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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?

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up vote 4 down vote accepted

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.

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Distribute and then recollect terms. $$(2x - 3)\sum_n \frac{(5x)^n}{n!} = 2x \sum_n \frac{(5x)^n}{n!} - 3 \sum_n \frac{(5x)^n}{n!}$$$$ = \sum_{n=1}^\infty \frac{2 \cdot 5^{n-1}}{(n-1)!} x^n - \sum_n \frac{3 \cdot 5^n}{n!} x^n$$ $$ = 3 + \sum_{n=1}^\infty \frac{2n \cdot 5^{n-1} + 5^n}{n!} x^n$$

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how did you get from step 1 to step 2, its a bit confusing where the n-1 is coming from – Math_Illiterate Apr 30 '12 at 11:54
Why it is $5^{n-1}$ in the second line, rather than $5^{n+1}$? – Jay Wong Jan 16 at 16:27

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