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Taylor Polynomial

  1. Apr 8, 2009 #1
    1. The problem statement, all variables and given/known data
    The Taylor polynomial of degree 100 for the function f about x=3 is given by
    p(x)= (x-3)^2 - (x-3)^4/2! +... + (-1)^n+1 [(x-3)^n2]/n! +... - (x-3)^100/50!
    What is the value of f^30 (3)?

    D) 1/15! or E)30!/15!


    2. Relevant equations



    3. The attempt at a solution
    I know the bottom of the answer is 15! because n=15 (for the exponent to be 30) but i'm not sure what the top does.
     
  2. jcsd
  3. Apr 8, 2009 #2
    By (-1)^n+1 [(x-3)^n2]/n!, do you mean:

    [tex]\frac{(-1)^{n+1}}{n!}(x-3)^{2n}[/tex]

    In any case, ask yourself this: What does the term with [itex]f^{(30)}(3)[/itex] in the Taylor expansion look like?
     
  4. Apr 8, 2009 #3
    how do you get the math problem to look like that -.-

    and that is the question I'm asking for help on...

    i think the term looks like
    (x-3)^30
    15!
    ?

    but how does that answer the question?
    Thanks!
     
  5. Apr 8, 2009 #4
    See here.

    That's what it looks like in p(x), but if you didn't know of p(x), what would the term look like?
     
  6. Apr 8, 2009 #5
    I'm not sure what you mean..?

    like f'(30) (x-3)^30 kind of thing?
    30!
     
  7. Apr 8, 2009 #6
    Yes, that kind of thing. So now you know what the general form of the term looks like and what it actually is. I leave the rest to you.
     
  8. Apr 8, 2009 #7
    i still don't understand =/


    I know the bottom of the term is 15! but I don't know how to get what the top is...
     
  9. Apr 8, 2009 #8
    You wrote, sort of, that the term that contains [itex]f^{(30)}(3)[/itex] is

    [tex]\frac{f^{(30)}(3)}{30!}(x-3)^{30}[/tex]

    and this should equal

    [tex]\frac{(x-3)^{30}}{15!}[/tex]

    right? So what is [itex]f^{(30)}(3)[/itex]?
     
  10. Apr 8, 2009 #9
    so it is E 30!
    15!
    Thanks so much!

    just one last question..
    I don't understand how you got
    [tex]
    \frac{f^{(30)}(3)}{30!}(x-3)^{30}
    [/tex]

    only because I thought that it would be the 30th derivative of f(3) not f^30 (3)
     
  11. Apr 8, 2009 #10
    I got that from the definition of the Taylor polynomial.
     
  12. Apr 8, 2009 #11
    okay. thanks so much again!
     
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