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Math Help - [SOLVED] Taylor polynomial question?

  1. #1
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    Talking [SOLVED] Taylor polynomial question?

    Hi =)


    Use a suitable Taylor polynomial for e^x to evaluate the 24th derivative of f(x) = exp (x^8) at x = 0 (express your answer in terms of factorials).

    Can someone teach me how to do so?
    The answer is 24!/3!


    thanks in advance =)
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  2. #2
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    \exp(x) = 1+x+{x^2\over2!}+{x^3\over3!}\cdots so \exp(x^8) = 1+x^8+{x^{16}\over2!}+{x^{24}\over3!}\cdots. The coefficient of x^n in this power series is equal to {1\over n!}\;{\mathrm{d}^n\exp(x^8)\over\mathrm{d}x^n}(0) by Taylor's theorem. Equating these for n=24 gives {1\over 3!} = {1\over 24!}\;{\mathrm{d}^{24}\exp(x^8)\over\mathrm{d}x^{2  4}}(0).
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  3. #3
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    Quote Originally Posted by maddas View Post
    ...The coefficient of x^n in this power series is equal to {1\over n!}\;{\mathrm{d}^n\exp(x^8)\over\mathrm{d}x^n}(0) by Taylor's theorem. Equating these for n=24 gives {1\over 3!} = {1\over 24!}\;{\mathrm{d}^{24}\exp(x^8)\over\mathrm{d}x^{2  4}}(0).
    I don't understand what the d stands for. could you explain? thanks!
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  4. #4
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    {\mathrm{d}^nf(x)\over\mathrm{d}x^n}(x) is Leibniz's notation for the nth derivative of f at the point x. Perhaps you write it f^{(n)}(x)?
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  5. #5
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    oh! yea, we write it like that.
    thank you!!
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