A polynomial of degree N (greater or equal to one) can be written as:
P(Z) = C(Z - Z1)(Z - Z2)(Z - Z3)…(Z - ZN)
Assume a polynomial has no repeated roots. Then the derivative of the polynomial taken at a root Z1 will be equal to what you get when you remove the (Z - Z1) from the polynomial and input Z = Z1.
Why is this?
To clarify the question:
Let P(Z) = Z^2 + 3Z + 2 = (Z + 1)(Z + 2)
P’(Z) = 2Z + 3
P’(-2) = -4 + 3 = -1
But if you remove the (Z + 2) from P(Z), you are left with just Z + 1
And Z1 + 1 = -2 + 1 = -1
Can anyone explain why this is? It seems like it should be obvious, but I’m just not seeing it.
You don’t actually need a polynomial. Take any function of the form
f(z) = (z - z[sub]0[/sub])g(z) ,
with g(z) arbitrary. Distributing the g(z) into the binomial factor gives
f(z) = z g(z) - z[sub]0[/sub] g(z) .
The derivative of f(z), then, can be written (product rule)
f’(z) = z g’(z) + g(z) - z[sub]0[/sub] g’(z) .
For z=z[sub]0[/sub], the first and third terms cancel, leaving:
f’(z[sub]0[/sub]) = g(z[sub]0[/sub]) .
More intuitively, perhaps, you can think of the problem in terms of Taylor series about (z-z[sub]0[/sub]). The function g(z) near z[sub]0[/sub] can be written as a constant term g(z[sub]0[/sub]) plus higher-order terms. The (z-z[sub]0[/sub]) piece is its own Taylor series. When you multiply the two, the coefficient of the resultant first (and lowest) order term is simply the above constant term: g(z[sub]0[/sub]). And, this term’s coefficient is the derivative at z[sub]0[/sub].
A final approach: The line (z-z[sub]0[/sub]) grabs ahold of g(z) at z[sub]0[/sub] and says, “I’m a line, so you’re not going to have any derivative less that a first derivative here, and your value here will set the size of that first derivative since I have unit slope.”
Okay, so perhaps anthropomorphizing a binomial factor is a little silly, but I did it anyway.
The other posters have this from the calculus viewpoint. From algebraic geometry, it’s technical, but basically saying that there are no repeated roots means that you’re intersecting the variety in the projective plane described by the polynomimal transversally by a (projective) line. The derivative is the jacobian of the projection map at the appropriate point, which… is probably more intricate than you want to know, so just read the other post.
As a nitpick, though, your first assertion is only true if you’re working over an algebraically closed field. Just try working over the real numbers and try writing x[sup]2[/sup]+1 in that form. The upshot is that technically the other posts don’t work in general, since there are technicalities about what it means to differentiate over other fields than the real numbers. So, either you have to change and say “assume a polynomial can be written like so with no repeated roots” or the answer needs to get a lot more complicated.
I may be getting myself in trouble here, but it seems we’re given that derivative, more or less. Given the usual assumptions that allow us to take the derivative of P, I think it still works, over the domain that it is valid. Pasta and my approach did not assume that there were no repeated roots, and I don’t think there are any problems with extending the approach.
For your example, P(x) = x[sup]2[/sup]+1, we can write it f(z) = z[sup]2[/sup]+1, or f(z) = (z + i)(z - i) = (z - i)g(z)
So, one of the roots is i, and g(z) would be (z + i)
f’(z) = 2z, so f’(i) = 2i, but that is the same as g(i) = (i + i)
With repeated roots, f’(z0) = 0, but that’s equal to g(z0) as well.
I just said to work over the real numbers. My nitpick wasn’t with the “no repeated roots”, but with “a polynomial can be written like this”.
To factor z[sup]2[/sup]+1 (as a polynomial in R) you need to adjoin a square root of -1, which moves you to C. In C, the notion of “derivative” is not nearly so simple as it is in R. In particular, you’re making the (highly non-canonical) choice of which square root of -1 is i and which is -i, which makes a (non-canonical distinction) between the holomorphic and anholomorphic components of the actual deRham differential.
In the end, yes you get the same answer, but the reasoning from earlier posts is all wrong in between if you’re not working over R.
Mathochist is correct, I should have mentioned that I was working over the field of complex numbers.
Given that, I think the methods in the other posts work.
The product rule works for functions of complex variables, and Pasta’s Taylor series approach is a nice solution. Thanks, by the way!
That was my nitpick though. Whether the root was repeated didn’t seem to make a difference. I’m not sure where you were going with “From algebraic geometry, it’s technical, but basically saying that there are no repeated roots means that you’re intersecting the variety in the projective plane described by the polynomimal transversally by a (projective) line.”
The previous posts just assumed that there was a root, called z0, or z1.
The algebraic geometry is a particularly elegant solution, and goes right to the heart of the fact that this is a polynomial. The fuzziness is accessible even without that.
Basically, when you talk about a function on the complex plane you no longer have a canonical vector field around into which to plug the function to “take the derivative”. You can derive in any direction you want. What is usually done is to pick (non-canonical choice!) a given square root of -1 and call it i. Then we can define
z = x + iy
{\bar z} = x - iy
and define the holomorphic and anholomorphic derivative operators
Holomorphic and anholomorphic functions are those for which the anholomorphic or holomorphic part (respectively) of the differential vanishes. Note that if we had chosen the other square root of -1 the holomorphic and anholomorphic functions are interchanged. (n.b.: An algebraic geometer would say that the (Spec R)-scheme Spec C has a nontrivial involution)
Now, on the complex plane with a choice of complex structure (square root of -1 and notion of (an)holomorphic functions), the analysis posted earlier applies to holomorphic functions and \frac{\partial}{\partial z}, as well as to anholomorphic functions and \frac{\partial}{\partial {\bar z}}
So, given a polynomial in R we extend our base field by tensoring with Cand choosing a square root of -1. Now the polynomial factors into linear terms, and non-real terms occur in conjugate pairs, so switching to the other choice of i leaves everything invariant and it now doesn’t matter which choice we make to generalize the polynomial, the variable, the base field, and the differential operatos (as long as we choose them consistently).
If we started with a polynomial in C[Z], we really do need to have in mind a choice of i. There is no canonical choice of “the derivative”, as showed up in the OP.
I was asking about the “From algebraic geometry, it’s technical, but basically saying that there are no repeated roots means that” part. The repeated roots don’t seem to affect the problem, as stated.
Polynomials in z, as in the OP, are holomorphic. I don’t see anything wrong with Pasta’s analysis then. Even the Taylor series part where he goes intuitive. I do draw the line at his anthropomorphizing.
For this fixed class of functions one choice of derivative makes them holomorphic and the other makes them all have derivative zero. It’s still a choice, even if only one is interesting. There’s still no canonical choice of “the derivative”.
And, please note that I didn’t say that there was anything incorrect with the other analyses, but just that they were incomplete and charging ahead with “the derivative” without justifying and clarifying what exactly was being used.
I’m not sure what your point is, as far as Pasta’s post, but I got to ask about this: are you saying that when you have a function of x, f(x), and you take the derivative with respect to x, there is no reason to choose that derivative over taking the derivative with respect to y? The derivative with respect to y is zero, sure…but so is the derivative with respect to w, and v, and u. Is that what you mean by “no canonical choice”?
The complex plane is a complex manifold of complex dimension 1. That is, it’s a manifold if real dimension 2 with a complex structure – a section of the endomorphism bundle of the tangent bundle with square -1. What the geometry defines for any smooth function from the plane to itself is a map from the tangent bundle to itself, the differential of the function.
The thing that gets glossed over is that there are two complex structures related by conjugation. Picking one decomposes the sheaf of smooth functions into the sheaves of holomorphic and anholomorphic functions with respect to that complex structure. “The derivative” is only specified once you pick one or the other complex structure and restrict to the apropriate sheaf of holomorphic functions.
That’s not a function, it’s a 1-form. Its two component functions (in the z-{\bar z} basis) are the holomorphic and anholomorphic derivatives. It’s just like the real derivative f’(x) is the (only) component of the 1-form f’(x)dx.
I grabbed a few books, and looked them over, and I really don’t understand your objections.
It would seem that you would disagree with this webpage, correct? If you do not disagree, please explain what assumptions were being made in the above posts that are not explicit there.
To the extent that I think the term “complex differentiable” is misleading, yes. The Cauchy-Riemann equations are exactly the condition that the anholomorphic derivative vanishes. This presupposes a choice of which direct summand of the sheaf of smooth complex-valued functions is holomorphic and which is anholomorphic, which is what I’m saying isn’t a canonical choice.
If we’ve picked a complex structure, then we have the following calculation of the total differential.
Now the Cauchy-Riemann equations are equivalent to the vanishing of the second term
df = \frac{\partial f}{\partial z}dz
So there’s only one function to specify the differential of a holomorphic function, and that function is misleadingly called “the derivative”, sweeping the fact that a non-canonical choice has been made all under the rug.