[Math] Proving Lagrange’s Remainder of the Taylor Series

calculustaylor expansion

My text, as many others, asserts that the proof of Lagrange's remainder
is similar to that of the Mean-Value Theorem.

To prove the Mean-Vale Theorem, suppose that f is differentiable over $(a, b)$
and continuous over $[a, b]$. Then, for $x \in (a, b)$, define
$$ F(x) = f(x) – f(a) – {{f(b)-f(a)} \over {b-a}}(x-a)$$
$F(b) = 0, F(a) = 0$. So apply Rolle's Theorem and you'll have a $\zeta \in (a, b)$ such that
$$F'(\zeta) = f'(\zeta) – {{f(b)-f(a)} \over {b-a}} = 0$$
$$f'(\zeta)(b-a) = f(b) – f(a)$$
I was able to find a graphical interpretation of $F(x)$,
which helped me a great deal in understanding (and memorizing) the proof.
Namely, $F(x)$ represents the difference between $f(x) – f(a)$ and
the height of the triangle created by the secant line between $f(a)$ and $f(b)$.

Is there something similar with the proof of Lagrange's Remainder?

Let $f$ be a real-value function on an open interval $I$, $c \in I$.
Furthermore, $f^{(n+1)}(t)$ exists for every $t \in I$.
Let $x \in I$ be fixed and $M$ be a value such that $f(x) = T_n(c, x) + M(x-c)^{n+1}$

Now, the text defines
$$g(t) = f(t) – T_n(c, t) – M(t-c)^{n+1}$$
I was able to confirm that $g(c) = 0, g(x) = 0$, and $g'(x) = g''(x) \space … \space = g^{(n)}(c) = 0$.
The question is, what does $M(t-c)^{n+1}$ represent?
Is it the entire difference between $f(t)$ and the Taylor expansion of $f$,
or only a part of the difference?

Thanks for taking the time to read this and response.

Best Answer

Old question, but...

The short answer is that $M(t-c)^{n+1}$ represents the entire difference between $f(t)$ and the Taylor series of $f$ about $c$ at the specific point $t=x$. In other words, $M(t-c)^{n+1}=R_n(c, x)$, or the error when using $f(x)\approx T(c, x)$ to estimate $f(x)$ for some specific $x$ near $c$. (The choice is made in order to turn the proof into a problem of solving for $M$.)

(It's easy to think that in that case the function $g(t) = f(t) - T_n(c, t) - M(t-c)^{n+1}$ must be zero for all $t$, but this is not (necessarily) the case since $M$ is defined in terms of the error when approximating $f$ at a specific value $x$ as opposed to the general $t$.)

It is clear after a bit of exploration that $g^\prime(x) = g^{\prime\prime}(x) \space ... \space = g^{(n)}(c) = 0$ (as you point out). Rolle's theorem can be applied to this (which is where the proof is similar to that of the mean value theorem) repeatedly (unlike in the proof of the MVT, where it's only applied once).

As for a geometric interpretation, it is a bit tricky in general, but it can be done with a clever choice of $f(t)$. For example, $f(t)=e^t$ is convenient because of its property $f^{(k)}(t)=f(t)$ for all $k\in \mathbb Z^{\ge 0}$. Now:

  • Plot $f(t)=e^t$.
  • Take the first few terms $T_n(t)$ from the Taylor series about some point $c$. (Choose a low $n$ such as 2 or 3 so that the error can be easily seen.)
  • Use polynomial $T(t)$ to approximate $f(x)$ for some $x$ near $c$.
  • Plot the point $(x, T(x))$ so that the gap between the estimate and the true value is apparent.
  • Now you know that there is some point $\nu\in]x, c[$ such that $\displaystyle f^{(n+1)}(\nu)\cdot\frac{(x-c)^{n+1}}{(n+1)!}$ is equal to the gap between the point and the curve.
  • Since we chose $f(t)=e^t$, the value of the $f^{(k)}$ is the value of $f$, i.e. the curve.
  • Since $x$, $c$ and $n$ are of our own choosing, the constant $\frac{(x-c)^{n+1}}{(n+1)!}$ can be determined. The gap between the point and the curve can now be interpreted as the dilation of this factor of one of the values of the curve in the region $]x,c[$.

This geometric interpretation can also be explored with choices of $f$, like $f(t)=\sin t$, where $f^{(k)}(t)$ is easily determined for arbitrary $k$. (Actually, this might be a little less misleading, because the gap is seen as a dilation of some value of the derivative, a separate graph (unless $4|k$ in the case of $\sin t$, of course)).

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