Taylor Series of Products Here's a taylor series problem I've been working on. I'll list a few steps to the problem and tell you guys where I'm getting stuck. Thanks in advance for the help. 
So my questions builds off the fact that 
$ e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}$
and we are asked to find the taylor series of the following function: 
$f(x) = (2x-3)\cdot e^{5x}$ around a = 0
So I first decided to calculate the taylor series for $e^{5x}$ by generating a few terms and noticing the pattern. I then found the following series to represent $e^{5x}$
$ e^{5x} = \sum_{n=0}^{\infty}\frac{5^n}{n!} \cdot x^n$
Next I know I must multiply this series by (2x-3) somehow so I begin like this: 
$(2x-3) \cdot \sum_{n=0}^{\infty}\frac{5^n}{n!} \cdot x^n$
$\sum_{n=0}^{\infty}\frac{(2x-3)5^n}{n!} \cdot x^n$
My problem with this answer is that it's not in the correct form for a taylor series and must be in the form:
$\sum b_n \cdot x^n$
Does anyone know the type of manipulations I must do to convert my result to the correct form?
 A: The Taylor series for $e^{5x}$ can be obtained without generating a few terms and noticing a pattern. Just use the fact that $e^t$ has Taylor series 
$$e^t=1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+\cdots$$
and let $t=5x$.
As for the Taylor series of $(2x-3)e^{5x}$, suppose that you know that the Taylor series of $f(x)$ is $a_0+a_1x+a_2x^2+\cdots$. Then the coefficient of $x^n$ in the Taylor series for $(2x-3)f(x)$ is $2a_{n-1}-3a_n$. We need to make a minor adjustment for the constant term.
A: Distribute and then recollect terms.
$$(2x - 3)\sum_n \frac{(5x)^n}{n!} = 2x \sum_n \frac{(5x)^n}{n!} - 3 \sum_n \frac{(5x)^n}{n!}$$$$ = \sum_{n=1}^\infty \frac{2 \cdot 5^{n-1}}{(n-1)!} x^n - \sum_n \frac{3 \cdot 5^n}{n!} x^n$$ $$ = 3 + \sum_{n=1}^\infty \frac{2n \cdot 5^{n-1} + 5^n}{n!} x^n$$
A: The taylor series for $f(x) = (2x-3)$ is itself. 
The taylor series for $g(x) = \exp(-5x) = \sum_{n = 0}^{\infty} \frac{5^n}{n!}x^n$.
How can we find the Taylor series of the product $f(x) \cdot g(x)$? The Taylor series will always have a form of the following:
$$f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + ... \\
g(x) = b_0 + b_1 x + b_2 x^2 + b_3 x^3 + ...$$
That means that the product of those will be:
$$
(a_0 + a_1 x + a_2 x^2 + a_3 x^3 + ...)(b_0 + b_1 x + b_2 x^2 + b_3 x^3 + ...) = a_0 b_0 + (a_1 b_0 + a_0 + b_1)x + (a_2 b_0 + a_1 b_1 + a_0 b_2) x^2 + (a_3 b_0 + a_2 b_1 + a_1 b_2 + a_0 b_3)x^3 + ...
$$
This shows that this can be reordered into a taylor series with different coefficients. Let's call this new coefficient $c_j$. Then the new taylor series becomes 
$$
\sum c_j (x - x_0)^n.
$$
Where $c_j$ is given by 
$$
c_j = a_j b_0 + a_{j-1} b_1 + a_{j-2} b_2 + a_{j-3} b_3 + ... + a_1 b_{j-1} + a_0 b_{j} = \sum_{k = 0}^{j} a_{j-k} \; b_k
$$
(Simply check the first term where $j = 0$, we obtain $a_0 b_0$. For $j = 1$ we obtain $a_1 b_0 + a_0 b_1$, exactly the same.)
So, back to the problem. The product $P(x) = f(x) \cdot g(x)$ can be written as:
$$
P(x) = 2x \sum_{n = 0}^{\infty} \frac{5^n}{n!} x^n - 3 \sum_{n = 0}^{\infty} \frac{5^n}{n!} x^n = 2x \sum_{n = 0}^{\infty} \frac{5^n}{n!} x^n - \sum_{n = 0}^{\infty} 3 \frac{5^n}{n!} x^n
$$
We are left with the first term. The coefficient of $2x$ are $a_1 = 2$ and the rest is zero. This will simplify the expression. For $g(x) = \sum_{n = 0}^{\infty} \frac{5^n}{n!} x^n$ the coefficients are $b_j = \frac{5^j}{j!}$.:
$$
c_j = a_j b_0 + a_{j-1} b_1 + ...= a_1 b_0 = 2 \frac{5^{(n-1)}}{(n-1)!}
$$
Therefore $P(x) = f(x) \cdot g(x)$ is (with $x_0 = 0$)
$$
P(x) = \sum_{n = 0}^{\infty} 2 \frac{5^{(n-1)}}{(n-1)!} x^n - \sum_{n = 0}^{\infty} 3 \frac{5^n}{n!} x^n
$$
