Why is the polyval command giving two different answers?
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Why does the polyval operator not work as expected. Is the ans variable not stored as a column vector? Why aren't the second, fifth, and sixth results equal?
>> roots([1,-8,17,2,-24])
ans =
4.0000
3.0000
2.0000
-1.0000
>> polyval([1.-8,17,2,-24],ans)
ans =
-192.0000
-54.0000
-8.0000
-2.0000
>> roots([1,-8,17,2,-24])
ans =
4.0000
3.0000
2.0000
-1.0000
>> x=ans
x =
4.0000
3.0000
2.0000
-1.0000
>> polyval([1,-8,17,2,-24],x)
ans =
1.0e-13 *
0.8882
0.3197
0.0355
0.1421
>> polyval([1,-8,17,2,-24],[2.0000;3.0000;-1.0000;3])
ans =
0
0
0
0
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Risposta accettata
Alberto
il 22 Set 2014
Instruction roots uses an iterative numeric method to approximate the solution in float arithmetic. What you get is an excellent approximation.
If you need the exact solution you should try a symbolic method:
g = x^4-8*x^3 + 17*x^2 +2*x -24
g =
x^4 - 8*x^3 + 17*x^2 + 2*x - 24
>> sol=solve(g==0)
sol =
2
3
4
-1
1 Commento
Matt J
il 23 Set 2014
You also may need a symbolic version of polyval, even when you have the exact roots:
>> polyval([1,-8,17,2,-24]/3,[4 3 2 -1])
ans =
1.0e-14 *
0.8882 0.1776 0.1776 0.1776
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