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May 18th, 2019, 08:38 AM  #1 
Senior Member Joined: Aug 2018 From: România Posts: 112 Thanks: 7  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, 02:17 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,855 Thanks: 744 
Simplest solution f''(x)=0.

May 18th, 2019, 09:53 PM  #3 
Senior Member Joined: Aug 2018 From: România Posts: 112 Thanks: 7  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 10:57 PM. 
May 18th, 2019, 10:55 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 21,124 Thanks: 2332 
Yes.


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