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Re: Newbie question solving lin sys
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From: |
John B. Thoo |
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Subject: |
Re: Newbie question solving lin sys |
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Date: |
Fri, 27 Dec 2002 16:55:47 -0800 |
On Friday, December 27, 2002, at 03:51 PM, William Lash wrote:
**Three follow-up questions. (I hope that you don't mind.)
1. What does the dot after the s in s.^[0:11] mean in the "for"
loop?
In Octave (and Matlab), usually things like multiplication and
exponentiation are matrix type operations. Putting a "." in front
of the operation makes it do things in an "element by element"
fashion. To be honest, I wasn't sure that s.^[0:11] would give me
a row vector with the elements s^0 through s^11 untill I tried
it.
I see. Neat.
2. What in the "for" loop tells Octave to add the terms?
The multiplication of the row vector containing the powers of s
and the column vector containing the values of x in matrix
arithmetic give you the sum of the products as above. Think of
it as S*X where S is 1x12, and X is 12x1. You could do the same
using:
sum(x'.*s.^[0:11])
Oh, right, Octave treats everything as vectors. (I've played a little
bit with Maple in the past; this is different here.) Duh!
3. How do I plot the graph on [0:11] without a loss of precision? I
tried this without success:
octave:84> gplot [0:11] "x(1) + x(2)*x + x(3)*x**2 + x(4)*x**3 +
x(5)*x**4 + x(6)*x**5 + x(7)*x**6 + x(8)*x**7 + x(9)*x**8 +
x(10)*x**9 +
x(11)*x**10 + x(12)*x**11"
line 0: undefined function: x
I would probably do something like:
for s = [0:11]
z(s+1) = s.^[0:11] * x;
end
plot([0:11],z)
That works to give me a polygonal curve that connects the 12 points.
That's good, but how would I plot the polynomial with more grid points
(smoother)? It works reasonably when I do "format long" and then copy &
paste the coefficients to plot
octave:139> gplot [0:11] "22 + 129.316052352359*x -
330.008191451222*x**2 + 356.743251568416*x**3 - 211.743557625161*x**4 +
77.5116682813202*x**5 - 18.4455900615644*x**6 + 2.90919225949428*x**7 -
0.301909718420775*x**8 + 0.0198178458639753*x**9 -
0.000745701048375786*x**10 + 0.0000122504808365439*x**11"
but there must be an easier way. :-)
Once again, thanks very much for all your help.
---John.
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