积分法的误差

Before learning what the integral form of the error of a Taylor series is, we will first learn, through a true anecdote, why it is important for young mathematicians to learn this concept.

In the early 1920's, Igor Tamm was a professor at the Odessa Polytechnic University in what is now Ukraine. This was during the Russian Civil War, when getting food was difficult, so Igor one day went to a nearby village to trade silver spoons for chickens. There, he was captured by an invading anti-communist army, accused of being a communist agent, and sent to the army commander to be killed. Igor confessed to the commander that he was just a mathematics professor trying to get food in the village. The commander, being skeptical, wanted Igor to prove he was a mathematician, so he ordered him to correctly derive the error of the n-term Taylor approximation or be shot. Igor worked out the solution at gunpoint and finally was set free for having the correct solution. Afterwards, he became a prominent physicist and a Nobel Laureate.

Therefore, a thorough understanding of the error, or remainder, term in the Taylor series approximation is crucial not only in mathematics but also in everyday life.

We will prove here that the error of an $n$-term Taylor approximation around $a$ is equal to $\int_{a}^{x}\frac{f^{(n+1)}(z)(x-z)^n}{n!}dz$, or $f(x)=\sum_{j=0}^{n}{\frac{f^{(j)}(a)(x-a)^{j}}{j!}}+\int_{a}^{x}\frac{f^{(n+1)}(z)(x-z)^n}{n!}dz$.

Base case: $n=0$.

Then it is clear that $f(x)=\sum_{j=0}^{n}{\frac{f^{(j)}(a)(x-a)^{j}}{j!}}+\int_{a}^{x}\frac{f^{(n+1)}(z)(x-z)^n}{n!}dz=f(a)+\int_{a}^{x}\frac{f^{(1)}(z)}{1}dz$.

Inductive step:

Assume that $f(x)=\sum_{j=0}^{n-1}{\frac{f^{(j)}(a)(x-a)^{j}}{j!}}+\int_{a}^{x}\frac{f^{(n)}(z)(x-z)^{(n-1)}}{(n-1)!}dz$.

We will prove that then $f(x)=\sum_{j=0}^{n}{\frac{f^{(j)}(a)(x-a)^{j}}{j!}}+\int_{a}^{x}\frac{f^{(n+1)}(z)(x-z)^n}{n!}dz$.

Try proving this on your own and then click "Continue" for the solution. (Hint: integration by parts is good)

Integrating by parts, we can replace the indefinite integral $\int\frac{f^{(n)}(z)(x-z)^{(n-1)}}{(n-1)!}dz=-\int{\frac{f^{(n)}(z)}{(n-1)!}d\frac{(x-z)^n}{n}}$ with ${-\frac{f^{(n)}(z)}{(n-1)!}\frac{(x-z)^n}{n}}+\int{\frac{(x-z)^n}{n}d\frac{f^{(n)}(z)}{(n-1)!}}$, which simplifies to ${-\frac{f^{(n)}(z)(x-z)^n}{n!}}+\int\frac{f^{(n+1)}(z)(x-z)^n}{n!}dz$.

Now, we set the limits of integration as $a$ and $x$ to get $\int_{a}^{x}\frac{f^{(n)}(z)(x-z)^{(n-1)}}{(n-1)!}dz=({-\frac{f^{(n)}(x)(x-x)^n}{n!}}+{\frac{f^{(n)}(a)(x-a)^n}{n!}})+\int_{a}^{x}\frac{f^{(n+1)}(z)(x-z)^n}{n!}dz\\={\frac{f^{(n)}(a)(x-a)^n}{n!}}+\int_{a}^{x}\frac{f^{(n+1)}(z)(x-z)^n}{n!}dz$.

Replacing the error integral for the $n-1$ term approximation with this new expression, we see that $f(x)=\sum_{j=0}^{n}{\frac{f^{(j)}(a)(x-a)^{j}}{j!}}+\int_{a}^{x}\frac{f^{(n+1)}(z)(x-z)^n}{n!}dz$, so the induction is complete.

If $T_n(x)$ is the degree $n$ Taylor approximation of $f(x)$ at $x=a$, then the approximation error $R_n(x) = f(x) - T_n(x)$ (sometimes called remainder), in integral form, is $$R_n(x) = \frac{1}{n!}\int_a^x (x-t)^n f^{(n+1)}(t)\, dt.$$ (Technical hypotheses: this formula holds if $f^{(n+1)}$ is continuous on an open interval $I$ containing $a$, and $x$ lies in $I$.)