# Taylor polynomial of $\int_{0}^{x}\sin(t^2)dt$

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?

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The last term is $\frac{F^{(3)}(0)}{3!}$, which is $\frac 13$. You can estimate the accuracy using an integral remainder. –  Davide Giraudo Dec 26 '11 at 23:12
Oh, how silly of me! $cos(0)=1$ of course. Thank you. You can post this as an answer? –  ro44 Dec 26 '11 at 23:13

We have $$f(x)=f(0)+xf'(0)+\frac{x^2}2f''(0)+\frac{x^3}{3!}f^{(3)}(0)+\int_0^x\frac{(x-t)^3}{3!}f^{(4)}(t)dt,$$ so \begin{align*} \sin x^2&=x^2+\int_0^x\frac{(x-t)^3}{3!}(8\cos(t^2)(1-5t)+16t^4\sin(t^2))dt\\ &=x^2+\frac 43\int_0^x(x-t)^3(\cos(t^2)(1-5t)+2t^4\sin(t^2))dt, \end{align*} and for $x\geq 0$ \begin{align*} \left|\int_0^x\sin t^2dt-\frac{x^3}3\right|&=\left|\int_0^x\sin t^2dt-\int_0^xt^2dt\right|\\ &\leq \frac 43\left|\int_0^x\int_0^t(t-s)^3(\cos(s^2)(1-5s)+2s^4\sin(s^2))dsdt\right|\\ &\leq \frac 43\int_0^x\int_0^t(t-s)^3(1+5s+2s^4)dsdt\\ &\leq \frac 43\int_0^x\int_0^tt^3(1+5s+2s^4)dsdt\\ &=\frac 13\left(\left[t^4\int_0^t(1+5s+2s^4)ds\right]_0^x-\int_0^xt^4(1+5t+2t^4)dt\right)\\ &=\frac 13x^4\left(x+\frac 52x^2+\frac 25x^5\right)-\frac 13\left(\frac{x^5}5+\frac 56x^6+\frac 29x^9\right)\\ &=\frac{x^5}3\left(\frac45+\frac 53x+\frac 25x^4-\frac 29x^4\right). \end{align*}

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Thanks! I have also learned the Lagrange form of the remainder, which I think you can use to show the remainder is smaller than $+-10^-2$. –  ro44 Dec 26 '11 at 23:38
@AndréNicolas Thanks! I will correct it. –  Davide Giraudo Dec 27 '11 at 11:07

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.

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