How do you find the Maclaurin Series for #f(x)= x / (1-x^4)#?

Question

Ezekiel Alexander · Accepted Answer

I (well, Wolfram Alpha) got:

#x + x^5 + x^9 + . . . #.

A Maclaurin series is just the Taylor series expansion around #a = 0#.

A Taylor series is written as:

#= sum_(n=0)^(oo) (f^((n))(a))/(n!)(x-a)^n#

where:

Since you will have to take many derivatives, let's just do them right now. We can try three of them, but even that's a lot, as there will be a lot of quotient rule and chain rule!

#color(green)(f^((0))(x)) = f(x) = color(green)(x/(1-x^4))#

#color(green)(f'(x)) = ((1-x^4)(1) - (x)(-4x^3))/(1-x^4)^2#

#= color(green)((1 + 3x^4)/(1-x^4)^2)#

And I'll skip the work for the second and third derivatives because you shouldn't need to do this on a test. I just did this in Wolfram Alpha in this case.

#color(green)(f''(x)) = ((1-x^4)^2(12x^3) - (1 + 3x^4)(2(1-x^4)*-4x^3))/(1-x^4)^4#

#[ . . . ] = color(green)((-12x^11 - 8x^7 + 20x^3)/(1-x^4)^4)#

#color(green)(f'''(x)) = ((1-x^4)^4(-132x^(10) - 56x^6 + 60x^2) - (-12x^11 - 8x^7 + 20x^3)(4(1-x^4)^3(-4x^3)))/(1-x^4)^8#

#[ . . . ] = color(green)((60x^26 + 24x^22 - 636x^18 + 1104x^14 - 636x^10 + 24x^6 + 60x^2)/(1-x^4)^8)#

Alright, and now the Taylor series centered around #a = 0#, or the Maclaurin series, truncated at #n = 3#, is:

#color(blue)(sum_(n=0)^(N) (f^((n))(a))/(n!)(x-a)^n)#

#= (f^((0))(0))/(0!)x^0 + (f'(0))/(1!)x^1 + (f''(0))/(2!)x^2 + (f'''(0))/(3!)x^3 + . . . #

#= f(0) + f'(0)x + (f''(0))/(2)x^2 + (f'''(0))/(6)x^3 + . . . #

#= cancel(0/(1-0^4))^(0) + (1 + cancel(3(0)^4)^(0))/(1-cancel((0)^4)^(0))^2 x + cancel((-12(0)^11 - 8(0)^7 + 20(0)^3)/(1-(0)^4)^4)^(0) x^2 + cancel((60(0)^26 + 24(0)^22 - 636(0)^18 + 1104(0)^14 - 636(0)^10 + 24(0)^6 + 60(0)^2)/(1-(0)^4)^8)^(0) x^3 + . . . #

So, naturally, we needed to do more derivatives... Great. Anyways, you would (eventually) get:

#= color(blue)(x + x^5 + x^9 + . . . )#

Isaiah Allen · Answer

undefined To find the Maclaurin series for the function $ f(x) = \frac{x}{1-x^4} $, we first express the function as a geometric series.

$$ f(x) = x(1 + x^4 + (x^4)^2 + (x^4)^3 + \ldots) $$

Now, we recognize that this is a geometric series with first term $ a = x $ and common ratio $ r = x^4 $.

The sum of an infinite geometric series is given by:

$$ S = \frac{a}{1 - r} $$

Substituting the values for $ a $ and $ r $, we get:

$$ f(x) = \frac{x}{1 - x^4} = x(1 + x^4 + (x^4)^2 + (x^4)^3 + \ldots) $$

$$ = x + x^5 + x^9 + x^{13} + \ldots $$

This is the Maclaurin series for $ f(x) = \frac{x}{1 - x^4} $.