Prove that the error term in the Taylor expansion of 
satisfies the following inequality.
![\[ \arctan(x) = \sum_{k=0}^{n-1} \frac{(-1)^k x^{2k+1}}{2k+1} + E_{2n} (x), \qquad |E_{2n}(x)| \leq \frac{x^{2n+1}}{2n+1}, \quad \text{if } 0 \leq x \leq 1. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-109163600b9e50e97b428c9960ae13a6_l3.png "Rendered by QuickLaTeX.com")
Proof. To prove this we will work directly from the definition of the error as an integral,
![\[ E_n (x) = \frac{1}{n!} \int_a^x (x-t)^n f^{(n+1)}(t) \, dt. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-bb2061dc2c51382b2bb5b411fd822ef1_l3.png "Rendered by QuickLaTeX.com")
We know for 
we have, (we need for the expansion of 
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*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-5e53ec8ca0552d226904aa41d2d1a9eb_l3.png "Rendered by QuickLaTeX.com")
Therefore we have
![\[ E_{2n}(x) = \int_0^x \frac{(-1)^n t^{2n}}{1+t^2}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-0bf9048ab160d4e863cc2fabe947d620_l3.png "Rendered by QuickLaTeX.com")
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*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-7801a028d4b8594e005fcb1ef4ce10ef_l3.png "Rendered by QuickLaTeX.com")
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.
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!
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?
So, perhaps the restriction |x| infinity.
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)).
I’ll need to think about this. You’re right that the expansion is valid for all
(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 
).