Second-degree Taylor Polynomial for $e^{4x}$ centered at $x=3$

Question

Find the second-degree polynomial about the given point. $$e^{(4x)}, x = 3$$

${e^{4x} = }$ ______________

---

I'm stuck on how to find the approximate value for $e^{4x}$. This is what I first tried doing but couldn't find the appropriate answer.

$$f(x) = P_2(x) = f(0) + f'(0)x + {f''(0)\over2}{x^2}$$ $$P_2(x) = e^{(4)(0)} + 4e^{(4)(0)}(3) + {16e^{(4)(0)}\over2}(9)$$ $$P_2(x)= 1 + 12 + 72$$ $$P_2(x) = 85$$

I don't know what I'm doing wrong because I believe that I am following the formula correctly. Any help would be great, thanks.

Answer 1

You are following the correct formula... except your whole series is based around x=0!

Also note that when trying to find the polynomial, you do not evaluate the $x,$ $x^2,$ etc. terms. It should be in your notes, as Ted Shifrin seems to have said.

What you should do instead is work at x=3 ; note that this changes $x$ into $x-3$ and $x^2$ to $(x-3)^2$ as a result of the 'transformation' from $x=0$ to $x=3.$

So it would be...

$P_2(x)=f(3)+f'(3)(x-3)+f''(3)\frac{(x-3)^2}{2}=e^{12}+4e^{12}(x-3)+8e^{12}(x-3)^2=\boxed{e^{12}(8x^2-44x+61)}.$

Answer 2

You can use the classical formula if you start making $4x=4(x-3)+12$ and then define $y=4(x-3)$. So $$e^{4x}=e^{4(x-3)+12}=e^{12}e^{4(x-3)}=e^{12} e^y$$ $$e^y=1+y+\frac{y^2}{2}+O\left(y^3\right)$$ Replace $y$ by $4(x-3)$ to get $$e^{4x}=e^{12}\left(1+4(x-3)+\frac 12 4^2(x-3)^2+O\left((x-3)^3\right)\right)$$ No, expand $$1+4(x-3)+\frac 12 4^2(x-3)^2=8 x^2-44 x+61$$ to end with pie314271's result.