Writing a function as a power series?? 3x² / (x-6)²?

Question

f(x) = 3x² / (x-6)²
How can we re-write this into the form a / (1 - r) ?
So that we can then re-write the above into Σ arⁿ?

3x² ∫ 1 / (x-6)²

3x² d/dx (-1 / (x-6))

3x² (-1/6) d/dx Σ (1 / 1-(x/6))

(1 / 1-(x/6)) is in the form a / 1 - r, so Σ arⁿ is:

3x²/6 d/dx (Σ(1)(x/6)ⁿ

d/dx Σ (-3)xⁿ⁺² / 6ⁿ⁺¹

Now, if this is correct, which probably isn't lol, what do I do with the d/dx? :c And I'm not sure how to evaluate the thing now...

I am to find the first 5 terms of the power series, at x = 0?? :c
and the radius of convergence :c I guessed 6 and I was right :o

Answer 1

　

You need to rewrite 1/(x−6) in the form a/(1−bx)

which has power series: a + a(bx) + a(bx)² + a(bx)³ + ...

then differentiate both sides, and finally multiply by Cx²



1/(x−6) = −1/6 * 1/(1−x/6)

1/(x−6) = −1/6 [1 + x/6 + (x/6)² + (x/6)³ + (x/6)⁴ + ...]

1/(x−6) = −1/6 [1 + x/6 + x²/6² + x³/6³ + x⁴/6⁴ + ...]



Differentiate:



−1/(x−6)² = −1/6 [1/6 + 2x/6² + 3x²/6³ + 4x³/6⁴ + ...]



Multiply both sides by −3x²



3x²/(x−6)² = x²/2 [1/6 + 2x/6² + 3x²/6³ + 4x³/6⁴ + ...]

3x²/(x−6)² = x²/(2·6) + 2x³/(2·6²) + 3x⁴/(2·6³) + 4x⁵/(2·6⁴) + ...

3x²/(x−6)² = 3·1x²/(6·6) + 3·2x³/(6·6²) + 3·3x⁴/(6·6³) + 3·4x⁵/(6·6⁴) + ...

3x²/(x−6)² = 3·1(x/6)² + 3·2(x/6)³ + 3·3(x/6)⁴ + 3·4(x/6)⁵ + ...



f(x) = 3x²/(x−6)² = Σ [n = 1 to ∞] 3n(x/6)ⁿ⁺¹





This is true for:

|x/6| < 1

|x| < 6



——————————————————————————————



Alternate method



We can use generalized form of binomial expansion for (1+a)ⁿ

when n is negative or rational:



(1+a)ⁿ = C(n,0) + C(n,1) a + C(n,2) a² + C(n,3) a³ + ...

(1+a)ⁿ = 1 + n!/(1! (n−1)!) a + n!/(2! (n−2)!) a² + n!(3! (n−3)!) a³ + ...

(1+a)ⁿ = 1 + n/1! a + n(n−1)/2! a² + n(n−1)(n−2)/3! a³ + ...



3x²/(x−6)²

= 3x²/(−6(1−x/6))²

= 3x²/(36(1−x/6)²)

= 3x²/36 (1−x/6)⁻²

= 3(x/6)² [1 + (−2)/1! (−x/6) + (−2)(−3)/2! (−x/6)² + (−2)(−3)(−4)/3! (−x/6)³

........... + (−2)(−3)(−4)(−5)/4! (−x/6)⁴ + . . .]

= 3(x/6)² [1 + 2(x/6) + 3(x/6)² + 4(x/6)³ + 5(x/6)⁴ + . . .]

= 3·1(x/6)² + 3·2(x/6)³ + 3·3(x/6)⁴ + 3·4(x/6)⁵ + 3·5(x/6)⁶ + . . .

　

Answer 2

I don't understand the last step in line 1 where you seem to replace 1/(x-6)² with the integral of that.



What I'd do is write:

1/(x - 6)² as - d/dx [1/(x-6)] = d/dx [1/(6-x) = (1/6) d/dx [1/(1 - x/6) ]



Okay so far? The 1/(1-x/6) at the end is what you can use the geometric series expansion on:



1/(x - 6)² = (1/6) d/dx [1 + x/6 + (x/6)² + (x/6)³ + ...



...and then take term-by-term derivatives:



= (1/6)[1/6 + 2x/36 + 3x²/216 + .... + (n+1)x^n/6^(n+1) + ...]



Multiplying that by 3x² and converting to sigma notation should be pretty easy from here out. The geometric series expansion required |x/6| < 1, and multiplying by 3x² didn't change that.

Answer 3

The given expression equals (by partial fractions):

3 + 108/(x-6)^2 + 36/(x-6)

You can find the power series for each of those 3 terms and combine them.

For the second term, its integral is -108/(x-6), so you could write down the power series of that and differentiate term by term.

Answer 4

You can do so only when 6>> x. Otherwise not.