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

Find $$(fofofo...ofOgogo...og)=\frac{1}{x}$$ {fofo... are 101 times and gogo.. are 100 times}

Then Find $x$?

I calculated as follows Since $$(fog)=\frac{\frac{2x+1}{2-x}+2}{1-2{\frac{2x+1}{2-x}}}$$ $$(fog)=\frac{2x+1+4-2x}{2-x-4x-2}$$ $$(fog)=\frac{5}{-5x}=-\frac{1}{x}$$

and $$(fog)o(fog)=x$$

So 100th term as $(fog)o(fog)o...o(fog) = x$

Hence $$(fofofo...ofOfofo...og)=\frac{1}{x}$$ $$f(fog)o(fog)o(fog)o....o(fog)=1/x$$ $$f(x)=1/x$$ $$f(x)=\frac{x+2}{1-2x}=\frac{1}{x}$$ $$x^2+2x=1-2x$$ $${x^2}+4x-1=0$$

$$x=\frac{-4\pm\sqrt{4^2+4}}{2(1)}$$ $$x=-2\pm\sqrt{5}$$

Is the method and Answer Correct? Need Your Suggestion.

  • 1
    $\begingroup$ FYI you should use $\texttt{\circ}$ instead of $o$. $\endgroup$ – kccu May 4 at 13:54
  • $\begingroup$ Your calculation of $f \circ g$ is correct but not very useful. The calculation of $(f \circ g) \circ (f \circ g)$ is not needed because that expression does not appear in the problem. What would be more useful would be to calculate $f\circ f$ and $g \circ g$, then $f \circ f \circ f$ and $g \circ g \circ g$. I was guessing that the fourth iterate of each is the identity, but it seemed not to be so when I tried to put it in Alpha. Then you could say that the whole chain of $100\ g$'s is the identity, as is the chain of $100\ f$'s and you are left with $f$ $\endgroup$ – Ross Millikan May 4 at 14:06
  • $\begingroup$ @RossMillikan finding fofof.. is a hard process so I choose fog method. is my answer wrong? $\endgroup$ – Pankaj Solanki May 4 at 14:13
  • $\begingroup$ I didn't understand what you did in the step right after the word "Hence". Did you use that $(f\circ g)\circ(f\circ g)=f \circ f \circ g \circ g$ ? Because this is usually wrong. $\endgroup$ – Fareed AF May 4 at 14:27

Your method is almost correct. The result is definitely correct.

You only need to show that $f$ and $g$ commute (which is true):

  • $(f\circ g)(x) = (g\circ f)(x) = -\frac{1}{x}$

Hence, you have

  • $((f\circ g)\circ(f\circ g))(x) = x$

Since $f$ and $g$ commute using the symbols $f^n = \underbrace{f\circ \cdots \circ f}_{n-fold\; composition\; of \;f}$ etc., you get your result

$f^{101}\circ g^{100} = ((f\circ g)^2)^{50}f = f$

The remaining part of your solution is exactly as you did.


Composition of functions is not -- in general -- a commutative process, so what you've computed is apparently different from what the problem specifies. You are not to rearrange the order of composition for different functions unless you've first shown that it is allowable.

What you need may be a little tenacity. Composing functions of the types above with themselves -- that is $ffff\cdots f(x)$ -- usually ends one up in the identity function after a sufficient number of iterated compositions. Thus, you need not compute after some point on.

  • $\begingroup$ In the present example $f$ and $g$ do commute and their composition is $-\frac{1}{x}$. So, rearranging the composition leads directly to the solution in this case. $\endgroup$ – trancelocation May 4 at 14:58
  • $\begingroup$ @trancelocation Right. However, I think OP was not aware of the general non commutativity of composition; hence he should have proved that they did in this case before manipulating with abandon. $\endgroup$ – Allawonder May 4 at 18:11

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