Harmonic Cantilever
===================
References:
http://www.astc.org/resource/exhibits/goldwater.htm
http://www.newtrier.k12.il.us/academics/math/Connections/curves/harmo...
http://mps.uchicago.edu/docs/2005/LabelsText/HARMONIC%20CANTILEVER.doc
But why?
By mathematical induction. Let the length of the blocks on the stack
be two (= 2 in a certain physical unit).
Then the center of gravity of a single block is given by: Z_1 = 1 .
Suppose the center of gravity of the first (n) blocks is given by Z_n ,
then add one block and find the new center of gravity Z_(n+1) . Place
the new block with its left side underneath the old blocks, such that
they cannot collapse, i.e. left side exactly at Z_n .
Picture (with non-proportional font):
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1234567890123456789012345678901234567890123456789012345678901234567890
Z is given by the mass of the previous (n) blocks times the old center
of gravity, plus a distance equal to the old center of gravity (so that
the blocks will not collapse), plus one times one (= center of gravity
of the newly added block times its mass). The whole must be devided by
the new number of blocks. Hence the formula becomes:
Z_(n+1) = (n.Z_n + Z_n + 1)/(n+1) .
Or:
Z_(n+1) = Z_n + 1/(n+1)
Starting with n = 0 and Z_0 = 0 this becomes:
Z_1 = 1 , Z_2 = 1 + 1/2 , Z_3 = 1 + 1/2 + 1/3 ,
Z_4 = 1 + 1/2 + 1/3 + 1/4 , .. which is precisely the harmonic series.
Where Z_4 = 25/12 and subsequent terms > 2 (= length of single block):
see above picture. Yet the whole thing doesn't collapse.
Nice huh?
Han de Bruijn
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Han de Bruijn wrote:
> References:
> http://www.astc.org/resource/exhibits/goldwater.htm
> http://www.newtrier.k12.il.us/academics/math/Connections/curves/harmo...
> http://mps.uchicago.edu/docs/2005/LabelsText/HARMONIC%20CANTILEVER.doc
We seek a _continuous_ approximation of the Harmonic Cantilever, which
was presented in a previous poster:
y = 0 x = 0
y = -1 x = 1
y = -2 x = 1 + 1/2
y = -3 x = 1 + 1/2 + 1/3
y = -4 x = 1 + 1/2 + 1/3 + 1/5
...........................................
y = -n x = 1 + 1/2 + 1/3 + ... + 1/n
y = -(n+1) x = 1 + 1/2 + 1/3 + ... + 1/n + 1/(n+1)
It is seen that y(n+1) - y(n) = [ x(n+1) - x(n) ] y(n+1)
For large (n), the finite differences [ x(n+1) - x(n) ] become smaller
and smaller and the quotient [ y(n+1) - y(n) ] / [ x(n+1) - x(n) ] is
an approximation of the differential quotient dy/dx . Hence dy/dx = y .
The solution of this differential equation is: y = c.exp(x) . From the
above table, we read that c < 0 . A kind of best fit between the stack
of bricks in the Harmonic Cantilever and this function has been used to
determine the constant (c), experimentally. The result is shown in the
following picture, for 50 bricks.
http://hdebruijn.soo.dto.tudelft.nl/jaar2007/stapelen.jpg
Question: are these exponential curves relevant for the way cathedrals
are built. Or do we have to expect other shapes? And why then?
Han de Bruijn
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Han de Bruijn wrote:
> determine the constant (c), experimentally. The result is shown in the
> following picture, for 50 bricks.
So far so good. But the experimental guess can be replaced by an exact
least squares minimalization procedure:
sum(k=1..N) [ y_k - c.exp(x_k) ]^2 = minimum(c)
Differentiating to (c) and equating the result to zero then results in:
c = [ sum(k=1..N) y_k.exp(x_k) ] / [ sum(k=1..N) exp(2.x_k) ]
Where y_k = -1, -2, -3, .. , -N
And x_k = 1, 1 + 1/2 , 1 + 1/2 + 1/3 , .. , 1 + 1/2 + 1/3 + .. + 1/N
The result is shown again in:
http://hdebruijn.soo.dto.tudelft.nl/jaar2007/stapelen.jpg
Han de Bruijn
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On 1 feb, 16:41, Han de Bruijn wrote:
> Han de Bruijn wrote:
> > determine the constant (c), experimentally. The result is shown in the
> > following picture, for 50 bricks.
> So far so good. But the experimental guess can be replaced by an exact
> least squares minimalization procedure:
Surprisingly enough, if we replace this least squares (c) by the Euler-
Mascheroni constant gamma = 0.5772156649015328606065120900824024310 ..
then the result is "almost" a best fit as well: gamma.e^x . Why oh why?
Han de Bruijn
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On 2 feb, 18:24, Han.deBru...@DTO.TUDelft.NL wrote:
> On 1 feb, 16:41,Han de Bruijn wrote:
> >Han de Bruijn wrote:
> > > determine the constant (c), experimentally. The result is shown in the
> > > following picture, for 50 bricks.
> > So far so good. But the experimental guess can be replaced by an exact
> > least squares minimalization procedure:
> Surprisingly enough, if we replace this least squares (c) by the Euler-
> Mascheroni constant gamma = 0.5772156649015328606065120900824024310 ..
> then the result is "almost" a best fit as well: gamma.e^x . Why oh why?
And an even better result is obtained with c = exp(-gamma) instead of
c = gamma. Part of the explanation is that these numbers don't differ
too much:
exp(-E) = 5.61459483566885E-0001 = E = 5.77215664901533E-0001
It is thus _conjectured_ that the best continuous fit to the Harmonic
Cantilever is (left hand side, as defined before):
y = - exp(x - gamma)
But .. can it be proved?
More or less. We know that y = -N . Then taking the logarithm at both
sides results in:
ln(N) = x - gamma <==> gamma = sum(k=1..N) 1/k - ln(N)
However, the limit of the latter expression, for N -> oo, is precisely
the definition of the Euler-Mascheroni constant gamma. So, that's why!
In practice, this means that the fit is best for large N and worse for
small N, quite in agreement with experience:
http://hdebruijn.soo.dto.tudelft.nl/jaar2007/stapelen.jpg
Nice, huh?
Han de Bruijn
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On 2 feb, 21:06, Han.deBru...@DTO.TUDelft.NL wrote:
> It is thus _conjectured_ that the best continuous fit to the Harmonic
> Cantilever is (left hand side, as defined before):
> y = - exp(x - gamma)
Meanwhile, this has become a theorem, not a conjecture.
When combined with the least squares method for finding the best fit,
we find another result (quite surprising to me, at least).
Let the function E(k) be defined by:
E(k) = exp[ sum(m=1..k) 1/m ]
Then: exp(-gamma) =
lim(N->oo) [ sum(k=1..N) k.E(k) ] / [ sum(k=1..N) E(k)^2 ]
And yes, this also is a theorem, not a conjecture. Is it well known?
Here comes a program snippet that confirms the theorem experimentally.
program HdB;
const
gamma : double = 0.577215664901532860606512090082402431042;
function konstante : double;
{
Least Squared Best Fit
}
const
N : integer = 10000000;
var
y : integer;
x,u,v,p : double;
begin
x := 0;
u := 0; v := 0;
for y := 1 to N do
begin
p := exp(x);
x := x + 1/y;
u := u + y*p;
v := v + p*p;
end;
konstante := u/v;
end;
begin
Writeln(exp(-gamma),' = ',konstante);
end.
Output:
5.61459483566885E-0001 = 5.61459525677516E-0001
Han de Bruijn
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On 2 feb, 21:58, Han.deBru...@DTO.TUDelft.NL wrote:
> Let the function E(k) be defined by:
> E(k) = exp[ sum(m=1..k) 1/m ]
> Then: exp(-gamma) =
> lim(N->oo) [ sum(k=1..N) k.E(k) ] / [ sum(k=1..N) E(k)^2 ]
= lim(N->oo) [ sum(k=1..N) E(k)^2 exp{- sum(m=1..k) 1/m + ln(k)}]
/ [ sum(k=1..N) E(k)^2 ]
This is a weighted mean of functions exp{- sum(m=1..k) 1/m + ln(k)}]
where the weights E(k)^2 are positive & uniformly increasing with (k)
and eventually approaching infinity. Meaning that the weights E(k)^2,
for large (k), are predominant. But exp{- sum(m=1..k) 1/m + ln(k)}]
for large (k) is approximately exp(- gamma) . Thus it is seen that
the above limit is mainly a mean of exp(- gamma) approximations.
I'm pretty sure that this sloppy argument can be made more rigorous
when needed.
Han de Bruijn