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Math Help - Compute the 9th derivative

  1. #1
    s3a
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    Compute the 9th derivative

    How do I do this the way I am supposed to? (without differentiating 9 times)

    Any help would be greatly appreciated!
    Thanks in advance!
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  2. #2
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    Quote Originally Posted by s3a View Post
    How do I do this the way I am supposed to? (without differentiating 9 times)

    Any help would be greatly appreciated!
    Thanks in advance!
    Use the hint. The Taylor series for \tan^{-1}x is

     <br />
\tan^{-1}x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots.<br />

    Subs in for x \rightarrow \frac{x^3}{4}. You really only need to consider powers up to x^9.
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  3. #3
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    Quote Originally Posted by s3a View Post
    How do I do this the way I am supposed to? (without differentiating 9 times)

    Any help would be greatly appreciated!
    Thanks in advance!
    did you use the hint?

    f(x) = \arctan\left(\frac{x^3}{4}\right) = \frac{x^3}{4} - \frac{x^9}{4^3 \cdot 3} + \frac{x^{15}}{4^5 \cdot 5} - \frac{x^{21}}{4^7 \cdot 7} + ...

    f^{(9)}(0) = -\frac{9!}{4^3 \cdot 3}
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  4. #4
    s3a
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    For arctan(x) I get = sigma from 0 to inf of (-x^(2n)) which gives - x - x^3 /3 - x^5 /5 etc without the alternating. What did I do wrong so far? I did d/dx arctan(x) = 1/(1-(-x^2)) then integrate the sum.
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  5. #5
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    Quote Originally Posted by s3a View Post
    For arctan(x) I get = sigma from 0 to inf of (-x^(2n)) which gives - x - x^3 /3 - x^5 /5 etc without the alternating. What did I do wrong so far? I did d/dx arctan(x) = 1/(1-(-x^2)) then integrate the sum.
    \frac{1}{1-r} = 1 + r + r^2 + r^3 + ...

    \frac{1}{1-(-x^2)} = 1 + (-x^2) + (-x^2)^2 + (-x^2)^3 + ... = 1 - x^2 + x^4 - x^6 + ...
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  6. #6
    s3a
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    Ok I see how to get the summation of arctan(x^3 /4) but then I don't know what to do. Do I just write down the polynomials up to the x^9 power and do something there? What about f^(n) (a) x^n /n! formula ? What I was trying to do before my current attempt was to just do the summation of f^(n) (a) x^n /n! and to replace f^(n) with it's appropriate algebraic expression but it was tough to compute the second derivative and on.

    Sorry for being stupid
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  7. #7
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    So you're OK with this, correct?

    f(x) = \arctan \left(\frac{x^3}{4}\right) = \frac{x^3}{4} - \frac{x^9}{4^3 \cdot 3} + \frac{x^{15}}{4^5 \cdot 5} - \frac{x^{21}}{4^7 \cdot 7} + \cdots

    Now, consider what happens when you take 9 derivatives of this expression.

    The first term \frac{x^3}{4} will eventually become zero (beyond 3 derivatives).

    The second term -\frac{x^9}{4^3 \cdot 3} will become a constant. What constant does it become?

    The remaining terms will turn into the form C \cdot x^n where C is a constant and n is a positive integer. When you substitute x=0 into a term of this form, does it make sense that these terms will all become zero?

    Therefore, there is only one nonzero term, which is the second term.
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  8. #8
    s3a
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    OMG! Yes, I get it now, thanks!
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