Maclaurin Series using Substitution

[alanwhite]

Homework Statement

Use a known Maclaurin series to compute the Maclaurin series for the function: f(x) = x/(1-4(x^2))

Homework Equations

1/(1-x) = ∑x^n

The Attempt at a Solution

I tried removing x from the numerator for: x ∑ 1/(1-4(x^2)), which would end up through substitution as x ∑ (4^n)(x^2n). Not too sure this is correct use of substitution however.

 

[szynkasz]

Yes, it is correct.

 

[Mark44]

Homework Statement

Use a known Maclaurin series to compute the Maclaurin series for the function: f(x) = x/(1-4(x^2))

Homework Equations

1/(1-x) = ∑x^n

The Attempt at a Solution

I tried removing x from the numerator for: x ∑ 1/(1-4(x^2)), which would end up through substitution as x ∑ (4^n)(x^2n). Not too sure this is correct use of substitution however.

What you did makes sense, but how you described what you did doesn't make sense. If you can find the series for 1/(1 - 4x²), just multiply term-by-term to get the series for x/(1 - 4x²). Pulling a variable out of a summation that involves x isn't a valid operation.
For example,
$$\sum_{n = 1}^k n^2 \neq n \cdot \sum_{n = 1}^k n $$

 

[alanwhite]

So in essence, I would write the terms of the series ∑ (4^n)(x^2n) and multiply each term by x? Alright, is there no way of writing the series so that there is no x variable outside of the summation?

 

[SammyS]

So in essence, I would write the terms of the series ∑ (4^n)(x^2n) and multiply each term by x? Alright, is there no way of writing the series so that there is no x variable outside of the summation?

Mark did say to take the sum, then multiply through by x, term by term. (Basically, that's the distributive law, and the extra x will be inside of the sum.For example,

$\displaystyle x\left(\sum _{n=1}^ k x^n \right) = \sum _{n=1}^ k x^{n+1}$