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a(n) is a sequence. I can't find an example :/

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can someone check my way to solve it?

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It should work because $f(x) = x^{1/3}$ is continuous over $\mathbb{R}$. The trick is, you don't have to use the same $\epsilon$. In other words, if you have some $\epsilon$ for the cubic convergence, create a function of that $\epsilon$ and $a$ for the linear conversion, and it would work... –  gt6989b Nov 15 '12 at 21:29
    
@Haskell: How will the OP learn if you make the corrections for him without even leaving a comment? I would have rejected the edit had I note been preempted. Alon: what you have here is a sequence, not a series. –  Brian M. Scott Nov 16 '12 at 0:37

1 Answer 1

up vote 2 down vote accepted

Observe that $1\le |a_n^2+a_na+a^2|$ is only sufficient to show $|a_n-a|\le|a_n^3-a^3|$. Your example with $a_n=a$ is not affected by this. I assume the sequence is a sequence of real numbers (not complex numbres). Then the implication is true:

If $a=0$ then for given $\epsilon>0$ there exists $N$ such that $n>N$ implies $|a_n^3|<\epsilon^3$, hence $|a_n|<\epsilon$.

Do you see what you can do in case $a\ne 0$? Assume we can find a number $c>0$ (not depending on $a_n$) and $n_1\in\mathbb N$ such that $|a_n^3-a^3|\ge c\cdot|a_n-a|$ holds for all $n>n_1$. Then we have already won because that allows to proceed as follows: Given $\epsilon>0$ there is an $n_0$ such that $|a_n^3-a_n|<c\epsilon$ for all $n>n_0$, hence $|a_n-a|\le\frac1c|a_n^3-a^3|<\epsilon$ for all $n>\max\{n_0,n_1\}$.

How can we find such $c$? You were close, but one needs to be careful with signs (that's also why I extracted the case $a=0$ separately). From $a_n^3\to a^3\ne0$ we see that there is an $n_1$ such that $n>n_1$ implies $|a_n^3-a^3|<a^3$, hence for such $n$, the numbers $a_n^3$ and $a^3$ have the same sign and the same holds for the signs of $a_n$ and $a$. As you noted, we have $a_n^3-a^3=(a_n-a)(a_n^2+a_na+a^2)$. As $a_n$ and $a$ have the same sign for $n>n_1$, all summands in $a_n^2+a_na+a^2$ are positive, hence $a_n^2+a_na+a^2\ge a^2>0$ and we can simply take $c=a^2$.

Remark: In general, for $a_m\approx a$ we'd have $a_n^3-a^3\approx 3a^2\cdot (a_n-a)$. Thus taking $c=a^2$ when actually $c\approx 3a^2$ would be possible is a somewhat wasteful estimate, but who cares?

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I'm sorry, but I don't know what to do if a is not zero. is it not the case that I wrote? :/ and you are right! the sequence is a sequence of real numbers.. –  Alon Shmiel Nov 15 '12 at 21:46
    
I am deleting my answer about nonzero a case, since I assumed continuity of cuberoot, which you are trying to prove. –  coffeemath Nov 16 '12 at 4:46

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