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 
Senior Member Joined: Sep 2015 From: USA Posts: 2,496 Thanks: 1371  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 
Member Joined: Aug 2018 From: România Posts: 84 Thanks: 4  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  
Member Joined: Aug 2018 From: România Posts: 84 Thanks: 4  Quote:
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?

June 25th, 2019, 08:55 PM  #8 
Member Joined: Aug 2018 From: România Posts: 84 Thanks: 4  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  
Member Joined: Aug 2018 From: România Posts: 84 Thanks: 4  Quote:
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  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Derivatives, trignometric functions and exponential functions  Nij  Calculus  2  November 25th, 2015 06:20 AM 
functions  kfarnan  Calculus  5  September 10th, 2011 03:51 AM 
fog functions  UltraMath  Algebra  3  September 4th, 2011 01:43 PM 
functions  panky  Calculus  0  December 31st, 1969 04:00 PM 