Limits Plus: IV

Suppose $f(x)$ is three times differentiable in the domain (some hints how do I prove this supposition always holds?). Now differentiating the given equation three times individually with respect to $x$ and to $y$ results in: $f'''(x+y)=f'''(x)=f'''(y)$ Let this quantity be $A$ so $f'''(x)=A$ . Then integrating three times we get: $f(x)=\frac{A}{6}x^3+\frac{B}{2}x^2+Cx+D$ for some unknown constants $A, B, C, D$ . Substituting back in the original equation gives: $\frac{A}{2}\left(x^2y+xy^2\right)+Bxy=D+x^2y+xy^2$ This hold for all reals so we can equate the coefficients of the similar monomials from each side. This gives the equations: $\text{For } x^2y: \frac{A}{2}=1 \Rightarrow A=2$ $\text{For } xy^2: \frac{A}{2}=1 \Rightarrow A=2$ $\text{For } xy: B=0$ $\text{For free terms}: 0=D$ However we have no information for $C$ . So for now $f(x)=\frac{1}{3}x^3+Cx$ Using the limit: $1=\lim_{x\to 0}\frac{f(x)}{x}=\lim_{x\to 0}\frac{1}{3}x^2+C=C$ So $C=1$ and $f(x)=\frac{1}{3}x^3+x$ . We easily find that: $f(0)+f'(0)+f'(x)-x^2=0+1+x^2+1-x^2=\fbox{2}$

From the required expression we can guess that f'(x) - x^2 would be some constant independent of x..... So let f'(x) = x^2 +c So f(x) = (x^3)/3 + cx + K where K is the constant of integration....now as f(0) = 0 as evident from the given relation...K is equal to 0. So f(x) =(x^3)/3 + cx. We can clearly see that it satisfies the relation given..... Again as limit x->0 f(x)/x is 1. Then we can say that c= 1. Again we can also say that f'(0) = 1. So f(0) + f'(0) + c = 2