I got one real and 2 imaginary roots. But by Descartes' Rule of signs, there are three sign changes in this equation and we are suppose to have 3 real roots. However, MATLAB gives one real and two imaginary roots. How can it be possible or how can it be derived from Descartes' rule?
Actually, Descarte's Rule of signs gives you the number of positive and negative roots.
So, if you have a function with three sign changes you may have either three positive roots OR (and that is also important) one positive root. This is also a possibility of Descartes: You can have also a number of positive roots which is your number of sign changes minus an even natural number, so in your example: 3 - 2 = 1
You can go now one step further and try to find negative roots by making f(x) --> f(-x) and reapply descartes rule and you get your number of negative roots.
If you know the number of negative and positive roots, you can check for the number of complex roots with the following method:
you know that an n-th degree polynomial (yours has n = 3) must have n roots. Thus you obtain the number of complex roots by taking the total number of roots and substract the positive and negative ones (but check before if there is no root at x = 0)!
So, n - (p+q) = number of complex roots
Here p = number of positive roots and q = number of negative ones.
For your function that means:
number of complex roots = 3 - (1 + 0) = 2 complex roots (as there are no negative ones and there is none at 0.
And this is in agreement with your solution from matlab.
Extremum points of the function are 1-1/3^(1/2) and 1+1/3^(1/2). 1-1/3^(1/2) is local maximum point, other local minimum point and f(1-1/3^(1/2))
Actually, Descarte's Rule of signs gives you the number of positive and negative roots.
So, if you have a function with three sign changes you may have either three positive roots OR (and that is also important) one positive root. This is also a possibility of Descartes: You can have also a number of positive roots which is your number of sign changes minus an even natural number, so in your example: 3 - 2 = 1
You can go now one step further and try to find negative roots by making f(x) --> f(-x) and reapply descartes rule and you get your number of negative roots.
If you know the number of negative and positive roots, you can check for the number of complex roots with the following method:
you know that an n-th degree polynomial (yours has n = 3) must have n roots. Thus you obtain the number of complex roots by taking the total number of roots and substract the positive and negative ones (but check before if there is no root at x = 0)!
So, n - (p+q) = number of complex roots
Here p = number of positive roots and q = number of negative ones.
For your function that means:
number of complex roots = 3 - (1 + 0) = 2 complex roots (as there are no negative ones and there is none at 0.
And this is in agreement with your solution from matlab.
Thank you very much Johannes Gruenwald and Kemal Aydin for your suggestions on this question.
No, Descartes' rule says that the number of positive real roos equals the number of changes in sign or less thatn that by 2 or 4 or .... Therefore, since there are 3 sign changes then you should get 3 or 1 real root(s).
A. Umar
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