What is the purpose of the second variable in Taylors remainder theorem?
In textbooks and online tutorials I see that the remainder is calculated by using a new unknown variable on the same interval. For example we take the Taylor polynomial $T_n(a)$ but find the remainder $R(x)$ with a new variable $z$ inside it. See this tutorial for an example.
Maybe I'm dense but I haven't seen an explanation for this other variable being used. Or at least not one I can understand. I originally thought the remainder was the difference between the value of $f(a)$ and $T_n(a)$ but it appears instead that it is a function like the derivative that can find the error at any point. But I'm not even sure if that's the case.
What is the purpose of $z$ in that example? And why do we find the remainder at an arbitrary value instead of the value we are estimating?
I'm in Calc II if that's relevant. Thanks.
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$\begingroup$Consider the easy case where $f(x)=f(c)+f'(z)(x-c)$. If you are in Calc 2, you should have learned linear approximation. The linear approximation of $f(x)$ around point $c$ is $f(x)\approx f(c)+f'(c)(x-c)$. See the following picture:
If you use $x$ in that derivative, you only get an approximation of $f(x)$, but if you use $f'(z)$ where $z$ is some value such that the slope of the tangent line at that point is equal to the slope of the secant line between $c$ and $x$, then you get the exact value of $f(x)$. So that's why we have to use a new variable $z$. Because it depends on your $c$ and $x$.
Edit: This is to answer the question from the comment. The remainder, in this linear case, is the error when you use $f(c)$ to approximate $f(x)$. It is $f(x)-f(c)=f'(z)(x-c)$ for some $z$. You can see from the picture that $z$ exists somewhere between the two points.
In second order case, we write $$f(x)=f(c)+f'(c)(x-c)+\frac{f''(z)}{2}(x-c)^2$$
Considering this as linear approximation with a remainder $\frac{f''(z)}{2}(x-c)^2$. So $f(x)$ is approximated by $f(c)+f'(c)(x-c)$, but the error is the remainder term.
The following picture shows the remainder (or error) when you approximate $f(x)$ with $0$-th order $f(c)$ or with $1$st order $f(c)+f'(c)(x-c)$:
Taylor's theorem with remainder term is not meant to allow the exact computation of $f(x)$ for some $x$ near $a$, using a hell of a detour. The remainder term is just intended to be a help in estimating the error when you replace the exact value $f(x)$ by its $n$th Taylor approximation $j_{\>a}^nf(x)$. The size of the error is connected with the distance $|x-a|$ and the size of $f^{n+1}$ in the neighborhood of $a$ in a particular way. Writing $f^{n+1}(\xi)$, whereby $\xi$ is some unknown point between $a$ and $x$, is just a handy way of referencing the dependence of the error on $f^{(n+1)}$.
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