View Full Version : Evaluating the product of two Taylor series
05-7-2012, 02:36 AM
I work as an online calculus tutor, and one of the problems that I was helping a student with today seemed a little too difficult to do by brute force. I was wondering if there was a more efficient method, or any other way to solve the problem by hand.
The problem asks to find the taylor series representation of:
Usually, for functions that have nasty derivatives, there are shortcuts in finding their taylor series which the problem is highlighting for practice. For example, to find it for x^2/(1-x^3), you would take the series for 1/(1-x), the geometric series, replace the x's with x^3 and multiply it by x^2.
Does anyone know how to find an analytic solution using the taylor series of ln(x+1) and/or 1/(1-x)?
05-7-2012, 03:18 AM
The Taylor series of ln (1+x) is
And the Taylor series of 1/(1-x) is
Are you asking for http://www4b.wolframalpha.com/Calculate/MSP/MSP46551a19d75g0ei345770000193964ehc1537gg4?MSPSto reType=image/gif&s=59&w=67&h=37 or http://www4b.wolframalpha.com/Calculate/MSP/MSP132861a19d60d20c8gafd00001fhd39a75hccgh06?MSPSt oreType=image/gif&s=47&w=69&h=36? (log x is ln x in WolframAlpha lmfao)
EDIT: For http://www4b.wolframalpha.com/Calculate/MSP/MSP46551a19d75g0ei345770000193964ehc1537gg4?MSPSto reType=image/gif&s=59&w=67&h=37:
You would need to multiply the expansion term by term. I like to keep the Maclaurin series of ln x grouped as one factor, and distribute that in the expansion of the geometric series. From this you should get something like:
Where the coefficient of x^n is the nth partial sum of the alternating harmonic series.
EDIT2: For http://www4b.wolframalpha.com/Calculate/MSP/MSP132861a19d60d20c8gafd00001fhd39a75hccgh06?MSPSt oreType=image/gif&s=47&w=69&h=36:
Look at the Maclaurin series of ln(1+x). We can get the Maclaurin series of ln(1-x) by replacing every x with a -x in the series expansion. When we subtract the two series, half of the terms cancel in the expansion. You are left with the series expansion of ln(1+x)-ln(1-x) or http://www4b.wolframalpha.com/Calculate/MSP/MSP132861a19d60d20c8gafd00001fhd39a75hccgh06?MSPSt oreType=image/gif&s=47&w=69&h=36. The final answer you should get is:
Hope this helps (:
05-7-2012, 08:10 PM
That does help, I was thinking that multiplying infinite series out wouldnt be possible, but after some thought i realized that you can solve for the first few coefficients by multiplying and adding together the terms that gave the corresponding order of x, and then finding a pattern.
Edit: I was just checkin out your skill determining system, and it looks pretty well thought out. I was wondering about the reason for the square in ( Xp/Q(p) )^2, it seems to me that a linear system would make more sense.
Heres an EXTREMELY simplified example to show what i mean... suppose theres 10 FGOs in the game, player 1 has AAAd 4 and player 2 has AAAd 6. An estimate of X12 ignoring the almost AAA scores would be 4 and 6 respectively, giving ratings of 12.16 and 12.36. Seems to me that people who have AAAd a little more or less than half of FGOs should have a rating of around 12.5?
Im thinking the square takes into account the difficulty range of 12's, assuming the AAAs one person has are easier songs than their un-AAAd.
What is your reason behind it?
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