Re: [math-fun] Quadratic equations --- still getting it wrong!
<< I have never seen this done in a textbook. But Bob Churchhouse in his Numerical Analysis course at Cardiff always used to discuss the behaviour of the coefficients as functions of the roots. I thought it a most useful exercise, and did not hesitate to appropriate it for my own course when the time came. WFL
Isn't this relationship -- for a monic polynomial P(x) = (x-r_1)...(x-r_n) = x^n + c_(n-1) x^(n-1) + ... + c_0 just given by the elementary symmetric functions: c_(n-k) = (-1)^k S_k(r_1,...,r_n) where S_k is the sum of all products of the r_j for k distinct indices j ??? But also I wonder how much light this would shed on numerical problems of calculating the reverse relationship: finding the roots from the coefficients. --Dan ________________________________________________________________________________________ "Outside of a dog, a book is man's best friend. Inside of a dog, it's too dark to read." --Groucho Marx
On 5/19/10, Dan Asimov <dasimov@earthlink.net> wrote:
<< I have never seen this done in a textbook. But Bob Churchhouse in his Numerical Analysis course at Cardiff always used to discuss the behaviour of the coefficients as functions of the roots. I thought it a most useful exercise, and did not hesitate to appropriate it for my own course when the time came. WFL
Isn't this relationship -- for a monic polynomial
P(x) = (x-r_1)...(x-r_n) = x^n + c_(n-1) x^(n-1) + ... + c_0
just given by the elementary symmetric functions:
c_(n-k) = (-1)^k S_k(r_1,...,r_n)
where S_k is the sum of all products of the r_j for k distinct indices j ???
Umm --- going to have to withdraw that gratuitously half-baked posting --- maybe go away to work out just exactly what it was that Bob did do --- well, it was 40-odd years back! WFL
But also I wonder how much light this would shed on numerical problems of calculating the reverse relationship: finding the roots from the coefficients.
--Dan
________________________________________________________________________________________ "Outside of a dog, a book is man's best friend. Inside of a dog, it's too dark to read." --Groucho Marx
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On 5/19/10, Fred lunnon <fred.lunnon@gmail.com> wrote:
On 5/19/10, Dan Asimov <dasimov@earthlink.net> wrote:
<< I have never seen this done in a textbook. But Bob Churchhouse in his Numerical Analysis course at Cardiff always used to discuss the behaviour of the coefficients as functions of the roots. I thought it a most useful exercise, and did not hesitate to appropriate it for my own course when the time came. WFL
... Umm --- going to have to withdraw that gratuitously half-baked posting --- maybe go away to work out just exactly what it was that Bob did do --- well, it was 40-odd years back! WFL
But also I wonder how much light this would shed on numerical problems of calculating the reverse relationship: finding the roots from the coefficients.
--Dan
Yes, I'd got it the wrong way about. Let the polynomial equation be f(x) == a_0 + a_1 x + ... + a_n x^n = 0, where the degree n is actually irrelevant. Then the differential of root x with respect to coefficient a_k is given by d x / d a_k = - x^k / f'(x), showing that a multiple root is inevitably ill-conditioned. Which of course is obvious from the picture. WFL
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Dan Asimov -
Fred lunnon