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#1




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! 
#2




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.

#3




Aren't y = 2 x^2 and y = 3 x^2 two different polynomials with the same roots and same values at zero?

#4




Quote:
Quote:
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; 04192014 at 05:17 PM. 


#5




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. 
#6




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/∏)(1x/2∏)(1+x/2∏) . . . and I wasn't sure such an equivalence was kosher.
Last edited by KarlGauss; 04192014 at 05:27 PM. 
#7




Quote:
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 nowherevanishing function f. 
#8




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.

#9




Quote:
Consider, for example, a polynomial f(x) with roots at 1, 2, and 3, such that f(0)=6. Then f(x)=(x+1)(x2)(x3) f(x)=(x+1)^2(x2)(x3) both satisfy these conditions but are manifestly not the same polynomial. The root x=1 appears with different multiplicity in the two polynomials. 


#10




And we also have to make sure we're talking about complex roots, not just real ones.

#11




Quote:
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) = (e^{ix}  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 f_{N}(x). Note that f_{N} 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 f_{N}(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 f_{N}(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 kth factor to be 1 when N/2 <= k), which is sufficient justification for commuting the limits.] Last edited by Indistinguishable; 04202014 at 02:43 PM. 
#12




There will be those who excoriate me for writing "/2i" when they would prefer "/(2i)"...
Last edited by Indistinguishable; 04202014 at 03:15 PM. Reason: Wasn't "/[kπ]" barbaric enough? 
#13




Quote:
Although, I'm pretty sure the statements in that post all remain true in either interpretation, as long as you're consistent about it. 
#14




Except for "sin(x) = (e^{ix}  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.



#15




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 nonroot). And of course, you must include all complex roots.

#16




Oh, I ended up phrasing things such that this restriction is unnecessary.
Last edited by Indistinguishable; 04202014 at 08:01 PM. 
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