@Carl Love 

This was my first thought too. But sometimes it is better to keep some RootOfs.
Or, make the alias only if they repeat.

Probably the best solution would be an interactive program with suggested sub-expressions to be "covered" and the possibility to use your own names for them.

@Carl Love 

Sorry, I simply forgot to do it, not because I do not appreciate your professional contribution. I really do; and thank you, of course!

V.A.

@Carl Love 

Interesting approach.
It should also consider situations like

eval(g(x),x=2*x);

where it fails.

@ecterrab 

My point is that in this situation eval acts blindly just like subs (in other situations).
After all, eval already does a "corruption of the programming language" when executing eval(int(f(x),x),x=1);
As I said, it is not a big deal, but the user must be informed about it.

@Kitonum 

Nice, vote up!

But what would be the simplest workaround for g(x) ?
I mean simpler than eval(g(_x), _x=x)

@ecterrab 

Correct? For g(x),  I would expect e.g.

int(f(_x+x), _x=x..2*x):

Edit. I know that in int, the dummy variable is not local to int.
But mathematically, g has nothing to do with x, so, g(x) should be as above.
I did not insert "bug" as a keyword, because I don't see this as a bug,
but mathematically it's incorrect.

P.S. I also know that the "correct" g should be
g:= proc(a) local x; int(f(x+a),x=a..2*a) end:

@Markiyan Hirnyk 

This is an excellent idea!
It makes possible to
1. Prove easily the theorem
2. Obtain easily animations even for n>50 circles.

V.A.

@Axel Vogt 

I think that sqrt is not important (exept probably for the initial "division by 0" because sqrt in not differentiable at 0).

method=Laplace works because Int is a convolution.

@Axel Vogt 

Do you mean y(s)=sqrt(s)  and the corresponding F(s)?

I just took it randomly. intsolve solves it wrt invlaplace but is not smart enough to use linearity and retrive sqrt(s).

@Kitonum 

Yes, and for n circles,

0 < x <= m(n) := (1-sin(Pi/n))/(1+sin(Pi/n)) = tan(Pi/4 - Pi/(2*n))^2.

Do you have a reference for this problem?

If you want to test intsolve you should choose some equations with known solutions.
E.g. take y(s)=sqrt(s) , compute F(s) and call intsolve(...).
And note that invlaplace could be in a distributional (generalized) sense.

@Markiyan Hirnyk 

@Markiyan Hirnyk 

It worked for me, but invlaplace could not be computed symbolically.

Your F(z) is arbitrary (generic) so you cannot hope for more.
Try e.g. replacing F(z) with a concrete expression e.g. z^2.

@Ferdinand

The theorem seems to be this:

Let C be a circle of radius 1 and center O.
For each 0 < x <= 1/3 there exists 0 <= d(x) < 1 such that the following holds.

For any circle C' having the radius x and having its center at the distance d(x) from O
and for any point P on C', denote by C(1) the unique circle which is outer-tangent to C' at P and is also inner-tangent to C.
For each k=2..6 define successively the unique (Apollonius) circle C(k) inner-tangent to C, and outer-tangent to C' and C(k-1).
Then C(6) is outer-tangent to C(1).



As the OP noticed, d(1/3)=0 and so, for x=1/3 all the circles C(k) have radius 1/3.
Is the exact expression of d(x) known as a function of x? [it should be]


Note also that the geometry package has a command Apollonius which constructs the Apollonius circles but it fails if their number is not maximal (i.e. 8). (see http://www.mapleprimes.com/questions/205718-Apollonius-geometry-Problem)

I tried to understand the statement of the theorem behind your construction.

Let C and C' be two fixed circles, C' inside C such that radius(C)=1, radius(C')=x, 0<x<=1/3.

Consider n in N and the circles C(1),...,C(n) inside C such that C(k) is tangent to
C, C', C(k+1) and C(k-1) for k=1..n where C(0)=C(n) and C(n+1)=C(1).

Then there exists x such that the existence of such circles is guaranteed. Furthermore,
given P a point on C' and imposing that the tangency point between C(1) and C' is P then n=6 and the positions of the circles C(1),...,C(n) are unique.

Is it correct? Or maybe this is valid only for a special position of center(C').
[Probably the position of center(C') is essential because of the Apollonius construction].