| Topic: |
Science > Physics |
| User: |
"Tobin Fricke" |
| Date: |
22 Jul 2005 02:21:46 PM |
| Object: |
Legendre Transform |
I'm trying to get a better understanding of the Legendre Transform. The
Wikipedia entry[1] for Legendre Transform mentions that it is related to
integration by parts--what is the relationship? Also, I've heard that
there is a relation to the Laplace transform--what is it?
I was introduced to the Legendre transform in thermodynamics as a
transform from convex functions of the form y = f(x) to 'tangent lines' of
the form b = g(m), where b is the intercept of a tangent line through
(x,y) and m is its slope. But apparently the Legendre transform is also
equivalent to taking a derivative, computing a functional inverse, and
then integrating. How do you show the equivalence?
The Wikipedia page also says that the Legendre Transform of (1/2) x^t A x
is (1/2) y^t A^-1 y (where x and y are vectors, x^t represents the
transpose of x, and A is an invertible matrix). How is this result
obtained and what does it mean?
thanks,
Tobin
[1] http://en.wikipedia.org/wiki/Legendre_transform
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| User: "Zigoteau" |
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| Title: Re: Legendre Transform |
23 Jul 2005 04:04:27 AM |
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Hi, Tobin,
I'm trying to get a better understanding of the Legendre Transform.
Join my club.
The
Wikipedia entry[1] for Legendre Transform mentions that it is related to
integration by parts--what is the relationship? Also, I've heard that
there is a relation to the Laplace transform--what is it?
Neither relationship has ever struck me.
I was introduced to the Legendre transform in thermodynamics as a
transform from convex functions of the form y = f(x) to 'tangent lines' of
the form b = g(m), where b is the intercept of a tangent line through
(x,y) and m is its slope.
That sounds like the blind men's description of an elephant
Yes the Legendre transform pops up frequently in thermodynamics and is
associated with the richness of e.g. the phase diagrams of mixtures.
Thermodynamic systems evolve until they reach equilibrium. For a
totally isolated system, the internal energy U remains constant =U_0.
The system evolves until its entropy S is maximized subject to the
condition U=U_0. Mathematically, by Lagrange's method of undetermined
multipliers, that problem is equivalent to minimizing U subject to
S=S_0.
No system is every totally isolated, and all interesting systems
interact with their surroundings. Gas in a cylinder of constant volume
V can exchange heat. The condition for equilibrium is that the
Helmholtz free energy A=U-TS is minimized for constant temperature
T=T_0 defined by the environment.
Now at constant volume TdS=dU, so that T = (dU/dS)|V. Hence A =
U-S(dU/dS)|V is the Legendre transform of U, considering the latter to
be a function of S and V.
Since TdS=dU, it also follows that dA = -SdT and S = -(dA/dT)|V. Hence
U = A-T(dA/dT)|V is a Legendre transform, if A is considered as a
function of T and V.
In fact in general, the inverse of a Legendre transform is always a
Legendre transform, but the point you must notice is that you have to
change the independent variables. The natural independent variables for
U are S and V, The natural independent variables for A are T and V.
In just the same way, you can now deal with the case of thermodynamic
systems consisting of condensed phases (liquids and solids), where the
volume cannot be adequately kept constant, and a much better
experimental variable is the pressure P. The condition for equilibrium
now involves the Gibbs' free energy G = A + PV = A-V(dA/dV)|T, which is
the Legendre transform of A wrt the variable V. The natural independent
variables for G are P and T. Just as in the previous case, the
relationship for getting back from G to A is also a Legendre transform.
But apparently the Legendre transform is also
equivalent to taking a derivative, computing a functional inverse
In general, the inverse of a Legendre transform is a Legendre
transform. I don't see a crucial connection with differentiation,
although of course the new independent variable is the derivative of
the stating function wrt the old independent variable.
, and
then integrating. How do you show the equivalence?
The Wikipedia page also says that the Legendre Transform of (1/2) x^t A x
is (1/2) y^t A^-1 y (where x and y are vectors, x^t represents the
transpose of x, and A is an invertible matrix). How is this result
obtained and what does it mean?
A is a matrix, x is a vector and x^t is its transpose. This result is
the result of Legendre transformation of (1/2) x^t A x wrt all the
components of x. Just apply the basic formula. Can you see how to do it
now?
You can clearly see the self-inverse nature of this result , as
(A^-1)^-1 = A
Cheers,
Zigoteau.
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| User: "Zigoteau" |
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| Title: Re: Legendre Transform |
23 Jul 2005 04:35:19 AM |
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PS
I see that the Wikipedia definition differs from mine in the sign of
the result. Not sure about that. Certainly in thermodynamics it only
occurs with the convention I have used. This explains their claim of a
connection to integration by parts.
The example Wikipedia gives is absolutely correct, but glosses over a
complexity of the application to thermodynamic systems. The derivative
of a second-order function is monotonic, i.e. its derivative takes on a
particular value just once. In thermodynamic systems you can get
potentials which have points of inflection, so that a particular value
of the gradient, say P, can occur for two distinct values of the
original independent variable V. This happens in e.g. a van der Waals
gas, and when it happens you get a first-order phase transition. For a
vdW gas it is the liquid-gas transition.
Cheers,
Zigoteau.
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| User: "Eric Gisse" |
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| Title: Re: Legendre Transform |
22 Jul 2005 06:56:42 PM |
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Tobin Fricke wrote:
I'm trying to get a better understanding of the Legendre Transform. The
Wikipedia entry[1] for Legendre Transform mentions that it is related to
integration by parts--what is the relationship? Also, I've heard that
there is a relation to the Laplace transform--what is it?
I would guess that it is 'related' to integration by parts in that you
can undo integration by parts by swapping the derivative and
anti-derivative and applying the integration by parts rule.
I have no idea how it relates to the Laplace transform, I don't see a
direct relationship.
<derail> Transforms in general, and specifically integral transforms,
oddly fascinate me...I wish I knew what makes them work so well but all
the books I read are more about method and existance theorems, rather
than the theory on what makes them tick. </derail>
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| User: "" |
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| Title: Re: Legendre Transform |
23 Jul 2005 05:01:07 PM |
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Eric Gisse wrote:
<derail> Transforms in general, and specifically integral transforms,
oddly fascinate me...I wish I knew what makes them work so well but all
the books I read are more about method and existance theorems, rather
than the theory on what makes them tick. </derail>
Integral transforms are cool things. What follows is strongly coloured
by the context in which I use them.
It's all about doing stuff in infinite-dimensional vector spaces. OK,
take a function with infinite degrees of freedom - that's easy, just a
function on the real line will do. This is the general case in field
theory; you have something like F(x,y,z), where F is some scalar or
vector valued function. Let us just consider the 1D scalar case, F(x).
The "obvious" way to specify F is to simply give the value of F for all
values of x. Fine, except for the fact that we have an infinite number
of possible values of x. Well, in practice, you might be able to get
away with a discrete subset of x, but sometimes you can't. What to do?
OK, a diversion to Cartesian vectors. Take 3 Cartesian unit vectors, u
v w, and you can write any vector V in terms of x,y,z components as
V = Vx u + Vy v + Vw w
which is all kinda nice, but not very useful unless you have a method
for finding Vx etc. The recipe is simple and is:
Vx = V.u
which gives you the projection of V onto u. If u is not a unit vector,
then you had better use V.u/sqrt(u.u)
Back to F(x)! As long as F(x) isn't too obnoxiously behaved (piecewise
continuous is sufficient), you know you can write F(x) as a Fourier
integral:
F(x) = integral( A(k)exp(ikx) )
which is nice if A(k) is better behaved than F(x), and in many other
cases too. Now, what this is is simply the infinite-dimensional
equivalent of V = Vx u + ... . In a vector space of functions, the
scalar product is
F(x).g(x) = integral( w(x)F(x)g*(x) )/sqrt(integral( w(x)g(x)g*(x) ))
where w(x) is a weighting function (effectively the metric) which in
useful cases is often equal to 1 (just as the Cartesian metric tensor =
I), and * is the complex conjgate. If the denominator equals 1, then
you have the convenient case of g(x) being a unit vector. So, choose an
orthogonal unit basis (ie integral(w(x)g_n(x)g*_m(x)) = delta(n,m)),
and you're doing pretty much the Cartesian vector thing. Note how
F(x).g(x) is simply the usual integral transform formula. The
relationship to V.u/sqrt(u.u) is not made explicit often enough.
And now for a rant about Fourier transforms! Basically, apart from the
conveniences of (a) exp(x) being the mathematically simplest special
function, with the super-convenient d/dx exp(ax) = a exp(ax) and (b)
plane waves being simple and all that, Fourier integrals suck. Unless
you can do them (especially the inverse transform) analytically, you
WILL get failure to converge, as a result of a finite discrete
representation of the continuous A(k). Computationally, the discrete
bases are much better.
--
Timo
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| User: "Eric Gisse" |
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| Title: Re: Legendre Transform |
24 Jul 2005 03:44:45 AM |
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wrote:
Eric Gisse wrote:
<derail> Transforms in general, and specifically integral transforms,
oddly fascinate me...I wish I knew what makes them work so well but all
the books I read are more about method and existance theorems, rather
than the theory on what makes them tick. </derail>
Integral transforms are cool things. What follows is strongly coloured
by the context in which I use them.
It's all about doing stuff in infinite-dimensional vector spaces. OK,
take a function with infinite degrees of freedom - that's easy, just a
function on the real line will do. This is the general case in field
theory; you have something like F(x,y,z), where F is some scalar or
vector valued function. Let us just consider the 1D scalar case, F(x).
The "obvious" way to specify F is to simply give the value of F for all
values of x. Fine, except for the fact that we have an infinite number
of possible values of x. Well, in practice, you might be able to get
away with a discrete subset of x, but sometimes you can't. What to do?
OK, a diversion to Cartesian vectors. Take 3 Cartesian unit vectors, u
v w, and you can write any vector V in terms of x,y,z components as
V = Vx u + Vy v + Vw w
which is all kinda nice, but not very useful unless you have a method
for finding Vx etc. The recipe is simple and is:
Vx = V.u
which gives you the projection of V onto u. If u is not a unit vector,
then you had better use V.u/sqrt(u.u)
Back to F(x)! As long as F(x) isn't too obnoxiously behaved (piecewise
continuous is sufficient), you know you can write F(x) as a Fourier
integral:
F(x) = integral( A(k)exp(ikx) )
which is nice if A(k) is better behaved than F(x), and in many other
cases too. Now, what this is is simply the infinite-dimensional
equivalent of V = Vx u + ... . In a vector space of functions, the
scalar product is
F(x).g(x) = integral( w(x)F(x)g*(x) )/sqrt(integral( w(x)g(x)g*(x) ))
where w(x) is a weighting function (effectively the metric) which in
useful cases is often equal to 1 (just as the Cartesian metric tensor =
I), and * is the complex conjgate. If the denominator equals 1, then
you have the convenient case of g(x) being a unit vector. So, choose an
orthogonal unit basis (ie integral(w(x)g_n(x)g*_m(x)) = delta(n,m)),
and you're doing pretty much the Cartesian vector thing. Note how
F(x).g(x) is simply the usual integral transform formula. The
relationship to V.u/sqrt(u.u) is not made explicit often enough.
The actual relationships and theory are not made explicit at all,
actually. This is a big personal bugbear. I think it is also why I am
so interested in this subject. The interest will probably wane once I
figure this stuff out to my satisfaction, as with everything else to
date.
Oh sure, things like existance for functions with suitable properties
are covered lightly if at all. But the actual honest-to-god theory that
makes integral transforms work is damn hard to find. Either I don't
know what I am looking for, or it simply isn't covered enough.
Lets look at my bookshelf.
My complex analysis text covers integral transforms (Fourier, Laplace,
Hankel). It talks about how to use them, how to find them, and if you
are really lucky - how to invert them. But no actual thought is put to
the theory behind them.
My ODE text covers Laplace transforms, and some existance theorems, but
not the theory. Inversion is limited to 'looking up from a table'. Now
that I think about it, having the inversion of the Laplace transform
hidden from me so completely in that book was probably what got me
interested in these in the first place.
My other PDE text is a Dover, which covers a decent spread of
techniques which include integral transforms. As with the rest of my
collection of books, the actual meat of integral transforms is hidden
from view.
This is something that annoys the ***** out of me, actually. Integral
transforms have an insane amount of utility, in my opinion. But a
problem, to me, beyond the occasional existance theroem, what makes
them work is rarely shown. To me, it is like not explaining limits and
then teaching people how to differentiate and integrate.
I am finding that what makes them work is hidden from view to the point
where they become a black box whose properties are overestimated or
unknown. Hell, even in the book "The use of integral transforms" by Ian
Sneddon, the actual theory is not touched. Though, to be fair, I should
have gotten the idea that he just wanted to cover application and not
theory - especially from the title of the book and the introduction.
Then again, I could be entirely wrong about all of this. Perhaps there
is a real kick-***** book out there that covers integral transform theory
until even the author got sick of it. I just haven't seen it yet. It
would probably help if I had access to a decent library again...
And now for a rant about Fourier transforms! Basically, apart from the
conveniences of (a) exp(x) being the mathematically simplest special
function, with the super-convenient d/dx exp(ax) = a exp(ax) and (b)
plane waves being simple and all that, Fourier integrals suck. Unless
you can do them (especially the inverse transform) analytically, you
WILL get failure to converge, as a result of a finite discrete
representation of the continuous A(k). Computationally, the discrete
bases are much better.
One of these days I really want to play with numerically inverting the
Laplace transform, among others. It would probably help if I had an
idea how to invert a contour integral, though. That would help a lot.
The thing about the transforms that catches my eye is their sheer
utility. Some ODEs or simple PDEs that are nigh-impossible or at least
a real pain get reduced to next to nothing in terms of complexity. From
my own playing though, more the Laplace than the Fourier transform.
Another thing is how they are related. I find it written a lot that the
Fourier and Laplace transforms are the same thing, only operating on
different parts of the complex plane. I agree, if how the Laplace
inversion integral is any indication.
But what I would like to see is transforms that are based off of
different functions. As I said, the black-boxish nature of the
transforms conveys a bit of mystery and a little bit of overconfidence
in their use. My pet hobby, for the moment are partial differential
equations. I only have a small set of tools at my disposal, and I find
myself coming to transforms really quickly. So the only really-useful
transforms I know of are the Laplace and Fourier.
I'd love to be able to create my own and play with them, but I don't
really understand how they work and I have ZERO idea how to invert my
new creation. However, I do know there is more to transforms than the
exponential. For example, the kernel for the Hankel transform is a
bessel function. For the Mellin transform, it is something different.
So on, and so forth.
Holy ***** that came out to be long. This isn't for a class - I wouldn't
even know which class would touch this stuff so I have no professor to
talk to about this stuff. For awhile, at least. Neat stuff though.
--
Timo
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| User: "Edward Green" |
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| Title: Re: Legendre Transform |
30 Jul 2005 11:35:48 PM |
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Eric Gisse wrote:
This is something that annoys the ***** out of me, actually. Integral
transforms have an insane amount of utility, in my opinion. But a
problem, to me, beyond the occasional existance theroem, what makes
them work is rarely shown. To me, it is like not explaining limits and
then teaching people how to differentiate and integrate.
Well, I'm not sure what you want from "what makes them work", but have
you considered that integral transforms are, AFAIK, all species of
changes of bases in vector spaces?
You may be aware of this abstractly, but I wonder if it has ever hit
you how concrete the identity is. Viz., any finite or countably
infinite dimensional vector can be represented by a row of numbers --
i.e., a sequence A(n). Well, what is a function f(x) but such a
sequence in the limit as the n are spaced closer and closer together on
a line? A graph of a function is nothing but a representation of the
components of an infinite dimensional vector ... the independent
variable is the index! An inner product in a function space is defined
like any other inner product -- except for the little matter of
infinities, of course.
What makes this so useful? Here I'm just thinking aloud, but I'm
thinking linearity + useful properties of particular bases. I.e., say
we want to evolve a vector under some operator F, and it happens that
this evolution is particularly simple is some one basis. So, we
transform to this basis (integral transform), write down the simple
evolution equation, and retransform. Example: analysis of vibration as
a sum of principal modes. Linearity would be essential.
Another thing -- just a check for myself here -- aren't the Laplace and
the Fourier transform really the same transform, with merely a switch
in what variable we consider to be the real element of the complex
transform?
As to Legendre transforms, I too know them only from a course in thermo
where they were represented as the transformation of a convex function
to its intercept via tangents. Damn. At some time I knew what was a
Legendre transform of what, and maybe even had an idea why that was
useful.
<...>
Another thing is how they are related. I find it written a lot that the
Fourier and Laplace transforms are the same thing, only operating on
different parts of the complex plane.
Well, that's what I suggested. That makes it sound a little mysterious
though... I think it is really very simple. In both cases the real
version can be extended into complex numbers, in which case they become
identical up to some identification of variables. They are only
distinct if you confine yourself to the real versions.
I agree, if how the Laplace
inversion integral is any indication.
But what I would like to see is transforms that are based off of
different functions. As I said, the black-boxish nature of the
transforms conveys a bit of mystery and a little bit of overconfidence
in their use. My pet hobby, for the moment are partial differential
equations. I only have a small set of tools at my disposal, and I find
myself coming to transforms really quickly. So the only really-useful
transforms I know of are the Laplace and Fourier.
I'd love to be able to create my own and play with them, but I don't
really understand how they work and I have ZERO idea how to invert my
new creation. However, I do know there is more to transforms than the
exponential. For example, the kernel for the Hankel transform is a
bessel function.
Oh. I didn't know that. Thanks. :-)
That makes sense, since the Bessel functions are widely libeled as
being orthogonal. Or are they...
A thought: since integral transforms are really changes of basis in
vector spaces, we have a distinction between orthogonal basis and,
well, non-orthogonal bases. The orthogonal kind are obviously much
easier to deal with ... for example, we are enabled to ignore the
distinction between "the things we form inner product with" and "the
things we multiply our inner products by to reconstruct the original
vector. In general these things are not the same things. And the form
of the inverse of orthogonal transformations are also particularly
simple: ditto not non-orthogonal cases. The two ideas (things we form
inner products with vs. things we mutliply the results by, and direct
vs. inverse) are closely related -- which is a cheap way of saying I
always have to go back and think about this, it's never become second
nature.
It is a useful way to think about things though, since I was at least
temporarily able to grok the transformation properties of covariant and
contravariant vectors in a non-standard way: they are _also_ examples
of the duality between "things you dot with and things you construct
with".
Anyway, transformation pairs in which the things you integrate with are
non-orthogonal would be much more messy. Hmm... but isn't this true of
the Laplace transform viewed as a real transform?
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| User: "Zigoteau" |
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| Title: Re: Legendre Transform |
25 Jul 2005 04:09:44 AM |
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Hi, Eric,
<snip>
This is all very interesting, but may I note:
1. the Legendre transform is not a Laplace transform, or vice versa.
No-one has yet come up with a sensible explanation of Wikipedia's claim
2. Yes, the Laplace transform is closely connected to the Fourier
transform, and so is the Hankel transform, which is the 2D Fourier
transform reexpressed in terms of rotational eigenfunctions.
One of these days I really want to play with numerically inverting the
Laplace transform, among others. It would probably help if I had an
idea how to invert a contour integral, though. That would help a lot.
Don't do it. The matrix of the discrete Laplace transform always has an
inverse, but AFAIK there is no simple expression for it corresponding
to the inverse of the DFT, and it's horrendously ill-conditioned.
Cheers,
Zigoteau.
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| User: "Eric Gisse" |
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| Title: Re: Legendre Transform |
25 Jul 2005 10:41:32 AM |
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Zigoteau wrote:
Hi, Eric,
<snip>
This is all very interesting, but may I note:
1. the Legendre transform is not a Laplace transform, or vice versa.
No-one has yet come up with a sensible explanation of Wikipedia's claim
I had never seen that before, and I still think it is odd. Cute if
true.
2. Yes, the Laplace transform is closely connected to the Fourier
transform, and so is the Hankel transform, which is the 2D Fourier
transform reexpressed in terms of rotational eigenfunctions.
At what point is all this stuff explained? The key ideas seem to be
hiding in the land of differential geometry, if my weekend reading is
any indication. Which would neatly explain why the base concepts are so
hard to find.
One of these days I really want to play with numerically inverting the
Laplace transform, among others. It would probably help if I had an
idea how to invert a contour integral, though. That would help a lot.
Don't do it. The matrix of the discrete Laplace transform always has an
inverse, but AFAIK there is no simple expression for it corresponding
to the inverse of the DFT, and it's horrendously ill-conditioned.
So the Laplace transform isn't terribly useful if you can't analyticaly
invert what you find?
This is why im curious to see what makes them work. Perhaps find
something better...or, if I actually understood what was going on, find
the limited range in which you can numerically invert the transform
without much grief.
Cheers,
Zigoteau.
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| User: "Zigoteau" |
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| Title: Re: Legendre Transform |
25 Jul 2005 03:55:07 PM |
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Hi, Eric,
2. Yes, the Laplace transform is closely connected to the Fourier
transform, and so is the Hankel transform, which is the 2D Fourier
transform reexpressed in terms of rotational eigenfunctions.
At what point is all this stuff explained?
It was probably all explained in detail a century ago, and people have
moved on to more exciting things. If you rediscover it, you can't get
it published in a journal that people read because it's already known.
In the early 1900s people wrote worthy books full of all the results of
functional analysis, and you can still find them on library shelves.
They're still useful, but they are long out of print and no-one is
going to republish them because the market for them is not enormous.
I in fact rediscovered it when thinking about the eigenvalue problem
for the Laplace operator del^2 in two dimensions with circular boundary
conditions (including the case of no boundary conditions). Since del^2
commutes with rotation, the angular part must be an eigenvalue of
d/dtheta, i.e. exp(i*n*theta). The radial part is then the Bessel
function J_n(r). If you Fourier transform to momentum space, del^2
becomes r^2, so the eigenfunctions are nonzero only on a circular
ring.. del^2 also commutes with d/dx and d/dy, so it is also possible
to express the solution as a sum of plane waves, i.e. a 2D Fourier
transform. Skipping a few steps, it follows that the 2D Fourier
transform of J_n(r)*exp(i*n*theta) has the form
a*delta(r-b)*exp(i*n*theta).
The key ideas seem to be
hiding in the land of differential geometry, if my weekend reading is
any indication. Which would neatly explain why the base concepts are so
hard to find.
?? Not sure about that. The operator del^2 is relevant to spaces that
are as flat as a pancake.
One of these days I really want to play with numerically inverting the
Laplace transform,
Don't do it. The matrix of the discrete Laplace transform always has an
inverse, but AFAIK there is no simple expression for it corresponding
to the inverse of the DFT, and it's horrendously ill-conditioned.
So the Laplace transform isn't terribly useful if you can't analyticaly
invert what you find?
Make up your mind - you just said you wanted to invert it numerically.
There is a list of analytical functions for which you can write down
the analytical inverse Laplace transform. It's the numerical inverse
which is ill-conditioned. The analytical inverse Laplace transform is
involved in one of the methods for solving linear differential
equations with constant coefficients, such as are encountered in
electric wave filters . There are a number of other methods, but some
people prefer Laplace transforms.
This is why im curious to see what makes them work. Perhaps find
something better...or, if I actually understood what was going on, find
the limited range in which you can numerically invert the transform
without much grief.
I'm all in favor of understanding what is going on.
Cheers,
Zigoteau.
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| User: "Eric Gisse" |
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| Title: Re: Legendre Transform |
25 Jul 2005 11:11:17 PM |
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Zigoteau wrote:
Hi, Eric,
2. Yes, the Laplace transform is closely connected to the Fourier
transform, and so is the Hankel transform, which is the 2D Fourier
transform reexpressed in terms of rotational eigenfunctions.
At what point is all this stuff explained?
It was probably all explained in detail a century ago, and people have
moved on to more exciting things. If you rediscover it, you can't get
it published in a journal that people read because it's already known.
In the early 1900s people wrote worthy books full of all the results of
functional analysis, and you can still find them on library shelves.
They're still useful, but they are long out of print and no-one is
going to republish them because the market for them is not enormous.
I don't want to rediscover the details, but I do want to know what they
are. Someone already did the work at least once, long before I was
born. No need to reinvent the wheel. "a week in the lab saves a day in
the library", and all that.
I in fact rediscovered it when thinking about the eigenvalue problem
for the Laplace operator del^2 in two dimensions with circular boundary
conditions (including the case of no boundary conditions). Since del^2
commutes with rotation, the angular part must be an eigenvalue of
d/dtheta, i.e. exp(i*n*theta). The radial part is then the Bessel
function J_n(r). If you Fourier transform to momentum space, del^2
becomes r^2, so the eigenfunctions are nonzero only on a circular
ring.. del^2 also commutes with d/dx and d/dy, so it is also possible
to express the solution as a sum of plane waves, i.e. a 2D Fourier
transform. Skipping a few steps, it follows that the 2D Fourier
transform of J_n(r)*exp(i*n*theta) has the form
a*delta(r-b)*exp(i*n*theta).
The key ideas seem to be
hiding in the land of differential geometry, if my weekend reading is
any indication. Which would neatly explain why the base concepts are so
hard to find.
?? Not sure about that. The operator del^2 is relevant to spaces that
are as flat as a pancake.
I am talking about the mechanics of integral transforms. You make it
sound like they have simply faded into the past due to not being
interesting.
One of these days I really want to play with numerically inverting the
Laplace transform,
Don't do it. The matrix of the discrete Laplace transform always has an
inverse, but AFAIK there is no simple expression for it corresponding
to the inverse of the DFT, and it's horrendously ill-conditioned.
So the Laplace transform isn't terribly useful if you can't analyticaly
invert what you find?
Make up your mind - you just said you wanted to invert it numerically.
There is a list of analytical functions for which you can write down
the analytical inverse Laplace transform. It's the numerical inverse
which is ill-conditioned. The analytical inverse Laplace transform is
involved in one of the methods for solving linear differential
equations with constant coefficients, such as are encountered in
electric wave filters . There are a number of other methods, but some
people prefer Laplace transforms.
No...you misunderstood me. I do not dispute the utility of the Laplace
transform. That is one of the things that fascinates me about it, its
unparalleled utility.
It just appears that the Laplace transform loses much of its' utility
the moment you obtain a transformed equation that you can't invert
manually either via manually slugging out the Bromwhich integral or
looking up the function in a table.
For example, it is fairly easy to make a PDE so that the transform of
one of the variables has a form of an ODE which you cannot analytically
solve. That is where numerically inverting a Laplace transform would
come in. Though from the sound of it, a better idea would semm to be to
skip the transform and simply numerically wail on the PDE.
I'm not sure what makes me think that numericaly inverting an integral
transform as a way of solving a PDE is a neat idea, but its interesting
to think about. This is why I want to get back to school.
This is why im curious to see what makes them work. Perhaps find
something better...or, if I actually understood what was going on, find
the limited range in which you can numerically invert the transform
without much grief.
I'm all in favor of understanding what is going on.
Cheers,
Zigoteau.
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| User: "Zigoteau" |
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| Title: Re: Legendre Transform |
26 Jul 2005 02:21:20 AM |
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Hi Eric,
In the early 1900s people wrote worthy books full of all the results of
functional analysis, and you can still find them on library shelves.
I don't want to rediscover the details, but I do want to know what they
are. Someone already did the work at least once, long before I was
born. No need to reinvent the wheel. "a week in the lab saves a day in
the library", and all that.
Right, then go to the reference shelves of your library and dig out
those worthy tomes. They're a bit hard going, though.
?? Not sure about that. The operator del^2 is relevant to spaces that
are as flat as a pancake.
I am talking about the mechanics of integral transforms. You make it
sound like they have simply faded into the past due to not being
interesting.
It is not that they are not intrinsically interesting, just that these
days not many people get excited by them.
Make up your mind - you just said you wanted to invert it numerically.
There is a list of analytical functions for which you can write down
the analytical inverse Laplace transform.
No...you misunderstood me. I do not dispute the utility of the Laplace
transform. That is one of the things that fascinates me about it, its
unparalleled utility.
I'm not sure if I can go along with that. If I have a problem like that
to solve, I usually do it by the operator method. I end up with a set
of simultaneous equations to solve for the coefficients in terms of the
boundary conditions. The Laplace transform method can't handle all
forms of boundary condition, although when it can you obviously end up
with the same solution in the end. Laplace transforms are for people
with neat desks, and my desk has always been a riot.
It just appears that the Laplace transform loses much of its' utility
the moment you obtain a transformed equation that you can't invert
manually either via manually slugging out the Bromwhich integral or
looking up the function in a table.
For example, it is fairly easy to make a PDE so that the transform of
one of the variables has a form of an ODE which you cannot analytically
solve. That is where numerically inverting a Laplace transform would
come in. Though from the sound of it, a better idea would semm to be to
skip the transform and simply numerically wail on the PDE.
Or Fourier transforms. They're the Laplace transform along the
imaginary-s axis, and usually they're just as closely related to your
original differential equation as the LT. The inversion of an FT to get
the original function is not totally free of complications, but it is a
lot less temperamental than inverting an LT and there's a fast
algorithm that is even available in Excel.
I'm not sure what makes me think that numericaly inverting an integral
transform as a way of solving a PDE is a neat idea, but its interesting
to think about. This is why I want to get back to school.
Well I'm the last one who would want to stop you.
Cheers,
Zigoteau.
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| User: "James Dolan" |
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| Title: Re: Legendre Transform |
28 Jul 2005 05:50:17 AM |
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in article <1122306092.767356.147960@o13g2000cwo.googlegroups.com>,
eric gisse <jowr.pi@gmail.com> wrote:
|> This is all very interesting, but may I note: 1. the Legendre
|> transform is not a Laplace transform, or vice versa. No-one has
|> yet come up with a sensible explanation of Wikipedia's claim
|
|I had never seen that before, and I still think it is odd. Cute if
|true.
the legendre transform is the limit as t goes to zero of the conjugate
of the laplace transform by the transformation f |-> f#, where
f#(x) = log_t(f(x)).
(in other words legendre transform is a sort of "low temperature
limit" of laplace transform.)
some fiddling with conventions might be needed to get this to work,
but it's roughly correct.
it's actually rather simpler than that, though; when the formulas for
laplace transform and legendre transform are written in sensible
language, it's obvious that they're really the same formula.
--
[e-mail address jdolan@math.ucr.edu]
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| User: "James Dolan" |
|
| Title: Re: Legendre Transform |
30 Jul 2005 10:40:58 PM |
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|
in article <dcad99$cg7$1@glue.ucr.edu>,
james dolan <jdolan@math-cl-n03.math.ucr.edu> wrote:
|in article <1122306092.767356.147960@o13g2000cwo.googlegroups.com>,
|eric gisse <jowr.pi@gmail.com> wrote:
|
||> This is all very interesting, but may I note: 1. the Legendre
||> transform is not a Laplace transform, or vice versa. No-one has
||> yet come up with a sensible explanation of Wikipedia's claim
||
||I had never seen that before, and I still think it is odd. Cute if
||true.
|
|the legendre transform is the limit as t goes to zero of the
|conjugate of the laplace transform by the transformation f |-> f#,
|where f#(x) = log_t(f(x)).
|
|(in other words legendre transform is a sort of "low temperature
|limit" of laplace transform.)
|
|some fiddling with conventions might be needed to get this to work,
|but it's roughly correct.
|
|it's actually rather simpler than that, though; when the formulas for
|laplace transform and legendre transform are written in sensible
|language, it's obvious that they're really the same formula.
to me it's funny that there was a lively discussion about the
unlikeliness of a relationship between laplace transform and legendre
transform until someone showed how they're actually related, upon
which everyone apparently lost interest. but maybe one reason for it
is that some people don't know what it means for a transformation u to
be a conjugate of a transformation v by a transformation w, so i'll
try presenting the relationship between laplace transform and legendre
transform in a different way here:
take a handful (f,g,h,...) of functions (to make the situation
especially simple and clear you should pick them to be quadratic
polynomials, but the basic relationship is still there even if they're
some other kind of functions), and calculate their legendre
transforms:
legendre transform(f) = ?
legendre transform(g) = ?
legendre transform(h) = ?
.
.
.
then write down the exponentials of the negatives of the original
functionals, including a parameter t as follows:
f$(x) = exp(-f(x)*t)
g$(x) = exp(-g(x)*t)
h$(x) = exp(-h(x)*t)
.
.
.
so for example if f,g,h,... are quadratic polynomials bounded below,
the above exponentials are the corresponding so-called "gaussian"
functions. then calculate the laplace transforms of these new functions:
laplace transform(f$)
laplace transform(g$)
laplace transform(h$)
.
.
.
(the laplace transform should be taken with respect to the variable x;
the variable t is just a parameter. a parameter is about halfway
between a variable and a constant.)
then look for a pattern. if you used quadratic polynomials then the
pattern should be very obvious, as long as there are no problems with
sign conventions or other notational conventions.
finally, understand _why_ the pattern occurs. basically it's because,
as i indicated earlier, when the formulas for legendre transform and
laplace transform are written in sensible language, it's obvious that
they're really the same formula.
i wouldn't really describe this as "cute", though; rather i'd say that
it's a key to the secret foundation of a huge body of mathematical
physics.
--
[e-mail address jdolan@math.ucr.edu]
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| User: "" |
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| Title: Re: Legendre Transform |
24 Jul 2005 02:00:44 AM |
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In article <1122153297.015822.300730@z14g2000cwz.googlegroups.com>, writes:
Eric Gisse wrote:
<derail> Transforms in general, and specifically integral transforms,
oddly fascinate me...I wish I knew what makes them work so well but all
the books I read are more about method and existance theorems, rather
than the theory on what makes them tick. </derail>
Integral transforms are cool things. What follows is strongly coloured
by the context in which I use them.
It's all about doing stuff in infinite-dimensional vector spaces. OK,
take a function with infinite degrees of freedom - that's easy, just a
function on the real line will do. This is the general case in field
theory; you have something like F(x,y,z), where F is some scalar or
vector valued function. Let us just consider the 1D scalar case, F(x).
The "obvious" way to specify F is to simply give the value of F for all
values of x. Fine, except for the fact that we have an infinite number
of possible values of x. Well, in practice, you might be able to get
away with a discrete subset of x, but sometimes you can't. What to do?
OK, a diversion to Cartesian vectors. Take 3 Cartesian unit vectors, u
v w, and you can write any vector V in terms of x,y,z components as
V = Vx u + Vy v + Vw w
which is all kinda nice, but not very useful unless you have a method
for finding Vx etc. The recipe is simple and is:
Vx = V.u
which gives you the projection of V onto u. If u is not a unit vector,
then you had better use V.u/sqrt(u.u)
Back to F(x)! As long as F(x) isn't too obnoxiously behaved (piecewise
continuous is sufficient), you know you can write F(x) as a Fourier
integral:
F(x) = integral( A(k)exp(ikx) )
which is nice if A(k) is better behaved than F(x), and in many other
cases too. Now, what this is is simply the infinite-dimensional
equivalent of V = Vx u + ... . In a vector space of functions, the
scalar product is
F(x).g(x) = integral( w(x)F(x)g*(x) )/sqrt(integral( w(x)g(x)g*(x) ))
where w(x) is a weighting function (effectively the metric) which in
useful cases is often equal to 1 (just as the Cartesian metric tensor =
I), and * is the complex conjgate. If the denominator equals 1, then
you have the convenient case of g(x) being a unit vector. So, choose an
orthogonal unit basis (ie integral(w(x)g_n(x)g*_m(x)) = delta(n,m)),
and you're doing pretty much the Cartesian vector thing. Note how
F(x).g(x) is simply the usual integral transform formula. The
relationship to V.u/sqrt(u.u) is not made explicit often enough.
And now for a rant about Fourier transforms! Basically, apart from the
conveniences of (a) exp(x) being the mathematically simplest special
function, with the super-convenient d/dx exp(ax) = a exp(ax) and (b)
plane waves being simple and all that, Fourier integrals suck. Unless
you can do them (especially the inverse transform) analytically, you
WILL get failure to converge, as a result of a finite discrete
representation of the continuous A(k). Computationally, the discrete
bases are much better.
Aye, no argument about it. But, look at the bright side, those
*****-poor convergence properties of Fourier transforms (and the series
aren't that much better) gave rise to a hell of a lot of first class
work in mathematics. Basically it forced the mathematicians to put
the notions of limits and convergence (which till then were treated
in a quite cavalier fashion) on a solid footing.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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