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Thread: Creating generic formula

  1. #1
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    Question Creating generic formula

    Hi,

    I'm trying to create a formula for the function $g^n(x)$ which has the following derivative pattern:

    $$g^1(x) = \frac{e^x(x-1) + 1}{x^2}$$
    $$g^2(x) = \frac{e^x(x^2 - 2x + 2) - 2}{x^4}$$
    $$g^3(x) = \frac{e^x(x^3 - 3x^2 + 6x - 6) + 6}{x^4}$$
    $$g^4(x) = \frac{e^x(x^4 - 4x^3 +12x^2 -24x + 24) - 24}{x^5}$$
    $$g^5(x) = \frac{e^x(x^5 - 5x^4 + 20x^3 - 60x^2 +120x - 120) + 120}{x^6}$$
    etc.

    What I have so far is $$g^n(x) = \frac{e^x(x^n - [SUMMATION HERE])}{x^{n+1}}$$

    I can't seem to figure out what sort of summation is required for the expression inside the brackets that says SUMMATION HERE. Help would be very much appreciated!
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  2. #2
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    Re: Creating generic formula

    $\displaystyle g^{(n)}(x)=\frac{e^xp_n(x)-(-1)^nn!}{x^{n+1}}$

    where $p_n$ is a monic polynomial of degree $n$ satisfying the condition

    $\displaystyle p_n(x)+ \left(p_n(x)\right)'=x^n$

    Prove this by induction.

    the coefficients of the polynomial $p_n$ are of the form $\frac{n!}{k!}$
    Last edited by Idea; Jun 3rd 2019 at 04:16 AM.
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  3. #3
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    Re: Creating generic formula

    Thanks for your reply.

    With your expression, can you show how this is equal to $g^{\prime}(1)$ and $g^{\prime}(2)$ ? I ask since I am a bit unsure how $p_n(x) - (-1)^n n!$ would work. It is also unclear to me where the value k came from.
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  4. #4
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    Re: Creating generic formula

    Here is an example

    $$p_5(x)=x^5-5x^4+20x^3-60x^2+120x-120=x^5-\frac{5!}{4!}x^4+\frac{5!}{3!}x^3-\frac{5!}{2!}x^2+\frac{5!}{1!}x-\frac{5!}{0!}$$

    in summation form it would be

    $$p_n(x)=\sum _{k=0}^n (-1)^{k+n}\frac{n!}{k!}x^k$$
    Last edited by Idea; Jun 4th 2019 at 06:45 AM.
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