Suppose I have the function $f(x)=\frac{1}{x-1}$ and its Taylor polynomials that model it. The Taylor polynomials are centred at $x=0$ and so the $n^{th}$ order Taylor polynomial would be: $$T_n\left(x\right)=-1-x-x^2-x^3-x^4-...-x^n$$

Also, as the order of the Taylor polynomial increase, it gets closer and closer to approximating $f(x)$. However, I have tried inputting larger and larger orders of Taylor polynomials and still come out short from actually modelling the function perfectly. So, is there a way $T_n$ can perfectly model $f(x)$?

I was thinking this could work if the domain of the whole thing was just at that one point where $T_n$ intercepts with $f(x)$. If this would be the case then $T_n$ = $f(x)$. would this be true?

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    $\begingroup$ If the $n$th Taylor polynomial ever matched its function perfectly (i.e. equal in value on its entire domain), then the function would be the polynomial. Unless the function you're approximating is a polynomial, there will always be some inaccuracy. $\endgroup$ – Christian May 14 '17 at 1:50
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    $\begingroup$ @Christian That should really be an answer . . . $\endgroup$ – Noah Schweber May 14 '17 at 1:52
  • $\begingroup$ @Christian So under no circumstance is this possible then? $\endgroup$ – S.Sharifi May 14 '17 at 1:54

No, unless your original function $f(x)$ happens to already be a polynomial, you'll never have a large enough $n$ so that $f(x) = T_n(x)$ for all $x$. This is because $f(x)$ is not a polynomial and $T_n(x)$ is.


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