1. ## 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!

2. ## 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 $\frac{n!}{k!}$

3. ## Re: Creating generic formula

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.

4. ## 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

$$p_n(x)=\sum _{k=0}^n (-1)^{k+n}\frac{n!}{k!}x^k$$