Why is the polyval command giving two different answers?
显示 更早的评论
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
采纳的回答
Alberto
2014-9-22
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 个评论
Matt J
2014-9-23
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
更多回答（1 个）
Because you have a typo in your call to polyval: a period appears where a comma should be.
2 个评论
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