2
$\begingroup$

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

$\endgroup$
7
  • $\begingroup$ Your "power series expansion" is not a power series. $\endgroup$
    – fqq
    May 11 '16 at 0:31
  • 1
    $\begingroup$ For the series expansion centered on $0$, you will need to evaluate the derivatives at $0$. But there is a quicker way, using the known expansion of $e^t$. $\endgroup$ May 11 '16 at 0:34
  • $\begingroup$ @fqq What did I do then? $\endgroup$
    – Speakmore
    May 11 '16 at 0:35
  • $\begingroup$ I would review the definition of a Taylor power series. Notice that the power series is simply the normal Taylor series for $e^x$ with$ x=2x$. Remember that you need to expand the Taylor series around a given point, so you evaluate all the derivatives at that point. Your professor has naturally given the expansion around $x=0$. If you want to do this manually (I.e. Without using the known series for the exponential function) then review the definition of a Taylor series. Does this help, at least? $\endgroup$
    – KR136
    May 11 '16 at 0:36
  • $\begingroup$ @Αδριανός I know what a Taylor Series is, as well as that when centered at 0 it's actually a Maclaurin Series (more specifically). The equation being: $$f(x)=f(a) + f^'(a)(x-a) + f^{''}(a)/(2!)(x-a)^2 ...$$ $\endgroup$
    – Speakmore
    May 11 '16 at 0:39
3
$\begingroup$

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.

$\endgroup$
-1
$\begingroup$

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
  • $\begingroup$ So I'm not entirely grasping the concept of what I do to come up with the $x^n$ part after plugging in 0 to my derivatives. Plug 0 into the first derivative, get 2. Plug 0 into the second derivative, get 2. Plug 0 into the third derivative, get $8/6$ = $4/3$ Plug 0 into the fourth derivative, get $16/24$ = $2/3$ $\endgroup$
    – Speakmore
    May 11 '16 at 0:45
  • $\begingroup$ You don't come up with $x^n$, they're just there. When you plug in $0$ to your derivatives, you calculate the $a_n$'s (with $n!$ in the denominator). You just write the $a_n$ followed by $x^n$ in the series. $\endgroup$ May 11 '16 at 0:48
  • $\begingroup$ So is $x^n$ after every summation for all problems. That's the given and you have to find what else you have to manipulate it by..? So $f(x) = cos(x)$ has some variation of $x^n$? $\endgroup$
    – Speakmore
    May 11 '16 at 0:53
  • $\begingroup$ Yes, every power series has some $x^n$. In fact, the power series of the cosine is $$\cos(x) = \sum\limits_{n=0}^{\infty}{(-1)}^n\frac{x^{2n}}{(2n)!}$$ (so in the previous formulation, $a_n=0$ for all odd $n$). $\endgroup$ May 11 '16 at 0:56
  • $\begingroup$ I guess I just don't grasp the concept of how you'd turn $cos(x)$ into the correct power series expansion on the right hand side. Or any function for that matter. $\endgroup$
    – Speakmore
    May 11 '16 at 1:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.