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Taylor Polynomials Questions

  1. Apr 25, 2006 #1
    Hi

    Given a function f(x) = sqrt(x) is the Taylor Polynomial of degree 2 for that function:

    [tex]\frac{x^2}{2} - 99x + 4901[/tex] where x = 100 ?

    Sincerely Fred
     
  2. jcsd
  3. Apr 25, 2006 #2
    Do you mean a=100, as in it's centered at a=100? If so, use Taylor's formula for approximating polynomials.

    [tex]f(x)=\sum_{n=0}^{\infty}\frac{f^{n}(a)*(x-a)^n}{n!}[/tex]

    I get the first two terms of the series are [tex]10+\frac{x-100}{20}[/tex]
     
  4. Apr 26, 2006 #3

    HallsofIvy

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    Staff Emeritus
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    The Taylor's polynomial of degree 2 for a given function at a given point must match the function's value, first derivative and second derivative at that point.
    Is [itex]\sqrt{100}= 100^2/2- 99(100)+ 4901[/itex]?

    Is the derivative of [itex]\sqrt{x}[/itex] at x= 100 equal to the derivative of that polynomial at x= 100?

    Is the second derivative of [itex]\sqrt{x}[/itex] at x= 100 equal to the second derivative of that polynomial at x= 100?

    If the answer to all three questions is correct, then that must be the
    Taylor polynomial.
     
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