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’.
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.
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!
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.
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.
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.
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) = (e[sup]ix[/sup] - e[sup]-ix[/sup])/2i = [(1 + ix/N)[sup]N[/sup] - (1 - ix/N)[sup]N[/sup]]/2i for infinitely large N, so that a factorization of sin(x) can be extracted from a factorization of the polynomial [(1 + ix/N)[sup]N[/sup] - (1 - ix/N)[sup]N[/sup]]/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)[sup]N[/sup] - (1 - ix/N)[sup]N[/sup]]/2i by f[sub]N/sub. Note that f[sub]N[/sub] 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[sub]N/sub/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[sub]N/sub/x = the product of 1 - (x/[N tan(kπ/N)])[sup]2[/sup] 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.]
Except for “sin(x) = (e[sup]ix[/sup] - e[sup]-ix[/sup])/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.
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.
