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Define g(x):=1-1/x. Then gg(x)=1/(1-x) and ggg(x)=x. Here we use the short notation fgh(x):=f(g(h(x))) for the composition of functions f, g, h. ggg(x)=x is really doing the trick for us. Rewrite the original functional equation as
(1) f(x)+fg(x)=1+x
Substitute x by g(x) and gg(x) respectively and obtain
(2) fg(x)+fgg(x)=1+g(x)
and
(3) fgg(x)+fggg(x)=1+gg(x)
The latter can be rewritten as
(3*) fgg(x)+f(x)=1+gg(x)
Solve (3*) for fgg(x) and insert into (2):
(2*) fg(x)+1+gg(x)-f(x)=1+g(x)
Solve (2*) for fg(x) and insert into (1):
(1*) f(x)+g(x)+f(x)-gg(x)=1+x
Rewrite (1*) and expand g:
2f(x)=1+x+gg(x)-g(x)=1+x+1/(1-x)-1+1/x=(x^3-x^2-1)/(x^2-x)
finally
f(x)=1/2 * (x^3-x^2-1)/(x^2-x)
The solution is very similar to the solution of atheron's first problem, only here we have to go one step further. In atheron's problem, one would set g(x):=1/x and utilize gg(x)=x. Credits to Steve and Richard who have spotted the right approach in no time at all. |
