Functional equeation

**ns1954** · January 21st 2010, 08:12 AM

Is it possible to find nonzero solution of functional equation

$<br /> (1+4x)f(2x)=2f(x+2x^2)<br />$

defined and analitic in some neighbourhood of $0$

**NonCommAlg** · January 24th 2010, 01:49 AM

Originally Posted by ns1954

Is it possible to find nonzero solution of functional equation

$<br /> (1+4x)f(2x)=2f(x+2x^2)<br />$

defined and analitic in some neighbourhood of $0$

never mind!

**ns1954** · January 24th 2010, 04:36 AM

After taking first derivate in FE, by shain rule, we have relation

$4f(2x)+2(1+4x)[f'(2x)-f'(x+2x^2)]=0$ or $4f(2x)+2(1+4x)[g(2x)-g(x+2x^2)]=0$ , by your notion.

I can't see how you derived $g(2x)=g(x+2x^2)$ from this.

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