manipulate well known Taylor series

Question

How can you manipulate a known power series when $x ≠ 0$ ?

For example $f(x)=3x e^x$ What is the Taylor series to the first 5 terms generated by $f$ at $x=4$ ?

I know $$3xe^x = 3x + 3x^2 + \frac{3x^3}{2} + \frac{3x^4}{3!} + \frac{3x^5}{4!} +... \frac {3x^{n+1}}{n!}$$ because of the well known Taylor series for

$$ e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$

but not sure how to rework this at $x=4$

Answer 1

Let $t = x - 4$. Then $f(x) = f(t+4) = 3e^4 (t+4) e^t$. Can you take it from there?

Answer 2

Taylor theorem states that $\dots$

$$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)(x-a)^n}{n!}$$

So what you want to do is take $f(x)$ to be $3xe^x$ and $a = 4$. From there it is just a plug and chug to find the polynomial expansion.

Answer 3

You can do it like this.

$3xe^x=3(x-4+4)e^{x-4+4}=3e^{4}(x-4)e^{x-4}+12e^4e^{x-4}$

Then use the series that you know $e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$ to instead write $e^{x-4}=\sum_{n=0}^{\infty}\frac{(x-4)^n}{n!}$

Putting this into the expression of your function above yields

$$\begin{align}3e^{4}(x-4)e^{x-4}+12e^4e^{x-4}&=3e^4(x-4)\sum_{n=0}^{\infty}\frac{x^n}{n!}+12e^4\sum_{n=0}^{\infty}\frac{x^n}{n!}\\&=12e^4+\sum_{n=1}^{\infty}3e^4\left(\frac{1}{(n-1)!}+\frac{4}{n!}\right)(x-4)^n\end{align}$$

Answer 4

If the radius of convergence of the development of the function and its derivatives around $x=0$ contains the new point ($x=4$), you can evaluate the new Taylor coefficients from the series:

$$f(4)=3\sum_{n=0}^\infty\frac{4^{n+1}}{n!}, \\f'(4)=3\sum_{n=0}^\infty(n+1)\frac{4^n}{n!}=3\sum_{n=1}^\infty\frac{4^n}{(n-1)!}+3\sum_{n=0}^\infty\frac{4^n}{n!}, \\f''(4)=3\sum_{n=2}^\infty\frac{4^{n-1}}{(n-2)!}+3\sum_{n=1}^\infty\frac{4^{n-1}}{(n-1)!}, \\\cdots$$