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Taylor approximations; the error term; convergence

The $ n$-th Taylor approximation $ T_{n}(f, a)(x)$ based at $ a$ to a function $ f(x)$ is the $ (n+1)$-th partial sum of the Taylor series:

$\displaystyle T_n(f,a)(x) = \sum_{k = 0}^n \frac{f^{k}(a)}{k!}(x - a)^k $

$\displaystyle = f(a) + f^{(1)}(a)(x -a) + \frac{f^{(2)}(a)}{2}(x - a)^2 + \frac{f^{(3)}(a)}{6}(x - a)^3 + \dots + \frac{f^{(n)}(a)}{n!}(x - a)^n.$

Note that $ T_n(f, a)(x)$ is a sum of $ n + 1$ terms and is a polynomial of degree at most $ n$ in $ x$.
Then $ T_n(f, a)(x)$ has the characteristic property that its derivatives agree with those of the function $ f(x)$, when both are evaluated at $ x = a$, up to and including the $ n$-th derivative.

Consider now the difference $ E_n(f, a)(x) = \vert f(x) - T_n(f, a)(x)\vert$.
Intuitively this should be small.
There are various estimates of the size of this difference: one is the following:

$\displaystyle E_n(f, a)(x) \le \frac{K_{n+1}}{(n+1)!} \vert x - a\vert^{n+1} .$

This estimate is valid throughout the interval $ a - r \le x \le a+r $, for a fixed positive $ r$, where the quantity $ K_{n+1}$ is the maximum of $ \vert f^{(n+1)}(x)\vert$ on that interval.
So for example, for the function $ f(x) = e^x$, we have $ K_{n+1} = e^{a + r}$ and

$\displaystyle E_n(f, a)(x) \le \frac{e^{a +r}}{(n+1)!} \vert x - a\vert^{n+1} .$

For the functions $ \sin(x)$ and $ \cos(x)$, we know that $ K_{n+1}$ is a value of one of the two functions $ \vert\sin(x)\vert$ or $ \vert\cos(x)\vert$, somewhere on the interval $ [a-r, a + r]$, which can never be larger than $ 1$, so we always have the following estimate:

$\displaystyle E_n(f, a)(x) \le \frac{1}{(n+1)!} \vert x - a\vert^{n+1} .$

For each of these functions, we notice that as $ n \rightarrow \infty$, the error goes to zero, since the denominator $ (n+1)!$ grows much faster than any power of the form $ u^n$ for fixed $ u$.

When the error goes to zero as $ n$ goes to infinity, we get two by-products: So we can conclude as stated earlier, that the Taylor series for the functions $ e^x$, $ \sin(x)$ and $ \cos(x)$ always represents the function, on any interval $ [a-r, a + r]$, for any reals $ a$ and $ r$, with $ r > 0$.
Since this is true for any real $ r > 0$, these Taylor series represent the functions on the entire real line.

As another example consider the function $ f(x) = \frac{1}{1 + x}$ and its expansion based at 0.
We have $ \vert f^{(n+1)}(x)\vert = \frac{(n+1)!}{(1 + x)^{n+1}}$, so, on the interval $ [-r, r]$, where $ 0 < r < 1$, we get $ K_{n+1} = \frac{(n+1)!}{(1 -r)^{n+1}}$ and then we have:

$\displaystyle E_n(f, 0)(x) \le \left(\frac{\vert x\vert}{1 - r}\right)^{n+1} .$

This goes to zero as $ n \rightarrow \infty$, provided $ \vert x\vert < 1 - r$.
Note that $ r$ must be restricted to the range $ 0 < r < 1$, since the function and its derivatives blow up as $ x \rightarrow -1^+$.
We conclude that the Taylor series represents the function $ \frac{1}{1 + x}$ on the interval $ [-r, r]$, for any $ 0 < r < 1$, so therefore also on the interval $ (-1, 1)$.

Finally, if a Taylor series converges on an open interval $ (p, q)$, then it converges absolutely on that interval.


next up previous
Next: Tricks with Taylor series Up: 23014convergence Previous: Taylor series based at
George A. J. Sparling 2003-12-08