I have a function $$f(x) = e^{-x(2a-x)}$$ It is easy to write out the Taylor expansion for this function in $x=0$ point. But.. I want to get a formula for the Taylor coefficients. I tried a lot. I tried to take derivatives, perform some simple transformations and used the computer, trying to calculate them. Nothing much happened. Maybe there is some clever trick here? I would be very grateful for that.
I got the following expression: $$f(x) = e^{-x(2a-x)} = \sum_{k=0}^{\infty}\frac{(-1)^kx^k(2a-x)^k}{k!}=\sum_{k=0}^{\infty}\frac{(-1)^kx^k}{k!}\sum_{m=0}^{k}\frac{k!(2a)^{k-m}(-1)^mx^m}{m!(k-m)!}=$$ $$=\sum_{k=0}^{\infty}\sum_{m=0}^{k}\frac{(-1)^{m+k}(2a)^{k-m}x^{k+m}}{m!(k-m)!}$$ I thought to designate $m + k = s$ and rewrite the sums in a more convenient form for determining the coefficients, but nothing happened.
Thank you in advance!
As you said, this is not the most simple problem.
Using brute force, developed at $x=0$ $$e^{-x(2a-x)}=1-2 a x+\left(2 a^2+1\right) x^2-\left(\frac{4 a^3}{3}+2 a\right) x^3+\left(\frac{2 a^4}{3}+2 a^2+\frac{1}{2}\right) x^4-\left(\frac{4 a^5}{15}+\frac{4 a^3}{3}+a\right) x^5+\left(\frac{4 a^6}{45}+\frac{2 a^4}{3}+a^2+\frac{1}{6}\right) x^6+O\left(x^7\right)$$
Now, searching for a pattern in $$e^{-x(2a-x)}=\sum_{k=0}^\infty c_k\,x^k$$ the $c_k$'s correspond to the recurrence equation $$c_k=-\frac 2 k(a\,c_{k-1}-c_{k-2})\qquad \text{with}\qquad c_0=1\qquad \text{and}\qquad c_1=-2a$$ I do not think that we could have anything more explicit.