Getting the hang of Taylor's formula

Question

As we know, Taylor's formula is a method which allows us to find power series for specific functions in a systematical way. I am trying to understand it, not just apply it. Let $f(x)$ be the power series:

$$f(x)=\sum_{i=0}^{\infty} a_{i}(x-x_{0})^i$$

My book states when $f(x)$ is convergent for $(x-x_{0})$ small enough, we can find the coefficient $a_{0}$ by setting $x=x_{0}$. Therefore:

$$f(x_{0})=\sum_{i=0}^{\infty} a_{i}(x_{0}-x_{0})^i= a_{0}$$

But, would not it be the indetermination $0^0$? How could we solve this equation then?

Thank you

It's a standard mathematical convention (widely discussed on this website, and back by several reasons) that $0^0=1$. — Clement C., Apr 24 '18 at 16:50
See e.g https://math.stackexchange.com/questions/11150/zero-to-the-zero-power-is-00-1 — Clement C., Apr 24 '18 at 16:51
This is one of many reasons why it is right to take $0^0=1$. — Angina Seng, Apr 24 '18 at 17:01

score 0 · Accepted Answer · answered Apr 24 '18 at 16:54

0

$0^0$ can be seen as

$$\lim_{x\to 0^+}x^x=\lim_{x\to0^+}e^{x\ln (x)}=e^0=1$$

answered Apr 24 '18 at 16:54

hamam_Abdallah

62,951

For Taylor series, the important limit is $\lim_{x\to0}x^0=1$. – Angina Seng Apr 24 '18 at 17:05
@LordSharktheUnknown If $x\to 0$ then $\lim_0x^x=\lim_0x^0$. – hamam_Abdallah Apr 24 '18 at 17:11
@Salahamam_Fatima a silly question, why does $x^x=e^{xln(x)}$? – JD_PM Apr 24 '18 at 18:42
It has to be a log property but I do not see it – JD_PM Apr 24 '18 at 18:43
@JD_PM In general $x^a=e^{a\ln (x)} $ for $x>0$ and $a $ real. – hamam_Abdallah Apr 25 '18 at 16:09

Getting the hang of Taylor's formula

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