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Taylor polynomial-problem

  1. Nov 6, 2013 #1
    Hi,

    I would really appreciate it if someone can help me with the following problem, regarding a taylor polynomial:

    A 2nd degree taylor polynomial to the function f around x = 1, is given by:

    T_2(x) = x + x^2

    Question:
    What is f´(1) ?

    Answer: 3

    (Btw: the question is from a multiple choice test, but the answer should be evident without consulting the possible answers)



    2. Relevant equations

    I usually solve this kind of problem simply by considering the general form of a taylorpolynomial:

    f(a) + f´(a)(x-a) + (f´´(a)/2!) * (x-a)^2

    and comparing it to the taylor polynomial given in the problem statement. From a little rearrangement, the answer is usually self-evident. But in this case the rearranging seems very elaborate, so im hoping that im missing some clever way to solve it.


    3. The attempt at a solution

    In this case I really dont know how go about it. I´ve only managed to simply make sense of the answer, by the following argument:

    f(1) + f´(1)(x-1) + f´´(1)(x-1)]^2 = x + x^2 ⇔

    f(1) + f´(1)x - f´(1) + f´´(1)(x^2 + 1 - 2x) = x + x^2 ⇔

    f(1) + f´(1)x - f´(1) + f´´(1)x^2 + f´´(1) - f´´(1)2x = x + x^2

    Inserting f´(1) = 3 evaluates to x + x^2 , if f´´(1) = 2 and f(1) = 2 .
    But there´s no way that I would have seen that, not knowing that f´(1) = 3 .

    There must be some nice way of solving this problem?

    Any help would be truly appreciated..!
     
  2. jcsd
  3. Nov 6, 2013 #2

    vanhees71

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    You should just expand your Taylor polynom in powers of [itex]x[/itex] and then compare the coefficients on both sides, which leads to a linear system of equations for f(1), f'(1), and f"(1).
     
  4. Nov 6, 2013 #3

    HallsofIvy

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    The "second degree Taylor polyomial" of f, about x= 1, is, by definition, [itex]f(1)+ f'(1)(x- 1)+ (f''(1)/2)(x- 1)^2[/itex]. You are told that the second order Taylor polynomial for this function is [itex]x+ x^2[/itex].

    I think Vanhees71 meant to say "expand your polynomial in powers of x-1". That is, find a, b, and c so that [itex]a(x- 1)^2+ b(x- 1)+ c= x+ x^2[/itex]. Then you can just "read off" the value of f'(1).
     
  5. Nov 6, 2013 #4
    Hi,
    I know that I need to find the coefficients, but not how to do it. It becomes an equation with three unknowns? How do i read off the value?

    I know how to solve this type of problem typically. When the derivative in question is simply the coefficient to x^n times the nth factorial. But thats hard do in this problem, i think.

    Thanks..!
     
  6. Nov 6, 2013 #5
    eipeplusone - as Hallsofivy mentioned - expand his equation on the left hand side in terms of a polynomial in x making sure you multiply a,b, and c through. Now collect terms of a, b, and c into coefficients to this polynomial. Equate those coefficients on left to like powers of known coefficients on right (start with highest term). You'll see an easy solution to yield a, b, and c. From there, you can compute f(1), f'(1) and f''(1)
     
  7. Nov 9, 2013 #6
    Great, I understand it now. Thanks a lot!
     
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