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?


3 Answers 3


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.


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

  • $\begingroup$ how did you get from step 1 to step 2, its a bit confusing where the n-1 is coming from $\endgroup$ Apr 30, 2012 at 11:54
  • $\begingroup$ Why it is $5^{n-1}$ in the second line, rather than $5^{n+1}$? $\endgroup$
    – Jay Wang
    Jan 16, 2016 at 16:27
  • $\begingroup$ @JayWong The sum was $\frac{(5x)^0}{0!}+\frac{(5x)^1}{1!}+\frac{(5x)^2}{2!}+\ldots$ for $n=0,1,2,\ldots$ and was rewritten as $\frac{(5x)^{(1-1)}}{(1-1)!}+\frac{(5x)^{(2-1)}}{(2-1)!}+\frac{(5x)^{(3-1)}}{(3-1)!}+\ldots$ for $n=1,2,3,\ldots$. Since the first index was increased by $1$, the terms $a_n$ had to be changed to $a_{n-1}$ to accomodate. $\endgroup$
    – Jam
    Dec 6, 2019 at 13:41

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 $x_0 = 0$)

$$ P(x) = \sum_{n = 0}^{\infty} 2 \frac{5^{(n-1)}}{(n-1)!} x^n - \sum_{n = 0}^{\infty} 3 \frac{5^n}{n!} x^n $$


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