Solving f(x) Without Derivatives, Limits, or Integrals

[Omid]

If :
f( [x-1]/[x+1]) + f(-1/x) + f( [x+1]/[1-x]) = x
Find f(x) without using derivatives nor limits nor Integrals.

Thanks

 

[Hurkyl]

Have you tried plugging in values for x?

 

[Omid]

Let's do it now.
x = 2 :
f(1/3) + f( -1/2) + f (-1/3) = 2.
Now what does it mean? I see a pattern but... I can't find f(x) anyway.
Can you give me some hints?

 

[James R]

f(x) = log(1/x)

 

[Hurkyl]

I mean lots of values... try to pick values of x so that you get f(y) terms where the y's are the same between different equations.

For instance, since that equation has a f(1/3) in it, you might like to plug in x=-3 next.

 

[Omid]

I plugged in so many xs but I just don't get it.
Consider that this problem must be solved in 2 or 3 minutes in the exam.
Anyway thanks for your help Hurkyl.

 

[Omid]

f(x) = log(1/x)

I don't know wether it's right or wrong. Anyway, can you tell me how did you get it?

 

[James R]

Omid,

There's no easy way to do this, because f could be anything. You really just have to guess.

My previous answer doesn't work, now that I've checked it. It's close, but if you apply it to the left hand side above, you get log(x) instead of x, so it's not quite there.

We can convert your expression to another form, which may be easier to work with.

Suppose we put

$$y=\frac{x-1}{x+1}$$

Then we have

$$\frac{x+1}{1-x} = -1/y$$

and

$$\frac{-1}{x} = -\frac{(y+1)}{y-1}}$$

Putting this back into your original expression, and changing the name of the variable y back to x, we get an alternative form of your original expression:

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

It still remains to find f(x), though.

 

[Omid]

Tell me how was I soppused to carry out?

Check it out:
>If :
>f( (x-1)/(x+1)) + f(-1/x) + f( (x+1)/(1-x)) = x >Find f(x) without using derivatives nor limits nor Integrals.
Cute.
Left side is not defined for x = -1, 0 or 1, but I guess we should
ignore that.
Let h(x) = (x-1)/(x+1), and let h[n] be h iterated n times.
Notice that h[2](x) = -1/x, h[3](x) = (x+1)/(1-x) and h[4](x) = x.
So the equation says f(h(x)) + f(h[2](x)) + f(h[3](x)) = x.
Replacing x by h(x), we have f(h[2](x)) + f(h[3](x)) + f(x) = h(x).
Similarly, f(h[3](x)) + f(x) + f(h(x)) = h[2](x), and
f(x) + f(h(x)) + f(h[2](x)) = h[3](x).
Solve this system of four equations for f(x), f(h(x)), f(h[2](x))
and f(h[3](x)) and you get your answer, which is a rational function
with numerator of degree 4 and denominator of degree 3.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
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I think what Robert said is somehow like what James_R has written.
But I don't understand how was I soppused to find or guess this ugly fraction by plugging in diffrent x values? .