2016.3.8 Question 8

1. If we replace $x$ with $-x$ in the original equation, we get

   $$f(-x)+(1-(-x))f(-(-x))={(-x)}^2,$$

   which simplifies to

   $$f(-x)+(1+x)f(x)=x^2$$

   as desired.

   Therefore, we have a pair of equations in terms of $f(x)$ and $f(-x)$:

   $$\{\begin{array}{cc}f(x)+(1-x)f(-x)\hfill & =x^2\hfill \end{array}$$

   Multiplying the second equation by $(1-x)$ gives us

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

   and subtracting the first equation from this

   $$-x^{2}f(x)=-x^3,$$

   which gives $f(x)=x$.

   Plugging this back, we have

   $$\begin{array}{llll}\hfill \text{LHS}& =f(x)+(1-x)f(-x)& \hfill & \\ \hfill & =x+(1-x)(-x)& \hfill & \\ \hfill & =x-x+x^2& \hfill & \\ \hfill & =x^2& \hfill & \\ \hfill & =\text{RHS}& \hfill & \end{array}$$

   which holds. Therefore, $f(x)=x$ is the solution to the functional equation.

2. For $x\ne 1$, we have

   $$\begin{array}{llll}\hfill K(K(x))& =\frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1}& \hfill & \\ \hfill & =\frac{(x+1)+(x-1)}{(x+1)-(x-1)}& \hfill & \\ \hfill & =\frac{2x}{2}& \hfill & \\ \hfill & =x,& \hfill & \end{array}$$

   for $x\ne 1$, as desired.

   The equation on $g$ is

   $$g(x)+\mathit{xg}(K(x))=x,$$

   and if we substitute $x$ as $K(x)$, we have

   $$g(K(x))+K(x)g(K(K(x)))=K(x),$$

   which simplifies to

   $$g(K(x))+K(x)g(x)=K(x).$$

   Multiplying the second equation by $x$, we have

   $$\mathit{xK}(x)g(X)+\mathit{xg}(K(x))=\mathit{xK}(x),$$

   and subtracting the first equation from this gives

   $$(\mathit{xK}(x)-1)g(x)=x(K(x)-1),$$

   which gives

   $$\begin{array}{llll}\hfill g(x)& =\frac{x\left(K(x)-1\right)}{\mathit{xK}(x)-1}& \hfill & \\ \hfill & =\frac{x\left(\frac{x+1}{x-1}-1\right)}{x\cdot \frac{x+1}{x-1}-1}& \hfill & \\ \hfill & =\frac{x\left[(x+1)-(x-1)\right]}{x(x+1)-(x-1)}& \hfill & \\ \hfill & =\frac{2x}{x^2+1},& \hfill & \end{array}$$

   for $x\ne 1$.

   If we plug this back to the original equation, we have

   $$\begin{array}{llll}\hfill \text{LHS}& =\frac{2x}{x^2+1}+x\frac{2\cdot \frac{x+1}{x-1}}{{\left(\frac{x+1}{x-1}\right)}^2+1}& \hfill & \\ \hfill & =\frac{2x}{x^2+1}+\frac{2x\cdot (x+1)\cdot (x-1)}{{(x+1)}^2+{(x-1)}^2}& \hfill & \\ \hfill & =\frac{2x}{x^2+1}+\frac{2x(x^2-1)}{2x^2+2}& \hfill & \\ \hfill & =\frac{2x}{x^2+1}+\frac{x(x^2-1)}{x^2+1}& \hfill & \\ \hfill & =\frac{x^3-x+2x}{x^2+1}& \hfill & \\ \hfill & =\frac{x(x^2+1)}{x^2+1}& \hfill & \\ \hfill & =x& \hfill & \\ \hfill & =\text{RHS},& \hfill & \end{array}$$

   so

   $$g(x)=\frac{2x}{x^2+1}$$

   is the solution to the original functional equation.

3. Let $H(x)=\frac{1}{1-x}$. Notice that

   $$\begin{array}{llll}\hfill H(H(x))& =\frac{1}{1-\frac{1}{1-x}}& \hfill & \\ \hfill & =\frac{1-x}{1-x-1}& \hfill & \\ \hfill & =\frac{x-1}{x}& \hfill & \\ \hfill & =1-\frac{1}{x}& \hfill & \end{array}$$

   and

   $$\begin{array}{llll}\hfill H(H(H(x)))& =\frac{1}{1-\left(1-\frac{1}{x}\right)}& \hfill & \\ \hfill & =\frac{x}{1}& \hfill & \\ \hfill & =x.& \hfill & \end{array}$$

   Now, if we replace all the $x$ with $\frac{1}{1-x}$, we will get

   $$h\left(\frac{1}{1-x}\right)+h\left(1-\frac{1}{x}\right)=1-\frac{1}{1-x}-\left(1-\frac{1}{x}\right),$$

   and doing the same replacement again gives us

   $$h\left(1-\frac{1}{x}\right)+h(x)=1-\left(1-\frac{1}{x}\right)-x.$$

   Summing these two equations, together with the original equation, gives us that

   $$2\cdot \left[h\left(\frac{1}{1-x}\right)+h\left(1-\frac{1}{x}\right)+h(x)\right]=3-2\cdot \left[x+\frac{1}{1-x}+\left(1-\frac{1}{x}\right)\right],$$

   and therefore

   $$h\left(\frac{1}{1-x}\right)+h\left(1-\frac{1}{x}\right)+h(x)=\frac{3}{2}-\left[x+\frac{1}{1-x}+\left(1-\frac{1}{x}\right)\right].$$

   Subtracting the second equation from this, gives that

   $$\begin{array}{llll}\hfill h(x)& =\left(\frac{3}{2}-\left[x+\frac{1}{1-x}+\left(1-\frac{1}{x}\right)\right]\right)-\left[1-\frac{1}{1-x}-\left(1-\frac{1}{x}\right)\right]& \hfill & \\ \hfill & =\frac{1}{2}-x.& \hfill & \end{array}$$

   Plugging this back to the original equation, we have

   $$\begin{array}{llll}\hfill \text{LHS}& =\frac{1}{2}-x+\frac{1}{2}-\frac{1}{1-x}& \hfill & \\ \hfill & =1-x-\frac{1}{1-x}& \hfill & \\ \hfill & =\text{RHS},& \hfill & \end{array}$$

   which satisfies the original functional equation. Therefore, the original equation solves to

   $$h(x)=\frac{1}{2}-x.$$