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Taylor's Formula to derive higher order derivatives

  1. Nov 30, 2011 #1
    The question asks the student to use Taylor's formula to calculate the exact values of higher derivatives

    f '[0], f '' [0], f ''' [0], ... , f^6'[0]

    of the function f[x] defined by the power series

    x/2 + x^2/12 + x^3/240 +x^4/10080 + ... +((k x^k)/(2 k)!) + ....

    My first attempt at a solution resulted in {0, 0, 0, 0 ...} which I know is not correct.

    I do know that Taylor's Fromula says that the expansion of f[x] in powers of (x-b) is given by

    f + f ' (x-b) + (f '' (x-b)^2)/2! + (f ''' (x-b)^2)/3! ... However, I just can't seem to figure out how to use it when I am given some given expansion such as above. I feel a little silly not being able to figure this out, but can anyone help me think about this? Maybe I need someone to explain it in a slightly different way. Thanks to all.
     
  2. jcsd
  3. Nov 30, 2011 #2

    SammyS

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    If f(x) = x/2 + x2/12 + x3/240 +x4/10080 + ... +((k xk)/(2 k)!) + ... ,

    then what is f '(x) ? , f ''(x) ? , ...

    Then let x = 0 for each derivative.
     
  4. Nov 30, 2011 #3
    Okay I got it. Looking at it from that approach is a great way to see, conceptually, what is going on. Thanks for taking the time. Of course, it seems that using Taylor Serires could be a rather inefficient method. I'm sure we will be learning other methods as the class advances.

    Thanks again
     
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