SOLUTION TO PROBLEM 13

This solution is by Andrew Gacek

Setting x=0, we get f(0)+f(0)=f(0). Thus f(0)=0.

Take the derivative with respect to x:

$$f'(x)=f'\left(\frac{x+y}{1-xy}\right)\left(\frac{1+y^2}{(1-xy)^2}\right).$$

Setting x=0 this becomes

f'(0)=f'(y)(1+y²).

Letting C=f'(0), we get the differential equation

$$f'(y)=\frac{C}{1+y^2},\qquad\qquad f(0)=0.$$

Integration leads to the solution $f(y)=C\cdot \arctan(y)$. This function is differentiable everywhere and satisfies the constraint for all values of C.

To see this take the following identity

$$\tan(a+b)=\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}$$

and let $a=\arctan(x)$ and $b=\arctan(y)$, to get

$$\tan(\arctan(x)+\arctan(y))=\frac{x+y}{1-xy},$$

$$\arctan(x)+\arctan(y)=\arctan\left(\frac{x+y}{1-xy}\right),$$

$$C\cdot \arctan(x)+C\cdot\arctan(y)=C\cdot\arctan\left(\frac{x+y}{1-xy} \right).$$

Also note that $\frac{d}{dx}(C\cdot\arctan(x))=\frac{C}{1+x^2}$.

Therefore, the functions in question are of the form $f(x)=C\cdot \arctan(x)$ for a real C.

Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Sat Dec 1 13:25:54 CST 2001