Rekna

If anyone can give me some hints beyond the ones I already have i'd be greatful.

Notation:
int--> stands for integreal

Here is the problem

Let P be a real polynomial of degree n such that:

int[a,b] P(x)*X^k dx = 0 where k=0,...,n-1.

a) Show that P has n real zeros (simple) in (a,b).

Hint: Show that p must have at least one zero in [a,b], and that assumption tha tp has less than n zeros yields a contradiction; (consider int[a,b] P(x)Qn-1(x) dx, where Qn-1(x)=(x-x1)...(x-xn-1) and Xi i=1,..,n-1 are the zeros of P in (a,b).

b) Show that the n-point interpolating rule on [a,b] based on zeros of p has degree 2n-1.

Hint: Consider the quotient and remainder polynomials whena given polynomial of degree 2n-1 is divided by p.


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Showing a zero exists in P is easy. Take the case where k=0 then we have the int[a,b] P(x) = 0. This means that P(x) must change signs at least once and hence has a zero.

But this is as far as I can get. I think my problem is I have very little background in this subject matter especially when it gets into the orthogonal inner products occuring when we multiply by X^k. Any suggestions of would be great.

Rekna

Found a link to the question. It is question 2 in this pdf.

http://www.cs.utah.edu/classes/cs521...ontent/hw5.pdf

Karkaboosh1

i didnt understand most of the problem... only at calc1 atm.. what cource level is that? 2? meh.

well for "Hint: Show that p must have at least one zero in [a,b]"
try using fund. theorem of calc. where:

y' for y = int[v(x),u(x)] f(t) dt = f(v(x)) v'(x) - f(u(x) u'x

you get f(t), which would be the function. find the roots/zeroes of the function to prove that a zero does exist, and find teh derivative of that to prove taht the function is constantly increasing or decreasing to show taht only 1 zero exists. im preaty sure there is something wrong with my logic here since the problem is of hgiher level than what i am doing now.

Stompy

Ugh, I *HATE* PDFs. Hate hate hate The damn thing, being as small as it is, locked my machine up for a minute.. [/anti-PDF rant]

Anyway..

I think that's beyond Calc 2. I just took that final today and I have a pretty high grade in the class... and the first post made almost absolutely no sense to me... (unless it was worded improperly)

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stonegrody

Why the hell did I click on this? I don't know anything about math...

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hilbert25

Ah.... was going to say seems like algebra crap, but just looked at the pdf... now that integral there changes everything. That looks a heck of a lot more like multi-v which I've tried hard to kill all the brain cells for remembering it with alchohol.

FloydianOne

Pretty sure its a "Constructing Proofs" class... Im a CS major and i had to take a course CS1050 which was very similar to that class. Im pretty sure I can do that problem, but its 4am and xmas so ill look at it later...

Rekna

It is a masters level advanced scientific computing course. I ended up figuring it out thanks to a book that had it in it. The first part is actually pretty easy.

Since we know that the integral of P(x)*X^k is equal to zero for k=0..n-1 we know that there is at least 1 zero, look at the case where k=0. We then have the integral of P(x) is equal to zero (thus either P(x) is the zero polynomial or it changes sign at least once).

Now assume P(x) has r < n-1 zeros. And let Q(x) be the interpolating polynomial of P(x) on those zeros. Then P(x)*Q(x) shares zeros with P(x). Thus if they share signs then the integral of P(x)*Q(x)>0 or if they have opposite signs the integral of P(x)*Q(x)<0. But if you expand Q(x) out you get a linear combination of X^k and thus the integral must be zero from the given statements above thus we have a contradiction and P(x) must have more than n-2 zeros.