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

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*}


  1. hteica says:

    Hi, x bigger or equal to 0 is a sufficient condition for the inequalities to hold. But the case that x bigger than 1 is almost useless , since abs(E2n(x)) will have a large value, we won’t be able to use it to approximate things, I think that’s why Apostol mentioned the important part, namely x smaller or equal to 1. That’s become one of the motivation for the last problem of this exercise where we are to approximate the value of pi.

  2. tom says:

    Yeah, that’s what I suspected- that you were thinking of an infinite series. I figured it out after browsing thru Courant. I understand your thinking, but we haven’t officially gotten to infinite series yet!

  3. tom says:

    You stated we need x —> {0,1} for the expansion of 1/1+x^2. Can you please elaborate on this? Apostol expands 1/1-x with restriction x ≠ 1, then substitutes -x^2 for x in the polynomial – but then x will never be 1. Expanding 1/1+x^2 directly also seems reasonable. When |x|>1 I see the error term becomes unreasonable. Did I miss something?

      • tom says:

        My comment seems to have gotten messed up… What I said was perhaps the restriction on x is to insure the convergence if the Taylor series went to infinity? (I’m assuming the decomposed polynomial acts like 1-r^n/1-r when n —> infinity, rather then 1/1-r + En(r)).

      • RoRi says:

        I’ll need to think about this. You’re right that the expansion is valid for all x \neq 1 (since it is just finitely many terms… I think when I wrote this I was thinking about the expansion for the geometric series which is only valid for |x| < 1).

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