Eikonal Blog

2010.01.05

Lie Derivative

Filed under: mathematics — Tags: — sandokan65 @ 16:14

Definition: For an abstract object ${\cal T}(x)$ at the spot $x$ of the manifold $M$, the Lie derivative ${\cal L}_{\xi}$ along the vector $\xi$ is defined as:

${\cal L}_{\xi}{\cal T}(x):\equiv \lim_{\epsilon\rightarrow 0} \frac1{\epsilon} \left[{\cal T}(x+\epsilon\xi)-{\cal T}(x)\right]$.

For the tangent tensor fields defined as
${\cal T}(x) = T_{\mu_1\cdots\mu_p}{}^{\nu_1\cdots\nu_q}(x) dx^{\mu_1} \otimes \cdots \otimes dx^{\mu_p} \otimes \partial_{\nu_1} \otimes \cdots \otimes \partial_{\nu_q}$
we can use the following basic formulas:
${\cal L}_{\xi} dx^{\mu} = (\partial_{\alpha}\xi^\mu) dx^\alpha$,
${\cal L}_{\xi} \partial_{\nu} = - (\partial_{\nu}\xi^{\beta}) \partial_{\beta}$.

These lead to following expressions for the components of the $(p,q)$-tensor field ${\cal T}$:

$({\cal L}_{\xi} T)_{\mu_1\cdots\mu_p}{}^{\nu_1\cdots\nu_q}(x) = \xi^\alpha\partial_\alpha T_{\mu_1\cdots\mu_p}{}^{\nu_1\cdots\nu_q} + \sum_{i=1}^{p} T_{\mu_1\cdots\hat{\mu_i}\alpha\mu_p}{}^{\nu_1\cdots\nu_q}(x) (\partial_{\mu_i}\xi^\alpha) - \sum_{j=1}^{q} T_{\mu_1\cdots\mu_p}{}^{\nu_1\cdots\hat{\nu_j}\beta\nu_q}(x) (\partial_{\beta} \xi^{\nu_j})$.

Note: the imprecise/incorrect but universally accepted notation is ${\cal L}_{\xi} T_{\mu_1\cdots\mu_p}{}^{\nu_1\cdots\nu_q}(x)$.

Examples:

• ${\cal L}_{\xi} \varphi = \xi^\alpha\partial_\alpha \varphi,$
• ${\cal L}_{\xi} V_{\mu} = \xi^\alpha\partial_\alpha V_{\mu} + V_{\alpha} (\partial_\mu \xi^\alpha),$
• ${\cal L}_{\xi} A^{\nu} = \xi^\alpha\partial_\alpha A^{\nu} - A^{\alpha} (\partial_\alpha \xi^\mu)$,
• ${\cal L}_{\xi} S_{\mu\nu} = \xi^\alpha\partial_\alpha S_{\mu\nu} +S_{\alpha\nu} \partial_\mu\xi^\alpha + S_{\mu\alpha} (\partial_\nu \xi^\alpha) = \partial_\mu (S_{\alpha\nu} \xi^\alpha) + \partial_\nu (S_{\alpha\mu} \xi^\alpha) - (\partial_\mu S_{\alpha\nu} + \partial_\nu S_{\alpha\mu} - \partial_\alpha S_{\mu\nu} )$.

A consequence of this last example is the case when $S_{\mu\nu}$ are components of the metric tensor $g_{\mu\nu}$, where one gets:

${\cal L}_{\xi} g_{\mu\nu} = \partial_\mu \xi_\nu + \partial_\nu \xi_\mu - 2 \Gamma_{\mu\alpha\nu}\xi^{\alpha} = \nabla_\mu \xi_\nu + \nabla_\nu \xi_\mu$.

Note: in all above expressions for the Lie derivatives of the tensor components, it is possible to replace the ordinary partial derivatives $\partial_\mu$ with the corresponding covariant derivatives $\nabla_\mu$.

—-
References: T1249 (1997.05.04)

Squares with just two different decimal digits

Filed under: mathematics, number theory, puzzles — Tags: — sandokan65 @ 16:00

I used to state here

There exist only finitely many squares with just two decimal digits” providing a really short list

• $38^2 = 1,444$
• $88^2 = 7,744$
• $109^2 = 11,881$
• $173^2 = 29,929$
• $212^2 = 44,944$
• $235^2 = 55,255$
• $3,114^2 = 9,696,996$

plus infinite classes:

• $10^{2n}$,
• $4\cdot 10^{2n}$,
• $9\cdot 10^{2n}$.

This is not correct. Thanks to Bruno Curfs for pointing this out, as well as for finding the reference [2] cited below.

• Namely, one can see that squares of all single digit numbers satisfy targeted condition:
• $1^2 = 01$,
• $2^2 = 04$,
• $3^2 = 09$,
• $4^2 = 16$,
• $5^2 = 25$,
• $6^2 = 36$,
• $7^2 = 49$,
• $8^2 = 64$,
• $9^2 = 81$,
• Then, there are squares of following two-digit numbers:
• $10^2 = 100$,
• $11^2 = 121$,
• $12^2 = 144$,
• $15^2 = 225$,
• $20^2 = 400$,
• $21^2 = 441$,
• $22^2 = 484$,
• $26^2 = 676$,
• $30^2 = 900$,
• $38^2 = 1,444$,
• $88^2 = 7,744$,
• For squares of 3-digit numbers one finds:
• $100^2 = 10,000$,
• $109^2 = 11,881$
• $173^2 = 29,929$
• $212^2 = 44,944$
• $235^2 = 55,255$
• $264^2 = 69,696$ (thanks to Bruno Curfs)

Of the higher cases, there are known only infinitely long classes:

• $10^{2n}$
• $4\cdot 10^{2n}$
• $9\cdot 10^{2n}$

• $3,114^2 = 9,696,996$
• $81,619^2 = 6,661,661,161$.

Author of page “1. Squares of 2 different digits” (http://www.asahi-net.or.jp/~KC2H-MSM/mathland/math02/three01.htm) provides an implementation of the algorithm to perform exhaustive check. According to his numerical experiments, there are no other sporadic occurrences up to at least $y=10^{48}$ (i.e. $x=10^{24}$).

Reference:

Similar here: More simple math wonders – https://eikonal.wordpress.com/2012/03/14/more-simple-math-wonders/ | Mental calculation of cube root of a six-digit number – https://eikonal.wordpress.com/2010/01/14/mental-calculation-of-cube-root-of-a-two-digit-number/ | Squares with just two different decimal digits – https://eikonal.wordpress.com/2010/01/05/squares-with-just-two-different-decimal-digits/ | Number theory finite concidental sums – https://eikonal.wordpress.com/2010/01/05/number-theory-finite-considental-sums/

Number theory finite concidental sums

Filed under: mathematics, number theory — Tags: — sandokan65 @ 15:43

$1+3+3^n+3^{n+1}+3^{2n} = (2+3^n)^2$

$1+3^n+3^{n+1}+3^{2n} +3^{2n+1}= (1+2 \cdot 3^n)^2$

$1+7+7^2+7^3=20^2$

$1+12^2+12^3+12^4+12^5=521^2$

Reference: T1277

Similar here: More simple math wonders – https://eikonal.wordpress.com/2012/03/14/more-simple-math-wonders/ | Mental calculation of cube root of a six-digit number – https://eikonal.wordpress.com/2010/01/14/mental-calculation-of-cube-root-of-a-two-digit-number/ | Squares with just two different decimal digits – https://eikonal.wordpress.com/2010/01/05/squares-with-just-two-different-decimal-digits/ | Number theory finite concidental sums – https://eikonal.wordpress.com/2010/01/05/number-theory-finite-considental-sums/

Sums

Filed under: mathematics, number theory — Tags: , — sandokan65 @ 15:39

$\sum_{k=1}^\infty \frac1{k^k} = 1.291,285,997,...$ (see Prudnikov 5.1.30.1)

$\sum_{k=1}^\infty \frac{(-)^k}{k} = -0.783,430,051,...$

$\sum_{k=1}^\infty \frac{k!}{k^k} = 1.879,853,86...$

$\sum_{k=1}^\infty \frac{(-)^k k!}{k^k} = -0.655,831,60...$

Reference: T1277

Pi calculations

Filed under: mathematics — Tags: , , , — sandokan65 @ 12:00

Bailey-Borwein-Plouffe formula:
$\pi = \sum_{n=0}^\infty \frac1{16^n} \left( \frac{4}{8n+1} - \frac{2}{8n+4} - \frac1{8n+5} - \frac1{8n+6} \right)$. (Sources: [2], [3]).

Plouffe:
$\frac{\pi}{8} = \sum_{n=0}^\infty \frac1{n 2^{\left[\frac{n+1}2\right]}} \left( \left[\frac{n+7}8\right] - \left[\frac{n+6}8\right] + \left[\frac{n+1}8\right] - \left[\frac{n+4}8\right]\right)$. (Source: [3])

Belard:
$\pi = \frac1{2^6} \sum_{n=0}^\infty \frac{(-)^n}{2^{10n}} \left( -\frac{2^5}{4n+1} - \frac1{4n+3} + \frac{2^8}{10n+1} - \frac{2^6}{10n+3} - \frac{2^2}{10n+5} - \frac{2^2}{10n+7} + \frac1{10n+9} \right)$. (Source: [1])

Belard:
$\pi = \frac1{740,025}\left(\sum_{n=1}^\infty \frac{3P(n)}{\binom{7n}{2n} 2^{n-1}} - 20,379,280 \right)$ where:
$P(n) = - 885,673,181 n^5 + 3,125,347,237 n^4 - 2,942,969,225 n^3 + 1,031,962,795 n^2 - 196,882,274 n + 10,996,648$. (Source: [1])

Chudnovsky:
$\frac1\pi = \frac{\sqrt{10005}}{4270934400} \sum_{k=0}^\infty (-)^k \frac{(6k)!}{(k!)^3 (3k)!} \frac{(13591409+545140134 k)}{640320^{3k}}$

Related at this blog: Calculating e – https://eikonal.wordpress.com/2010/08/06/calculating-e/

Vlatko Stefanovski

Filed under: music — sandokan65 @ 04:20

Meiko Kaji

Filed under: music — Tags: — sandokan65 @ 00:44

Ryuichi Sakamoto

Filed under: music — Tags: — sandokan65 @ 00:39