It's been too long ago, I can't remember how to reduce difference
equations.

Suppose I have...

f(N+1) = 0.97*f(N)

Now I know, off the seat of my pants (I think :-), that...

f(N) = Const*(0.97^N)

But I can't remember how to rigorously get to that conclusion.

...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

On 2/5/2013 11:05 AM, Jim Thompson wrote:
It's been too long ago, I can't remember how to reduce difference
equations.

Suppose I have...

f(N+1) = 0.97*f(N)

Now I know, off the seat of my pants (I think :-), that...

f(N) = Const*(0.97^N)

But I can't remember how to rigorously get to that conclusion.

...Jim Thompson


Mathematical induction. It's true for N=0, by construction, and if it's
true for N, it's true for N+1 by the use of your recurrence relation.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510 USA
+1 845 480 2058

hobbs at electrooptical dot net
http://electrooptical.net

On Tue, 05 Feb 2013 11:13:45 -0500, Phil Hobbs
wrote:

On 2/5/2013 11:05 AM, Jim Thompson wrote:
It's been too long ago, I can't remember how to reduce difference
equations.

Suppose I have...

f(N+1) = 0.97*f(N)

Now I know, off the seat of my pants (I think :-), that...

f(N) = Const*(0.97^N)

But I can't remember how to rigorously get to that conclusion.

...Jim Thompson


Mathematical induction. It's true for N=0, by construction, and if it's
true for N, it's true for N+1 by the use of your recurrence relation.

Cheers

Phil Hobbs

"Truth" isn't "solution" ;-)

Isn't there some rigorous way to derive the "continuous" equation from
the difference equation?

...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

On 05/02/2013 16:21, Jim Thompson wrote:
On Tue, 05 Feb 2013 11:13:45 -0500, Phil Hobbs
wrote:

On 2/5/2013 11:05 AM, Jim Thompson wrote:
It's been too long ago, I can't remember how to reduce difference
equations.

Suppose I have...

f(N+1) = 0.97*f(N)

Now I know, off the seat of my pants (I think :-), that...

f(N) = Const*(0.97^N)

But I can't remember how to rigorously get to that conclusion.

...Jim Thompson

Mathematical induction. It's true for N=0, by construction, and if it's
true for N, it's true for N+1 by the use of your recurrence relation.

Cheers

Phil Hobbs

"Truth" isn't "solution" ;-)

He is right though you have it for N=0 F(0), F(1) = 0.97*F(0)
and then the recurrence relation gives you F(N+1) and if paranoid
you can check it out formally by substituting your guess into it
explicitly

F(N) = k*(0.97^N)
F(N+1) = 0.97*k*(0.97^N) == k*(0.97^(N+1)) QED

which as Phil says is proof by induction.

This one is just a dull harmonic series.

Isn't there some rigorous way to derive the "continuous" equation from
the difference equation?

...Jim Thompson

Not necessarily - the finite difference equation is always a discrete
approximation to the true full differential equation.

How well the finite difference equation reflects reality and what other
approximations you have made along the way affects whether or not you
can get back to the continuous version unambiguously.

Numerical approximations to derivatives are notoriously fickle.

--
Regards,
Martin Brown

On 2/5/2013 11:05 AM, Jim Thompson wrote:
It's been too long ago, I can't remember how to reduce difference
equations.

Suppose I have...

f(N+1) = 0.97*f(N)

Now I know, off the seat of my pants (I think :-), that...

f(N) = Const*(0.97^N)

But I can't remember how to rigorously get to that conclusion.

...Jim Thompson


You've received good answers so far, but another way to do it is through
generating functions. The differential equation for the "exponential
generating function" of this recurrence can be written by inspection;
it's y' = 0.97*y.

Assuming F(0) = 1, the solution to this differential equation is y =
e^(0.97*x). The solution to your recurrence is then the coefficient of
x^k/k! in the Taylor series expansion of e^(0.97*x), or 0.97^k.

On Tue, 05 Feb 2013 13:41:05 -0500, bitrex
wrote:

On 2/5/2013 11:05 AM, Jim Thompson wrote:
It's been too long ago, I can't remember how to reduce difference
equations.

Suppose I have...

f(N+1) = 0.97*f(N)

Now I know, off the seat of my pants (I think :-), that...

f(N) = Const*(0.97^N)

But I can't remember how to rigorously get to that conclusion.

...Jim Thompson


You've received good answers so far, but another way to do it is through
generating functions. The differential equation for the "exponential
generating function" of this recurrence can be written by inspection;
it's y' = 0.97*y.

Assuming F(0) = 1, the solution to this differential equation is y =
e^(0.97*x). The solution to your recurrence is then the coefficient of
x^k/k! in the Taylor series expansion of e^(0.97*x), or 0.97^k.

Aha! Thanks, bitrex, you rang the right chime :-}

I'm slow this morning, that should have been obvious to me :-(

...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

On 2/5/2013 1:49 PM, Jim Thompson wrote:
On Tue, 05 Feb 2013 13:41:05 -0500, bitrex
wrote:

On 2/5/2013 11:05 AM, Jim Thompson wrote:
It's been too long ago, I can't remember how to reduce difference
equations.

Suppose I have...

f(N+1) = 0.97*f(N)

Now I know, off the seat of my pants (I think :-), that...

f(N) = Const*(0.97^N)

But I can't remember how to rigorously get to that conclusion.

...Jim Thompson


You've received good answers so far, but another way to do it is through
generating functions. The differential equation for the "exponential
generating function" of this recurrence can be written by inspection;
it's y' = 0.97*y.

Assuming F(0) = 1, the solution to this differential equation is y =
e^(0.97*x). The solution to your recurrence is then the coefficient of
x^k/k! in the Taylor series expansion of e^(0.97*x), or 0.97^k.

Aha! Thanks, bitrex, you rang the right chime :-}

I'm slow this morning, that should have been obvious to me :-(

...Jim Thompson


You can also "solve" it just by plugging in a trial solution r^n into
the equation - any linear recurrence with constant coefficients can be
solved this way. See:
http://en.wikipedia.org/wiki/Recurre...eneral_methods

On Tue, 05 Feb 2013 09:21:34 -0700, Jim Thompson wrote:

On Tue, 05 Feb 2013 11:13:45 -0500, Phil Hobbs
wrote:

On 2/5/2013 11:05 AM, Jim Thompson wrote:
It's been too long ago, I can't remember how to reduce difference
equations.

Suppose I have...

f(N+1) = 0.97*f(N)

Now I know, off the seat of my pants (I think :-), that...

f(N) = Const*(0.97^N)

But I can't remember how to rigorously get to that conclusion.

...Jim Thompson


Mathematical induction. It's true for N=0, by construction, and if it's
true for N, it's true for N+1 by the use of your recurrence relation.

Cheers

Phil Hobbs

"Truth" isn't "solution" ;-)

Isn't there some rigorous way to derive the "continuous" equation from
the difference equation?

I'm not sure what you mean by "continuous" equation -- a difference
equation lives in discrete-time, and has no direct relationship to a
differential equation in continuous time. (You can _make_ direct
relationships, but to do so you have to specify how sampling and
reconstruction are carried out).

I was taught in differential equations class that when you dig right down
to the bottom of things, you solve a differential equation by guessing an
answer, then proving that you were right. Finding symbolic solutions to
difference equations is the same general procedure. So I think that's as
much rigor as you're going to find in this.

Yes, there are recipes for these solutions, but all of of them (including
the use of the z transform) are just the results of the guess-then-prove
technique being carried out for a whole class of difference equations,
rather than any one specific one.

In the case of a linear, shift-invariant difference equation, the recipe
is to find the auxiliary polynomial of the difference equation, and
"posit" that the solution is the sum of A_k * d_k^N, where d_k is the k'th
root of the auxiliary polynomial and A_k is a constant that goes with it.

In your case (assuming that f(N) is the N'th element in f, which is a
possibly infinitely long vector of values), then your auxiliary
polynomial is z - 0.97, your "posited" values of d are just d = 0.97, and
your "posited" solution is

f(N) = A * (0.97)^N

Note, too, that just like differential equations, linear difference
equations can be homogeneous or non-homogeneous (yours is homogeneous).
You can find the non-homogeneous solutions to difference equations the
same way as you do for differential equations.

And, finally, if you do this a lot with linear, shift-invariant
difference equations, it pays to learn how to use the z-transform. It
simplifies things almost as much as the Laplace transform does for linear
time-invariant differential equations, and makes all of this folderol
much easier to remember.

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com

On Tue, 05 Feb 2013 13:16:39 -0600, Tim Wescott
wrote:

On Tue, 05 Feb 2013 09:21:34 -0700, Jim Thompson wrote:

On Tue, 05 Feb 2013 11:13:45 -0500, Phil Hobbs
wrote:

On 2/5/2013 11:05 AM, Jim Thompson wrote:
It's been too long ago, I can't remember how to reduce difference
equations.

Suppose I have...

f(N+1) = 0.97*f(N)

Now I know, off the seat of my pants (I think :-), that...

f(N) = Const*(0.97^N)

But I can't remember how to rigorously get to that conclusion.

...Jim Thompson


Mathematical induction. It's true for N=0, by construction, and if it's
true for N, it's true for N+1 by the use of your recurrence relation.

Cheers

Phil Hobbs

"Truth" isn't "solution" ;-)

Isn't there some rigorous way to derive the "continuous" equation from
the difference equation?

I'm not sure what you mean by "continuous" equation -- a difference
equation lives in discrete-time, and has no direct relationship to a
differential equation in continuous time. (You can _make_ direct
relationships, but to do so you have to specify how sampling and
reconstruction are carried out).

I was taught in differential equations class that when you dig right down
to the bottom of things, you solve a differential equation by guessing an
answer, then proving that you were right. Finding symbolic solutions to
difference equations is the same general procedure. So I think that's as
much rigor as you're going to find in this.

Yes, there are recipes for these solutions, but all of of them (including
the use of the z transform) are just the results of the guess-then-prove
technique being carried out for a whole class of difference equations,
rather than any one specific one.

In the case of a linear, shift-invariant difference equation, the recipe
is to find the auxiliary polynomial of the difference equation, and
"posit" that the solution is the sum of A_k * d_k^N, where d_k is the k'th
root of the auxiliary polynomial and A_k is a constant that goes with it.

In your case (assuming that f(N) is the N'th element in f, which is a
possibly infinitely long vector of values), then your auxiliary
polynomial is z - 0.97, your "posited" values of d are just d = 0.97, and
your "posited" solution is

f(N) = A * (0.97)^N

Note, too, that just like differential equations, linear difference
equations can be homogeneous or non-homogeneous (yours is homogeneous).
You can find the non-homogeneous solutions to difference equations the
same way as you do for differential equations.

And, finally, if you do this a lot with linear, shift-invariant
difference equations, it pays to learn how to use the z-transform. It
simplifies things almost as much as the Laplace transform does for linear
time-invariant differential equations, and makes all of this folderol
much easier to remember.

Yep, It's all coming back to me... guess a solution and prove it fits
:-(

...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

On Tue, 05 Feb 2013 12:18:20 -0700, Jim Thompson wrote:

On Tue, 05 Feb 2013 13:16:39 -0600, Tim Wescott
wrote:

On Tue, 05 Feb 2013 09:21:34 -0700, Jim Thompson wrote:

On Tue, 05 Feb 2013 11:13:45 -0500, Phil Hobbs
wrote:

On 2/5/2013 11:05 AM, Jim Thompson wrote:
It's been too long ago, I can't remember how to reduce difference
equations.

Suppose I have...

f(N+1) = 0.97*f(N)

Now I know, off the seat of my pants (I think :-), that...

f(N) = Const*(0.97^N)

But I can't remember how to rigorously get to that conclusion.

...Jim Thompson


Mathematical induction. It's true for N=0, by construction, and if
it's true for N, it's true for N+1 by the use of your recurrence
relation.

Cheers

Phil Hobbs

"Truth" isn't "solution" ;-)

Isn't there some rigorous way to derive the "continuous" equation from
the difference equation?

I'm not sure what you mean by "continuous" equation -- a difference
equation lives in discrete-time, and has no direct relationship to a
differential equation in continuous time. (You can _make_ direct
relationships, but to do so you have to specify how sampling and
reconstruction are carried out).

I was taught in differential equations class that when you dig right
down to the bottom of things, you solve a differential equation by
guessing an answer, then proving that you were right. Finding symbolic
solutions to difference equations is the same general procedure. So I
think that's as much rigor as you're going to find in this.

Yes, there are recipes for these solutions, but all of of them
(including the use of the z transform) are just the results of the
guess-then-prove technique being carried out for a whole class of
difference equations, rather than any one specific one.

In the case of a linear, shift-invariant difference equation, the recipe
is to find the auxiliary polynomial of the difference equation, and
"posit" that the solution is the sum of A_k * d_k^N, where d_k is the
k'th root of the auxiliary polynomial and A_k is a constant that goes
with it.

In your case (assuming that f(N) is the N'th element in f, which is a
possibly infinitely long vector of values), then your auxiliary
polynomial is z - 0.97, your "posited" values of d are just d = 0.97,
and your "posited" solution is

f(N) = A * (0.97)^N

Note, too, that just like differential equations, linear difference
equations can be homogeneous or non-homogeneous (yours is homogeneous).
You can find the non-homogeneous solutions to difference equations the
same way as you do for differential equations.

And, finally, if you do this a lot with linear, shift-invariant
difference equations, it pays to learn how to use the z-transform. It
simplifies things almost as much as the Laplace transform does for
linear time-invariant differential equations, and makes all of this
folderol much easier to remember.

Yep, It's all coming back to me... guess a solution and prove it fits
:-(

The guy who taught my second term of diff eqs clearly wanted us to
remember this for all time, because he made this a mantra. Nearly every
class meeting he would put some new form of differential equation up on
the board, and he'd say "Now, how do we solve this differential
equation?" then (because we didn't all shout it out in unison) he'd
answer himself: "We guess, and prove that we're right!"

That was 30 years ago. It's stuck with me, so I guess he met his goal in
my case.

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com

On Tue, 05 Feb 2013 13:26:23 -0600, Tim Wescott
wrote:

On Tue, 05 Feb 2013 12:18:20 -0700, Jim Thompson wrote:

On Tue, 05 Feb 2013 13:16:39 -0600, Tim Wescott
wrote:

On Tue, 05 Feb 2013 09:21:34 -0700, Jim Thompson wrote:

On Tue, 05 Feb 2013 11:13:45 -0500, Phil Hobbs
wrote:

On 2/5/2013 11:05 AM, Jim Thompson wrote:
It's been too long ago, I can't remember how to reduce difference
equations.

Suppose I have...

f(N+1) = 0.97*f(N)

Now I know, off the seat of my pants (I think :-), that...

f(N) = Const*(0.97^N)

But I can't remember how to rigorously get to that conclusion.

...Jim Thompson


Mathematical induction. It's true for N=0, by construction, and if
it's true for N, it's true for N+1 by the use of your recurrence
relation.

Cheers

Phil Hobbs

"Truth" isn't "solution" ;-)

Isn't there some rigorous way to derive the "continuous" equation from
the difference equation?

I'm not sure what you mean by "continuous" equation -- a difference
equation lives in discrete-time, and has no direct relationship to a
differential equation in continuous time. (You can _make_ direct
relationships, but to do so you have to specify how sampling and
reconstruction are carried out).

I was taught in differential equations class that when you dig right
down to the bottom of things, you solve a differential equation by
guessing an answer, then proving that you were right. Finding symbolic
solutions to difference equations is the same general procedure. So I
think that's as much rigor as you're going to find in this.

Yes, there are recipes for these solutions, but all of of them
(including the use of the z transform) are just the results of the
guess-then-prove technique being carried out for a whole class of
difference equations, rather than any one specific one.

In the case of a linear, shift-invariant difference equation, the recipe
is to find the auxiliary polynomial of the difference equation, and
"posit" that the solution is the sum of A_k * d_k^N, where d_k is the
k'th root of the auxiliary polynomial and A_k is a constant that goes
with it.

In your case (assuming that f(N) is the N'th element in f, which is a
possibly infinitely long vector of values), then your auxiliary
polynomial is z - 0.97, your "posited" values of d are just d = 0.97,
and your "posited" solution is

f(N) = A * (0.97)^N

Note, too, that just like differential equations, linear difference
equations can be homogeneous or non-homogeneous (yours is homogeneous).
You can find the non-homogeneous solutions to difference equations the
same way as you do for differential equations.

And, finally, if you do this a lot with linear, shift-invariant
difference equations, it pays to learn how to use the z-transform. It
simplifies things almost as much as the Laplace transform does for
linear time-invariant differential equations, and makes all of this
folderol much easier to remember.

Yep, It's all coming back to me... guess a solution and prove it fits
:-(

The guy who taught my second term of diff eqs clearly wanted us to
remember this for all time, because he made this a mantra. Nearly every
class meeting he would put some new form of differential equation up on
the board, and he'd say "Now, how do we solve this differential
equation?" then (because we didn't all shout it out in unison) he'd
answer himself: "We guess, and prove that we're right!"

That was 30 years ago. It's stuck with me, so I guess he met his goal in
my case.

That's probably what surprised me most... start the class with, "There
are these few basic equations we can solve exactly..." :-(

...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

On 2/5/2013 2:16 PM, Tim Wescott wrote:
On Tue, 05 Feb 2013 09:21:34 -0700, Jim Thompson wrote:

On Tue, 05 Feb 2013 11:13:45 -0500, Phil Hobbs
wrote:

On 2/5/2013 11:05 AM, Jim Thompson wrote:
It's been too long ago, I can't remember how to reduce difference
equations.

Suppose I have...

f(N+1) = 0.97*f(N)

Now I know, off the seat of my pants (I think :-), that...

f(N) = Const*(0.97^N)

But I can't remember how to rigorously get to that conclusion.

...Jim Thompson


Mathematical induction. It's true for N=0, by construction, and if it's
true for N, it's true for N+1 by the use of your recurrence relation.

Cheers

Phil Hobbs

"Truth" isn't "solution" ;-)

Isn't there some rigorous way to derive the "continuous" equation from
the difference equation?

I'm not sure what you mean by "continuous" equation -- a difference
equation lives in discrete-time, and has no direct relationship to a
differential equation in continuous time. (You can _make_ direct
relationships, but to do so you have to specify how sampling and
reconstruction are carried out).

I was taught in differential equations class that when you dig right down
to the bottom of things, you solve a differential equation by guessing an
answer, then proving that you were right. Finding symbolic solutions to
difference equations is the same general procedure. So I think that's as
much rigor as you're going to find in this.

There are exceptions to this rule, especially transform methods and
variation of parameters. It's much more true with PDEs, where e.g.
there are 7 types of coordinate systems in which the Laplacian
separates, including such intuitive ones as parabolic cylindrical
coordinates.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510 USA
+1 845 480 2058

hobbs at electrooptical dot net
http://electrooptical.net

On Tue, 05 Feb 2013 23:32:24 -0500, Phil Hobbs wrote:

On 2/5/2013 2:16 PM, Tim Wescott wrote:
On Tue, 05 Feb 2013 09:21:34 -0700, Jim Thompson wrote:

On Tue, 05 Feb 2013 11:13:45 -0500, Phil Hobbs
wrote:

On 2/5/2013 11:05 AM, Jim Thompson wrote:
It's been too long ago, I can't remember how to reduce difference
equations.

Suppose I have...

f(N+1) = 0.97*f(N)

Now I know, off the seat of my pants (I think :-), that...

f(N) = Const*(0.97^N)

But I can't remember how to rigorously get to that conclusion.

...Jim Thompson


Mathematical induction. It's true for N=0, by construction, and if
it's true for N, it's true for N+1 by the use of your recurrence
relation.

Cheers

Phil Hobbs

"Truth" isn't "solution" ;-)

Isn't there some rigorous way to derive the "continuous" equation from
the difference equation?

I'm not sure what you mean by "continuous" equation -- a difference
equation lives in discrete-time, and has no direct relationship to a
differential equation in continuous time. (You can _make_ direct
relationships, but to do so you have to specify how sampling and
reconstruction are carried out).

I was taught in differential equations class that when you dig right
down to the bottom of things, you solve a differential equation by
guessing an answer, then proving that you were right. Finding symbolic
solutions to difference equations is the same general procedure. So I
think that's as much rigor as you're going to find in this.

There are exceptions to this rule, especially transform methods and
variation of parameters. It's much more true with PDEs, where e.g.
there are 7 types of coordinate systems in which the Laplacian
separates, including such intuitive ones as parabolic cylindrical
coordinates.

I would contend, though, that all those methods basically came about
either because someone guessed an answer to a general class of problems
and proved it's validity for that class (e.g. the Laplace transform as
used for solving LTI differential equations), or because someone found a
trick (variation of parameters, separation of variables, etc.) that
applies to a number of different problems.

So in those cases the "guess and prove" has been done for you, but you
still have to make sure that your problem fits into the class of things
that the method solves for (and heaven knows, I've seen people trying to
cram nonlinear or time varying system descriptions into the Laplace
transform, and being all confused when their nonsensical formulations
give nonsensical results).

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com

On 05/02/2013 19:26, Tim Wescott wrote:
On Tue, 05 Feb 2013 12:18:20 -0700, Jim Thompson wrote:

On Tue, 05 Feb 2013 13:16:39 -0600, Tim Wescott
wrote:

And, finally, if you do this a lot with linear, shift-invariant
difference equations, it pays to learn how to use the z-transform. It
simplifies things almost as much as the Laplace transform does for
linear time-invariant differential equations, and makes all of this
folderol much easier to remember.

Yep, It's all coming back to me... guess a solution and prove it fits
:-(

The guy who taught my second term of diff eqs clearly wanted us to
remember this for all time, because he made this a mantra. Nearly every
class meeting he would put some new form of differential equation up on
the board, and he'd say "Now, how do we solve this differential
equation?" then (because we didn't all shout it out in unison) he'd
answer himself: "We guess, and prove that we're right!"

That was 30 years ago. It's stuck with me, so I guess he met his goal in
my case.

I still recall my first serious brush with exotic differential equations
in the freshman year although for different reasons.

The lecturer was an internationally famous astronomer and brilliant
analytical solver of novel differential equations. The snag was that he
could not teach for toffee and merely demonstrated pulling rabbit out of
hat again and again and again. His coursework was impossible.

This particular course was so incomprehensible that after a while the
best of us went to the other stream of maths on group theory since it
was so much easier and the exam questions were likely to be possible to
solve in finite time. Some from the other course which then became badly
overcrowded then went to the ODE course so they could sit down.

Indirectly he probably contributed to the increased use of computers to
solve differential equations as we later moved into research.

--
Regards,
Martin Brown

On Wed, 06 Feb 2013 07:56:07 +0000, Martin Brown
wrote:

On 05/02/2013 19:26, Tim Wescott wrote:
On Tue, 05 Feb 2013 12:18:20 -0700, Jim Thompson wrote:

On Tue, 05 Feb 2013 13:16:39 -0600, Tim Wescott
wrote:

And, finally, if you do this a lot with linear, shift-invariant
difference equations, it pays to learn how to use the z-transform. It
simplifies things almost as much as the Laplace transform does for
linear time-invariant differential equations, and makes all of this
folderol much easier to remember.

Yep, It's all coming back to me... guess a solution and prove it fits
:-(

The guy who taught my second term of diff eqs clearly wanted us to
remember this for all time, because he made this a mantra. Nearly every
class meeting he would put some new form of differential equation up on
the board, and he'd say "Now, how do we solve this differential
equation?" then (because we didn't all shout it out in unison) he'd
answer himself: "We guess, and prove that we're right!"

That was 30 years ago. It's stuck with me, so I guess he met his goal in
my case.

I still recall my first serious brush with exotic differential equations
in the freshman year although for different reasons.

The lecturer was an internationally famous astronomer and brilliant
analytical solver of novel differential equations. The snag was that he
could not teach for toffee and merely demonstrated pulling rabbit out of
hat again and again and again. His coursework was impossible.

This particular course was so incomprehensible that after a while the
best of us went to the other stream of maths on group theory since it
was so much easier and the exam questions were likely to be possible to
solve in finite time. Some from the other course which then became badly
overcrowded then went to the ODE course so they could sit down.

Indirectly he probably contributed to the increased use of computers to
solve differential equations as we later moved into research.


I recall struggling through Diffy-Q, which was widely
regarded as a "bag of tricks" subject... you just had to
figure out which trick to pull out of the bag for each
special case.

Then next term came Laplace Transforms, where we learned
that everything that needed doing could be done with simple
algebra via Laplace... all that Diffy-Q torture had been
just for background information and building character!


Bob Masta

DAQARTA v7.21
Data AcQuisition And Real-Time Analysis
www.daqarta.com
Scope, Spectrum, Spectrogram, Sound Level Meter
Frequency Counter, Pitch Track, Pitch-to-MIDI
FREE Signal Generator, DaqMusic generator
Science with your sound card!

On Wed, 06 Feb 2013 13:29:16 +0000, Bob Masta wrote:

On Wed, 06 Feb 2013 07:56:07 +0000, Martin Brown
wrote:

On 05/02/2013 19:26, Tim Wescott wrote:
On Tue, 05 Feb 2013 12:18:20 -0700, Jim Thompson wrote:

On Tue, 05 Feb 2013 13:16:39 -0600, Tim Wescott

wrote:

And, finally, if you do this a lot with linear, shift-invariant
difference equations, it pays to learn how to use the z-transform.
It simplifies things almost as much as the Laplace transform does
for linear time-invariant differential equations, and makes all of
this folderol much easier to remember.

Yep, It's all coming back to me... guess a solution and prove it fits
:-(

The guy who taught my second term of diff eqs clearly wanted us to
remember this for all time, because he made this a mantra. Nearly
every class meeting he would put some new form of differential
equation up on the board, and he'd say "Now, how do we solve this
differential equation?" then (because we didn't all shout it out in
unison) he'd answer himself: "We guess, and prove that we're right!"

That was 30 years ago. It's stuck with me, so I guess he met his goal
in my case.

I still recall my first serious brush with exotic differential equations
in the freshman year although for different reasons.

The lecturer was an internationally famous astronomer and brilliant
analytical solver of novel differential equations. The snag was that he
could not teach for toffee and merely demonstrated pulling rabbit out of
hat again and again and again. His coursework was impossible.

This particular course was so incomprehensible that after a while the
best of us went to the other stream of maths on group theory since it
was so much easier and the exam questions were likely to be possible to
solve in finite time. Some from the other course which then became badly
overcrowded then went to the ODE course so they could sit down.

Indirectly he probably contributed to the increased use of computers to
solve differential equations as we later moved into research.


I recall struggling through Diffy-Q, which was widely regarded as a "bag
of tricks" subject... you just had to figure out which trick to pull out
of the bag for each special case.

Then next term came Laplace Transforms, where we learned that everything
that needed doing could be done with simple algebra via Laplace... all
that Diffy-Q torture had been just for background information and
building character!

Using the Laplace transform is just a really versatile trick for
approximately solving real-world problems. Here's the reasoning that you
should keep in mind whenever you use it (or the z transform):

1: All real-world systems are nonlinear and time varying.
2: The Laplace transform only works on systems that are linear and time-
invariant.
3: Thus, I cannot use the Laplace transform to solve this problem.
4: But I can come _close_ by linearizing this here system
5: And now I can use Laplace!

This works great a whole lot of the time -- but it doesn't always, and
engineers who are steeped in Laplace (or the z transform) tend to forget
that they've skipped over steps 1-3, and did 4 without questioning why,
or when it is not valid to do so.

--
Tim Wescott
Control system and signal processing consulting
www.wescottdesign.com

On 2/6/2013 12:59 AM, Tim Wescott wrote:
On Tue, 05 Feb 2013 23:32:24 -0500, Phil Hobbs wrote:

On 2/5/2013 2:16 PM, Tim Wescott wrote:
On Tue, 05 Feb 2013 09:21:34 -0700, Jim Thompson wrote:

On Tue, 05 Feb 2013 11:13:45 -0500, Phil Hobbs
wrote:

On 2/5/2013 11:05 AM, Jim Thompson wrote:
It's been too long ago, I can't remember how to reduce difference
equations.

Suppose I have...

f(N+1) = 0.97*f(N)

Now I know, off the seat of my pants (I think :-), that...

f(N) = Const*(0.97^N)

But I can't remember how to rigorously get to that conclusion.

...Jim Thompson


Mathematical induction. It's true for N=0, by construction, and if
it's true for N, it's true for N+1 by the use of your recurrence
relation.

Cheers

Phil Hobbs

"Truth" isn't "solution" ;-)

Isn't there some rigorous way to derive the "continuous" equation from
the difference equation?

I'm not sure what you mean by "continuous" equation -- a difference
equation lives in discrete-time, and has no direct relationship to a
differential equation in continuous time. (You can _make_ direct
relationships, but to do so you have to specify how sampling and
reconstruction are carried out).

I was taught in differential equations class that when you dig right
down to the bottom of things, you solve a differential equation by
guessing an answer, then proving that you were right. Finding symbolic
solutions to difference equations is the same general procedure. So I
think that's as much rigor as you're going to find in this.

There are exceptions to this rule, especially transform methods and
variation of parameters. It's much more true with PDEs, where e.g.
there are 7 types of coordinate systems in which the Laplacian
separates, including such intuitive ones as parabolic cylindrical
coordinates.

I would contend, though, that all those methods basically came about
either because someone guessed an answer to a general class of problems
and proved it's validity for that class (e.g. the Laplace transform as
used for solving LTI differential equations), or because someone found a
trick (variation of parameters, separation of variables, etc.) that
applies to a number of different problems.

So in those cases the "guess and prove" has been done for you, but you
still have to make sure that your problem fits into the class of things
that the method solves for (and heaven knows, I've seen people trying to
cram nonlinear or time varying system descriptions into the Laplace
transform, and being all confused when their nonsensical formulations
give nonsensical results).


Well, the question of how mathematical theorems are discovered is an
interesting one--George Polya wrote a series of books on the subject
back in (iirc) the 1950s, of which the best known is the elementary one,
"How To Solve It". Great book.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510 USA
+1 845 480 2058

hobbs at electrooptical dot net
http://electrooptical.net

On 2/6/2013 10:51 AM, Tim Wescott wrote:
On Wed, 06 Feb 2013 13:29:16 +0000, Bob Masta wrote:

On Wed, 06 Feb 2013 07:56:07 +0000, Martin Brown
wrote:

On 05/02/2013 19:26, Tim Wescott wrote:
On Tue, 05 Feb 2013 12:18:20 -0700, Jim Thompson wrote:

On Tue, 05 Feb 2013 13:16:39 -0600, Tim Wescott

wrote:

And, finally, if you do this a lot with linear, shift-invariant
difference equations, it pays to learn how to use the z-transform.
It simplifies things almost as much as the Laplace transform does
for linear time-invariant differential equations, and makes all of
this folderol much easier to remember.

Yep, It's all coming back to me... guess a solution and prove it fits
:-(

The guy who taught my second term of diff eqs clearly wanted us to
remember this for all time, because he made this a mantra. Nearly
every class meeting he would put some new form of differential
equation up on the board, and he'd say "Now, how do we solve this
differential equation?" then (because we didn't all shout it out in
unison) he'd answer himself: "We guess, and prove that we're right!"

That was 30 years ago. It's stuck with me, so I guess he met his goal
in my case.

I still recall my first serious brush with exotic differential equations
in the freshman year although for different reasons.

The lecturer was an internationally famous astronomer and brilliant
analytical solver of novel differential equations. The snag was that he
could not teach for toffee and merely demonstrated pulling rabbit out of
hat again and again and again. His coursework was impossible.

This particular course was so incomprehensible that after a while the
best of us went to the other stream of maths on group theory since it
was so much easier and the exam questions were likely to be possible to
solve in finite time. Some from the other course which then became badly
overcrowded then went to the ODE course so they could sit down.

Indirectly he probably contributed to the increased use of computers to
solve differential equations as we later moved into research.


I recall struggling through Diffy-Q, which was widely regarded as a "bag
of tricks" subject... you just had to figure out which trick to pull out
of the bag for each special case.

Then next term came Laplace Transforms, where we learned that everything
that needed doing could be done with simple algebra via Laplace... all
that Diffy-Q torture had been just for background information and
building character!

Using the Laplace transform is just a really versatile trick for
approximately solving real-world problems. Here's the reasoning that you
should keep in mind whenever you use it (or the z transform):

1: All real-world systems are nonlinear and time varying.
2: The Laplace transform only works on systems that are linear and time-
invariant.
3: Thus, I cannot use the Laplace transform to solve this problem.
4: But I can come _close_ by linearizing this here system
5: And now I can use Laplace!

This works great a whole lot of the time -- but it doesn't always, and
engineers who are steeped in Laplace (or the z transform) tend to forget
that they've skipped over steps 1-3, and did 4 without questioning why,
or when it is not valid to do so.


You can often use it to derive a perturbation series for those times
when linearizing isn't quite enough.

I highly recommend Bender & Orszag's "Advanced Mathematical Methods for
Scientists and Engineers", which has all that sort of stuff--boundary
layer theory, steepest descents, and a whole lot of other asymptotic
methods. Arfken's applied math book is good as well--it used to be the
standard textbook for physics undergraduates.

It's often possible to derive a series solution, which may or may not
converge, and then convert to a continued fraction or apply convergence
tricks such as Shanks's algorithm or Richardson extrapolation. When
those work, which isn't always, they can be practically supernatural.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510 USA
+1 845 480 2058

hobbs at electrooptical dot net
http://electrooptical.net

On 2/6/2013 4:51 PM, Tim Wescott wrote:
On Wed, 06 Feb 2013 13:29:16 +0000, Bob Masta wrote:

On Wed, 06 Feb 2013 07:56:07 +0000, Martin Brown
wrote:

On 05/02/2013 19:26, Tim Wescott wrote:
On Tue, 05 Feb 2013 12:18:20 -0700, Jim Thompson wrote:

On Tue, 05 Feb 2013 13:16:39 -0600, Tim Wescott

wrote:

And, finally, if you do this a lot with linear, shift-invariant
difference equations, it pays to learn how to use the z-transform.
It simplifies things almost as much as the Laplace transform does
for linear time-invariant differential equations, and makes all of
this folderol much easier to remember.

Yep, It's all coming back to me... guess a solution and prove it fits
:-(

The guy who taught my second term of diff eqs clearly wanted us to
remember this for all time, because he made this a mantra. Nearly
every class meeting he would put some new form of differential
equation up on the board, and he'd say "Now, how do we solve this
differential equation?" then (because we didn't all shout it out in
unison) he'd answer himself: "We guess, and prove that we're right!"

That was 30 years ago. It's stuck with me, so I guess he met his goal
in my case.

I still recall my first serious brush with exotic differential equations
in the freshman year although for different reasons.

The lecturer was an internationally famous astronomer and brilliant
analytical solver of novel differential equations. The snag was that he
could not teach for toffee and merely demonstrated pulling rabbit out of
hat again and again and again. His coursework was impossible.

This particular course was so incomprehensible that after a while the
best of us went to the other stream of maths on group theory since it
was so much easier and the exam questions were likely to be possible to
solve in finite time. Some from the other course which then became badly
overcrowded then went to the ODE course so they could sit down.

Indirectly he probably contributed to the increased use of computers to
solve differential equations as we later moved into research.


I recall struggling through Diffy-Q, which was widely regarded as a "bag
of tricks" subject... you just had to figure out which trick to pull out
of the bag for each special case.

Then next term came Laplace Transforms, where we learned that everything
that needed doing could be done with simple algebra via Laplace... all
that Diffy-Q torture had been just for background information and
building character!

Using the Laplace transform is just a really versatile trick for
approximately solving real-world problems. Here's the reasoning that you
should keep in mind whenever you use it (or the z transform):

1: All real-world systems are nonlinear and time varying.
2: The Laplace transform only works on systems that are linear and time-
invariant.
3: Thus, I cannot use the Laplace transform to solve this problem.
4: But I can come _close_ by linearizing this here system
5: And now I can use Laplace!

This works great a whole lot of the time -- but it doesn't always, and
engineers who are steeped in Laplace (or the z transform) tend to forget
that they've skipped over steps 1-3, and did 4 without questioning why,
or when it is not valid to do so.

These discussion brings back some memories.
I hope i don't have bad dreams tonight.

Following to Laplace we where tortured with Hilbert transformations.

After university i never had to solve a differential- or integral equations.

On Wed, 06 Feb 2013 23:14:04 +0100, tuinkabouter
wrote:

On 2/6/2013 4:51 PM, Tim Wescott wrote:
On Wed, 06 Feb 2013 13:29:16 +0000, Bob Masta wrote:

On Wed, 06 Feb 2013 07:56:07 +0000, Martin Brown
wrote:

On 05/02/2013 19:26, Tim Wescott wrote:
On Tue, 05 Feb 2013 12:18:20 -0700, Jim Thompson wrote:

On Tue, 05 Feb 2013 13:16:39 -0600, Tim Wescott

wrote:

And, finally, if you do this a lot with linear, shift-invariant
difference equations, it pays to learn how to use the z-transform.
It simplifies things almost as much as the Laplace transform does
for linear time-invariant differential equations, and makes all of
this folderol much easier to remember.

Yep, It's all coming back to me... guess a solution and prove it fits
:-(

The guy who taught my second term of diff eqs clearly wanted us to
remember this for all time, because he made this a mantra. Nearly
every class meeting he would put some new form of differential
equation up on the board, and he'd say "Now, how do we solve this
differential equation?" then (because we didn't all shout it out in
unison) he'd answer himself: "We guess, and prove that we're right!"

That was 30 years ago. It's stuck with me, so I guess he met his goal
in my case.

I still recall my first serious brush with exotic differential equations
in the freshman year although for different reasons.

The lecturer was an internationally famous astronomer and brilliant
analytical solver of novel differential equations. The snag was that he
could not teach for toffee and merely demonstrated pulling rabbit out of
hat again and again and again. His coursework was impossible.

This particular course was so incomprehensible that after a while the
best of us went to the other stream of maths on group theory since it
was so much easier and the exam questions were likely to be possible to
solve in finite time. Some from the other course which then became badly
overcrowded then went to the ODE course so they could sit down.

Indirectly he probably contributed to the increased use of computers to
solve differential equations as we later moved into research.


I recall struggling through Diffy-Q, which was widely regarded as a "bag
of tricks" subject... you just had to figure out which trick to pull out
of the bag for each special case.

Then next term came Laplace Transforms, where we learned that everything
that needed doing could be done with simple algebra via Laplace... all
that Diffy-Q torture had been just for background information and
building character!

Using the Laplace transform is just a really versatile trick for
approximately solving real-world problems. Here's the reasoning that you
should keep in mind whenever you use it (or the z transform):

1: All real-world systems are nonlinear and time varying.
2: The Laplace transform only works on systems that are linear and time-
invariant.
3: Thus, I cannot use the Laplace transform to solve this problem.
4: But I can come _close_ by linearizing this here system
5: And now I can use Laplace!

This works great a whole lot of the time -- but it doesn't always, and
engineers who are steeped in Laplace (or the z transform) tend to forget
that they've skipped over steps 1-3, and did 4 without questioning why,
or when it is not valid to do so.

These discussion brings back some memories.
I hope i don't have bad dreams tonight.

Following to Laplace we where tortured with Hilbert transformations.

After university i never had to solve a differential- or integral equations.

Pure Laplace is a PITA, but the Heaviside short-hand version is a
beautiful engineering tool.

...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

On 06/02/2013 22:14, tuinkabouter wrote:
On 2/6/2013 4:51 PM, Tim Wescott wrote:
On Wed, 06 Feb 2013 13:29:16 +0000, Bob Masta wrote:

On Wed, 06 Feb 2013 07:56:07 +0000, Martin Brown
wrote:


This particular course was so incomprehensible that after a while the
best of us went to the other stream of maths on group theory since it
was so much easier and the exam questions were likely to be possible to
solve in finite time. Some from the other course which then became
badly
overcrowded then went to the ODE course so they could sit down.

Indirectly he probably contributed to the increased use of computers to
solve differential equations as we later moved into research.


I recall struggling through Diffy-Q, which was widely regarded as a "bag
of tricks" subject... you just had to figure out which trick to pull out
of the bag for each special case.

Then next term came Laplace Transforms, where we learned that everything
that needed doing could be done with simple algebra via Laplace... all
that Diffy-Q torture had been just for background information and
building character!

Using the Laplace transform is just a really versatile trick for
approximately solving real-world problems. Here's the reasoning that you
should keep in mind whenever you use it (or the z transform):

1: All real-world systems are nonlinear and time varying.
2: The Laplace transform only works on systems that are linear and time-
invariant.
3: Thus, I cannot use the Laplace transform to solve this problem.
4: But I can come _close_ by linearizing this here system
5: And now I can use Laplace!

This works great a whole lot of the time -- but it doesn't always, and
engineers who are steeped in Laplace (or the z transform) tend to forget
that they've skipped over steps 1-3, and did 4 without questioning why,
or when it is not valid to do so.

These discussion brings back some memories.
I hope i don't have bad dreams tonight.

Following to Laplace we where tortured with Hilbert transformations.

After university i never had to solve a differential- or integral
equations.

I still have serious hatred of Green's functions arising from that
particular course. Lucky we no longer do it analytically any more.

--
Regards,
Martin Brown

On 06/02/2013 17:42, Phil Hobbs wrote:

You can often use it to derive a perturbation series for those times
when linearizing isn't quite enough.

I highly recommend Bender & Orszag's "Advanced Mathematical Methods for
Scientists and Engineers", which has all that sort of stuff--boundary
layer theory, steepest descents, and a whole lot of other asymptotic
methods. Arfken's applied math book is good as well--it used to be the
standard textbook for physics undergraduates.

Our local choice was Matthews & Walker Mathematical Methods of Physics.

It's often possible to derive a series solution, which may or may not
converge, and then convert to a continued fraction or apply convergence
tricks such as Shanks's algorithm or Richardson extrapolation. When
those work, which isn't always, they can be practically supernatural.

Always used to be popular with the turbulent flow brigade - often gave
divergent power series that could only be tamed with Shank's. The pure
mathematicians used to cringe at the abuse of method but they could not
deny that the results it predicted matched experimental data well!

Works best on alternating series with poor convergence or divergence!

I was once very interested in extended convergence tricks and have used
the rational approximation for Log(1+x) in anger several times.

Log(1+x) = x(6+x)/(6+4x) where -1/2 x 1

The other common one sometimes useful is

Sqrt(x) = (1+3x)/(3+x) where -1/2 x 2

Only really any good if you have fast hardware divide.

These tricks sometimes allow a rough polynomial approximation soluble
analytically to provide a much better initial input guess for more rapid
convergence of an iterative method like N-R.

--
Regards,
Martin Brown

Martin Brown wrote:
On 06/02/2013 22:14, tuinkabouter wrote:
On 2/6/2013 4:51 PM, Tim Wescott wrote:
On Wed, 06 Feb 2013 13:29:16 +0000, Bob Masta wrote:

On Wed, 06 Feb 2013 07:56:07 +0000, Martin Brown
wrote:


This particular course was so incomprehensible that after a while the
best of us went to the other stream of maths on group theory since it
was so much easier and the exam questions were likely to be possible to
solve in finite time. Some from the other course which then became
badly
overcrowded then went to the ODE course so they could sit down.

Indirectly he probably contributed to the increased use of computers to
solve differential equations as we later moved into research.


I recall struggling through Diffy-Q, which was widely regarded as a "bag
of tricks" subject... you just had to figure out which trick to pull out
of the bag for each special case.

Then next term came Laplace Transforms, where we learned that everything
that needed doing could be done with simple algebra via Laplace... all
that Diffy-Q torture had been just for background information and
building character!

Using the Laplace transform is just a really versatile trick for
approximately solving real-world problems. Here's the reasoning that you
should keep in mind whenever you use it (or the z transform):

1: All real-world systems are nonlinear and time varying.
2: The Laplace transform only works on systems that are linear and time-
invariant.
3: Thus, I cannot use the Laplace transform to solve this problem.
4: But I can come _close_ by linearizing this here system
5: And now I can use Laplace!

This works great a whole lot of the time -- but it doesn't always, and
engineers who are steeped in Laplace (or the z transform) tend to forget
that they've skipped over steps 1-3, and did 4 without questioning why,
or when it is not valid to do so.

These discussion brings back some memories.
I hope i don't have bad dreams tonight.

Following to Laplace we where tortured with Hilbert transformations.

After university i never had to solve a differential- or integral
equations.

I still have serious hatred of Green's functions arising from that
particular course. Lucky we no longer do it analytically any more.


I recall the materials course where the prof explained everything using
tensor calculus. Unfortunately, nobody had previously bothered to teach us
tensor calculus. All the lectures were pure gibberish. I'm not at all sure
how I passed that course.

On 2/7/2013 5:40 AM, Martin Brown wrote:
On 06/02/2013 17:42, Phil Hobbs wrote:

You can often use it to derive a perturbation series for those times
when linearizing isn't quite enough.

I highly recommend Bender & Orszag's "Advanced Mathematical Methods for
Scientists and Engineers", which has all that sort of stuff--boundary
layer theory, steepest descents, and a whole lot of other asymptotic
methods. Arfken's applied math book is good as well--it used to be the
standard textbook for physics undergraduates.

Our local choice was Matthews & Walker Mathematical Methods of Physics.

It's often possible to derive a series solution, which may or may not
converge, and then convert to a continued fraction or apply convergence
tricks such as Shanks's algorithm or Richardson extrapolation. When
those work, which isn't always, they can be practically supernatural.

Always used to be popular with the turbulent flow brigade - often gave
divergent power series that could only be tamed with Shank's. The pure
mathematicians used to cringe at the abuse of method but they could not
deny that the results it predicted matched experimental data well!

Works best on alternating series with poor convergence or divergence!

I was once very interested in extended convergence tricks and have used
the rational approximation for Log(1+x) in anger several times.

Log(1+x) = x(6+x)/(6+4x) where -1/2 x 1

The other common one sometimes useful is

Sqrt(x) = (1+3x)/(3+x) where -1/2 x 2

Only really any good if you have fast hardware divide.

These tricks sometimes allow a rough polynomial approximation soluble
analytically to provide a much better initial input guess for more rapid
convergence of an iterative method like N-R.


Good stuff. Did you ever read Forman Acton's "Numerical Methods That
Work"?

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510 USA
+1 845 480 2058

hobbs at electrooptical dot net
http://electrooptical.net

On Fri, 08 Feb 2013 02:23:34 GMT, the renowned Ralph Barone
wrote:

Martin Brown wrote:
On 06/02/2013 22:14, tuinkabouter wrote:
On 2/6/2013 4:51 PM, Tim Wescott wrote:
On Wed, 06 Feb 2013 13:29:16 +0000, Bob Masta wrote:

On Wed, 06 Feb 2013 07:56:07 +0000, Martin Brown
wrote:


This particular course was so incomprehensible that after a while the
best of us went to the other stream of maths on group theory since it
was so much easier and the exam questions were likely to be possible to
solve in finite time. Some from the other course which then became
badly
overcrowded then went to the ODE course so they could sit down.

Indirectly he probably contributed to the increased use of computers to
solve differential equations as we later moved into research.


I recall struggling through Diffy-Q, which was widely regarded as a "bag
of tricks" subject... you just had to figure out which trick to pull out
of the bag for each special case.

Then next term came Laplace Transforms, where we learned that everything
that needed doing could be done with simple algebra via Laplace... all
that Diffy-Q torture had been just for background information and
building character!

Using the Laplace transform is just a really versatile trick for
approximately solving real-world problems. Here's the reasoning that you
should keep in mind whenever you use it (or the z transform):

1: All real-world systems are nonlinear and time varying.
2: The Laplace transform only works on systems that are linear and time-
invariant.
3: Thus, I cannot use the Laplace transform to solve this problem.
4: But I can come _close_ by linearizing this here system
5: And now I can use Laplace!

This works great a whole lot of the time -- but it doesn't always, and
engineers who are steeped in Laplace (or the z transform) tend to forget
that they've skipped over steps 1-3, and did 4 without questioning why,
or when it is not valid to do so.

These discussion brings back some memories.
I hope i don't have bad dreams tonight.

Following to Laplace we where tortured with Hilbert transformations.

After university i never had to solve a differential- or integral
equations.

I still have serious hatred of Green's functions arising from that
particular course. Lucky we no longer do it analytically any more.


I recall the materials course where the prof explained everything using
tensor calculus. Unfortunately, nobody had previously bothered to teach us
tensor calculus. All the lectures were pure gibberish. I'm not at all sure
how I passed that course.

Bell curve.


Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com

On 08/02/2013 02:32, Phil Hobbs wrote:
On 2/7/2013 5:40 AM, Martin Brown wrote:

Works best on alternating series with poor convergence or divergence!

I was once very interested in extended convergence tricks and have used
the rational approximation for Log(1+x) in anger several times.

Log(1+x) = x(6+x)/(6+4x) where -1/2 x 1

The other common one sometimes useful is

Sqrt(x) = (1+3x)/(3+x) where -1/2 x 2

Only really any good if you have fast hardware divide.

These tricks sometimes allow a rough polynomial approximation soluble
analytically to provide a much better initial input guess for more rapid
convergence of an iterative method like N-R.


Good stuff. Did you ever read Forman Acton's "Numerical Methods That
Work"?

Yes - but along time ago. I don't possess a copy any more.

One memorable fun version of 3 term Shanks with terrible numerical
stability (but you can divide it out to get something much better) is
difference of geometric mean over arithmetic mean

given the partial sums a, b, c

x' = (a*c-b^2)/(a+c-2*b)

And if you feed it 1,3,7 it gives

x' = (1*7-3^2)/(1+7-2*3) = (7-9)/(8-6) = -1

Users of twos compliment arithmetic take note!

A numerically stable version that is formally equivalent is

x' = (a+2b+c)/4 - (a-c)^2/(a+c-2b)/4

Subject to typos and lapses of memory...

--
Regards,
Martin Brown