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I have been trying to solve an exercise based on the relations and functions. Right now I had stuck to a question based on functions. The question says:

A real valued function $f(x)$ satisfies the functional equation

$$f(x-y)=f(x).f(y)-f(a-x).f(a+y)$$

where $a$ is given constant and also $f(0)=1$. Then $f(2a-x) =$

So far I know in these kind of questions we have to replace a specific variable with another value (maybe any constant or another variable). Like I had done an example previously,

- Number of onto functions
- expansion of $\cos^k(\theta)$
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Here is an example:

Questionsays, $f(x)+2f(1-x)=x^2$, Then what is $f(x)$?

Answer:Given, $$f(x)+2f(1-x)=x^2$$

Let is be equation $(1)$.Replace $x$ with $1-x$.

So now equation is $$f(1-x)+2f(x)=(1-x)^2$$

Let is be equation $(2)$.

So after multiplying equation $(2)$ with $2$ (name it equation $(3)$) and subtracting equation $(1)$ and $(3)$ we can get the $f(x)$.

I know This one is quite easier than that I had asked because my question would need much more substitutions.

Thanks in advance!

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Let $x = y= 0$. This gives $f(0) = f(0)^2-f(a)^2$ or $f(a) = 0$. Then let $x = 0$.

This gives $f(-y) = f(0)f(y)-f(a)f(a+y)$ or $f(-y) = f(y)$. Then let $x = a$.

This gives $f(a-y) = f(a)f(y)-f(0)f(a+y)$ or $f(a-y)+f(a+y) = 0$.

In this equation, set $y = x-a$. This gives $f(a-(x-a))+f(a+(x-a)) = 0$ or $f(a-x+a) = f(2a-x) = -f(x)$ as desired.

I dont think that finding a closed form is easy.

$\bullet \quad x=y=0$ $\rightarrow $ $f(a)=0.$

$\bullet \quad x=0 \rightarrow f(y)=f(-y)$. i.e., $f$ is an even function.

$\bullet \quad x=y=a \rightarrow f(2a)=-1$

$\bullet \quad y=2a \rightarrow f(x-2a)=f(2a-x)=-f(x)$

Writing out $f(a-0)$ one realizes that $f(a)=0$. Knowing that it is easy to check that $f(-a)=f(0-a)=0$ and that writing out $1=f(a-a)$ leads to $f(2a)=-1$. With this we can write out

$$

f(2a-x)=\underbrace{f(2a)}_{-1}f(x)-\underbrace{f(a-2a)}_0f(a+x)=-f(x)

$$

But then I do not know much more by now.

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