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$S^0:\mathbb N\to \Bbb R$ is a function. For any $m\in \Bbb N$, we define

$$S^m:\Bbb N\to \Bbb R$$ $$S^m(n)=\sum_{k=1}^n S^{m-1}(k)$$

For each $m\in \Bbb N$, we have: $$\lim_{n \to \infty}S^m(n)=0$$

Can we deduce $S^0\equiv 0?$

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@julien: I denote the nested partial sum of $f:\Bbb N \to \Bbb R$ by $S^m_f$. Now index $f$ is not needed. nested sums are like $S^{m-1}\circ S =S^m$. However you can edit my post. –  user59671 Mar 5 '13 at 19:13
    
Ah, right, there was that good question, still unanswered, +1. –  1015 Apr 8 '13 at 21:46
    
Have you tried to use a power series and Abel's like results? –  Davide Giraudo Apr 14 '13 at 16:06
    
to use power series, I need a convergent sequence of roots of the power series of $S^0$. –  user59671 Apr 14 '13 at 16:14

2 Answers 2

No, there are functions such that all your nested partial sums converge to $0$, but that are nonzero :

Suppose you have constructed the first $k(n)$ terms for $S_0$, such that $S_1(k(n)) = S_2(k(n)) = \ldots = S_n(k(n)) = 0$, and let $\varepsilon > 0$. You want to pick the next $k(n+1)-k(n)$ terms so that the next partial sums stay bounded by $\varepsilon$, and $S_m(k(n+1)) = 0$ for $0 \le m \le n+1$.

This is always possible : in fact you can uniquely choose the next $n+2$ terms such that $S_m(k(n)+n+2) = 0$ for $0 \le m \le n+1$, by solving an invertible linear system of $n+2$ equations. Then, in order to not go over $\varepsilon$ in the intermediate partial sums, find $N$ such that they stay under $N\varepsilon$. Divide those $n+2$ numbers by $N$, and apply them $N$ times to get what you wanted.


You can also have a more explicit construction :
Pick $S_0(0,1) = +1, 0$, so that $S_0(1) = 0$
Then pick $S_0(2+n) = - S_0(n \mod 2)/2$ for $0 \le n < 4$, so that $S_0(5) = S_1(5) = 0$
Then pick $S_0(6+n) = - S_0(n \mod 6)/3$ for $0 \le n < 18$, so that $S_0(23) = S_1(23) = S_2(23) = 0$
Then pick $S_0(24+n) = - S_0(n \mod 24)/4$ for $0 \le n < 96$, so that $S_0(119) = \ldots = S_3(119) = 0$.
And so on.

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What is this function $k(n)$? –  Zach L. May 18 '13 at 16:11
    
@ Zach L. You choose $k(n)$ as you are building your counter-example. –  mercio May 18 '13 at 18:44
    
Suppose $\epsilon>0$ is fixed. Suppose for some $n$ I have some $k(n)$ such that $S^1(k(n))=...=S^n(k(n))\le \epsilon$ now how can I find $k(n+1)>k(n)$ with $S^1(k(n))=...=S^{n+1}(k(n+1))\le \epsilon$. How will this prove $S^n(k)\to 0$? How can I be sure there is a strictly increasing $k(n)$? a clear proof is appreciated. Or a more clear definition for a nozero $S^0$ satisfying the conditions. –  user59671 May 22 '13 at 11:13
    
I don't look at an index $k(n)$ where the $S^i(k(n))$ are less than $\varepsilon$, but at one where they are all $0$. Then you can control what happens to the next $S^i(k(n)+j)$ because they all depend linearly on the $S^0(k(n)+j)$. –  mercio May 22 '13 at 11:44

Hmm, it might be that I did not understand the question correctly, but I find the problem interesting so I'll give it a try and please correct me if I'm wrong.

Here is, what I understand from the definitions of the $S_m$ with an example at $n=4$: $$ \begin{eqnarray} S^1(4) &=& S^0(1)+ S^0(2)+ S^0(3)+ S^0(4) \\ &=& 1 S^0(1)+ 1 S^0(2)+ 1 S^0(3)+ 1 S^0(4) \\ S^2(4) &=& S^1(1)+ S^1(2)+ S^1(3)+ S^1(4) \\ &=& 4 S^0(1)+ 3 S^0(2)+ 2 S^0(3)+ 1 S^0(4) \\ S^3(4) &=& S^2(1)+ S^2(2)+ S^2(3)+ S^2(4) \\ &=& 10 S^0(1)+ 6 S^0(2)+ 3 S^0(3)+ 1 S^0(4) \\ \end{eqnarray} $$ and $$ S^m(n) = \sum_{k=1}^n \binom{k+m}{m}S^0(k)$$ where also $ \lim _{n \to \infty}S^m(n) = 0$ is assumed. I hope I've got this correct so far.

Now in the example, if we subtract $S^1(4)$ from $S^2(4)$ we get $$S^2(4)-S^1(4) = S^2(3)$$ which is generalizable in the obvious way $$S^m(n)-S^{m-1}(n) = S^m(n-1) $$ or $$S^m(n)-S^m(n-1) = S^{m-1}(n) $$ But as $S^m(n)$ goes to zero so does $S^m(n-1)$, and also $ \lim_{n \to \infty}S^m(n)-S^{m-1}(n) = 0 $
If that reasoning is correct, then we have also from $m=1$ only that $$ \lim_{n \to \infty}S^0(n) = 0 $$

Hmm, that's not the final solution I'm afraid. Did your question actually mean that $S^0(n)=0$ for all $n \in \mathbb N$ ?


[Update]: Ok, after that we can write things in matrix-notation. For shortness let's denote $S^0(n)$ as $a_n$ and $S^m(n)$ as $z_{m,n}$ and let us then rewrite the vector/matrix-product for the $n=4$ example: $$ \begin{array} {lll} & * &\left[ \begin{array} {rrrr} \quad 1 & \quad 4 & \quad 10 & \quad 20\\ 1 & 3& 6 &10\\ 1 & 2& 3 & 4\\ 1 & 1& 1 & 1 \end{array} \right] \\ \left[ \begin{array} {r} a_1 & a_2 & a_3 & a_4 \end{array} \right] & = &\left[ \begin{array} {rrrrr} \ \ z_{1,4} & \ z_{2,4} & \ z_{3,4} & \ \ z_{4,4} \end{array} \right] \end{array} $$ If this scheme is extended to infinite size ($ n \to \infty$) in the obvious way, the problem says that we expect that for all k we find $z_{k,n} \to 0$

To divide&conquer the problem, we can LR-factorize the coefficients matrix, let's call it $C_n$ with the index n for the size such that $C_n = L_n \cdot R_n$ with $L_n$ and $R_n$ triangular matrices. We have then the formal definition $$ \begin{eqnarray} A_n \cdot C_n &=& Z_n \\ A_n \cdot (L_n \cdot R_n) &=& Z_n \\ \end{eqnarray}$$ We get then the interesting effect, that $L_n$ becomes the lower triangular Pascal-matrix and its entries are fixed for any size; so we can look at the left dot-product $ A_n \cdot L_n = B_n $ first and this looks then like

$$ \begin{array} {lll} & * &\left[ \begin{array} {rrrr} \quad 1 & & & \\ 1 & 1& &\\ 1 & 2& 1 & \\ 1 & \ 3& \ 3 & \ 1 & \cdots\\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array} \right] \\ \left[ \begin{array} {r} a_1 & a_2 & a_3 & a_4 & \cdots \end{array} \right] & = &\left[ \begin{array} {rrrrr} \quad b_0 & b_1 & b_2 & b_3 & \cdots \end{array} \right] \end{array} $$ where if we understand the values $a_n$ as coefficients of a power series in x we have that the $b_0$ can expressed as a function $f(x)$ at $x=1$ by the power series $$ f(x) = \sum_{k=0}^\infty a_{1+k} x^k $$and the $b$'s of the following columns are then the derivatives such that $$ b_k= x^k {f^{(k)}(x)\over k!} \qquad \qquad \text{ at } x=1 $$

Now the right matrix $R$ is upper triangular; and its entries change with the increasing $n$; we get with the $n$ kept indeterminate $$ R = \left[ \begin{array} {ccccc} 1 & n & \binom{n+1}{2} &\binom{n+2}{3} &\binom{n+3}{4}& \cdots \\ . & -1 & -n & -\binom{n+1}{2} &-\binom{n+2}{3} & \cdots \\ . & . & 1 & n & \binom{n+1}{2} & \cdots \\ . & . & . &- 1 & -n & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots &\ddots\\ \end{array} \right] $$ and we want, that the matrix-product of $B_n \cdot R_n = Z_n $ in the limit $ n \to \infty$ arrives at the nullvector $ \lim_{n \to \infty} Z_n = [0,0,0,...]$.

From the first column of $R$ we see, that $b_0=f(1)=0$ is required to make $z_1=S^1(n)=0$.
From this seems(!) to follow, that for the second column to provide $z_2=0$ also $b_1=0$ is required and from this for all other columns too. However, because n goed to infinity it might be, that there is some more consideration of the limits required - but I'm not yet sure of this.
But given, that this considerations are all correct, then we see, that $f(x)$ must be a function, where all derivatives at $x=1$ become zero - and I know only one function, which is not constant and has still all derivatives at some (single (!)) point $x_0$ equal to zero, this is $g(x)=\exp(-{1 \over x^2})$ at $x_0=0$ .

Caution: if divergent expressions are involved this might look differently - I don't see this yet.

So, although this small study was a nice experience I still cannot answer this question to one side so far..., sorry...hmmm...

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you did understand the question correctly. and the question is whether $S^0(n)=0$ for all $n$. –  user59671 May 18 '13 at 15:44

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