5
$\begingroup$

So, given a Taylor series: $$f(x)=f(x_0)+f'(x_0)(x-x_0)+f''(x_0)\frac{(x-x_0)^2}{2!}+\cdot\cdot\cdot+f^{(n)}(x_0)\frac{(x-x_0)^n}{n!}+R_n$$ The error $R_n$ is given by: $$R_n=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1}$$ for some $\xi\in(x_0,x)$.

What does that mean? I don't understand what $\xi$ is and why the remainder is equal to that. Is there some geometrical proof?
Thank you for your time.

$\endgroup$
  • 2
    $\begingroup$ This is a very interesting piece of exposition on that theorem from a master. $\endgroup$ – Giuseppe Negro May 7 '16 at 10:23
1
$\begingroup$

I see that this is a pretty old question, but here goes anyway.

The main problem with a geometric approach is that we are often dealing with very high order derivatives. Just by looking at a graph we can easily get a sense of such geometric interpretations as value, gradient and concavity - corresponding respectively to $f^{(0)}(x)$, $f^{(1)}(x)$ and $f^{(2)}(x)$ - but after that it starts to become a struggle to interpret function behaviour visually.

Having said that, if we choose $f(x)=e^x$, for which $f^{(k)}(x) = e^x$ for all $k$, we can produce something of a geometric interpretation of the Lagrange error term, especially if we start off with low degree Taylor polynomials for the approximation and incrementally incorporate more terms from the series into the polynomial approximation.

We know that:

$$\begin{align} f(b) = \sum_{k=0}^\infty\frac{f^{(k)}(a)(b - a)^k}{k!}&= \sum_{k=0}^n\frac{f^{(k)}(a)(b - a)^k}{k!}+\sum_{k=n+1}^\infty\frac{f^{(k)}(a)(b - a)^k}{k!}\\\\ &= T_n(b:a) + R_n(b:a) \end{align}$$

And that:

$$\left(\exists c \in ]a, b[\right)\left(\frac{f^{(n+1)}(c)(b-a)^{n+1}}{(n+1)!}=R_n(b:a)=\sum_{k=n+1}^\infty\frac{f^{(k)}(a)(b - a)^k}{k!}\right)$$

Now, for example, let's say that we want to use a Taylor series of $f(x) = e^x$ about $a = 0$ to estimate $f(b = 2)$.

If we make the $n=0$ assertion (in other words, we say that $f(b)\approx f^{(0)}(a)$ independent of $b$, which generally is only a good idea for $b$ very close to $a$), we are effectively saying that there is some $c$ in the region $]a, b[$ such that:

$$\frac{f^{(0+1)}(c)(b-a)^{0+1}}{(0+1)!}=e^b - e^a$$

enter image description here

In this particular example, where we have specified the values, we can calculate $c$:

$$2e^c=e^2 - 1\\c = \ln\left(\frac{e^2 - 1}{2}\right)\approx 1.16$$

With $c$ and $f(x)$ translated down by twice the gradient of $f$ at $c$ (which, since we're using $f(x)=e^x$, is twice $f(c)$) shown:

enter image description here

Similarly, if we make the $n=1$ assertion (in other words we say that $f(b)\approx f^{(0)}(a)+ f^{(1)}(a)(b - a)$) we are effectively saying that there is some $c$ in the region $]a, b[$ such that:

$$\frac{f^{(1+1)}(c)(b-a)^{(1 + 1)}}{(1+1)!}=e^b - (e^a + e^a(b - a))$$

Since we have specified all values we can again calculate $c$, which, together with the graph of $f(x)$ translated down by twice the concavity of $f$ at $c$ (which, since we're using $f(x) = e^x$, is twice the value of $f$ at $c$) is illustrated:

enter image description here

Of course, we can continue on similarly from here (though we have to now pay a little more attention to the factorials in the denominators), incorporating successively more terms from the Taylor series into the Taylor polynomial. The nice thing about using $f(x)=e^x$ is that we will always be able to interpret derivatives evaluated at $c$ as the height of the curve above the $x$ axis at $c$, which would not be the case with other functions.

While a geometric interpretation of the Lagrange error term would be a lot more complicated with other functions, I found that when I could make sense of what was happening in the simple $e^x$ case, the Lagrange error term made a lot more sense in general.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.