2015_30 Simple and elegant Part 1

$\boxed{Solution 1:}$

$f\left( x+{ y }^{ n } \right) =f\left( x \right) +{ (f(y)) }^{ n }\quad \forall x,y\epsilon R$

Notice that from first principles we have $f'(0) = \lim_{x\rightarrow 0} \frac{f(0+h) - f(0)}{h}$ Now putting $(x,y) = (0,0)$ we get that $f(0) = 0$ we can use it to prove that a limit is of $0/0$ form.

Also we will denote $h = (h^{\frac{1}{n}})^n$ in order to keep conformity with the first equation.Plugging in all our changes gives us :

$f’(0) = \lim_{x\rightarrow 0} \frac{f(0) + (f(h^{\frac{1}{n}}))^n - f(0)}{h} = \lim_{x\rightarrow 0} \frac{ (f(h^{\frac{1}{n}}))^n}{(h^{\frac{1}{n}})^n} = [\lim_{x\rightarrow 0} \frac{f(h^{\frac{1}{n}})}{h^{\frac{1}{n}}}]^n$

Since the final limit is of the form $0/0$ we simply apply L Hospital 's to evaluate it and finally we get,

$f'(0) = [f'(0)]^n$ This holds true if $f’(0) = 1$ or $f’(0) = 0$ .Since $f’(0) > 0$ Hence we can conclude

$\boxed{f’(0) = 1}$ .

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$\boxed{Solution 2 :}$ $f\left( x+{ y }^{ n } \right) =f\left( x \right) +{ (f(y)) }^{ n }\quad \forall x,y\epsilon R$

This functional equation is satisfied for $f(x) = x$ and $f(x) = 0$ .

If $f(x) = x ; f'(x) = 1$ and if $f(x) = 0 ; f'(x) = 0$ .But since $f'(x) > 0$ we must have

$\boxed{f'(x) = 1}$