Power Series Expansion $e^{2x}$
I'm currently finding the power series expansion centered at 0 for a bunch of functions. My answers are starting not to add up with the 'correct' ones and I'm looking for some assistance as to what I'm doing wrong, or if my answers are correct but just written differently...
For the problem: $$f(x) = e^{2x}$$
I derived it 4 times:
$$\frac{f^1(x)}{n!} = \frac{2e^{2x}}{1!}$$
$$\frac{f^2(x)}{n!} = \frac{4e^{2x}}{2!}$$
$$\frac{f^3(x)}{n!} = \frac{8e^{2x}}{3!}$$
$$\frac{f^4(x)}{n!} = \frac{16e^{2x}}{4!}$$
That was enough for me to "come up with the pattern." So I wrote for my power series expansion:
$$\sum_{n=0}^\infty \frac{2^ne^{2n}}{n!}$$
However, my professors answer is:
$$e^{2x} = \sum_{n=0}^\infty \frac{(2x)^n}{n!} = \sum_{n=0}^\infty (2)^n\frac{x^n}{n!}$$
I'm not really sure where the $x$ comes from in his solution, replacing $e$.
Thanks for your help
2 Answers
$\begingroup$As you mentioned in your comment, and as others have pointed out, you know that the Taylor series about $0$, or the Maclaurin series for a function $f(x)$, is defined:
$f(x)=\displaystyle \sum_{n=0}^∞ \frac{x^n \cdot f^{(n)}(0)}{n!}$
Where $f^{(n)}(0)$ represents the $n$th derivative of $f$ evaluated at $0$.
You did right in noticing that the $n$th derivative of $f(x)=e^{2x}$ is:
$f^{(n)}(x)=2^n e^{2x}$
Now, notice, as you let $x=0$, the $n$th derivative evaluated at $0$ is:
$f^{(n)}(0)=2^n$
Now, using the definition you provided for the Maclaurin series, we have that:
$f(x)=e^{2x}=\displaystyle \sum_{n=0}^∞ \frac{2^{n} x^n}{n!}$
as req'd.
$\endgroup$ $\begingroup$A power series expansion of $f$ about a point $x=x_0$ is an expression of the form
$$\sum\limits_{n=0}^{\infty} a_n\cdot (x-x_0)^n$$
If $f$ is analytic, then the coefficient $a_n$ is given by
$$a_n =\frac{ f^{(n)}(x_0)}{n!}$$
In your case, $x_0=0$ and $f(x)=e^{2x}$. You've calculated the first few derivatives of $f$, but you have not evaluated them at $x_0=0$ (to calculate the $a_n$).
A different approach, which I recommend, is the following. The power series of $e^x$ about $0$ is well known and given by
$$\sum\limits_{n=0}^{\infty}\frac{x^n}{n!}$$
In fact, if you have never before calculated this power series, I encourage you to do so using the observations above.
Now, $e^x$ is a very wel-behaved function, and we have that it actually is equal to its power series at every $x$. In particular, if you substitute $x$ by $2x$ in the power series expansion of $e^x$, you will obtain the power sereis expansion for $e^{2x}$:
$$\sum\limits_{n=0}^{\infty}\frac{(2x)^n}{n!}= \sum\limits_{n=0}^{\infty}2^n \cdot \frac{x^n}{n!}$$
$\endgroup$ 6
