Polynomial into e^ix expansion ?
Nimo
f(x) be any polynomial,
I can go for Taylor's polynomial expansion
for that function at x values
say x = p,q,r , I can get Taylor's Polynomials
Is there any method or procedure to
express the same function in polynomial using
e^ix instead of using only x,
why I'm asking this particular one is
for higher powers of x its very hard to compute
where as if we use e^ix it will eventually
become e^inx for big 'n' values.
/*Question in a Single Line*/
How to go for a polynomial expansion for a function say F(x)
at given point using e^ix instead of using only x.?
Have you catch my_point.?
waiting..
galathaea
http://en.wikipedia.org/wiki/Fourier_series
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
galathaea: prankster, fablist, magician, liar
Robert Israel
> Let
>
> f(x) be any polynomial,
> I can go for Taylor's polynomial expansion
> for that function at x values
> say x = p,q,r , I can get Taylor's Polynomials
>
> Is there any method or procedure to
> express the same function in polynomial using
> e^ix instead of using only x,
> why I'm asking this particular one is
> for higher powers of x its very hard to compute
> where as if we use e^ix it will eventually
> become e^inx for big 'n' values.
Huh? What will become e^inx? What does whether
n is big or small have to do with it? And what's
hard to compute about powers of x?
> /*Question in a Single Line*/
>
> How to go for a polynomial expansion for a function say F(x)
> at given point using e^ix instead of using only x.?
Look up "Fourier series".
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Nimo
<isr...@math.MyUniversitysInitials.ca> wrote:
-----------
look if this is the problem,
Let
T(p), T(q) & T(r) be the Taylor's polynomials
for the function f(x) at x = p,q,r
I don't have the function f(x)
but data points as this
x = 1 2 3 4
y = 2 5 8 7
T[f(x)] at x = 1
f(1) + 1/1! df(1)/dx (x-1)^1 + 1/2! d^2f(1)/dx^2 x-1)^2----(1)
here I got a DDE.
here if you observe that I used x, but I'm thinking
other, let's see
x = c ===> (x-c)===> (x-1)
this is what we write for expansion
but, can't I write like this ?
e^ix =e^ic------(2)
coming to Taylor Polynomial expansion
how to use eq(2) in eq(1)
and more over how to solve this
Delay Differential equation(DDE)?
Is there any flaw in my method ?
Ray Vickson
There is exactly one polynomial y = f(x) that matches the data (the so-
called Lagrange polynomial---see http://en.wikipedia.org/wiki/Lagrange_polynomial
), but there are infinitely many non-polynomials that are an exact fit
to the data. No amount of wishful thinking or arm-waving or
incomprehensible lunatic raving will ever change that fact. You just
are not thinking, and worse still, you never seem to learn and never
seem to want to learn.
R.G. Vickson
Matt
> ), but there are infinitely many non-polynomials that are an exact fit
> to the data.
There are infinitely many polynomials too, surely? (Only one of lowest
order, but infinitely many of higher order.) Or am I misunderstanding
something?
Ray Vickson
No, you are right. I should not reply to anything before having my
morning coffee.
RGV
mike3
> ), but there are infinitely many non-polynomials that are an exact fit
> to the data. No amount of wishful thinking or arm-waving or
> incomprehensible lunatic raving will ever change that fact. You just
> are not thinking, and worse still, you never seem to learn and never
> seem to want to learn.
>
It seems to me that what he wants to find is an expansion for f that
looks
like this:
f(x) = a_0 + a_1 (e^(ix)) + a_2 (e^(ix))^2 + a_3 (e^(ix))^3 + ...
i.e. a "polynomial" with "e^ix" replacing x. That is a Fourier series,
and I
don't think it exists for non-periodic functions like he wants (i.e.
if f is a
polynomial function.).
Matt
But it could be an approximation on the interval spanning the given
data points, which is possibly all he wants.
mike3
That could be. One can pass an e^(ix) expansion through a given
set of data points.
Consider 2 points, (x_1, y_1) and (x_2, y_2). We have
a_0 + a_1 e^(ix_1) = y_1
a_0 + a_1 e^(ix_2) = y_2
Then this is just linear equations for a_0 and a_1, giving
a_0 = y_1 - (e^(ix_1) (y2 - y1))/(e^(ix_2) - e^(ix_1))
a_1 = (y_2 - y_1)/(e^(ix_2) - e^(ix_1)).
Then the interpolation is f(x) = a_0 + a_1 e^(ix).
Nimo
=-=-=-=-=-=-=-
apologized for late reply.
please discard
http://groups.google.com/group/sci.math/browse_thread/thread/f8a1ddbcd45e0578?hl=en#
"how to form a differential equation thread"
Thank you.
we will continue here.
-=-=-=-=-=-=-=-
Generally we will write a polynomial
by taking x
P = a0 + a1x + a2x^2 + a3x^3 + a4x^4+............+ an x^n__(1)
here we can form a polynomial depending up on our wish
may be fitting roots like this (x-a)(x-b)(x-c).....(x-n)
or an interpolation polynomial like 'Lagrangian' form
but the central part of everything is 'x'
"Whole_my doubt is instead of using 'x'.
I have to have use 'w'"
Let,
omega = w = e^ix = cisx.
In eq(1) instead of going for x, I need to go for omega
P = a0 + a1 (w) + a2 (w)^2 + a3 (w)^3 +......+...
here whether we can go for fitting roots
(w-a)(w-b)(w-c).........or
going for interpolation polynomial like Lagrangian
form, what ever it is,
/**/ Which ever polynomial I deal, replacing 'x'
with omega(w) that's exactly my problem /**/
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=--=-=-=-=-
In the course of asking the above, I asked all other stuff,
I'm doing paper work, when ever I got a
doubt I posted that thread. Other than
that I don't have any reasons for asking
several question about e^ix.
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
Thanks for those who helped
in previous posts.
Matt
> "how to form a differential equation thread"
> Thank you.
> we will continue here.
> -=-=-=-=-=-=-=-
>
> Generally we will write a polynomial
> by taking x
>
> P = a0 + a1x + a2x^2 + a3x^3 + a4x^4+............+ an x^n__(1)
>
> here we can form a polynomial depending up on our wish
> may be fitting roots like this (x-a)(x-b)(x-c).....(x-n)
> or an interpolation polynomial like 'Lagrangian' form
> but the central part of everything is 'x'
> "Whole_my doubt is instead of using 'x'.
> I have to have use 'w'"
>
> Let,
> omega = w = e^ix = cisx.
> In eq(1) instead of going for x, I need to go for omega
>
> P = a0 + a1 (w) + a2 (w)^2 + a3 (w)^3 +......+...
>
> here whether we can go for fitting roots
> (w-a)(w-b)(w-c).........or
> going for interpolation polynomial like Lagrangian
> form, what ever it is,
>
> /**/ Which ever polynomial I deal, replacing 'x'
> with omega(w) that's exactly my problem /**/
If your known data points are (x_1,y_1), (x_2,y_2), ... (x_n,y_n) then
you can set up simultaneous equations like this:
a_0 + a_1*e^(i*x_1) + a_2*e^(2*i*x_1) + ... + a_{n-1}*e^((n - 1)
*i*x_1) = y_1
a_0 + a_1*e^(i*x_2) + a_2*e^(2*i*x_2) + ... + a_{n-1}*e^((n - 1)
*i*x_2) = y_2
...
and solve for the a's in the usual way (here I'm writing e^(n*i*x) for
(e^(i*x))^n). The resulting function
y = a_0 + a_1*e^(i*x) + a_2*e^(2*i*x) + ... + a_{n-1}*e^((n - 1)
*i*x)
will pass through the data points, but there's no reason to expect it
will do exactly what you want between them, and in general it won't be
real-valued for real x other than x_1, x_2, ... x_n.
Alternatively -- amounting to exactly the same thing but not
explicitly giving you the values of the a's -- you can use Lagrange-
style interpolation except that everywhere you have an x or an x_i in
the Lagrange interpolation formula you use e^(i*x) and e^(i*x_i)
instead.
You realise that if instead you had
y = ... + b_2*e^(-2*i*x) + b_1*e^(-i*x) + a0 + a_1*e^(i*x) + a_2*e^
(2*i*x) ...
then you'd have a standard fourier series? Then you could find the
coefficients to approximate to a known function using standard
methods. I'm not sure if there's a way to adapt these methods to give
you a series without the negative powers. Maybe someone else will.
Matt
> You realise that if instead you had
>
> y = ... + b_2*e^(-2*i*x) + b_1*e^(-i*x) + a0 + a_1*e^(i*x) + a_2*e^
> (2*i*x) ...
>
I should have written it more tidily as
y = ... + a_{-2}*e^(-2*i*x) + a_{-1}*e^(-i*x) + a_0 + a_1*e^(i*x) +
a_2*e^(2*i*x) ...
Nimo
-----
especially @ matt
Thanks for the help
problem is OKAY.
and thanks to all those
who replied this thread.