1one function is f(x)=x.
Determine all functions (from R → R)
\boxed{xf(x+xy)=xf(x)+f(x^2)f(y)}
FOR ALL REALS X,Y.....
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5 Answers
11The qsn asks all.....Either prove ur function's uniqueness else go on fighting! [66]
1do we have to tell the number of functions or what the functions really are...?
1@arshad u gotta name the function
@shoumik i dont fight 4 nething i m apostle of non-voilence
341Setting y = 0, we see that f(0) = 0.
We see by setting x = 1, that f(1+y) = f(1) (1 + f(y)).
So if f(1) = 0, f(x) = 0 for all x.
Suppose f(1) ≠0:
Setting x = 1 and y = -1, we have
f(0) = 0 = f(1) (1 + f(-1)) and since f(1) ≠0, we must have f(-1) = -1.
Now in the original equation we set y = -1 and obtain the equation
x f(x) = f(x2)
We see that f(0) = 0. When x ≠0, we have by letting g(x) = \frac{f(x)}{x}
g(x) = g(x2)
Since g(x) is even, WLOG x>0
From the given functional equation, we have
g(x) = g(x^{\frac{1}{2}}) = .... = g(x^{\frac{1}{2n}})=...
For all x>0, the sequence \{x^{\frac{1}{2n}}\} converges to 1.
Since we assume that f(x) is continuous and so g(x) is continuous and by Heine's Theorem of continuity, we have the sequence \{g(x^{\frac{1}{2n}})\} converges to g(1). But since all the terms in the sequence are equal, we have g(x) = g(1) for all x
So that f(x) = kx.
Since x f(x) = f(x2), we have k= 1 and hence f(x) = x