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Very Easy Taylor Series Approximation Help

  1. Apr 16, 2007 #1
    1. The problem statement, all variables and given/known data

    Approximate f by a Taylor polynomial with degree n at the number a.

    f(x) = x^(1/2)

    a=4
    n=2

    4<x<4.2
    (This information may not be needed for this, there are two parts but I only need help on the first)

    2. Relevant equations

    Summation [tex]f^(i) (a) * (x-a)^i / i![/tex]

    3. The attempt at a solution

    Derivatives...

    [tex]x^(1/2)
    0.5 * x^(-1/2)
    -0.25 * x^(-3/2)[/tex]

    So the Taylor series of order n=2 should be...

    2 + 0.25(x-4) - (1/64)(x-4)(x-4)

    Now, my question...

    To find the approximation of square root of x, do I just plug in 4 to that?

    That would make most of the terms zero, leaving me with 2

    Does this mean that whenever the Taylor series is centered at a number, the first term is the approximation?


    Ugh, sorry for the failure using tex tags :( It's my first post here
     
  2. jcsd
  3. Apr 16, 2007 #2

    Dick

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    Yes. Exactly. If x=a the first term IS the approximation. The usual game is to approximate something like 4.1 using the series.
     
    Last edited: Apr 17, 2007
  4. Apr 17, 2007 #3

    HallsofIvy

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    I think Dick is interpreting the question differently than I am.

    What you have: 2+ (1/4)(x-4)- (1/64)(x- 4)2 alread is the quadratic approximation to [itex]\sqrt{x}[/itex] around x= 4.

    Putting x= 4 would just give you the obvious [itex]\sqrt{4}= 2[/itex].
     
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