I just learned about Taylor polynomials, and I am trying to estimate $\int_{0}^{1/2}\sin(x^2)dx$ using the 3rd degree Taylor polynomial of $F(x)=\int_{0}^{x}\sin(t^2)dt$ at $0$. I get the following:
$F'(x)=\sin(x^2)$,
$F''(x)=2x\cos(x^2)$,
$F^{(3)}(x)=2\cos(x^2)-4x^2\sin(x^2)$.
And so the estimate is:
$$\frac{F(0)}{0!}\cdot 1+\frac{F'(0)}{1!}\cdot \frac{1}{2}+\frac{F''(0)}{2!}\cdot \frac{1}{4}+\frac{F^{(3)}(0)}{3!}\cdot \frac{1}{8} = 0.$$
However this seems very weird (shouldn't it be closer to the actual value than $0$?), is my estimation correct?
Best Answer
We first write down the Taylor expansion of $\sin t$ about $t=0$. This has undoubtedly already been done in your course. If it hasn't, calculation of the derivatives at $0$ is easy. We get $$\sin t =t-\frac{t^3}{3!}+\frac{t^5}{5!}-\frac{t^7}{7!}+\cdots.$$ Put $t=x^2$. We get $$\sin(x^2)=x^2-\frac{x^6}{3!}+\frac{x^{10}}{5!}-\frac{x^{14}}{7!}+\cdots.$$ Integrating term by term, we pbtain $$F(x)=\frac{x^3}{3}- \frac{x^7}{7\cdot 3!}+\frac{x^{11}}{11\cdot 5!}-\frac{x^{15}}{15\cdot 7!}+\cdots.$$
The degree $3$ expansion is then simply $\dfrac{x^3}{3}$. (Of course we need theorems to justify the substitution and the term by term integration.)
Let $x=1/2$. To estimate the error when we truncate at the $x^3/3$ term, it is in this case most efficient to note that our series is (for $x=1/2$) an Alternating Series. So the truncation error has absolute value less than $(1/2)^7/(7\cdot 3!)$. The Lagrange form of the remainder, though useful for theoretical purposes, is often less useful when we want to produce good error estimates.