If Re[s] > 1 then,
\zeta(s) = \sum_{k=1}^{\infty} k^{-s}
Can't we just differentiate term by term?
\frac{\partial}{\partial s} k^{-s} = -k^{-s} \log k
Then we have:
\frac{\partial}{\partial s} \zeta(s) = -\sum_{k=1}^{\infty} k^{-s} \log k
If Re[s] <= 1 then we could use...
Cauchy-Riemann in R^2
One last thing regarding Cauchy-Riemann in R^2\;.
We can write the Cauchy-Riemann condition as:
\mathbf{e}_1 \nabla f = 0
or left-multiplying both sides by \mathbf{e}_1 we get
\nabla f = 0
But we saw that f = \mathbf{e}_1 F\;.
So if f is analytic...
So what the heck is the complex derivative anyhow?
We've seen how we can map vectors in R^2 to spinors in C using some simple algebraic rules defined for multiplying vectors by vectors. By selecting a unit vector lying on the "real axis" we found a unique bijective map R^2 \longleftrightarrow...
Vectors, Spinors, and Cauchy-Riemann
For what follows, I shall use the following typeface convention:
scalars: \mathit{x}, \mathit{y}
vectors: \mathbf{u}, \mathbf{v}
spinors: \mathtt{z}, \mathtt{w}
All scalars here are assumed to be real numbers. \mathbf{e}_1 and \mathbf{e}_2 are...
Vectors And Spinors
A related topic I wanted to bring up was how vectors in R^2 are related to points on the complex plane and what I think the complex plane can be intuitively considered to represent geometrically. Note that this approach is just one of potentially many interpretations we...
Ok...so consider then:
x = A\, r [\cos \varphi]^{\frac{1}{n}}
y = B\, r [\sin \varphi]^{\frac{1}{n}}
from which we then get:
\frac{x^{2n}}{A^{2n}} +\frac{y^{2n}}{B^{2n}} = r^{2n}
Let \{\mathbf{e}_1, \mathbf{e}_2} \} form an orthonormal basis for the rectangular coordinate system and...
I think you think correctly...but...
I guess my focus was on avoiding the use of z^\ast entirely in the derivation of CR to illustrate what I feel is the essence of CR...linear independence, and hence an invertible map (x,y) \leftrightarrow (z, w)\;.
Here's an interesting way to look at CR I feel is often overlooked:
Let:
z = x + i y
z^{\ast} = x - i y
One common form for the CR condition is to say that if some function f is analytic then it does not depend on z^{\ast}\;. That is,
\frac{\partial f}{\partial z^{\ast}} = 0
But...
Reciprocal Frame
Say we have a frame \{ \mathbf{e}_1, \mathbf{e}_2, \ldots, \mathbf{e}_n\}\;. Even if we have not defined an inner product,
\mathbf{e}_i \cdot \mathbf{e}_k = ?
we can still construct a reciprocal frame \{ \mathbf{e}^1, \mathbf{e}^2, \ldots, \mathbf{e}^n\}\;, no? We would...
Embedding Problem
It can be useful in practice to embed manifolds into (pseudo)Euclidean spaces of higher dimension since the tangent spaces at all points in our manifold will be (pseudo)Euclidean hyperplanes.
Curvature is intrinsic in the sense that these tangent hyperplanes are different...
OK - I feel stupid now. I suppose we could just complete the square in the original sum:
(2\ell+1)e^{-k\, \ell(\ell+1)} = (2\ell+1)e^{-k(\ell^2 + \ell + \frac{1}{4})+\frac{k}{4}} = (2\ell+1)e^{\frac{k}{4}}e^{-k(\ell+\frac{1}{2})^2}
Then letting \lambda = \ell + 1/2 we have:
2\lambda \...
I'm not exactly sure how to evaluate this last sum (I might be unnecessarily complicating things further) but I'll throw out some ideas anyhow...
Consider extending the real numbers by adding a number u that is neither 1 nor -1 such that
u^2 = 1
much the same way we extend the real...
Here's an idea:
Break up the 2 \ell +1 into \ell + (\ell + 1) \;. Then we have:
Z_\ell = F(\ell) + G(\ell)
where
F(\ell) = \ell \, e^{-k \, \ell(\ell+1)}
and
G(\ell) = (\ell+1) \, e^{-k \, \ell(\ell+1)}
then
G(\ell - 1) = \ell e^{-k \, \ell(\ell-1)}
And so
H(\ell) =...
Expansion of a function in terms of Legendre polynomials directly provides a polynomial function you can use to plot approximations to the function. So I'm not sure what you're referring to here.