# How to recast these two exponential of infinite power series as simple power series?

I have two exponential of infinite power series, with different expressions for coefficients $a_n$, that I would like to recast as two other power series without the exponential
$$\exp\left(\sum_{n=1}^{\infty}a_n x^n \right)=\sum_{n=1}^{\infty}b_n x^n$$ The two approaches I considered are :

• Expanding the exponential term as a Taylor series $$\exp\left(\sum_{n=1}^{\infty}a_n x^n \right) = \sum_{k=0}^{\infty} \frac{\left(\sum_{n=1}^{\infty}a_n x^n \right)^k}{k!}$$ in which case the numerator of the fraction is the Cauchy product of $k$ infinite power series;
• Using complete Bell polynomials $B_n$ $$\exp\left(\sum_{n=1}^{\infty}a_n x^n \right) = \sum_{n=0}^\infty \frac{B_n(a_1,\dots,a_n)}{n!} x^n$$

In both cases, and even if my $a_n$ coefficients are very (very) simple, I'm struggling to obtain an exploitable expression for $b_n$ coefficients. The $a_n$ coefficient of my two power series are

• $a_n=C_1$

and

• $a_n=(-1)^{n+1}(\frac{C_2}{n}+nC_3)$

where $C_1$, $C_2$ and $C_3$ are constants. I am sure that it is possible to reduce $b_n$ to polynomials of degree $n$, but I just can't make it.

Thank you in advance for you help.

## 1 Answer

Of course $$\sum_{n=1}^\infty C_1 x^n = \frac{C_1 x}{1-x}\ \text{for}\ |x|<1$$ so in your first case you want the Maclaurin series of $$\exp\left(\frac{C_1 x}{1-x}\right)$$ This is the solution of the differential equation $$(x-1)^2 y'(x) - C_1 y(x) = 0 \ \text{with}\ y(0) = 1$$ from which we can get the recurrence $$n b_n - (C_1 + 2 + 2 n) b_{n+1} + (n+2) b_{n+2} = 0,\ b_0 = 1,\; b_1 = C_1$$ but I don't know of a closed-form solution to this.

In your second case, $$\sum_{n=1}^\infty (-1)^{n+1} \left(\frac{C_2}{n}+n C_3\right) x^n = C_2 \ln(1+x) + \frac{C_3 x}{(1+x)^2}$$ so you want the Maclaurin series of $$(1+x)^{C_2} \exp \left(\frac{C_3 x}{(1+x)^2}\right)$$ This is a solution of the differential equation $$(1+x)^3 y'(x) - (C_2 x^2 + (2 C_2 - C_3) x + C_2 + C_3) y(x) = 0 \ \text{with}\ y(0)=1$$ which leads to the recurrence $$(n+3) b_{n+3} + (3n+6-C_2-C_3) b_{n+2} + (3n+3-2 C_2 + C_3) b_{n+1} + (n-C_2) b_n = 0,\; b_0 = 1,\; b_1 = C_2 + C_3,\; b_2 = \frac{(C_2+C_3)^2}{2} - \frac{C_2}{2} - 2 C_3$$

• Thank you very much. Your solution is far more simple than using the two approaches I tried. Unfortunately I don't understand how you managed to obtain the recurrence from the differential equation. – Eliam Dec 7 '17 at 19:32
• If $y(x) = \sum_n b_n x^n$, collect the terms of the left side of the differential equation involving $x^n$. Thus you get $(n+1) b_{n+1}$ from $y'$, $n b_n$ from $x y'$, etc. – Robert Israel Dec 7 '17 at 22:51
• I figured out in the meantime, but thank you very much for your help and your explanation. – Eliam Dec 9 '17 at 17:55