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Old 11-17-2003, 07:18 PM   #11 (permalink)
Rekna
Junkie
 
hmm

Edit used periods to help with spacing

Here is the way I did it but I came up with a different solution then the problem. Can anyone find my logic error?

First I make one assumption:

The maximum height of the entire bubble tree can be found by maximizing the height of every touching pair of bubbles.

Now if this is incorrect someone please explain to me why.

Ok the rest will be full of crappy notation that can be written very well in this text editor but I will try.

First we draw a line going horizontal across each radius. This breaks our tree into n+1 intervals (1 interval is the bottom of the base circle radius, 1 interval is the top circle radius, and the rest of the intervals are the distances between consecutive radiuses.

If we denote the distance between the Nth and Nth+1ís circles radius as D(n) and the radius of the Nth circle as R(n) we then have our height

H(n)=R(1)+R(n)+[D(1)+...+D(n-1)]

Ok I hope you are still following (a tablet PC would be nice right about now).

Letís now look at a simple example with 2 circles.

If we set the radius of our larger circle to be R(n) and the smaller circles radius to be R(n+1) then we can say the following.

If we draw a line between the radius of the larger circle and connect it to a point where the two circles touch that distance is R(n). Then if we draw from the bottom circleís radius straight up to the top circles radius and then complete the triangle we have the following triangle.

...... R(n+1)
...... -------
...... |...../
D(n)..|..../ R(n)
...... |../
...... |/
We then have D(n)=sqrt(R(n)^2-R(n+1)^2)

Back to our original height equation for 2 circles we now have

H(n+1)= R(n) + R(n+1) + D(n)

H(n+1)= R(n) + R(n+1) + sqrt(R(n)^2-R(n+1)^2)

We now want to maximize this using R(n+1) as our variable

R(n) is a constant

Hí(n)=1-1*R(n+1)/sqrt(R(n)^2-R(n+1)^2)

Setting this equal to zero and solving gives us

R(n+1)=R(n)/sqrt(2)

This is a FO linear recursion with which gives us a general solution of

R(n)=(1/sqrt(2))^(n-1)*R(1)

So we now know the maximized radius at every height. Our calc work is done and now all we need to do is a simplifying of our original series


H(n)=R(1)+R(n)+[D(1)+...+D(n-1)]

We know R(1)=1
We know R(n)=(1/sqrt(2))^(n-1)

So now letís look at our summation of the Ds

n-1
Sum sqrt[ ((1/sqrt(2))^(2k-2) Ė (1/sqrt(2))^(2k)) ]
k=1


working with what is inside of the outside sqrt we can factor out a (1/sqrt(2))^(2k-2) leaving us with

n-1
Sum sqrt[ ((1/sqrt(2))^(2k-2)*(1 Ė (1/sqrt(2))^2) ]
k=1

OR

n-1
Sum sqrt[ ((1/sqrt(2))^(2k-2)*(1/2) ]
k=1


By putting in the outer square root as a 1/2 power we have

n-1
Sum (1/sqrt(2))^(k-1)*(1/sqrt(2))
k=1

We can then place the last 1/sqrt(2) into our exponent by adding 1 to th power.

n-1
Sum (1/sqrt(2))^k
k=1

Now donít forget about our first 2 terms that we still have. We have the 1 + (1/sqrt(2))^(n-1)

We can add our 1 directly into our sum by lowering k by 1.


.......................n-1
(1/sqrt(2))^(n-1) + Sum (1/sqrt(2))^k
.......................k=0

The summation is now a geometric series with base 1/sqrt(2) giving us the final equation for H(n)

H(n)= (1/sqrt(2))^(n-1) + 1-(1/sqrt(2))^n
..............................--------------------
....................................1-1/sqrt(2)

Placing values into this match the equation 1+sqrt(n) for H(1) and H(2). After that it gets of a minor amount.

My guess is my initial assumption might be wrong.

Last edited by Rekna; 11-17-2003 at 07:25 PM..
Rekna is offline  
 

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