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[heart_over_head]

How do you find the power series for 1 / ((1 - x^2)^2)?

THanks a lot.

Comments

[astarcambiata]

Start with a geometric series and play around a bit ;)

[heart_over_head]

I just figured it out. Thanks. This is all too cool.

[astarcambiata]

You're welcome.
I do so miss ye olde series manipulations, and the mystery they once presented.

Another way (to check yourself)

[anonymous]

f (x) = 1/(1 - x^2)^2

f(0) = 1 * 0!
f'(0) = 0
f''(0) = 2 * 2!
f'''(0) = 0
f^{(4)} (0) = 3 * 4!
f^{(5)} (0) = 0
f^{(6)} (0) = 4 * 6!
f^{(7)} (0) = 0
f^{(8)} (0) = 5 * 8!
f^{(9)} (0) = 0
f^{(10)} (0) = 6 * 10!
...
f^{n} (0) = (n/2 + 1) * n!     for even n

f(x) = 1 + 2 x^2 + 3 x^4 + 4 x^6 + 5 x^7 + 6 x^10 + ... + (n+1) x^{2n}

Takie dela.
Privet.

Re: Another way (to check yourself)

[heart_over_head]

Okay, woah. How much math was required to know this? How did you pull that out of your ass?

Re: Another way (to check yourself)

[astarcambiata]

They are using the Taylor series method (specifically they computed a Maclaurin series), which you will undoubtedly learn this semester. Check out http://en.wikipedia.org/wiki/Taylor_Series

PS

[astarcambiata]

Finding the Taylor series expansion with derivatives is not as slick for this particular problem, but in time it becomes the more reliable and definite procedure for finding a local series expansion of of an arbitrary (well behaved) function. Much later on you'll discover, should you take Complex Analysis, the Laurent series expansion, of which Taylor is a special case (and Maclaurin a special case of that centered at zero). Much more cumbersome to compute in this fashion, but the point is that there is a definite process for determining the coefficients of the terms of a power series (in the case of Laurent series, it permits negative powers, and can be defined by contour integration). In general, it's more efficient to avoid computing the derivatives and instead combine known series, or take term by term derivatives of known series.