Power series question

**virtuoso735** · April 12th 2009, 10:43 AM

We just started power series in class and I'm somewhat confused about them. I'm stuck on this particular problem. It says to make a power series for f(x) = x/(2x^2+1). What I did was to base it on geometric series so I convered the equation to look like 1/(1+x) so now it looks like x/[1-(-2x^2)]. What I got for the power series was x multiplied by the sum of (-2x^2)^n from n=0 to infinity. Is this right?

**Aryth** · April 12th 2009, 01:14 PM

You're close, except you forgot to add a coefficient, rational functions have this relation:

$\frac{x}{2x^2 + 1} = \sum_{n=0}^{\infty} a_nx^n$

It's that simple, now you have to find the coefficients:

Multiply the denominator through:

$x = (2x^2 + 1)\sum_{n=0}^{\infty} a_nx^n$

Now we multiply the sum through:

$x = 2\sum_{n=0}^{\infty}a_nx^{n+2} + \sum_{n=0}^{\infty}a_nx^n$

Adjust the indices:

$x = 2\sum_{n=2}^{\infty}a_{n-2}x^n + \sum_{n=0}^{\infty}a_nx^n$

Now we combine like terms:

$x = a_0 + a_1x + \sum_{n=2}^{\infty}(2a_{n-2} + a_n)x^n$

Since there are no coefficients on the left side, then:

$a_0 = 0$

Since the coefficient of the x term on the left side is 1, then the coefficient of the x term on the right side is also 1:

$a_1 = 1$

This gives you:

$x = x + \sum_{n=2}^{\infty}(2a_{n-2} + a_n)x^n$

Which gives you an interesting result:

$a_n = -2a_{n-2}$

Now you have your power series:

$f(x) = \sum_{n=2}^{\infty}-2a_{n-2}x^n$

This gives you:

$f(x) = \sum_{n=0}^{\infty}-2a_nx^{n+2}$

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