How would I set up the Taylor's Inequality to prove that the function $f(x) = \frac{1}{x}$ is equal to its Taylor Series expansion centered at $x=1$?
I've done the Taylor series expansion, but I'm a bit unsure on how the function is equivalent to its Taylor Series Expansion. I understand that the Taylor's Inequality would satisfy: $f(x) = T(x) + R(x)$, but I don't know how to complete it.
Best Answer
So you are trying to prove that your function is equal to its power series representation, and as you said to do this you will use Taylors theorem and inequality.
$\mathbf{Thereom:}$ If $f(x)=T_n(x)+R_n(x) , $ where $T_n$ is the $n^{th}$ degree taylor polynomial of $f$ at $a$ and $\lim_{n\to \infty} R_n(x)=0$ for $|x-a| \lt R$ , then $f$ is equal to the sum of its Taylor series on the interval $|x-a| \lt R$ ( theorem as given in Stewart 7ed.)
And we are wanting to take advantage of the following
$\mathbf{Lemma}:$ If $| f^{n+1}(x)| \le M$ for $|x-a| \le d$ , then the remainder $R_n(x)$ of the Taylor series satisfies the inequality
$$|R_n(x)| \le \frac{M}{(n+1)!}|x-a|^{n+1}$$ for |$x-a| \le d$
So what is the general form of the taylor series for $f(x)=\frac{1}{x}$ centred at $x=1$?
As you said , it has the general form $\sum_{n=0}^{\infty}(-1)^{n}(x-1)^{n}$ and converges for $|1-x| \lt 1$
Now , $f(x)=(1/x)$ and we have that $f^{n}(1)=(-1)^{n}n!$
$f^{n+1}(1)=(-1)^{n+1}(n+1)! \le (n+1)! $
$|x-a| \le 1$
and by the lemma we have $| R_n(x)| \le \frac{(n+1)!}{(n+1)!}|x-a|^{n+1}$
Do you see anything we could do from here?