how to form a differential equation ?
Nimo
who helped me in my previous threads
1) polynomial into e^ix expansion
2) I don't want a tangent, but e^ix
3) Taylor's series, Lagrange polynomial, e^ix
4) How to calculate e^ix
instead of bringing all those thread into main page
I'm appreciating their help here, once again thank you.
/*=-=-=-=-=-=-=-=-==-=-*/
Now coming to the
Problem: -
for given data how to frame a Differential_Equation ?
for example see this, totally random
x = 1 2 3 4 5 6
y = 50 1 30 122 0 18
what's the procedure to do that ?
Robert Israel
It's not at all clear what you mean, but I suppose you
want a differential equation that has a solution going
through those points.
Well, you might start by specifying the general form of
differential equation you want. In general, the solutions
of a DE of order m will have m arbitrary parameters. You have
6 data points, so a fairly arbitrary DE of order 6 or more
might be able to fit those points; if it has order m < 6,
you'll need to fit 6-m arbitrary parameters in the DE.
For example, let's try a homogeneous linear DE of order 3
with 3 arbitrary parameters:
y''' + c1 y'' + c2 y' + c3 y = 0
If r1, r2, r3 are the roots of the polynomial r^3 + c1 r^2 + c2 r + c3,
the solutions of the DE are y = a1 exp(r1 x) + a2 exp(r2 x) + a3 exp(r3 x).
In Maple:
> S := unapply(add(a[i]*exp(r[i]*x),i=1..3),x);
Y := [50, 1, 30, 122, 0, 18];
eqs:= {seq(S(x) = Y[x], x=1..6)};
sols:= evalf([allvalues(solve(eqs))]);
It's somewhat surprising that "solve" works here...
There are 6 solutions, but they are actually equivalent (since you can
permute the three terms of the solution).
> S1:= eval(S(x),sols[1]);
S1 :=
20.35019430*exp(.2279070630*x)+(.675907739-10.07805992*I)*exp((.3228362982+2.007021168*I)*x)+(.675907741+10.07805991*I)*exp((.3228362984-2.007021168*I)*x)
Due to roundoff error there are numbers here that should
be equal but are not quite equal.
> S1:= op(1, S1) + 2*evalc(Re(op(2,S1)));
S1 := 20.35019430*exp(.2279070630*x)+1.351815478*exp(.3228362982*x)
*cos(2.007021168*x)+20.15611984*exp(.3228362982*x)*sin(2.007021168*x)
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Gib Bogle
Read a calculus text, then an introductory text on differential
equations. Otherwise you should be able to find plenty of good material
online, e.g. lecture notes from basic university courses. Before
"framing" a d.e., you have to know what a d.e. is.
Virgil
Gib Bogle <bo...@ihug.too.much.spam.co.nz> wrote:
> Problem: -
> for given data how to frame a Differential_Equation ?
> for example see this, totally random
>
> x = 1 2 3 4 5 6
> y = 50 1 30 122 0 18
>
> what's the procedure to do that ?
For this sort of data, a finite difference approach might be more
appropriate.
Nimo
<isr...@math.MyUniversitysInitials.ca> wrote:
> > First of all let me extend my thanks for all
> > who helped me in my previous threads
> > 1) polynomial into e^ix expansion
> > 2) I don't want a tangent, but e^ix
> > 3) Taylor's series, Lagrange polynomial, e^ix
> > 4) How to calculate e^ix
>
> > instead of bringing all those thread into main page
> > I'm appreciating their help here, once again thank you.
> > /*=-=-=-=-=-=-=-=-==-=-*/
> > Now coming to the
>
> > Problem: -
> > for given data how to frame a Differential_Equation ?
> > for example see this, totally random
>
> > x = 1 2 3 4 5 6
> > y = 50 1 30 122 0 18
>
> > what's the procedure to do that ?
" It's not at all clear what you mean, but I suppose you
want a differential equation that has a solution going
through those points."
yeah, that's right.
I want the equation for which will
give my y values for x,
I know there is Lagrange for that, but
can't I try something new ?
Actually what I'm saying is
lets see in brief :
At a particular point
to make linear or quadtratic or cubic
or any other higher degree for a function, there is
Taylor's expansion.
In the same way at that particular point
instead of going for linear or quadratic or any other
I want to go for e^ix
utilizing Taylor's power series formula
Have you catch my point ?
not Fourier's 2Pi peridic concepts
or integral tranforms
don't confuse this with that.
look this,for a function f(x) at x=a
I'm going for making that function
cubic at that point
T[f(x) = f(a) + f1(a)(x-a) + 1/2! f2(a)(x-a)^2 +1/3! f3(a)(x-a)^3
now I've gone for doing cubing
the function at that particular point
we can also go for doing linear ,
quadratic or any other depending up on our wish.
In the same way using Taylor's formula going for this 'e^ix'
again repeating not Fourier expansion, don't confuse.
I hope this time I clarified the confusion.
If there is any doubt,let me know that.
Thanks.
Nimo
----
any doubts here?
[Mr.] Lynn Kurtz
wrote:
Yes. I *doubt* anyone can make sense of your incoherent post.
--Lynn
Nimo
> On Fri, 9 Jan 2009 21:21:50 -0800 (PST), Nimo <azeez...@gmail.com>
so, where is your doubt ?
can you please share it,
I hope, I can clarify it.
[Mr.] Lynn Kurtz
wrote:
>On Jan 10, 11:06 pm, "[Mr.] Lynn Kurtz" <ku...@asu.edu.invalid> wrote:
No, I'm not even going to try. Why don't you just continue in the
"polynomial into e^ix expansion" thread that you started about this
subject and where there are several replies to which you haven't
responded? Like I told you before, if you don't acknowledge people's
replies, soon nobody will bother with your threads. It's already
starting to happen, isn't it?
--Lynn
Nimo
> On Sun, 11 Jan 2009 04:04:35 -0800 (PST), Nimo <azeez...@gmail.com>
apologized, I state everything in
'Polynomial into e^ix thread'
please discard this thread.
Thank you.