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May 10th, 2019, 08:18 AM   #1
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A functional equation

Hello all,

Resolve the functional equation $\displaystyle f(x^n-i\sqrt2)+f(x^n)+f(x^n+i\sqrt2)=a+bi$ where $\displaystyle i^2=-1$ , $\displaystyle n\in \mathbb N$* and $\displaystyle a,b\in \mathbb R$.

All the best,

Integrator

Last edited by Integrator; May 10th, 2019 at 08:22 AM.
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May 10th, 2019, 08:03 PM   #2
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Try to set x=1 and x=0 .
for x=1 , $\displaystyle f(1+i\sqrt{2} ) + f(1) +f(1-i\sqrt{2} )=a+bi$.
for x=0 , $\displaystyle f(-i\sqrt{2} )+f(0)+f(i\sqrt{2})=a+bi$.
Also $\displaystyle f(1+i\sqrt{2} ) + f(1) +f(1-i\sqrt{2} )=f(-i\sqrt{2} )+f(0)+f(i\sqrt{2})$.
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May 10th, 2019, 08:53 PM   #3
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Quote:
Originally Posted by idontknow View Post
Try to set x=1 and x=0 .
for x=1 , $\displaystyle f(1+i\sqrt{2} ) + f(1) +f(1-i\sqrt{2} )=a+bi$.
for x=0 , $\displaystyle f(-i\sqrt{2} )+f(0)+f(i\sqrt{2})=a+bi$.
Also $\displaystyle f(1+i\sqrt{2} ) + f(1) +f(1-i\sqrt{2} )=f(-i\sqrt{2} )+f(0)+f(i\sqrt{2})$.
Hello,

I do not understand!What is the general form of the functions $\displaystyle f(x)$?
----------------------
For example, resolve the following case:
If $\displaystyle n=2$, $\displaystyle a=2$ and $\displaystyle b=1$ then what is the general form of the function $\displaystyle f(x)$?

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