Solving Functional Equations: Tips & Examples

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To solve the functional equations f(x+y)+f(x-y)=2f(x)f(y) and f(x)+(x+1/2)f(1-x)=1, start by substituting specific values for x and y. Setting y=0 in the first equation leads to the conclusion that f(0)=1. Additionally, substituting x=0 reveals that f(-y)=f(y), indicating that f is an even function. These insights form the basis for further exploration of the equations. Understanding these properties is crucial for finding the general solution.
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Hello,

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

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

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.

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

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.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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