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Find the limit as x goes to 0 of (sin x) / (arctan x)

by RoRi

Evaluate the limit.

    \[ \lim_{x \to 0} \frac{\sin x}{\arctan x}. \]

We know (p. 287) the following expansions as x \to 0,

    \[ \sin x = x + o(x^2), \qquad \text{and} \qquad \arctan x = x + o(x^2). \]

(Note that these are the same expansion when we use only the first order terms. This tells us that \sin x and \arctan x behave similarly near 0. We would need to take higher order terms to differentiate between the two. For instance, if we wanted to include cubic terms we would have \sin x = x - \frac{1}{6}x^3 + o(x^4), but \arctan x = x - \frac{1}{3}x^3 + o(x^4).) From here we compute the limit,

    \[ \lim_{x \to 0} \frac{\sin x}{\arctan x} = \lim_{x \to 0} \frac{x + o(x^2)}{x+o(x^2)} = 1. \]

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*}

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

by RoRi
  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

Prove an inequality for the error of the Taylor polynomial of arctan x

by RoRi

Prove that the error term in the Taylor expansion of \arctan x satisfies the following inequality.

    \[ \arctan(x) = \sum_{k=0}^{n-1} \frac{(-1)^k x^{2k+1}}{2k+1} + E_{2n} (x), \qquad |E_{2n}(x)| \leq \frac{x^{2n+1}}{2n+1}, \quad \text{if } 0 \leq x \leq 1. \]

Proof. To prove this we will work directly from the definition of the error as an integral,

    \[ E_n (x) = \frac{1}{n!} \int_a^x (x-t)^n f^{(n+1)}(t) \, dt. \]

We know for 0 \leq x \leq 1 we have, (we need 0 \leq x \leq 1 for the expansion of \frac{1}{1+x^2} to be valid),

    \begin{align*} \arctan x &= \int_0^x \frac{1}{1+t^2} \, dt \\[9pt] &= \int_0^x \left( 1 - t^2 + t^4 - t^6 + \cdots + (-1)^{n-1} t^{2n-2} + \frac{(-1)^n t^{2n}}{1+t^2} \right) \, dt \\[9pt] &= \int_0^x \left( 1 - t^2 + t^4 - t^6 + \cdots + (-1)^{n-1} t^{2n-2} \right) \, dt + \int_0^x \frac{(-1)^n t^{2n}}{1+t^2} \, dt \\[9pt] &= \sum_{k=0}^{n-1} \frac{(-1)^k x^{2k+1}}{2k+1} + \int_0^x \frac{(-1)^n t^{2n}}{1+t^2} \, dt \\[9pt] &= \sum_{k=0}^{n-1} \frac{(-1)^k x^{2k+1}}{2k+1} + E_{2n}(x). \end{align*}

Therefore we have

    \[ E_{2n}(x) = \int_0^x \frac{(-1)^n t^{2n}}{1+t^2}. \]

So, we can bound the error term by bounding the integral,

    \begin{align*} | E_{2n} (x) | &= \left| \int_0^x \frac{(-1)^n t^{2n}}{1+t^2} \, dt \right| \\[9pt] &\leq \int_0^x \frac{t^{2n}}{1+t^2} \, dt \\[9pt] &\leq \int_0^x t^{2n} \, dt &(1+t^2 \geq 1) \\[9pt] &= \frac{x^{2n+1}}{2n+1}. \qquad \blacksquare \end{align*}

Prove an inequality for the error of the Taylor polynomial of sin x

by RoRi

Prove that the error of the Taylor expansion of \sin x satisfies the following inequality.

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

Proof. Since the derivatives of \sin x are always \cos x, -\sin x, -\cos x, or \sin x we know that for f(x) = \sin x we have |f^{(n+1)}(x)| \leq 1. (In other words, the n+1st derivative is bounded above by 1 and below by -1.) Therefore, we can apply Theorem 7.7 (p. 280 of Apostol) to estimate the error in Taylor’s formula at a= 0 with m = -1 and M = 1. For x > 0 this gives us

    \begin{alignat*}{3} & m \frac{(x-a)^{n+1}}{(n+1)!} &\leq \ \ E_n (x) &\leq M \frac{(x-a)^{n+1}}{(n+1)!} \\[9pt] \implies & -\frac{x^{n+1}}{(n+1)!} &\leq \ \ E_n (x) &\leq \frac{x^{n+1}}{(n+1)!} \\[9pt] \implies & -\frac{x^{2n+1}}{(2n+1)!} &\leq \ \ E_{2n} (x) &\leq \frac{x^{2n+1}}{(2n+1)!} \\[9pt] \implies & \phantom{-}|E_{2n}(x)| &\leq \frac{x^{2n+1}}{(2n+1)!}. \end{alignat*}

Next, (from the second part of Theorem 7.7) if x < 0 we have

    \begin{alignat*}{3} & m \frac{(a-x)^{n+1}}{(n+1)!} &\leq (-1)^{n+1} E_n (x) &\leq M \frac{(a-x)^{n+1}}{(n+1)!} \\[9pt] \implies & - \frac{(-x)^{n+1}}{(n+1)!} &\leq (-1)^{n+1} E_n (x) &\leq \frac{(-x)^{n+1}}{(n+1)!} \\[9pt] \implies & - \frac{(-x)^{2n+1}}{(2n+1)!} &\leq (-1)^{2n+1} E_{2n}(x) &\leq \frac{(-x)^{2n+1}}{(2n+1)!} \\[9pt] \implies & - \frac{(-x)^{2n+1}}{(2n+1)!} &\leq -E_{2n} (x) &\leq \frac{(-x)^{2n+1}}{(2n+1)!} \\[9pt] \implies & \phantom{-} |E_{2n} (x)| &\leq \frac{x^{2n+1}}{(2n+1)!}. \qquad \blacksquare \end{alignat*}