calculustaylor expansion
I was asked to use Taylor's Theorem to confirm that the Maclaurin polynomial $P_7(x)$ for $e^x$ is sufficient to give $e$ correct to $3$ decimal places. My calculus-book states:
"To get an approximation for $e=e^1$ correct to 3 decimal places, we
need to have $E_n(1)<0.0005$"
QUESTION: I don't see why an error of $0.0005$ is sufficiently small, since $e+0.0005=2.7187…$ which rounds to $2.719$ which is not correct to $3$ decimal places. Any thoughts?
We need to understand what Taylor's Theorem, Lagrange form of the remainder, says in this case.
We are using the Taylor polynomial $P_1(x)=x$ to approximate $\sin x$. We could therefore call the error term $E_1$. But in this case the second term in the Taylor expansion is $0$, so $P_1(x)=P_2(x)$, and therefore $E_1$ and $E_2$ are equal.
By the Lagrange form of the remainder, we have $$|E_2|=\left|\frac{-\cos \xi}{3!}x^3\right|\tag{1}$$ for some $\xi$ between $0$ and $x$. Since the absolute value of the cosine is $\le 1$, from (1) we obtain the estimate $$|E_2| \le \frac{|x|^3}{3!}.\tag{2}$$ It should now be straightforward to find the range of $x$ for which the right-hand side of (2) is $\lt 10^{-7}$.
Remark: If you have met alternating series, we can bypass the Lagrange form of the remainder. In this case we end up with basically the same estimate of the error. Usually, however, the Lagrange form of the remainder results in technically correct but excessively pessimistic estimates.
Your teacher's example cannot work in general as I present a counter example below. Nonetheless, I think that your teacher's approach is a reasonable way to explain the intuition behind what happens in a typical case, provided that the proper caveats are given.
I think a more reasonable stopping condition, for programming purposes, is to iterate until the value of $f$ is very small. If the first derivative is relatively large in a neighborhood of the last iterate, this might be enough to prove that there is definitively a root nearby. Of course, Christian Blatter has already provided sufficient conditions.
For a counter example, let's suppose that $$f(x) = x(x-\pi)^2 + 10^{-12}.$$ Then, the Newton's method iteration function is $$N(x) = x-f(x)/f'(x) = x-\frac{x (x-\pi)^2+10^{-10}}{3 x^2-4\pi x+\pi ^2}$$ and if we iterate $N$ 20 times starting from $x_0=3.0$, we get $$ 3., 3.07251, 3.10744, 3.12461, 3.13313, 3.13736, 3.13948, 3.14054, \ 3.14106, 3.14133, 3.14146, 3.14153, 3.14156, 3.14158, 3.14158, \ 3.14159, 3.14159, 3.14159, 3.14159, 3.14159, 3.14159 $$ Thus, your teacher's method implies there is a root at $x=3.14159$ when, of course, there is no root near here. There is, however, a root near zero to which the process eventually converges after several thousand iterates.
To place this in a broader context, let's examine the basins of attraction for this polynomial in the complex plane. There are three complex roots, one just to the left of zero and two at $\pi\pm\varepsilon i$ where $\varepsilon$ is a small positive number. In the picture below, we shade each complex initial seed depending on which of these roots Newton's method ultimately converges.
Now, it is a theorem in complex dynamics that, whenever two of these basins meet, there are points of the third basin arbitrarily near by. As a result, there is definitely a number whose decimal expansion starts with $3.14159$ that eventually converges to the root near zero under iteration of Newton's method.
Best Answer
You could use the Lagrange formula for the remainder in the Taylor Polynomial development of order $\;n\;$ around $\;x_0=0\;$ to evaluate the error :
$$\text{For}\;c\in (0,x)\;:\;\;\left|\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}\right|=\frac{e^c}{(n+1)!}|x|^{n+1}\le$$
$$\le\frac{e^c}{(n+1)!}\cdot\le\frac9{(n+1)!}$$
since we can choose $\;x\le1\;$ and estimating crudely $\;e^3\implies e^2<9\;$, and thus we want:
$$\frac9{(n+1)!}\cdot<10^{-3}\implies (n+1)!>9,000\implies n+1=8\implies n=7$$
and indeed:
$$e=1+1+\frac12+\frac16+\frac1{24}+\frac1{120}+\frac1{720}+\frac1{5,040}\approx2.71825$$
and then
$$2-2.71825=0.00002... <10^{-3}$$
It turns out that also $\;n=6\;$ would work, but we usually aren't that interested in the best or smallest possible $\;n\;$, but only in some value that will work and will be more or less close to the best one.