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- Thread starter olast1
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- #2

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Take a look at what its Taylor series does to its coefficients.

- #3

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Thank you for your answer, but I am not sure I understand what you mean. Can you explain?

- #4

Hurkyl

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In what class was this problem given?

- #5

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Master Level.

- #6

Hurkyl

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"Master level"?

You mean something like, say, a graduate algebra course?

You mean something like, say, a graduate algebra course?

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- #8

Hurkyl

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So, it wouldn't be fruitful to suggest trying to turn the problem into a zero-finding problem if Sturm's theorem wasn't something you'd be expected to use!

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- #10

Hurkyl

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- #11

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I do not know if it helps or overlap with what you are saying but here is what I tried to do at this point: if P(x) is a polynomial of degree n, then its derivative P'(x) is a polynomial of degree n-1. Therefore, I have tried to parametrize a polynomial of degre n-1 to guarantee that it is strictly greater than 0 for any x between [0,1]. The parametrization I found is

P'(x)=prod(i=1,...n-1){x-1/(1-Bi)} with Bi>0 for any i=1,...n-2 and Bn-1=exp[b*prod(i=1,...n-2){1-Bi}] and b>0 which I believe guarantees that P'(x)>0.

Now however, I am having problems relating the coefficients of P'(x) to the parameters of p(x).

Thank you for your time. I truly appreciate your help.

- #12

Hurkyl

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I am having problems relating the coefficients of P'(x) to the parameters of p(x).

D'oh, that should be the easy part! If

[tex]p(x) = \sum_{i = 0}^{n} a_i x^i[/tex]

Then you should be able to directly take a derivative, to get a formula for the coefficients of p'(x) in terms of that of p(x).

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- #14

Hurkyl

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Well, what's the derivative of a_{i} x^i?

(and don't forget about the constant terms...)

(and don't forget about the constant terms...)

- #15

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P'(x)=sum(i=1,...,n-1){Ci*x^i}

Then obviously I can easily relate the Ci to your Ai.

- #16

Hurkyl

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I guess I was still thinking about how the Sturm's theorem approach would work, since that uses the coefficients of the polynomial directly. (Maybe I'm thinking about something related to Sturm's theorem than Sturm's theorem itself -- I can never keep them all straight, but that keyword is enough for me to find it in my reference materials!)

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