Math problem thread

Find all differentiable functions f: R -> R+ for which the following is true: f'(x) = f(f(x))

Such a function does not exist. 

Since f(x)>0 for all x, f'(x)=f(f(x))>0 so f must be strictly increasing.  Then f(x)>0 implies f(f(x))>f(0), or f'(x)>f(0). So f(0) is a lower bound on the derivative.

Now consider the slope between f(0) and f(x) for x<0.  We have
f(0)<(f(0)-f(x))/(0-x), or f(x)<f(0)+xf(0)=(1+x)f(0).  But then f(x)<0 for any x<-1, which is a contradiction.

Of course, this is a matter of definitions, but usually R+ = [0,infinity), as opposed to (0,infinity), so f(x) = 0 would be valid.

This one's a bit tricky:

In the parallelogram ABCD, there is a point P on the line AB, and a point Q on the line BC, such that the areas of the triangles formed by APD = 7, PBQ = 28 and QCD = 14.

Find the area of triangle formed by PQD.

>>4
Might not be the fastest way, but anyway.
Let R be a point on CD and S a point on AD such that PR//AD and QS//AB. T the intersection point between PR and QS.
|APRD|=14, |SQCD|=28, |PBQT|=56, and let |STRD|=x, now it's easy to see that
|APTS|/|STRD|=|PBQT|/|TQCR|, that is (14-x)/x=56/(28-x)
then x=4,178 approx. So |PQD|=|ABCD|-(7+14+28)=(56+28+14-x)-49=44.822 approx.
(The result not being an integer makes me feel unsure, though)

>>3
I understand your point, but that's not what wolfram last said.  What notation would you use for positive definite functions?

f(x) = x
therefore f(f(x)) = x (substituting x for x)

>>7

That's a contradiction against the start conditions, which were that f'(x) = f(f(x)).
f(f(x)) = x, but f'(x) = 1, in this case.

Here's a nice one:


Before his death, an old pirate hid his treasure on an island with the following instructions for finding it later:
"On the island there is only one coconut palm tree, one banana tree and one big boulder. Start near the boulder, walk to the coconut tree, pick a coconut, turn 90 degrees counter-clockwise and walk then as many steps as you did from the boulder to the palm tree. Leave the coconut there. Return to the boulder, and walk this time to the banana tree, turning 90 degrees clockwise, and again walk as many steps as you did from the boulder to the banana tree. The treasure is now buried at the midpoint between you and the coconut."

You get your hands on these instructions and decide to get the treasure, so you set off to the island. You search around, but you can only find the coconut palm and the banana tree -- the boulder seems to be gone! The mission now seems impossible, but there is still a way to find the treasure. How?

>>9
If the boulder location is indeed irrelevant, then you can pretend it is wherever you want it to be (eg at one of the plants).  Proving that the boulder is irrelevant, however, is going to be much more challenging.

>>10
Place the island in the Cartesian plane with the coconut tree on (0,0) and the banana tree on (a,0), a>0. If the boulder is on (x,y), the first instructions correspond to rotate (x,y) -90º around the origin, which sends it to (y,-x) where you place the coconut. Now go back to the boulder, the second part of the instructions is rotating (x,y) +90º around (0,a), leaving you in (a-y,x-a). The midpoint between (y,-x) and (a-y,x-a) is (a/2,-a/2) which doesn't depend of (x,y). Showing that the location of the boulder is not relevant.

>>11
Sorry, it was "+90º around (a,0)" (in the fourth line).

ne^x, where can be any real number?

>>13

No. If f(x) = ne^x, f'(x) = ne^x = f(x).
One of the assumptions was that f'(x) = f(f(x)), but f(f(x)) = f(ne^x) = ne^(ne^x) != ne^x = f'(x)

>>2
>Then f(x)>0 implies f(f(x))>f(0)
why?

what about using the definition of a differentiable function ? Maybe using a relation like f'(x)*h=df(x)(h) for real functions...

>>15
Increasing function.

>>1
If f(x) is constant, it works.

>>15
Then f(x)>0 implies f(f(x))>f(0)
why?
f(x) > 0 implies f(f(x)) > 0 for all x in R as f(x) is in R+.

As f(f(x)) = f'(x), this means f(x) is a strictly increasing function.

Let y = f(x) for a given x. As f is strictly increasing, f(y) > f(y-z) for all z > 0. Also, we know that y = f(x) > 0. In particular, for z = y we have f(y) > f(y-y) = f(0) QED.

>>18

No.

If f(x) = c where c is a constant, f'(x) = 0
If c ≠ 0, f(f(x)) = f(c) = c ≠ 0 = f'(x), which contradicts the conditions the function has to meet.

>>15

>Then f(x)>0 implies f(f(x))>f(0)
why?

If f is strictly increasing, then x>y implies f(x)>f(y).  In other words, we can take f of both sides and keep the inequality.  Therefore, we can take f of both sides of f(x)>0 to get f(f(x))>f(0).

didn't read any single post but I may propose some ideas.

deriving the given equation gives : f''(x) = f'(f(x))*f'(x).

more like f''(x)/f'(x) = f'(f(x)) = f(f(f(x))) = f(f'(x)) amirite ? (assuming f'(x) is not zero, else is OH SHI- or more like f is a constant)

which is more like log(f'(x)) = int(fof') = int (f'of).

Nao I must go, so may the force be with you.