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

by RoRi
  1. Show that r = \frac{\sqrt{21}-3}{2} is an approximation of the nonzero root of the equation

        \[ \arctan x = x^2 \]

    using the cubic Taylor polynomial approximation to \arctan x.

  2. Given that

        \[ \sqrt{21} < 4.6 \qquad \text{and} \qquad 2^{16} = 65536 \]

    prove that the number r from part (a) satisfies

        \[ |r^2 - \arctan x| < \frac{7}{100}. \]

    Determine if (r^2 - \arctan r) is positive or negative and prove the result.

  1. Proof. From a previous exercise (Section 7.8, Exercise #3) we know

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

    So, to approximate the nonzero root of x^2 - \arctan x we have

        \begin{align*} x^2 - x + \frac{x^3}{3} \approx 0 && \implies && x^2 + 3x - 3 &\approx 0 \\ && \implies && x &\approx \frac{\sqrt{21}-3}{2}. \qquad \blacksquare \end{align*}

  2. We know from the same previous exercise we used in part (a) that the error term E_{2n}(x) for \arctan x satisfies the inequality

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

    Using the values for \sqrt{21} and 2^{16} given we have

        \begin{align*} \left|E_5 \left( \frac{\sqrt{21}-3}{2} \right)\right| &\leq \left| \frac{\left( \frac{\sqrt{21}-3}{2} \right)^5}{5} \right| \leq \left| \frac{0.8^5}{5} \right| = \frac{(4/5)^5}{5} \\[9pt] &= \frac{2^{10}}{5^6} = \frac{2^{16}}{10^6} = \frac{65536}{1000000} < \frac{7}{100}. \qquad \blaacksquare \end{align*}

2 comments

    • Mohammad Azad says:

      Something got messed up in my comment, what I said is: If you want to prove that the difference is positive or negative first prove that for x>0
      arctan(x)>x-x^3/3

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