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.


  • $\begingroup$ What did you get for the Taylor series? Hint: It is a very simple, so you won't need any general properties about Taylor series. $\endgroup$ – Thomas Andrews Apr 24 '15 at 20:22

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?

  • $\begingroup$ when it converges, should it be x-1 instead of 1-x? I'm confused here. $\endgroup$ – Brendon Apr 24 '15 at 21:32
  • $\begingroup$ and continuing from the lemma, we can just simplify the (n+1)!/ (n+1)! into 1 so that Rn(x) = lx-al^(n+1) $\endgroup$ – Brendon Apr 24 '15 at 21:36
  • $\begingroup$ If you do the ratio test you will see why, note that I included the absolute values $\endgroup$ – Quality Apr 24 '15 at 21:44
  • $\begingroup$ by using the Ratio test, I get that the interval of convergence would be 0<x<2 which is the same as function. Does that prove that Rn(x) is equal to 0? $\endgroup$ – Brendon Apr 24 '15 at 22:07

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