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 June 20th, 2019, 09:15 PM #1 Member   Joined: Aug 2018 From: România Posts: 84 Thanks: 4 Functions 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
 June 24th, 2019, 09:21 PM #2 Member   Joined: Aug 2018 From: România Posts: 84 Thanks: 4 Hello all, Does nobody have any idea? Thank you very much! All the best, Integrator Last edited by skipjack; June 24th, 2019 at 10:40 PM.
June 24th, 2019, 10:25 PM   #3
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Quote:
 Originally Posted by Integrator 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 skipjack; June 24th, 2019 at 10:40 PM.

 June 24th, 2019, 10:39 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,819 Thanks: 2158 How about $f(n) = n\ \forall n \in \mathbb N^{*}$?
June 25th, 2019, 05:58 AM   #5
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Quote:
 Originally Posted by skipjack 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 skipjack; June 25th, 2019 at 02:45 PM.

June 25th, 2019, 06:26 AM   #6
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Quote:
 Originally Posted by romsek 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 by Integrator; June 25th, 2019 at 06:41 AM.

 June 25th, 2019, 03:36 PM #7 Global Moderator   Joined: Dec 2006 Posts: 20,819 Thanks: 2158 What do "ascending" and $N^*$ mean? Thanks from Integrator
June 25th, 2019, 08:55 PM   #8
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Quote:
 Originally Posted by skipjack 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 skipjack; June 25th, 2019 at 09:10 PM.

 June 25th, 2019, 09:14 PM #9 Global Moderator   Joined: Dec 2006 Posts: 20,819 Thanks: 2158 That doesn't tell me what $\mathbb{N}^*$means. Also, is a constant function an ascending function?
June 26th, 2019, 09:02 PM   #10
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Quote:
 Originally Posted by skipjack 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 skipjack; June 26th, 2019 at 11:15 PM.

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