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Homework Help: Taylor Series

  1. Apr 5, 2007 #1
    1. The problem statement, all variables and given/known data

    Give the Taylor Series for exp(x^3) around x = 2.

    2. Relevant equations

    f(x) = Sum[f(nth derivative)(x-2)^n]/n!

    3. The attempt at a solution

    I know the solution for e^x but can't seem to find a formula for the nth derivative of exp(x^3) around x = 2.

    Thanks for any help.
  2. jcsd
  3. Apr 5, 2007 #2
    You know the formula for the taylor series expansion of e^y right?

    Just substitute y=x^3 into that expansion and voila you have the formula for e^(x^3)
  4. Apr 5, 2007 #3
    I do now that the trick you mentioned works around x = 0; however, I am highly suspicious that this works for x around other than 0. Please show me explicitely why if you still think as you said.
  5. Apr 5, 2007 #4
    It works just fine, but you get a Taylor series expansion in (x^3-2)^n, not (x-2)^n. It's still a Taylor series though.

    e^(x^3)=e^2 sum(n=0..infinity) (1/n!)(x^3-2)^n

    Edit- the correct expression about x=2 is
    e^(x^3)=e^8 sum(n=0..infinity) (1/n!)(x^3-8)^n
    Last edited: Apr 5, 2007
  6. Apr 5, 2007 #5
    I think your series would actually give the expansion for e^[(x-2)3] and not for e^[x3-2].
  7. Apr 5, 2007 #6
    Neither, it gives the T series expansion for e^(x^3)

    Edit- sorry I should have replaced 2 with 8 in the above equation. Then it works.
    Last edited: Apr 5, 2007
  8. Apr 5, 2007 #7
    christian how did you get the e^2?
  9. Apr 5, 2007 #8
    No it still does not (x-2)^3 is not the same as x3-8.
  10. Apr 5, 2007 #9
    d/dn exp(x) = exp(x)

    so when x=2, exp(x)=e^2

    However, when you substitute x=(x')^3, you have to change to e^8
  11. Apr 5, 2007 #10
    I never said it was!
  12. Apr 5, 2007 #11
    oh, I see. But then are saying that exp(2^3) is the nth derivative of f around x = 2? Seems strange since for instance, first f derivative = 3x^2exp(x^3). When you substitute x = 2, the above equation is unequal to exp(2^3) as you imply. Not to mention the other derivatives: all unequal to exp(2^3).
  13. Apr 5, 2007 #12
    Here- I'll go through the steps

    e^x = e^a sum (x-a)^n / n!


    e^(x^3)=e^a sum(x^3-a)^n / n!

    = e^(b^3) sum (x^3-b^3)^n / n!

    when x=b the RHS is equal to e^(b^3), thus the T series is evaluated about x=b. So for b=2 we have

    e^(x^3)= e^8 sum (x^3-8)^n / n!
  14. Apr 5, 2007 #13
    "The derivative of e^(x^3) is not e^(x^3) it is (3x^2)(e^(x^3)"

    Where did I say otherwise?
  15. Apr 5, 2007 #14
    I see what you are doing but doesn't the last equation you give has the (x^3-8)^n, when it should be somthing like (x^3 - 2)^n, for T be around x = 2?
  16. Apr 5, 2007 #15
    No, for the reasons I gave above. Look at the value of the expansion when x=2.

    The reason for all the confusion is because I'm doing a T series expansion in x^3, and not in x. To do a T series expansion in x might be much more difficult, but the Q doesn't require an expansion in powers of x, though it may be implied.
  17. Apr 5, 2007 #16
    yeah, you might be right, but it is definitely the first time I see it as
    (x^3 - 2^3)^n for T be about x = 2. Seems unusual.
  18. Apr 5, 2007 #17
    First time I've seen it too! It works though, looking at some graphing software.
  19. Apr 5, 2007 #18
    Thanks a lot for all the help. I am convinced.
  20. Apr 5, 2007 #19
    Aha, I had a brainwave!

    x'=x-2, and x' is small near x=2
    so x^3=(x'+2)^3

    so given that x^3 = approx= 12(x-2)+8 about x=2


    e^(x^3)=e^8 sum(n=0..infinity) (1/n!)(x^3-8)^n

    By substitution we have

    e^(x^3)=e^8 sum(n=0..infinity) (1/n!)(12(x-2))^n

    Which, again looking at my graphing program, works. So, this is a nicer answer, because the expansion is in (x-2). However, it would be nice to have a more concise proof.
    Last edited: Apr 5, 2007
  21. Apr 5, 2007 #20
    OK, here's the concise version....

    e^(x^3) about x=2
    let x'=x-2
    then e^(x^3)=e^[(x'+2)^3]
    =e^[12x'+8] (approximately)
    =e^8 e^12x'
    =e^8 sum{n=0..infinity} (1/n!)[12(x-2)]^n

    Edit: Aaarrgghh! it's not quite exact, because you also need to expand (x'/2+1)^3 to higher powers in x'. So you really end up with a sum over n and m, where m is a Taylor series expansion of (x'/2+1)^3 about x'=0 in powers of x'^m.

    It's messy, but doable.

    I wonder if there's a way to only sum over one index.
    Last edited: Apr 5, 2007
  22. Apr 5, 2007 #21
    OK, doing the full expansion of (x'/2+1)^3 I get this for the final? answer

    e^(x^3) about x=2

    = e^8 sum{l,m,n}12^n 6^m (x-2)^(n+2m+3l) / l!m!n!

    Edit: well I guess you can write it in powers of (x-2)^k by rewriting the above as

    = e^8 sum{k} [sum{l,m,n} delta(3l+2m+n,k) 12^n 6^m (x-2)^(n+2m+3l) / l!m!n!]
    Last edited: Apr 5, 2007
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