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I am stuck on two problems:

1) Prove that $$\int_0^1 \frac{1+x^{30}}{1+x^{60}}dx=1+\frac{c}{31}$$ where $0< c <1$.

2) Prove that $$0.493948<\int_0^{\frac{1}{2}}\frac{1}{1+x^4}dx<0.493958$$

Now for the first one I get Taylor expansion at $0$: $$\frac{1+x^{30}}{1+x^{60}}=1+E_0(x)$$ where $E_0(x)$ is the remainder. I suppose I need to use Lagrangian form of remainder, by differentianting the function and evaluating it at $c$ but I can't make any progress as the derivative is too clumsy.

For the second again I do not know how to incorporate the remainder in bounding the integral.

Hints are appreciated!


Considering the second problem, I just need help on bounding the remainder of Taylor expansion. I get:


How do I specify bound for the remainder?

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The first part is a duplicate. An approach using power series expansions appears in an answer there. – David Mitra Jan 9 '13 at 18:27
up vote 2 down vote accepted

For the first one, since $x \in (0,1)$ we have $1 < 1+x^{60} < 1 + x^{30}$. Hence, we get that $$1 < \dfrac{1+x^{30}}{1+x^{60}} < 1+x^{30}$$ Now you can get the answer.

For the second one, we have $$\dfrac1{1+x^4} = \sum_{k=0}^{\infty}(-x^4)^{k}$$ and hence $$\sum_{k=0}^{2n+1}(-x^4)^{k} < \dfrac1{1+x^4} < \sum_{k=0}^{2n}(-x^4)^{k}$$ Take $n=2$ to get what you want.

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note to myself - think out of the box. this way is indeed a simple one. but could I solve it using taylor expansion? if so, how? – Sarunas Jan 9 '13 at 18:22
@Sarunas What I have written is the first few terms in the Taylor expansion. – user17762 Jan 9 '13 at 18:23
@Sarunas: If you mean Taylor series for 1), note that the Taylor series for $\frac{1+x}{1+x^2} = 1+x-x^2+x^3\cdots$. If $x \in [0,1)$, you have $\frac{1+x}{1+x^2} \leq 1+x$. Now replace $x$ by $x^{30}$ and you have Marvis' estimate. – copper.hat Jan 9 '13 at 18:28
oh right! just brilliant! now the second problem... :) – Sarunas Jan 9 '13 at 18:35

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