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Reverse this taylor series



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?
(edited 7 years ago)
Reply 1
Avatar for Zacken
Zacken
22
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: an=n!(8)8n\displaystyle a_n = \frac{n!^{(8)}}{8^n} - not sure if that helps.
Reply 2
Avatar for Zacken
Zacken
22
Probably just simpler to look at is as:

an=(1+8)(1+8+8)⋯(1+8n)(8+8)(8+8+8)⋯(8+8n)\displaystyle a_n = \frac{(1+8)(1 + 8 + 8) \cdots (1 + 8n)}{(8+8)(8+8 + 8) \cdots ( 8 + 8n)}
Reply 3
Avatar for Y.X.
Y.X.
OP
1
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: an=n!(8)8n\displaystyle a_n = \frac{n!^{(8)}}{8^n} - not sure if that helps.
Reply 4
Avatar for TeeEm
TeeEm
19
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
Reply 5
Avatar for Zacken
Zacken
22
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. :facepalm:
Reply 6
Avatar for Y.X.
Y.X.
OP
1
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
(edited 7 years ago)
Reply 7
Avatar for TeeEm
TeeEm
19
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
Original post by TeeEm
Your series will converge to (1 - x) to the power of minus 1/8, valid for -1 <x<1


:eek:

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