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This is an exercise from Apostol (p.285) that I'm having trouble with (in fact, I'm having trouble with the whole section):

Prove that $\displaystyle{\int_0^1 \frac{1+x^{30}}{1+x^{60}} = 1 + \frac{c}{31}}, \qquad \text{where } 0 < c < 1.$

This comes from the section of Exercises following Taylor expansions, and Taylor's formula with error term. It seems like the approach should involve getting this as the error term of the Taylor expansion of a function that we know something about? I'm having trouble making much more progress than that.

Updated Progress: From the definition of the error term of the Taylor expansion we have:

$$E_n (x) = \frac{1}{n!} \int_a^x (x-t)^n f^{(n+1)} (t) dt$$

Alternatively, we may express this in the Lagrange form of the error term (derived from the weighted mean value theorem of integrals):

$$E_n (x) = \frac{f^{(n+1)}(c)}{(n+1)!} (x-a)^{n+1} \qquad \text{where } a < c < x.$$

Now, let $a = 0$, $x = 1$, $n = 30$, $f^{(n+1)}(t) = \dfrac{1+t^{30}}{(1+t^{60})(1-t)^{30}}$, and set the two alternative expressions of $E_n(x)$ equal to each other:

$$\begin{align*} \implies & \frac{1}{n!} \int_a^x (x-t)^n f^{(n+1)}(t) dt & = & \dfrac{f^{(n+1)}(c)}{(n+1)!} (x-a)^{n+1}\\ \implies & \frac{1}{30!} \int_0^1 (1-t)^{30} \dfrac{1+t^{30}}{(1+t^{60})(1-t)^{30}} & = & \dfrac{1}{31!} \dfrac{1+c^{30}}{(1+c^{60})(1-c)^{30}}\\ \implies & \int_0^1 \dfrac{1+t^{30}}{1+t^{60}} & = & \frac{1}{31} \frac{1+c^{30}}{(1+c^{60})(1-c)^{30}}. \end{align*} $$

I cannot seem to get from here to $=1 + \dfrac{c}{31}$.

If someone can show me pretty explicitly, step-by-step how to tackle this problem, I'd appreciate it. Dealing with the error term on Taylor expansions is giving me quite a bit of trouble, so hopefully seeing a full solution will help clear things up.

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3 Answers 3

up vote 6 down vote accepted

The approach of anon is better, but we can use the power series expansion of $\frac{1}{1+x^{60}}$ to conclude that for $-1\le x\le 1$, $$\frac{1+x^{30}}{1+x^{60}}=(1+x^{30})(1-x^{60}+x^{120}-x^{180}+\cdots).$$ Multiply out, and integrate term by term. We get that our integral $I$ is given by $$I=1+\frac{1}{31}-\frac{1}{61}-\frac{1}{91}+\frac{1}{121}+\cdots.$$ Group terms by twos, either starting at the beginning, or starting after the first term. From the two groupings, we can see that $$1<I<1+\frac{1}{31}.$$ We can use the same idea to get a string of increasingly more precise inequalities, such as $$1+\frac{1}{31}-\frac{1}{61}-\frac{1}{91}<I<1+\frac{1}{31}-\frac{1}{61}.$$

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Thanks for your answer. I just don't see how this follows from the definitions of the Taylor expansion or its error terms. This seems to be just expanding the integrand and integrating term by term. Likely I am missing the obvious connection between the two, but I am just learning this. Also, the book has not yet defined what is a series at all, only the definitions of Taylor polynomials and the error term on the Taylor polynomial expansion. –  user23784 Apr 5 '12 at 0:00
@rar: Then my answer is premature by a week or so. You will have material on alternating series and their error terms soon. The error term one gets is in general much better than the one we get from the Lagrange error formula, which for practical purposes is kind of useless. You will also get experience at finding series expansions by methods other than the direct application of the definition. That should happen in under two weeks. At this moment then, the obvious thing to use is that if $f(x)$ is your function, on $[0,1]$ we have $1\le f(x)\le 1+x^{30}$. –  André Nicolas Apr 5 '12 at 0:06
Depending on how quickly I read / how motivated I am to study... I see it is 2.5 chapters away now that I peak ahead. I'll have to revisit your answers then. I hope the methods currently available to me will be obsolete... these exercises are painful directly from the definitions of Taylor and the error term. Thanks again. –  user23784 Apr 5 '12 at 0:10

This doesn't use Taylor expansions, but I can't resist pointing it out:

$$\rm 1<\frac{1+x^{30}}{1+x^{60}}=1+x^{30}\frac{1-x^{30}}{1+x^{60}}<1+x^{30} \quad for~~0<x<1. $$

(Actually, as David Mitra points out, $\rm \large \frac{1+x^{30}}{1+x^{60}}\normalsize <1+x^{30}$ is more or less immediate...)

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I think you need $1$ on the left side... –  N. S. Apr 4 '12 at 21:20
@N.S. Right you are! –  anon Apr 4 '12 at 21:20
Can't you immediately say ${1+x^{30}\over 1+x^{60}}<1+x^{30}$ for $0<x<1$? –  David Mitra Apr 4 '12 at 21:27
@David: Eh, yeah, just by multiplying out I see. I'm so used to the adding-and-subtracting-in-the-numerator trick I do it reflexively without thinking. :/ –  anon Apr 4 '12 at 21:31
Thanks for the answer. Is there a way one could approach this with Taylor's formula? Or is this somehow related to a taylor expansion that I don't see? Honestly, I'm sufficiently confused with this section of the text, that it could be equivalent and I wouldn't realize it. –  user23784 Apr 4 '12 at 21:46

I think you are making the problem much harder than it is. It is easy to see (for $x \in (0,1)$) that $0 <x^{60} < x^{30}$, and so $1 < \frac{1+x^{30}}{1+x^{60}} < 1 + x^{30}$. Now integrate all three quantities, the integral will preserve the order, and you get $1 < \int_0^1 \frac{1+x^{30}}{1+x^{60}}dx < 1+\frac{1}{31}$, which is what you were asked to prove. There is no need to complicate matters with Taylor series, or remainder formulae.

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This is an Exercise to learn about Taylor expansions and the remainder formula (which I think is quite clear in my question)... so solving it by other means sort of defeats the purpose of the exercise. It's not that I want to prove this particular statement by any method available: I want to understand how to apply the error term in the Taylor formula, and understanding is aided by applying these methods to particular problems. –  user23784 Apr 5 '12 at 1:13

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