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Math Help - Taylor Polynomial

  1. #1
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    Taylor Polynomial

    hey..just wanted to check if im doing this correctly

    Find the Taylor polynomial of degree 4 for the function f(x) = x^(3/2) about the centre c=1

    the 1st 5 terms i got is...

    1, 3/2, 3/4, -3/8, 9/16

    Now im not quite sure if im writing the series correctly..

    1 + 3/2(3/2 - 1) + 3/4(3/4 -1) - 3/8(-3/8 -1)^2/2! + 9/16(9/16 - 1)^3/3!

    am i correct?? or should i just leave the bracket as (x - 1)^2 / 2! and so on......

    Thanks
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  2. #2
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    Quote Originally Posted by b00yeah05 View Post
    hey..just wanted to check if im doing this correctly

    Find the Taylor polynomial of degree 4 for the function f(x) = x^(3/2) about the centre c=1

    the 1st 5 terms i got is...

    1, 3/2, 3/4, -3/8, 9/16

    Now im not quite sure if im writing the series correctly..

    1 + 3/2(3/2 - 1) + 3/4(3/4 -1) - 3/8(-3/8 -1)^2/2! + 9/16(9/16 - 1)^3/3!

    am i correct?? or should i just leave the bracket as (x - 1)^2 / 2! and so on......

    Thanks
    If you have to find a polynomial approximation for f(x), your polynomial must have x's in it ...... !
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  3. #3
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    Leave the bracket as (x-1).

    1+\frac{3}{2}(x-1)+\frac{3}{16}(x-1)^{2}-\frac{1}{16}(x-1)^{3}+\frac{3}{128}(x-1)^{4}-\frac{3}{256}(x-1)^{5}+.....
    Last edited by galactus; May 25th 2008 at 05:40 PM. Reason: forgot my 1
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    Quote Originally Posted by b00yeah05 View Post
    Now im not quite sure if im writing the series correctly..

    1 + 3/2(3/2 - 1) + 3/4(3/4 -1) - 3/8(-3/8 -1)^2/2! + 9/16(9/16 - 1)^3/3!

    am i correct?? or should i just leave the bracket as (x - 1)^2 / 2! and so on......

    Thanks
    No, you should leave the factors of (x - 1) alone. The Taylor series expansion of a function f(x) about a point x = a is given by

    f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x - a)^n + \cdots

    In your case, we have the following:

    f(x) = x^{3/2}

    a = 1

    and

    \begin{array}{rclcrcl}<br />
f(x) & = & x^{3/2} & \Rightarrow & f(1) & = & 1\\<br />
f'(x) & = & \frac32x^{1/2} & \Rightarrow & f'(1) & = & \frac32\\<br />
f''(x) & = & \frac34x^{-1/2} & \Rightarrow & f''(1) & = & \frac34\\<br />
f'''(x) & = & -\frac38x^{-3/2} & \Rightarrow & f'''(1) & = & -\frac38\\<br />
f^{(4)}(x) & = & \frac9{16}x^{-5/2} & \Rightarrow & f^{(4)}(1) & = & \frac9{16}<br />
\end{array}

    So, you should get

    f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \frac{f^{(4)}(a)}{4!}(x - a)^4

    = 1 + \frac32(x - 1) + \frac38(x - 1)^2 - \frac3{48}(x - 1)^3 + \frac9{384}(x - 1)^4

    = 1 + \frac32(x - 1) + \frac38(x - 1)^2 - \frac1{16}(x - 1)^3 + \frac3{128}(x - 1)^4

    Note that the series expansion produces a polynomial in x and not a constant.
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