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Let $\Bbb{Z}^+$be the set of all non-negative integers where $n$ and $k$ are given natural numbers. We consider the following non-decreasing function,

$$f:\Bbb{Z}^+ \to \Bbb{Z}^+$$ such that

\begin{align} f\left(\sum_{i=1}^n a_i^n\right) =\frac 1 k \sum_{i=1}^n \left(f(a_i)\right)^n \end{align}

1.Find all functions for $n=2014$ and $k=2014^{2013}$

2.Find, how many functions and which satisfy the condition of the problem depending on the values ​​of parameters $n$ and $k$.

I have some ideas for this problem .
I know how to found $f(0)$ (just take $a_1=a_2=\cdots=a_n=0$) .
I also can found $f(1)$ (take $a_1=1$ and $a_2=\cdots=a_n=0$)
Then if $f(0)=0$ I've get that $f(a_n)+f(b_n)=f(a_n+b_n)$. I also think thas all functions will be as $f(x)=kx$ but I can't prove it .
Can you help me with some partical solution or some ideas which can help to find it.

And can someone tell me how in first example ($n=2014$ and $k=2014^{2013}$) found f(2015) or $f(2016)$ using that $f(i^n)=i\cdot f(1)$ ( for $2015>i>1$)and $f(1)=0$ or $f(1)=2014$

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    $\begingroup$ I understand that it's solution but I need find all solutions and prove that there are no other solutions $\endgroup$ – Antony Oct 9 '14 at 14:35
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    $\begingroup$ @YourAdHere $f(n)=0$ is a solution (the only one of the form $f(n)=cn$ with $c\ne 2014$) $\endgroup$ – Hagen von Eitzen Oct 14 '14 at 14:47
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    $\begingroup$ Meta discussion on this question, which is suspected to stem from a contest: meta.math.stackexchange.com/q/17040/163 $\endgroup$ – Tobias Kienzler Oct 19 '14 at 18:02
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    $\begingroup$ This question is almost surely taken from a contest-- it contains the current date and is completely arbitrary and unmotivated. Given that the date is this year, and the OP hasn't had the courtesy to link to the source, there is a definite risk that the contest is going on now. Until the OP shows us a link to this contest so that we can know it is over, I believe we should not answer it. Come on folks -- recognizing contest problems can be hard, and we don't have an obligation to always do so, but when they are this obvious and the OP has put in so little work, why risk helping someone cheat? $\endgroup$ – David E Speyer Oct 21 '14 at 14:52
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    $\begingroup$ I guess just like your colleague santoni7, you got this question from the Ukrainian school team maths competition. It would be better to clearly indicate this, currently the two of you appear to try and cheat the SE rep-system - despite the questions themselves actually sounding interesting $\endgroup$ – Tobias Kienzler Oct 21 '14 at 17:37
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I do not know how to define all solutions for all values, but I know how to find all solutions on $\{0,1,\dots,n\}$.

Depending on the choice of the parameters, there are up to $4$ possible solutions in $\{0,1,\dots,n\}$.

Defining $f(0)$

If all $a$s $= 0$, we have $f(0)=\frac{n}{k}f^n(0)$, thus either $f(0)=0$ or $f(0)=\left(\frac{k}{n}\right)^{\frac{1}{n-1}}$ if this number happens to be integer. For your numerical example, this number is not an integer.

Defining $f(1)$ given $f(0)$

If $a_1=1$ and the rest is $0$, we have $f(1)=\frac{n-1}{k}f^n(0)+\frac{1}{k}f^n(1)=\frac{n-1}{n}f(0)+\frac{1}{k}f^n(1)$. If $f(0)=0$, we have either $f(1)=0$ or $f(1)=k^{1/(n-1)}$; for your numerical example, the latter value is $2014$. If $f(0)\neq 0$, we can have between $0$ and $2$ possible values for $f(1)$ - we can use them as long as they are integer.

Defining $f(2), \dots, f(n)$ given $f(1)$ and$f(0)$

Once we have decided on the values for $f(0)$ and $f(1)$, we can define $f(m)$, $1<m\leq n$, by choosing $a_1=\dots=a_m=1$, and the rest is $0$.

We then have:

$f(m)=\frac{1}{k}\left(m f^n(1) + (n-m) f^n(1)\right) = \frac{1}{k}\left(m f^n(1) + (n-m) f^n(0)\right)=m\left(f(1)-\frac{n-1}{n}f(0)\right) + \frac{n-m}{n}f(0)=mf(1)+(1-m)f(0)$

This holds for any $f(1)$ and $f(0)$.

For your parameters, we have $f(m)=2014m$ and $f(m)=0$

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  • $\begingroup$ Don't you have any ideas how solve this problem for numbers bigger that n? $\endgroup$ – Antony Oct 9 '14 at 15:46
  • $\begingroup$ To be honest, I think it might be impossible. This is because for most $m$s, there will be no $n$ integers such as the sum of their $n$ power is $m$. You can define it on more intervals though, the next one will be starting from $2^n$. $\endgroup$ – Yulia V Oct 9 '14 at 16:24
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    $\begingroup$ I think we must somehow use fact that function is non-decreasing but I can't imagine how $\endgroup$ – Antony Oct 9 '14 at 16:30
  • $\begingroup$ I think you might need "strictly increasing" property. Why don't you try to compute it on $[2^n, \dots, 2^n+(n-1)]$ and publish the results? $\endgroup$ – Yulia V Oct 9 '14 at 16:45

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