Solving Functional Equations: Tips & Examples

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The discussion focuses on solving two functional equations: \( f(x+y) + f(x-y) = 2f(x)f(y) \) and \( f(x) + \left(x + \frac{1}{2}\right)f(1-x) = 1 \). The first equation reveals that \( f(0) = 1 \) and indicates that \( f \) is an even function, as derived from substituting specific values for \( x \) and \( y \). The second equation is linked to the first and requires similar analytical techniques for resolution.

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stanley.st
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Hello,

could you explain me what's the right way to solve these equations. I've never solved it before.

[tex]f(x+y)+f(x-y)=2f(x)f(y)\,\;\;\forall x,y\in\mathbb{R}[/tex]
[tex]f(x)+\left(x+\frac{1}{2}\right)f(1-x)=1\,\;\;\forall x\in\mathbb{R}[/tex]

thank you...
 
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stanley.st said:
Hello,

could you explain me what's the right way to solve these equations. I've never solved it before.

[tex]f(x+y)+f(x-y)=2f(x)f(y)\,\;\;\forall x,y\in\mathbb{R}[/tex]
[tex]f(x)+\left(x+\frac{1}{2}\right)f(1-x)=1\,\;\;\forall x\in\mathbb{R}[/tex]

thank you...

Since the equation is true for all values, use clever values of x and y to get a system of equations.
 
For example, if, in the first, we take y= 0, we get f(x+0)+ f(x- 0)= 2f(x)= 2f(x)f(0) so we must have f(0)= 1. But if we take x= 0, we get f(0+y)+ f(0- y)= f(y)+ f(-y)= 2f(0)f(y) = 2f(y). That is, f(-y)= f(y) for all y so f is an even function.
 

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