logarithmic trickery

Consider that since \log(a)\log(b)=\log(a^{\log b})=\log(b^{\log{a}}, and therefore, somewhat unexpectly: a^{\log b} = b^{\log a}
we can use it to construct new identities thusly

Let A = \prod_{n=1}^{\infty} \sqrt[2]{1+\frac{1}{n^{2}}}

Let B = \prod_{m=1}^{\infty} \sqrt[e^{m}]{1+\frac{1}{m^{3}}}

Therefore, since we know that A^{\log(B)}=B^{\log(A)} we can use that
to explicitly calculate a new identity thusly:

1. Compute the logarithm of A, \log(A):
\log(A) = \frac{1}{2} \sum_{v=1}^{\infty} \frac{(-1)^{v+1}\zeta(2v)}{v}

2. Compute the logarithm of B, \log(B):
\log(B) = \sum_{s=1}^{\infty} \frac{(-1)^{s+1}\mathrm{Li}_{(3s)}(1/e)}{s}

3. Combine them:

\prod_{n=1}^{\infty} \prod_{s=1}^{\infty} \sqrt[s]{\left(1+\frac{1}{n^{2}}\right)^{(-1)^{s+1}\mathrm{Li}_{(3s)}(1/e)}} = \prod_{m=1}^{\infty} \prod_{v=1}^{\infty} \sqrt[ve^{m}]{\left(1+\frac{1}{m^{3}}\right)^{(-1)^{v+1}\zeta(2v)}}

>>> A = fp.nprod(lambda n: sqrt(1+1/(n*n)), [1,inf])
>>> A
>>> B = fp.nprod(lambda m: (1+1/(m**3))**(1/exp(m)), [1,inf])
>>> B
>>> A**log(B)
>>> B**log(A)

dreaming about reading mathematics books

one: It is a thick, yellow, Springer-Verlag dealie, illustrated and about fractals. There is a color plate on nearly every page. I dreamt of being in a bookshop and considering buying it.

two: has lots of equations at the beginning, and they change as I read them. last page has a distorting/metamorphosing picture of einstein/feynman in green and blue hues.

three: also a text that changes, I remember a diagram of helix against a point scatter. Also changed.

last night: I remember something about the Euler-Mascheroni constant \gamma. Impenetrable references section. Also picture of fractals. 

Have you dreamt of reading mathematics books? If so, leave a comment about them.

the banshee

I had this thought in the shower, and it descends from my “symmetry may be a long term red herring” idea, and it goes like this:

Imagine that you have something quite like the Monster group M, except that there is a fixed pair of elements in this thing like M — let’s call it \mathcal{B} for banshee. Every time you take two elements from \mathcal{B} and calculate their product, there is a \frac{1}{|M|^{2}} chance that you’ll be given something which is not an element of M, like a real number or some other object.

Let’s say it transpires that the boogeyman actually has some utility: if you happen to have a hangup on symmetry, your apprehension of the Monster group is going to obscure the existence of the banshee to you.

Interesting, because of the sheer size of the order of the Monster, it is easier to confuse this banshee with a group than if you were to take Klein’s viergruppe or a cyclic group and replace one of it’s elements. The banshee \mathcal{B} is not a group