Functions

Aug 2018
137
7
România
Hello all,

Determine ascending functions \(\displaystyle f:\mathbb N^{*} \rightarrow \mathbb N^{*}\) with the property that \(\displaystyle \frac{f(1)+f(2)+\cdots +f(n)}{f(1)f(2)+f(2)f(3)+\cdots +f(n)f(n+1)}=\frac{3}{2f(n)+4}\) \(\displaystyle \forall n \in \mathbb N^{*}\).

All the best,

Integrator
 
Aug 2018
137
7
România
Hello all,

Does nobody have any idea? Thank you very much!

All the best,

Integrator
 
Last edited by a moderator:

romsek

Math Team
Sep 2015
2,969
1,676
USA
Hello all,

Does nobody have any idea? Thank you very much!

All the best,

Integrator
I suspect people are getting tired of putting a lot of time into your brain teasers. You clearly know the answers to your questions ahead of time.
 
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skipjack

Forum Staff
Dec 2006
21,482
2,472
How about $f(n) = n\ \forall n \in \mathbb N^{*}$?
 
Aug 2018
137
7
România
How about $f(n) = n\ \forall n \in \mathbb N^{*}$?
Hello,

Correct! And are there any other functions? How do we find all the functions? Thank you very much!

All the best,

Integrator
 
Last edited by a moderator:
Aug 2018
137
7
România
I suspect people are getting tired of putting a lot of time into your brain teasers. You clearly know the answers to your questions ahead of time.
Hello,

I do not know yet how to find all those functions ...I assumed that \(\displaystyle \sum_{k=1}^{n} f(k)=3\cdot g(n)\) and \(\displaystyle \sum_{k=1}^{n} f(k)f(k+1)=[2f(n)+4]\cdot g(n)\) where \(\displaystyle g(n) : \mathbb N^*\rightarrow \mathbb N^*\) \(\displaystyle \forall n\in \mathbb N^*\)...

All the best,

Integrator
 
Last edited:
Aug 2018
137
7
România
What do "ascending" and $N^*$ mean?
Hello,

That is clear! Apart from \(\displaystyle f(n)=n\), are other functions no longer? If there are no other functions, then how can we prove that \(\displaystyle f(n)=n\) is the only function? The function \(\displaystyle f(n):\mathbb N^* \rightarrow \mathbb N^*\) with \(\displaystyle f(n)=2n+3\) and \(\displaystyle n\in \mathbb N^*\), is ascending function? I think so....Thank you very much!

All the best,

Integrator
 
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skipjack

Forum Staff
Dec 2006
21,482
2,472
That doesn't tell me what $\mathbb{N}^*$means.

Also, is a constant function an ascending function?
 
Aug 2018
137
7
România
That doesn't tell me what $\mathbb{N}^*$means.

Also, is a constant function an ascending function?
Heloo,

I do not understand where you want to reach...
\(\displaystyle N^*\) is the set of natural numbers greater than zero.The function \(\displaystyle f(n)=C\) where \(\displaystyle C\) is an arbitrary constant and \(\displaystyle C\in \mathbb N^*\) is not a strictly ascending function.
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How do we prove that \(\displaystyle f(n)=n\) is the only function that complies with the condition from the problem?Thank you very much!

All the best,

Integrator
 
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