MATLAB Answers

Why is the polyval command giving two different answers?

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Baldemy
Baldemy on 22 Sep 2014
Commented: Matt J on 23 Sep 2014
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

Accepted Answer

Alberto
Alberto on 22 Sep 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 Comment
Matt J
Matt J on 23 Sep 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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More Answers (1)

Matt J
Matt J on 22 Sep 2014
Edited: Matt J on 22 Sep 2014
Because you have a typo in your call to polyval: a period appears where a comma should be.
  2 Comments
Matt J
Matt J on 22 Sep 2014
And what don't you like about it? The result is quite close to zero, as one would expect.

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