 (since n is an integer)
i.e. the 18th order of Taylor's polynomial.

10. Sep 5, 2011

### SammyS

Staff Emeritus
[Mod Note: Replaced large images with hyperlinks, Hoot]
This is the Original Post with the corrected Image links.

Hope that helps someone !

Last edited by a moderator: May 5, 2017
11. Sep 13, 2011

### jayman16

still dont understand how the coefficients c_18, 19, and 20 are obtained. can you put it in even simpler terms why c_18 is 1, 19 is -1 and 20 is 0?

12. Sep 13, 2011

### vela

Staff Emeritus
If you can't see it, follow Tomer's advice back in post #6.

13. Sep 13, 2011

### jayman16

i got it. thanks. would i be right in saying that the the only term with x^18 in the series would be coming from summation 0 to infinity (x^3n) when n = 6 on the right hand side. c_18x^18 = x^18 meaning c_18 = 1. Similarly the only term with x^19 on the right hand side is when n = 6 where it is -(x^3n+1) = - (x^19). Hence c_19x^19 = -(x^19) , c_19 is therefore -1 and since there are no integer values for n that can give me x^20 on the right hand side i can conclude that there is no x^20 term on the right. So c_20 must be 0 on the left removing the x^20 term on the left. Is this the reasoning?

14. Sep 13, 2011

### vela

Staff Emeritus
Yes, you got it.

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