From help-octave-request at bevo dot che dot wisc dot edu Fri Dec 27 17:52:24 2002
Subject: Re: Newbie question solving lin sys
From: "William Lash" <zlash at mail dot notwires dot com>
To: <zlash at mail dot notwires dot com>, "John B. Thoo" <jb2@ms.yuba.cc.ca.us>
CC: <help-octave at bevo dot che dot wisc dot edu>
Date: Fri, 27 Dec 2002 17:51:45 -0600




---------- Original Message ----------------------------------
From: John B. Thoo <jb2 at ms dot yuba dot cc dot ca dot us>
Date:  Fri, 27 Dec 2002 14:56:39 -0800

>Bill---
>
>Yowzer!  Your suggestion worked perfectly.
>
>octave:80> for s = [0:11]
> > s.^[0:11] * x
> > end
>ans = 22
>ans = 28.000
>ans = 31.000
>ans = 39.000
>ans = 46.000
>ans = 53.000
>ans = 56.000
>ans = 55.000
>ans = 49.000
>ans = 42.000
>ans = 33.000
>ans = 27.000
>
>I didn't realize there would be a loss of precision.  (Not only am I new 
>to Octave, but I'm also new to numerics in general.)
>
>**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.
>
>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])
>
>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)

>
>
>Thanks for your patience.
>
>---John.
>
>
>On Friday, December 27, 2002, at 02:15 PM, William Lash wrote:
>
>> Whenever you type in the numbers for you calculation in s you are
>> losing precision.  Instead of typing in:
>>
>>> octave:55> 22 + 129.32*s - 330.01*s**2 + 356.74*s**3 - 211.74*s**4 +
>>> 77.512*s**5 - 18.446*s**6 + 2.9092*s**7 - 0.30191*s**8 + 
>>> 0.019818*s**9 -
>>> 0.0007457*s**10 + 0.00001225*s**11
>>
>> Try the following:
>>
>> x(1)+x(2)*s+x(3)*s**2+x(4)*s**3+x(5)*s**4+x(6)*s**5+x(7)*s**6+x(8)*s**7+x(
>> 9)*s**8+x(10)*s**9+x(11)*s**10+x(12)*s**11
>>
>> There are better ways to write this in octave, one way is:
>>
>> s.^[0:11] * x
>>
>> and you could check all the values with a for loop:
>>
>> for s = [0:11]
>> s.^[0:11] * x
>> end
>>
>>
>> Bill
>
>
 
                 



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