# Functions

#### Integrator

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

#### Integrator

Hello all,

Does nobody have any idea? Thank you very much!

All the best,

Integrator

Last edited by a moderator:

#### romsek

Math Team
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.

Last edited by a moderator:

#### skipjack

Forum Staff
How about $f(n) = n\ \forall n \in \mathbb N^{*}$?

#### Integrator

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:

#### 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.
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:

#### skipjack

Forum Staff
What do "ascending" and $N^*$ mean?

1 person

#### Integrator

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

Last edited by a moderator:

#### skipjack

Forum Staff
That doesn't tell me what $\mathbb{N}^*$means.

Also, is a constant function an ascending function?

#### Integrator

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.
---------------------------------
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

Last edited by a moderator: