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May 18th, 2019, 07:38 AM  #1 
Member Joined: Aug 2018 From: România Posts: 54 Thanks: 3  An equation of recurrence
Hello all, Solve the recurrence equation $\displaystyle f''(x2)+f''(x)+f''(x+2)=2f''(x+1)$. All the best, Integrator 
May 18th, 2019, 01:17 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,763 Thanks: 697 
Simplest solution f''(x)=0.

May 18th, 2019, 08:53 PM  #3 
Member Joined: Aug 2018 From: România Posts: 54 Thanks: 3  Hello, Others say they are solutions and if $\displaystyle f''(x)=c_1\bigg(\frac{1\sqrt{14i}}{2}\bigg)^x+c_2\bigg(\frac{1+\sqrt{14i}}{2}\bigg)^x+c_3\bigg(\frac{1\sqrt{1+4i}}{2}\bigg)^x+c_4\bigg(\frac{1+\sqrt{1+4 i}}{2}\bigg)^x$ where $\displaystyle i^2=1$ and $\displaystyle c_1$ , $\displaystyle c_2$ , $\displaystyle c_3$ , $\displaystyle c_4$ are constants. Is it right what others say? Thank you very much! All the best, Integrator Last edited by skipjack; May 18th, 2019 at 09:57 PM. 
May 18th, 2019, 09:55 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133 
Yes.


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