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Calculus II Help : Taylor/Maclaurin Series

  1. Apr 27, 2009 #1
    Hey... This really sucks. I am in Calculus 2 and I have had 3 in-class exams, all 3 were A's. This last exam is take-home and it is entirely Maclaurin and Taylor series.. The only thing in the class to go over my head.

    Please help me out with these problems!

    1. The problem statement, all variables and given/known data


    Use the function : f(x) = 1 / x^2 to answer the following questions.

    #1
    a. Find a formula for the sequence of values given by f^n (2). Do this by computing enough derivatives of f(x) evaluated at 2 until you see a pattern.

    I got

    Σ ( (-1)^(n+1) * (n+1)! ) / ( -2 * 2^(n+1) )
    n=0

    b. Find formula for the sequence of values given by f^n (2) / n!

    I got like... ( (-1)^n (n+1)! ) / (4 * 2^n )

    c. What is the Taylor Series centered at a = 2 for the function f(x) = 1/x^2 ?


    Σ ( f^n*(2)*(x-2)^n ) / n!
    n=0


    d. What is interval of convergence for this Taylor series?

    No bueno.

    e. What is T4 (x) ?

    f. What are T4(3) and R4 (3) ?



    #2
    a. Find Maclaurin Series for the function:
    F(x) =
    x⌠ t^2 * e^ (-t^2) dt
    0⌡


    *Remember : e^x =

    Σ [ f^n * (a) * (x-a)^n ] / n!
    n=0

    I got... (something that didn't work)

    [ (-1)^n * t^2 (t^(2n) ] / n!


    b. Estimate value of
    1⌠ x^2 * e^(-x^2) dx
    0⌡
    by using M9(x), the Maclaurin polynomial of degree 9.


    #3

    a. Find the Maclaurin series for the function f(x) = arctan ( x^3 / 3 )


    b. What is the interval of convergence?


    c. Find the value of the first 10 coefficient terms: c0, c1, c2, c3, c4 ... c10 for this Maclaurin series.


    d. What is the value of f^21 (0), the 21st derivative evaluated at zero?



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 27, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You mean the nth derivative, not f to the nth power here, don't you? f(x)= x-2 so f(2)= 1/4; f'(x)= -2x-3 so f'(2)= -1/4; f"(x)= 6x-4 so f"(2)= 3/8; f"'(x)= -12x-5 so f"'(2)= 3/16, etc. I don't see where you got that sum.

    The only difference between (a) and (b) is that you have divided by n!. What happened to the sum? Why do you still have (n+1)!? (n+1)!/(n+1)= n+1.

    That's the formula, yes, but obviously you are expected to use your answers from (a) and (b)!

     
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