From help-octave-request at bevo dot che dot wisc dot edu Fri Dec 27 16:57:27 2002
Subject: Re: Newbie question solving lin sys
From: "John B. Thoo" <jb2 at ms dot yuba dot cc dot ca dot us>
To: <zlash at mail dot notwires dot com>
Cc: <help-octave at bevo dot che dot wisc dot edu>
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?

2. What in the "for" loop tells Octave to add the terms?

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


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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