Home » Blog » Use Taylor polynomials to approximate the nonzero root of x2=sin x

Use Taylor polynomials to approximate the nonzero root of x2=sin x

  1. Using the cubic Taylor polynomial approximation of \sin x, show that the nonzero root of the equation

        \[ x^2 = \sin x \]

    is approximated by r = \sqrt{15} - 3.

  2. Using part (a) show that

        \[ | \sin r - r^2  | < \frac{1}{200}, \]

    given that \sqrt{15} - 3 < 0.9. Determine whether (\sin r - r^2) is positive or negative and prove the result.


  1. Proof. The cubic Taylor polynomial approximation of \sin x is

        \[ \sin x = x - \frac{x^3}{3!} + E_{2n} (x). \]

    This implies

        \[ x^2 - \sin x \approx x^2 - x + \frac{x^3}{6}. \]

    Therefore, we can approximate the nonzero root by

        \begin{align*}  x^2 - \sin x = 0 && \implies && x^2 - x + \frac{x^3}{6} &\approx 0 \\  && \implies && 6x^2 + x - 1 &\approx 0 \\  && \implies && x &\approx \sqrt{15} - 3.  \qquad \blacksquare \end{align*}

  2. Proof. We know from this exercise (Section 7.8, Exercise #1) that for \sin x we have

        \[ |E_{2n} (x)| \leq \frac{|x|^{2n+1}}{(2n+1)!}. \]

    So, for n = 2, and using the given inequality \sqrt{15} - 3 < 0.9, we have

        \[ |E_{2n} (x)| \leq \frac{|0.9|^5}{120} = \frac{9^5}{10^5 (120)} < \frac{1}{200}. \]

    Furthermore, (\sin r - r^2) > 0 since

        \[ \sin r - r^2 = E_{2n}(x) = \frac{x^5}{5!} - \frac{x^7}{7!}+ \cdots \]

    with the absolute value of each term in the sum strictly less than the absolute value of the previous term (since x < 1 and (n+2)! > n!). Thus, each pair is positive, so the whole series is positive. \qquad \blacksquare

One comment

  1. Julio Vieira says:

    There is a little error when you calculate x^2-x+x^3/6 = 0, the next step is x^2+6x-6 = 0 and not 6x^2+x-1=0

    And thanks for solutions, it really helps me a lot!

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):