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 June 20th, 2019, 09:15 PM #1 Member   Joined: Aug 2018 From: România Posts: 88 Thanks: 6 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: 88 Thanks: 6 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: 20,969 Thanks: 2218 How about $f(n) = n\ \forall n \in \mathbb N^{*}$? June 25th, 2019, 05:58 AM   #5
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 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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 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: 20,969 Thanks: 2218 What do "ascending" and $N^*$ mean? Thanks from Integrator June 25th, 2019, 08:55 PM   #8
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 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: 20,969 Thanks: 2218 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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 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. Tags functions Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Nij Calculus 2 November 25th, 2015 06:20 AM kfarnan Calculus 5 September 10th, 2011 03:51 AM UltraMath Algebra 3 September 4th, 2011 01:43 PM panky Calculus 0 December 31st, 1969 04:00 PM

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