Let f(x) = (8+8x) / (x)
Find the general formula for the nth derivative of the Taylor series Σ(n = 0 to infinity) centered at x = 1
What I mean is the portion of the answer that asks for (f^(n) (1)) / n! when evaluating Taylor or McLaurin series, not the term carrying (x-a)^n.
The first numerator term coefficient is 16, which does not adhere to the pattern. However the successive coefficients do follow the pattern. Second coefficient is -8, then 16, then -48,192, and so on. I believed that the multiplier is
changing by ((-1)^n) * (8(n+1)), but the formula isn't working.
In the end I imagine the sum of the series to look something like 16 + Σ(n=0 to infinity) (((-1^n) * (8(n+1))/n!) * (x-1)^n, but obviously my formula for the nth derivative doesn't work. That's what I am looking for.
- ted sLv 78 months agoFavorite Answer
dⁿf / dxⁿ = (-1)ⁿ 8 ( n!) when x = 1.....f ' = - 8 x^(-2) ; f ' ' = - 8 ( -2 ) x^(-3)
f ' ' ' = - 8 ( -2 )(-3)x^(-4) , etc. now let x = 1
- husoskiLv 78 months ago
That simplifies to f(x) = 8 + 8x^-1, after which:
f'(x) = -8x^-2
f''(x) = 16x^-3,
f'''(x) = -48x^-4
... and it's pretty easy to see that the pattern for nth derivative is:
f^(n)(x) = 8 * (-1)^n * n! x^-(n+1)
...for n>=1. Applying that formula with n=0 gives 8 as a result, instead of 16, so you can write the Taylor series as
f(x) = f(1) + Σ f^(k)(1)/k! (x-1)^k [sum from k=1 to oo]
= 16 + 8 Σ (-1)^k (x-1)^k [after factoring out 8 and canceling k!/k!]
But the k=0 term of that sum would be 1, so you could just as easily write that as:
f(x) = 8 + 8 Σ (-1)^k (x-1)^k [sum from k=0 to oo]
...and that first 8 is the constant term in f(x) = 8 + 8x^-1.
Another way to get there, without derivatives, is to let u = x-1, x = u + 1, so that:
f(x) = 8 + 8/x = 8 + 8/(1 + u)
...and then recongize that 1/(1+u) is the sum of the geometric series
1/(1 + u) = 1 - u + u² - u³ + ... = Σ (-1)^k u^k [sum from k=0 to oo]