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Math Help - taylors series

  1. #1
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    taylors series

    write the first 4 terms of the taylor's series for f(x)=3^x centered at c=0
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jeph View Post
    write the first 4 terms of the taylor's series for f(x)=3^x centered at c=0
    recall that the Taylor series is of the form:

    \sum_{n=0}^{ \infty} \frac {f^{(n)}(x_0)}{n!} (x - x_0)^n where the series is centered at x_0

    the first four terms of the Taylor series is centered at 0 is given by:
    \sum_{n=0}^{3} \frac {f^{(n)}(0)}{n!} x^n

    f(x) = 3^x \Rightarrow f(0) = 1
    f'(x) = ln(3) \cdot 3^x \Rightarrow f'(0) = ln(3)
    f''(x) = (ln(3))^2 \cdot 3^x \Rightarrow f''(0) = (ln(3))^2
    f'''(x) = (ln(3))^3 \cdot 3^x \Rightarrow f'''(0) = (ln(3))^3

    So the first 4 terms of the Taylor series is:
    1 + ln(3) x + \frac {(ln(3))^2}{2!}x^2 + \frac {(ln(3))^3}{3!}x^3
    Last edited by Jhevon; May 23rd 2007 at 11:57 AM. Reason: I had x^2 where i should have x^3
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    when i differentiate 3^x i get 3^x(ln3)

    but when i differentiate that again i get

    [3^x]' ln3 + 3^x [ln3]'
    =3^x(ln3)(ln3) + 3^x 1/3
    3^x(ln3)^2 + 3^x 1/3
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jeph View Post
    when i differentiate 3^x i get 3^x(ln3)

    but when i differentiate that again i get

    [3^x]' ln3 + 3^x [ln3]'
    =3^x(ln3)(ln3) + 3^x 1/3
    3^x(ln3)^2 + 3^x 1/3

    ln3 is a constant! product rule is not needed here. even if you did the product rule, the derivative of ln3 would be zero, so the second part would dissappear anyway
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  5. #5
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    Quote Originally Posted by jeph View Post
    write the first 4 terms of the taylor's series for f(x)=3^x centered at c=0
    <br />
f(x) = 3^x = e^{\ln(3)x} = 1 + \ln(3)x + \frac{(\ln(3)x)^2}{2!} + \frac{(\ln(3)x)^3}{3!} + ...<br />

    from the Taylor series for \exp(x) about 0

    RonL
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