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Use Taylor polynomials to approximate an integral of sin (x2)

by RoRi
  1. Show that

        \[ \sin x = x - \frac{x^3}{3!} + r(x), \qquad \text{where} \quad |r(x)| \leq \frac{1}{2^5 \cdot 5!} \]

    when 0 \leq x \leq \frac{1}{2}.

  2. Using part (a) find an approximation for the integral

        \[ \int_0^{\frac{\sqrt{2}}{2}} \sin (x^2) \, dx. \]

  1. Proof. We know from this exercise that

        \[ \sin x = \sum_{k=1}^n \frac{(-1)^{k-1}x^{2k-1}}{(2k-1)!} + E_{2n}(x) \qquad \text{where} \quad |E_{2n}(x)| \leq \frac{|x|^{2n+1}}{(2n+1)!}. \]

    Therefore, when 0 \leq x \leq \frac{1}{2} we have

        \[ \sin x = x - \frac{x^3}{3!} + r(x) \qquad \text{where} \quad |r(x)| \leq \frac{1}{2^5 \cdot 5!}. \qquad \blacksquare \]

  2. From part (a) we know

        \begin{align*} \sin x = x - \frac{x^3}{3!} + r(x) && \implies && \sin (x^2) &= x^2 - \frac{x^6}{3!} + r(x^2)\\[9pt] && \implies && \int_0^{\frac{\sqrt{2}}{2}} \sin (x^2) \, dx &= \int_0^{\frac{\sqrt{2}}{2}} \left( x^2 - \frac{x^6}{3!} + r(x^2) \right) \, dx \\[9pt] &&&&&= \left( \frac{x^3}{3} - \frac{x^7}{7 \cdot 3!} \right) \Bigr \rvert_0^{\frac{\sqrt{2}}{2}} + \int_0^{\frac{\sqrt{2}}{2}} r(x^2) \, dx \\[9pt] &&&&&= \frac{55\sqrt{2}}{672} + R \end{align*}

    where

        \[ |R| \leq \int_0^{\frac{\sqrt{2}}{2}} \frac{1}{2^5 \cdot 5!} \, dx = \frac{1}{2^5 \cdot 5!} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{7680}. \]

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