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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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    $\begingroup$ It is like the $\xi$ in the mean value theorem $f(x) = f(y) + f'(\xi) (x-y)$, with $\xi \in (x,y)$. $\endgroup$
    – copper.hat
    Jun 18, 2015 at 16:15
  • $\begingroup$ It means that there exists a value of x in the given range such that when the remainder term is added to the value of Taylor polynomial, it will give the value of the function. That value of x is written as z. $\endgroup$
    – Bhaskar
    Jun 18, 2015 at 16:20
  • $\begingroup$ @L16H7 is the remainder (potentially) different for each x value? Or is it always the same? It seems analogous to the derivative function in that it let's you find the error at any point (like derivative let's you find slope at any point) but that's not clear to me yet. Thanks. $\endgroup$
    – Dave
    Jun 18, 2015 at 16:55
  • $\begingroup$ For different values of x, values of z may vary between x and a. So yes, value of remainder term may change with x. $\endgroup$
    – Bhaskar
    Jun 18, 2015 at 17:13

2 Answers 2

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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:

enter image description here

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 $f(c)+f'(c)(x-c)$:

enter image description here

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  • $\begingroup$ Thanks, the drawing helps a bit. So are you saying that we pick a z in order to "pull down" the tangent line (in this case) so it precisely hits f(x)? And we have to pick a z that MVT guarantees exists in that interval because we don't know the f(x) value that we are trying to find? This makes it seem like newton's method in a sense. But then why don't we find that specific z value in order to hit that f(x) with the tangent? $\endgroup$
    – Dave
    Jun 18, 2015 at 16:50
  • $\begingroup$ Also what exactly IS the remainder -- the difference between f(x) and the T_n(x) at that point? That's the visual I get from the drawing but my readings have been unclear on that point . Thanks. $\endgroup$
    – Dave
    Jun 18, 2015 at 16:51
  • $\begingroup$ @Dave: Yes, we use $z$ because we don't know what $z$ is, with variant $x$. The MVT only guarantees there is a $z$ but doesn't tell where it is. I will edit my answer for your other question. $\endgroup$
    – KittyL
    Jun 18, 2015 at 16:56
  • $\begingroup$ I wanted to let you know I didn't forget about this, I just let the answer gel in my head for a while before accepting. Your answer helped me understand that we are looking for the max error in the interval between x and c and that z is how we find the max on that interval. Before that it made no sense to me. So thanks for your thorough explanation and help. $\endgroup$
    – Dave
    Jun 25, 2015 at 19:00
  • $\begingroup$ @Dave: Glad to know that you understand it now. :) $\endgroup$
    – KittyL
    Jun 25, 2015 at 20:56
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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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