dydx
Lv 6
dydx asked in Science & MathematicsMathematics · 3 weeks ago

Taylor Series....?

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.

2 Answers

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  • ted s
    Lv 7
    3 weeks ago
    Best 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

  • 3 weeks 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]

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