You are Here: Home >< Maths

# Reverse this taylor series

Announcements Posted on
Four hours left to win £100 of Amazon vouchers!! Don't miss out! Take our short survey to enter 24-10-2016

1. Doing an online test, can't leave until I finish this.
I think this should be an application of reversing a Taylor series. There is no coefficient with the x, so this series should be about point x=0. And I figured out the sequences given below using the Taylor Series formula.

f(0) = 1

f'(0) = 1/8

f''(0) = 9/64

f'''(0) = 153/512

f''''(0) = 3825/4096
What I need to do next is find the equation f(x) which satisfy these sequences and I don't know how to continue it. Or this is not a Taylor series at all?
2. (Original post by Y.X.)

Doing an online test, can't leave until I finish this.
I think this should be an application of reversing a Taylor series. There is no coefficient with the x, so this series should be about point x=0. And I figured out the sequences given below using the Taylor Series formula.

f(0) = 1

f'(0) = 1/8

f''(0) = 9/64

f'''(0) = 153/512

f''''(0) = 3825/4096
What I need to do next is find the equation f(x) which satisfy these sequences and I don't know how to continue it. Or this is not a Taylor series at all?
It looks like the numerators are 8-fold factorials and the denominator is powers of 8.

i.e: - not sure if that helps.
3. Probably just simpler to look at is as:

4. I figured out the denominator to be (8^n)x(n!). As for the 8-fold factorial, I don't even know how to express that in the test lol. I think the question is asking me to write a formula to represent the sum of the infinite terms of the series.
(Original post by Zacken)
It looks like the numerators are 8-fold factorials and the denominator is powers of 8.

i.e: - not sure if that helps.
5. (Original post by Y.X.)
I figured out the denominator to be (8^n)x(n!). As for the 8-fold factorial, I don't even know how to express that in the test lol. I think the question is asking me to write a formula to represent the sum of the infinite terms of the series.
Your series will converge to (1 - x) to the power of minus 1/8, valid for -1 <x<1
6. (Original post by TeeEm)
Your series will converge to (1 - x) to the power of minus 1/8, valid for -1 <x<1
Urgh, how did I not spot that.
7. I checked this myself and it matched all the terms, thank you so much. Just can't find this type of question anywhere, maybe I need to search for conversion instead of Taylor series.
(Original post by TeeEm)
Your series will converge to (1 - x) to the power of minus 1/8, valid for -1 <x<1
8. (Original post by Y.X.)
I checked this myself and it matched all the terms, thank you so much. Just can't find this type of question anywhere.
you would in my undergrad resources under series, but it looks that someone very keen took them down while I was away.
all the best
9. (Original post by TeeEm)
Your series will converge to (1 - x) to the power of minus 1/8, valid for -1 <x<1

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: May 3, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

### Who is getting a uni offer this half term?

Find out which unis are hot off the mark here

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read here first

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams