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Old 04-19-2014, 04:30 PM
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Do the linear roots of a polynomial, and its value at zero, uniquely define that polynomial?


If P(x) is a polynomial of degree n and has roots a, b, c, . . . , d and P(0) = 1, then I see that if we create F(x) = (1 - x/a)(1 - x/b)(1 - x/c) . . . (1 - x/d), then F(x) will have a, b, c, . . . . , and d as roots, and that F(0) = 1.

What I'm not confident about is that it follows that F(x) equals P(x). Having specified its roots as above, I see that it is necessary that P(x) can be expressed as the product of the (1 - x/roots). But is expressing it that way sufficient to completely define P(x)?

It seems to me that it does completely define it but I don't know how to prove my 'intuition'.

Thanks!
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Old 04-19-2014, 04:50 PM
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Thinking graphically, it's trivial to draw a smooth curve in the real plane of any degree N with N real roots. Stretching that curve but keeping the roots the same would mean there are an infinity of polynomials with the same roots.
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Old 04-19-2014, 04:54 PM
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Aren't y = 2 x^2 and y = 3 x^2 two different polynomials with the same roots and same values at zero?
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Old 04-19-2014, 05:13 PM
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Quote:
Originally Posted by Napier View Post
Aren't y = 2 x^2 and y = 3 x^2 two different polynomials with the same roots and same values at zero?
Quote:
Originally Posted by friedo
Thinking graphically . . ."
Yes. I think I should have added "except for a constant multiplier". Likely there's more I took for granted.

ETA: I guess all that's specified is a 'family' of polynomials. Nothing unique. I will attempt to explain later where my question came from, and why I stupidly thought the polynomials were identical. No time just now. Thank you!

Last edited by KarlGauss; 04-19-2014 at 05:17 PM.
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Old 04-19-2014, 05:14 PM
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Polynomial interpolation requires values n+1 distinct points for a degree n (or less) polynomial.

Napier's example doubles down on your problem. The roots themselves are not distinct and the P(0) case is also at the roots.
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Old 04-19-2014, 05:26 PM
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My motivation: I was reading about Euler's solution to the Basel Problem (sum of the inverse squares) and it was explained how he equated the Taylor series for sin(x)/x with the polynomial given by (1 - x/∏)(1+x/∏)(1-x/2∏)(1+x/2∏) . . . and I wasn't sure such an equivalence was kosher.

Last edited by KarlGauss; 04-19-2014 at 05:27 PM.
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Old 04-19-2014, 06:15 PM
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Originally Posted by KarlGauss View Post
If P(x) is a polynomial of degree n and has roots a, b, c, . . . , d and P(0) = 1, then I see that if we create F(x) = (1 - x/a)(1 - x/b)(1 - x/c) . . . (1 - x/d), then F(x) will have a, b, c, . . . . , and d as roots, and that F(0) = 1.

What I'm not confident about is that it follows that F(x) equals P(x). Having specified its roots as above, I see that it is necessary that P(x) can be expressed as the product of the (1 - x/roots). But is expressing it that way sufficient to completely define P(x)?

It seems to me that it does completely define it but I don't know how to prove my 'intuition'.

Thanks!
A nonzero polynomial f(X) of degree n (over an arbitrary commutative ring) has at most n roots, since f(z) = 0 iff (X - z) divides f. Apply that result to F - P.

The technique of Euler you mentioned isn't kosher (as it stands, anyway; there are ways of making that sort of thing rigorous). For one thing, f(x) sin/x has the same set of zeros for any nowhere-vanishing function f.
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Old 04-19-2014, 09:31 PM
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A polynomial is uniquely determined by its roots and its value at any single point which is not a root. This is pretty easy to show and I'll leave it as an exercise for the reader.
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Old 04-20-2014, 02:43 AM
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Originally Posted by ultrafilter View Post
A polynomial is uniquely determined by its roots and its value at any single point which is not a root. This is pretty easy to show and I'll leave it as an exercise for the reader.
I feel the need to state explicitly what you already know that when you say "...determined by its roots..." you are including the multiplicity of said roots. Yes, I know you know this and that was implied in your statement. I'm making this explicit for the reader doing your exercise.

Consider, for example, a polynomial f(x) with roots at -1, 2, and 3, such that f(0)=6. Then

f(x)=(x+1)(x-2)(x-3)
f(x)=(x+1)^2(x-2)(x-3)

both satisfy these conditions but are manifestly not the same polynomial. The root x=-1 appears with different multiplicity in the two polynomials.
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Old 04-20-2014, 09:48 AM
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And we also have to make sure we're talking about complex roots, not just real ones.
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Old 04-20-2014, 02:41 PM
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Quote:
Originally Posted by KarlGauss View Post
My motivation: I was reading about Euler's solution to the Basel Problem (sum of the inverse squares) and it was explained how he equated the Taylor series for sin(x)/x with the polynomial given by (1 - x/∏)(1+x/∏)(1-x/2∏)(1+x/2∏) . . . and I wasn't sure such an equivalence was kosher.
Euler is often said to have glibly moved from "sin(x)/x has roots at nonzero integer multiples of π" to "sin(x)/x = the product of (1 - x/(kπ)) for each nonzero integer k" in solving the Basel problem, simply ignoring the problem of the fact that many other functions have precisely the same zeros.

But Euler was not, in fact, always so glib in presenting this argument! For example, in Volume 1, Chapter 9 of his Introductio in Analysin Infinitorum (translation by Ian Bruce available here), we see that Euler argues for the product formula by noting that sin(x) = (eix - e-ix)/2i = [(1 + ix/N)N - (1 - ix/N)N]/2i for infinitely large N, so that a factorization of sin(x) can be extracted from a factorization of the polynomial [(1 + ix/N)N - (1 - ix/N)N]/2i.

I'll write out in modern style the extraction of that factorization, but all the ideas are already present in the Euler. Let us denote [(1 + ix/N)N - (1 - ix/N)N]/2i by fN(x). Note that fN is a polynomial of degree either N - 1 (if N is even) or N (if N is odd), whose degree 1 term is 1. Furthermore, its roots occur where x/N is the tangent of a multiple of π/N. Putting these together and restricting attention to odd N, we get that fN(x)/x = the product of 1 - x/[N tan(kπ/N)] over the nonzero integers k in (-N/2, N/2).

As N approaches infinity, N tan(kπ/N) approaches kπ. Thus, we have that sin(x)/x = the product of 1 - x/[kπ] over the nonzero integers k.

[This last step is slightly glib, in that we've commuted limits without justification. We can rectify that by bundling together the factors where k differs only in sign, saying fN(x)/x = the product of 1 - (x/[N tan(kπ/N)])2 over k in (0, N/2). Now we note that the movement of the factors toward their limit is monotonic in N (considering the k-th factor to be 1 when N/2 <= k), which is sufficient justification for commuting the limits.]

Last edited by Indistinguishable; 04-20-2014 at 02:43 PM.
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Old 04-20-2014, 03:11 PM
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There will be those who excoriate me for writing "/2i" when they would prefer "/(2i)"...

Last edited by Indistinguishable; 04-20-2014 at 03:15 PM. Reason: Wasn't "/[kπ]" barbaric enough?
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Old 04-20-2014, 03:20 PM
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Quote:
There will be those who excoriate me for writing "/2i" when they would prefer "/(2i)"...
Huh, I didn't know that any of our members were TI-83 calculators.

Although, I'm pretty sure the statements in that post all remain true in either interpretation, as long as you're consistent about it.
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Old 04-20-2014, 05:12 PM
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Although, I'm pretty sure the statements in that post all remain true in either interpretation, as long as you're consistent about it.
Except for "sin(x) = (eix - e-ix)/2i" (rather, they will be each others negations on the schoolmarmish interpretation) and the corresponding problem with "...whose degree 1 term is 1" (rather, it will be -1). But, yes, it's pleasing to note that this error is rather more negligible with i than with most factors.
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Old 04-20-2014, 05:14 PM
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Although I am not adding to anything said above, let me summarize by saying that two polynomials of the same degree n with n+1 points in common are the same. If you want these points to be the roots, you need one more and if 0 is a root, having the same value at 0 won't give a new point. And if one or more is a double root, you won't have n+1 points, although in that case you have a root of the derivatives and beyond that it gets a bit complicated. But if they have the same roots with the same multiplicities, then you do need one more point (so that must be a non-root). And of course, you must include all complex roots.
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Old 04-20-2014, 08:00 PM
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...restricting attention to odd n...
Oh, I ended up phrasing things such that this restriction is unnecessary.

Last edited by Indistinguishable; 04-20-2014 at 08:01 PM.
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