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Science Forum Index » Math - Numerical Analysis Forum » Understanding linear operators
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| Guest |
 Posted: Wed Dec 20, 2006 3:16 pm |
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I have some basic problem with understanding linear operators (at least
in the area of numerical analysis) and I will be a very happy person if
you can help me out.
Let's say that I have a certain operator. Let's say that it's a simple
D operator, i.e. D[U(x)] = dU/dx. This far it's OK, and I understand
what I see. I even understand what D^4(U) means - it means a 4th
derivative by x of U. However, when I have an expression
Exp[dt*D^2]U(x), I'm lost. What's that suppose to mean? How it will
look in the combination of different forms of U(x)?
Same goes for such a basic operator as a shift operator E(U). E(U) =
U(x+h). It's fine, it's easy. However, what does ln(E) means is not
clear to me.
Can anyone please enlighten me? This is frustating, cause it doesn't
seem like a very hard thing. |
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| Olin Perry Norton |
| Posted: Wed Dec 20, 2006 4:18 pm |
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This raises a question in my mind:
The Taylor series for a function f(x) is an
expansion about a particular value of x.
Usually it's x=0, but it doesn't have to be.
But, when we're talking about an operator that
operates on a function, what "value" of the
function are we expanding about?
Olin Perry Norton |
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| Robert Israel |
| Posted: Wed Dec 20, 2006 6:28 pm |
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Guest
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In article <1166642189.282739.171240@a3g2000cwd.googlegroups.com>,
<Mollari@gmail.com> wrote:
Quote: I have some basic problem with understanding linear operators (at least
in the area of numerical analysis) and I will be a very happy person if
you can help me out.
Let's say that I have a certain operator. Let's say that it's a simple
D operator, i.e. D[U(x)] = dU/dx. This far it's OK, and I understand
what I see. I even understand what D^4(U) means - it means a 4th
derivative by x of U. However, when I have an expression
Exp[dt*D^2]U(x), I'm lost. What's that suppose to mean? How it will
look in the combination of different forms of U(x)?
Helmut has already mentioned one interpretation, in terms of Taylor
series. A more general notion is using a functional calculus, either
the holomorphic functional calculus (e.g. for closed operators on
a Banach space with nonempty resolvent set) or the Borel functional
calculus (for normal operators on a Hilbert space).
See e.g. <http://en.wikipedia.org/wiki/Holomorphic_functional_calculus>
and <http://en.wikipedia.org/wiki/Borel_functional_calculus>.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada |
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| Han de Bruijn |
| Posted: Thu Dec 21, 2006 3:48 am |
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Guest
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Mollari@gmail.com wrote:
Quote: I have some basic problem with understanding linear operators (at least
in the area of numerical analysis) and I will be a very happy person if
you can help me out.
Let's say that I have a certain operator. Let's say that it's a simple
D operator, i.e. D[U(x)] = dU/dx. This far it's OK, and I understand
what I see. I even understand what D^4(U) means - it means a 4th
derivative by x of U. However, when I have an expression
Exp[dt*D^2]U(x), I'm lost. What's that suppose to mean? How it will
look in the combination of different forms of U(x)?
Same goes for such a basic operator as a shift operator E(U). E(U) =
U(x+h). It's fine, it's easy. However, what does ln(E) means is not
clear to me.
Can anyone please enlighten me? This is frustating, cause it doesn't
seem like a very hard thing.
Here is an introductional writeup, which hopefully is an answer to some
of your questions:
http://hdebruijn.soo.dto.tudelft.nl/jaar2004/uitboek.pdf
Han de Bruijn |
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| Helmut Jarausch |
| Posted: Thu Dec 21, 2006 5:06 am |
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Guest
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Olin Perry Norton wrote:
Quote:
This raises a question in my mind:
The Taylor series for a function f(x) is an
expansion about a particular value of x.
Usually it's x=0, but it doesn't have to be.
But, when we're talking about an operator that
operates on a function, what "value" of the
function are we expanding about?
If the radius of convergence is infinity, then it
doesn't matter.
Otherwise one needs bounds on the singular values of
the operator (if it has a singular value decomposition).
If you expand around z with a radius of convergence rho,
all singular values should lie in (z-rho,z+rho) .
If the operator A is self-adjoint (think of a symmetric
matrix, e.g.) apply the taylor series to an eigenvector v
with eigenvalue lambda, like
sum a_k A^k v = (sum a_k lambda^k ) v
^^^^^^^^^^^^^^^^^^^
this conventional taylor series should converge
for all eigenvector/eigenvalue pairs!
Hope, this helps a bit.
--
Helmut Jarausch
Lehrstuhl fuer Numerische Mathematik
RWTH - Aachen University
D 52056 Aachen, Germany |
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| Olin Perry Norton |
| Posted: Thu Dec 21, 2006 11:31 am |
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Guest
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Helmut Jarausch wrote:
Quote: Olin Perry Norton wrote:
This raises a question in my mind:
The Taylor series for a function f(x) is an
expansion about a particular value of x.
Usually it's x=0, but it doesn't have to be.
But, when we're talking about an operator that
operates on a function, what "value" of the
function are we expanding about?
If the radius of convergence is infinity, then it
doesn't matter.
Otherwise one needs bounds on the singular values of
the operator (if it has a singular value decomposition).
If you expand around z with a radius of convergence rho,
all singular values should lie in (z-rho,z+rho) .
If the operator A is self-adjoint (think of a symmetric
matrix, e.g.) apply the taylor series to an eigenvector v
with eigenvalue lambda, like
sum a_k A^k v = (sum a_k lambda^k ) v
^^^^^^^^^^^^^^^^^^^
this conventional taylor series should converge
for all eigenvector/eigenvalue pairs!
Hope, this helps a bit.
Thanks or the answer. I will think about it. |
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