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computer the Gram-Schmidt orthogonal of (22,11,5),(13,6,3),(-5,-2,-1) belong to R^3.

I'm taking calculus and my professor introduced us to maple software. The professor asked us to plot the families of curves for this orthogonal equation:

dy/dx = (x^2) - (2y^2) - C = 0

This is what I had so far:

 

restart;

with(DEtools):

with(plots):

 

Function:=unapply(simplify(x^2-2y^2-C),(x,y)):

'Function'(x,y) = Function(x,y);

plotFunction:=C->implicitplot(eval(F(x,y),a=C),x=-5..5,y=5..5,scaling=constrained):

plot1:=display(seq(plotFunction(a),a=-5..5)):

display(plot1);

 

This is only display one family. How do I code for it plot the other families?

(The graph should look like curves converging from left, top and right sides toward to the origin of the axes)

Please help.

So I am working with the OrthgonalExpansions library (amazing library, by the way! Just what I needed!), and it is great, except it seems to be having trouble doing a 2d Fourier expansion.

I'll upload my document. transformApprox.mw

In this document, I am running the BesselSeries expansion as we speak, but the FourierSeries expansion (in 2d) can never seem to complete. It always say that it has an

Chebyshev polynomials have the form:

 

Tn(x) = cos(n*arccos(x))

 

orthogonality interval is (-1, 1) for n = (2, 3, 4,

1. I use your previous reply on V_G derive  on other characteristic function, but most are undefined

or complicated solution or can not evaluate, why?

for example

charc := 1+i*X/(i*X-1);
int(exp(-I*X*u)*charc, X = -infinity .. infinity);

it got this complicated thing
piecewise(Im(1/i) = 0, undefined, int(exp(-I*X*u)*(1+i*X/(i*X-1)), X = -infinity .. infinity, method = _UNEVAL))

Density := int(exp(-I*X*u)*charc, X = -infinity .. infinity);

Hi guys, I would like some help writing a procedure that checks whether the group of input vectors is an orthogonal basis. Any help would be great!

Sorry, this seems like a silly question. But is there an easy way to convert trig funcitons, or even non trig functions to orthogonal (in this case Legendre) polynomials? 

How can I show that the Least Square solution x = (A'.A)^-1.A'.y
Is different when A is an orthogonal matrix compared to an
overdetermined or underdetermind matrix.

Preferably transform the matrix using Singular Value Decomposition (SVD)
or something similar.

Thanx


Help!

I've used Eigenvectors to solve for eigenvalues & eigenvectors.  Eigenvalues works, no problem.

The eigenvectors are not normalized to unit magnitude (how would I do that for all eigenvectors?) and the usual matrix multiplication of the eigenmatrix by its transpose should give the identity matrix--and somehow it does not. 

Can someone point out the flaw in my thinking?  The file is attached.

 

Thanks for you insight! 

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