Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $F(x)=\displaystyle \int^x_0\frac{1}{1+t^3}dt$

1)Prove that F is well defined and differentiable for all $x\in\mathbb{R}$

2)If $n$ is a positive integer show that $F(x)=\displaystyle(\sum\limits_{k=0}^n (-1)^k\frac{x^{3k+1}}{3k+1})+(-1)^{n+1}\int_0^x\frac{t^{3n+3}}{1+t^3}dt$

3)Prove that $F(x)=\displaystyle\sum\limits_{k=0}^\infty (-1)^k\frac{x^{3k+1}}{3k+1}$ for all $|x|\le1$

For part one I'm assuming by well defined it means the integral exists which is direct since the function under integration is continuous, but the differentiability part is not clear I'm not sure how I should approach the limit. As for parts 2,3 I've never encountered any similar problems( I think I should take the limit as $n\to\infty$ in 3)

share|improve this question
    
Don't you mean for all $|x|\leq 1$? –  Pedro Tamaroff Dec 31 '12 at 16:07
    
@petertamaroff fixed –  user10444 Dec 31 '12 at 16:14
    
There is trouble at $x=-1$. Factoring $1+t^3$ as $(1+t)(1-t+t^2)$ and making the substitution $u=1+t$ one can see that $\int_0^{-1} \frac{dt}{1+t^3}$ diverges. And the series at the end diverges at $x=-1$, by comparison with the harmonic series. –  André Nicolas Dec 31 '12 at 16:19
    
See my answer. You want $|x|<1$. –  Pedro Tamaroff Dec 31 '12 at 16:29

2 Answers 2

up vote 1 down vote accepted

For 1) use the Fundumental Theorem of Calculus

For 2) and 3) use that $$\frac{1}{1+t^3}=\sum\limits_{k=0}^{n}(-1)^k t^{3k}+\frac{(-1)^{n+1}t^{3n+3}}{1+t^3}$$ after proving it (easy proof if you expand the summations).

EDIT: More detail:

1) Because $f(t)=\frac{1}{1+t^3}$ is continuous in $[0,x]$ by FCT, $F$ is differentiable in $[0,+\infty)$.

2) Taylor's Theorem.

share|improve this answer
    
Thank you for your reply and concerning the formula you posted is it valid for all t, and is there another proof besides working from right to left? –  user10444 Dec 31 '12 at 15:46
    
@user10444 I added some more detail in the edit. Are you sure the problem says $x\in \mathbb{R}$? –  Nameless Dec 31 '12 at 15:58
    
That's what's written, a restriction was only given in part 3 –  user10444 Dec 31 '12 at 16:01
    
@user10444 Well if $x\le -1$ the integrand is not continuous ($1+t^3$ becomes $0$). What do you know about improper integrals? –  Nameless Dec 31 '12 at 16:03
    
That I'm sure is a mistake since we covered improper integrals in calculus courses but skipped it in the analysis course –  user10444 Dec 31 '12 at 16:07

Careful about the first statement. The integral makes no sense if $x\leq -1$, since the integrand has an essential singularity at $x=-1$. So you can only work over $x>-1$.

For the third item, you need to determine for which $x$ we have $$R_n(x)=\int_0^x \frac{t^{3n+3}}{1+t ^3}dt\to 0$$

as $n\to \infty$.

Now, note that for $t=-1$ the function is undefined. Since we're expanding throughout the rigin, we can be sure we can't get past that singulatiry, that is, we can be sure that if $x\leq -1$, the series expansion will fail. Now, consider what happens for $-1<t<1$. In that case $t^k\to 0$, so we will exploit this fact to show that the error/remainder does vanish. Note that for $-1<t<1$ (so $-1<x<1$), $1+t^3>0$, so that

$$0\leq \left|\int_0^x \frac{t^{3n+3}}{1+t ^3}dt\right|\leq \int_0^x \frac{|t|^{3n+3}}{1+t ^3}dt$$

Now, if we fix an $-1<x<1$, the sequence of functions

$$f_n(t)=\frac{|t|^{3n+3}}{1+t ^3}$$

converges uniformly to $0$ over $[-x,x]$, in the sense that given any $\epsilon >0$, we can take $n$ large enough so that no matter what $t\in[-x,x]$ we take

$$\frac{|t|^{3n+3}}{1+t ^3}<\epsilon $$

This is merely because $|t|^{3n+3}\to 0$ and because the function is increasing in both directions of the axis. This only means that for any $-1<x<1$, and taking $n$ large enough, we will have $$\int_0^x \frac{|t|^{3n+3}}{1+t ^3}dt<x\epsilon $$ so that $R_n\to 0$, as desired. Now, try to show that for $x\geq 1$, the error term does not go to zero, so that the power series doesn't converge to $f$.

share|improve this answer
1  
It is not clear that there is a problem at $x=1$. Note that the series converges. –  André Nicolas Dec 31 '12 at 19:36
    
@AndréNicolas You're right. I'll address that later. –  Pedro Tamaroff Dec 31 '12 at 23:29

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.