# Remainder Term for a Taylor Series and Polynomials

If $$f(x)=e^x$$ and $$f_n(x)$$ is the $$n^{th}$$ order Taylor Polynomial of this function, then we know that the Taylor Theorem states there exists $$t\in [x_0, x]$$ such that the remainder term of is given by,

$$r_n(x)=\frac{f^{(n+1)}(t)}{(n+1)!}(x-x_0)^{n+1}=\frac{e^t}{(n+1)!}(x-x_0)^{n+1}$$

as $$f^{(n+1)}(t)=f(t)=e^t$$. Now we also know,

$$f=f_n+r_n$$

$$\Rightarrow e^x = f_n(x) + \frac{e^t}{(n+1)!}(x-x_0)^{n+1}$$

On the right hand side, $$f_n$$ is a polynomial and so is the expression for the remainder term. However $$e^x$$ is not a polynomial. How is it possible to write a non polynomial function in terms of a polynomial? We can use a similar procedure and arrive at this contradiction for other non polynomial functions as well. So what is the gap in my understanding?

• The second term on the RHS (the remainder term) is not a polynomial, because $t$ is a function of $x$, not a constant.
– user169852
Mar 15, 2021 at 5:23
• We aren't saying that the functions are equal, just that they take the same value at the particular number $x$. Mar 15, 2021 at 5:30

$$t$$ is just a value between $$x_0$$ and $$x$$ and $$e^t$$ is how to evaluate this constant. It does depend on $$x$$, but nobody said that Taylor polynomial is expressing the function in full. It has an error in its approximation.

The term with $$t$$ is that error term, so it talks about an error in Taylor polynomial approximation. It is not part of polynomial and of course that it depends on $$x$$, because typically more you are away from $$x_0$$ worse the error.

Just because it has $$(x-x_0)^{n+1}$$ in it does not make it a part of the polynomial, there are other expressions for the error term that look less like an additional polynomial term.

So, yes, besides $$(x-x_0)^{n+1}$$ the error term may have one additional dependency on $$x$$, or it may not, we do not know. All we know is that $$t$$ is between the two. In general it is movable regardless if this move is expressible as a function on $$x$$ or not.

Typically this $$t$$ is used to say that max error cannot be worse than, in your case, the maximum value of $$e^t$$ where $$t$$ is between $$x$$ and $$x_0$$ together with other values in that term. Some functions will have this max value fixed for quite some possible range between $$x$$ and $$x_0$$, some will not. But anyway $$t$$ is not really useful beside deciding this max error, because we know nothing else about $$t$$ except that it is between $$x$$ and $$x_0$$.

If $$t$$ would be a fixed value then the error term would be in form of $$k(x-x_0)^{n+1}$$, meaning every function would be expressible as a polynomial of $$n+1$$ degree, and unless the function is such a polynomial that is not the case.

• -1 vote and accepted answer. Funny. Mar 16, 2021 at 19:22