Continuously Under Construction
Continuously Under Construction
• This is some pre-formatted text:
Column1 Column2 Column3
-------------------------------
123.45 23.11 -3.1
===============================
• A little bit of pop art (using Google's monospaced webfont 'Cousine'):
airkreuzer airkreuzer
airkreuzer airkreuzer
airkreuzer airkreuzer
airkreuzer airkreuzer
airkreuzer airkreuzer
airkreuzer airkreuzer
airkreuzer airkreuzer
airkreuzer airkreuzer
airkreuzer airkreuzer
airkreuzer airkreuzer
• Here's a very nice example using pre-formatted text (found at Christian Hamann's page on the History of Slide Rules):
The Principle of the Slide Rule:
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Multiply = Add Logarithmic Scales, e.g. 2×2=4 or 2×3=6 or 2×4=8
Divide = Subtract Logarithmic Scales, e.g. 4÷2=2 or 6÷3=2 or 8÷4=2
1 2 3 4 5 6 7 8 9 1 ...
| | | | | | | | | |
____________________________________________________________
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1 2 3 4 5 6 ...
add ===>> <<=== subtract
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• Here are some lines of Pascal code:
program First;
begin
WriteLn ('Hello again, Martin.')
end.
• This is what the J interpreter has to say:
'Hello again, Martin' Hello again, Martin[More examples from various programming languages may be found on the ACM "Hello, world!" project page.]
• As the year draws to a close (this entry is from 2022-Dec-31) Christmas trees will soon be a memory again.
Here's a 12x5 array holding the first 60 Fibonacci numbers:
[[ 0 1 1 2 3] [ 5 8 13 21 34] [ 55 89 144 233 377] [ 610 987 1597 2584 4181] [ 6765 10946 17711 28657 46368] [ 75025 121393 196418 317811 514229] [ 832040 1346269 2178309 3524578 5702887] [ 9227465 14930352 24157817 39088169 63245986] [ 102334155 165580141 267914296 433494437 701408733] [ 1134903170 1836311903 2971215073 4807526976 7778742049] [ 12586269025 20365011074 32951280099 53316291173 86267571272] [139583862445 225851433717 365435296162 591286729879 956722026041]]The relevant Python code snippet looks like this:
import numpy as np def fibay(r,c): fs= [0,1] for k in range(r*c-2): fs.append(sum(fs[-2:])) return np.array(fs).reshape((r,c)) print(fibay(12,5))
Using a MathJax Content Delivery Network (CDN) by linking MathJax (JavaScript based) into the web pages that are to include mathematics; this solution calls the combined in-line configuration "TeX-MML-AM_CHTML" on the server to make simultaneous use of ASCIIMath and TeX/LaTeX coding possible:
❖ First, some ASCIIMath input:
The two possible solutions to a quadratic equation `ax^2 + bx + c = 0\ ,\ a!=0` are given by `x_(1;2) = (-b +- sqrt(b^2-4ac))/(2a) .`
❖ Second, using TeX/LaTeX input:
Given the quadratic equation \(a x^2 + b x + c = 0 \,, \,\, a \ne 0 \) then possible solutions are \( x_{1;2} = \cfrac{-b\pm\sqrt{b^2-4ac}}{2a}\,. \)
❖ Third, more TeX/LaTeX input (mixed at some stages with ASCIIMath):
• Porting PYTHAGORAS' theorem to the unit circle: \( sin^2(x)+cos^2(x)=1 \) .
• Probability for continued BERNOULLI trial \[ P(E) = {n \choose k}\, p^k\, (1-p)^{ n-k} \] where \( P(E) \) is the probability to find exactly \( k \) successes in \( n \) trials when the probability for success at each stage of the experiment is \( p \) and that of failure is \( 1 - p \) .
Imagine e.g. tossing a fair coin repeatedly; let \( E_{k<3} \) be "There will be less than three heads in five trials".
The probability is then given by \[ P(E_{k<3})
= \binom{5}{0} \left(\frac{1}{2}\right)^{0} \left(\frac{1}{2}\right)^{5}
+ \binom{5}{1} \left(\frac{1}{2}\right)^{1} \left(\frac{1}{2}\right)^{4}
+ \binom{5}{2} \left(\frac{1}{2}\right)^{2} \left(\frac{1}{2}\right)^{3}
= \left(\binom{5}{0} + \binom{5}{1} + \binom{5}{2}\right) \left(\frac{1}{2}\right)^{5}
= \left(1 + 5 +\frac{5\cdot4}{1\cdot2}\right) \left(\frac{1}{32}\right)
= \frac{16}{32} = 0.5 = 50\% \] (do not expect that result every time you try).
• The Golden Ratio \(\phi=(1+\sqrt{5})/2\) resp. \(\varphi=(\!-\!\,1+\sqrt{5})/2\)
] Phi=. -: >: %: 5 1.61803398875 ] phi=. -: <: %: 5 0.61803398875 Phi - phi NB. difference 1 Phi * phi NB. product 1 % Phi NB. reciprocal 1/Phi 0.61803398875 phi = % Phi NB. does phi equal 1/Phi ? (boolean) 1is the only number having a continued fraction representation this simple \[\phi=\frac{1+\sqrt{5}}{2}=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cdots}}}}\approx 1.61803\] consisting of 1s only (and therefore converging painfully slow). It is algebraic, as a root of the quadratic equation \[\,x^2-x-1=0\] and – as it can't be expressed by a fraction proper – irrational. Some call it the most irrational number.
10 # 1 NB. a list (vector) of 10 1s 1 1 1 1 1 1 1 1 1 1 (+ %)/ 1 #~ 10 NB. building the continued fraction (feeding 10) 1.61818181818 (+%)/\ 1 $~ 10 NB. showing progression 1 2 1.5 1.66667 1.6 1.625 1.61538 1.61905 1.61765 1.61818 (+ %)/ 1 #~ 15 NB. cf (feeding 15) 1.61803278689 (+ %)/ 1 #~ 25 NB. cf (feeding 25) 1.61803398867 (+ %)/ 1 #~ 50 NB. cf (feeding 50) 1.61803398875
As a reminiscence to the FIBONACCI series here is a related equation \[ \phi^{2}=\left(\frac{1+\sqrt{5}}{2}\right)^{2}=\frac{1+2\,\sqrt{5}+5}{4}=\frac{1+\sqrt{5}}{2}+1=\phi+1=\phi^{1}+\phi^{0} \] This can be generalised to \[\,\phi^{n+1}=\phi^{n}+\phi^{n-1}\]
• These I discovered by tinkering around, starting at the midth of intervall `[0;\pi]` and symmetrically spreading out to it's boundaries;
note `\phi` appearing in the fourth line:
\[
\begin{align}
\int\limits_{\pi/2}^{1\pi/2} \hspace{-0.5em}{sin(t)}dt &= 2\,cos(\frac{\pi}{2}) = 0\\
\int\limits_{\pi/3}^{2\pi/3} \hspace{-0.5em}{sin(t)}dt &= 2\,cos(\frac{\pi}{3}) = 1\\
\int\limits_{\pi/4}^{3\pi/4} \hspace{-0.5em}{sin(t)}dt &= 2\,cos(\frac{\pi}{4}) = \sqrt{2} \approx 1.41421\\
\int\limits_{\pi/5}^{4\pi/5} \hspace{-0.5em}{sin(t)}dt &= 2\,cos(\frac{\pi}{5}) = \frac{1}{2}(1+\sqrt{5}) = \phi \approx 1.61803\\
\int\limits_{\pi/6}^{5\pi/6} \hspace{-0.5em}{sin(t)}dt &= 2\,cos(\frac{\pi}{6}) = \sqrt{3} \approx 1.73205\\
&\vdots\\
\int\limits_{\pi/10}^{9\pi/10} \hspace{-0.5em}{sin(t)}dt &= 2\,cos(\frac{\pi}{10}) = \sqrt{\frac{1}{2}(5+\sqrt{5})} \approx 1.90211\\
\int\limits_{0}^{\pi} {sin(t)}dt &= 2\,cos(0) = 2
\end{align}
\]
sin SR3 1r10p1,9r10p1,400 1.902113032591 +: cos 1r10p1 1.90211303259 %:-:5+%:5 1.90211303259 bcf +: cos 1r10p1 NB. build continued fraction (bcf) 1 1 9 4 1 1 1 2 1 1 2 3 7 1 2 12 1 2 1 6 2 1 1 5 1 2 2 2 8 11 bcf %:-:(+%:)5 1 1 9 4 1 1 1 2 1 1 2 3 7 1 2 13 9 1 14 5 2 4 2 1 1 13 1 2 ecf 1 1 9 4 1 1 1 2 1 1 2 3 7 1 2 12 NB. evaluate continued fraction (ecf) up to first double-digit list item 1.90211303259 ecf 1 1 9 4 1 1 1 2 1 1 2 3 7 1 2 13 1.90211303259
• EULER's Product formula for calculating the Totient (number of co-prime integers below and including \(n\)) \[\varphi(n) = n\cdot\prod\limits_{i=1}^{i=k} (1-\frac{1}{p_{i}}) = \prod\limits_{i=1}^{i=k} (p_{i}-1)\cdot p_{i}^{(e_{i}-1)}=\bigg\lbrace \begin{equation}\begin{split} & n-1, \: \text{if}\: n\: \text{prime} \\ & \!\! < n-1, \: \text{if}\: n\: \text{composite} \end{split}\end{equation} \]
5 p: 11 NB. 11 is prime; Totient of 11 is 10 10 5 p: 12 NB. 12 is composite (non-prime); Totient of 12 is 4 4 5 p: 5039 NB. 5039 is prime 5038 5 p: !7 NB. 5040 is composite (non-prime) 1152 (- ~:) &. q: !7 NB. using Andrew Nikitin's translation of Product formula 1152
• EULER's Prime-Generating Polynomial `n^2 + n + 41` is known to generate (distinct) primes for `n = 0..39` and has less success for `n > 39` .
1 p: 41 1 1 p. i. 40 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 v=. 41 1 1 p. 39 + i. 12 (] ,:1&p:) v 1601 1681 1763 1847 1933 2021 2111 2203 2297 2393 2491 2591 1 0 0 1 1 0 1 1 1 1 0 1
• This is one of my favourites \[ \int_0^\infty e^{-x}\,\mathrm{d}x = 1 \] • Here is EULER's Identity (using the half-turn constant \(\pi\)) \[ e^{i \pi} = -1 \] or (using the full-turn constant \(\tau\)) \[ e^{i \tau} = 1 \] For further reading on \(\tau\) see The Tau Manifesto by Michael Hartl.
• This definite integral is used to calculate the area of one of the graph's spikes [the indefinite integral I looked up at the
Wolfram|Alpha site]. As the integrand contains \(\,sin\) and \(\,cos\) squared, the area is symmetrical to \(\,x=\frac{\pi}{2}\) and it will therefore suffice to calculate the left half and double the result:
\[\int_{0}^{\pi}\frac{sin^2(x)}{1 + cos^2(x)} dx
= 2\cdot\left[\sqrt{2}\:arctan\left(\dfrac{\tan(x)}{\sqrt{2}}\right)-x\right]_{0}^{\frac{\pi}{2}}
= 2\cdot\bigl(\sqrt{2}\,\frac{\pi}{2}-\frac{\pi}{2}-0\bigr)
= (\sqrt{2} - 1) \,\pi \approx 1.3\]
Calling NEWTON's Method (with seeds `-0.1` and `3`) seems to confirm the first roots being `0` and `pi`.
f VN^:1e5 (_0.1 3) 9.9984e_13 3.14159Numerical Integration using Trapezoid Rule and SIMPSON's Rule (9 resp 8 sections) yields
f=. ([: *: 1 o. ]) % [: >: [: *: 2 o. ] f TR 0 1p1 9 1.30129 f SR4 0 1p1 8 1.30129Here is the result from Calculus \((\sqrt{2} - 1) \,\pi\) for comparison:
1p1*<:%:2 1.30129
• Tetration of Infinite Height
Citing Brilliant [slightly modified Ed.]: "Tetration is, for the most part, another name for iterated exponentiation. Iterated exponentiation is when you raise a number to the power of itself several times. The formal definition of tetration is \[\large ^{n}x \equiv x^{x^{x^{^{.^{.^{.^{x}}}}}}},\] where there are \(\,n\) \(\,x\)'s on the right side of the equation. This definition allows tetration to represent huge numbers in a tiny amount of space."
Given this equation, what's the value of \(\,x\) ? \[x^{x^{x^{x^{x^{x\mathstrut^{\,.^{\,.^{\,.}}}}}}}}=2\]
Here we have infinite tetration !
Let's have a look at the top \(\; x^{x}=2\;\Longrightarrow\; x\cdot ld(x)=ld(2)=1\) : \[\Biggl\{ \begin{equation}\begin{split} & x=1:\:\:\: 1\cdot ld(1)=0\\ & x=2:\:\:\: 2\cdot ld(2)=2 \end{split}\end{equation}\] The solution should be found within the range of \(\,0\lt x\lt 2\) .
Setting the left side of the equation to \(f(x)\) , we may also set \(f(x)=x^{f(x)}=2\) .
Going to the logarithmic realm and back, we get
\[
\begin{align}
f(x)\cdot ln(x)&=ln(2)
\hspace{1.2em}\therefore\:
\frac{ln(2)}{ln(x)}=2
\hspace{1.2em}\:\therefore\:
\frac{ln(2)}{2}=ln(x)\\
x&=e^{\frac{ln(2)}{2}}=e^{ln(2)\cdot\frac{1}{2}}=\sqrt{2}
\end{align}
\]
expe 1.41421,1e_6 1.41421 1e_6 34 1.99997 expe (2%:2),1e_6 NB. <<<<< 1.41421 1e_6 34 2 expe 1.414219,1e_6 1.41422 1e_6 34 2.00005
The second (trivial) case is \(\,f(x)=1\) , as \(\,\sqrt[\leftroot{2}\uproot{2}\scriptstyle 1]{1}=1^{1}=1\) .
expe 1,1e_15 1 1e_15 1 1
If we broaden our view and accept not only integer but also real radicand/exponent pairs, we find convergence within the interval \(\,[\frac{1}{e};e]\) .
This has already been shown by Leonard EULER in 1783.
Upper interval boundary:
NB. expe (2.71%:2.71),1e_13 NB. 1.44466538 NB. 1e_13 NB. 6554 NB. 2.71 NB. expe (1x1%:1x1),1e_13 NB. 1.44466786 NB. 1.44466786··· (Wolfram|Alpha) NB. 1e_13 NB. 7358766 NB. number of steps 'soaring' ... NB. 2.71828109 NB. 2.718281828··· ≈ e (Wolfram|Alpha) NB. expe (2.72%:2.72),1e_13 NB. 1.44466775 NB. 1e_13 NB. 26727 NB. 2.71656546Lower interval boundary:
NB. expe (0.368%:0.368),1e_3 NB. 0.0661057 NB. 0.001 NB. 11687 NB. 0.3685 NB. expe (0.367%:0.367),1e_3 NB. below lower boundary ? NB. 0.0651334 NB. 0.001 NB. ?? NB. ???? NB. 1x_1 NB. 0.367879 NB. expe (1x_1%:1x_1),1e_3 NB. lower boundary seems to be 1/e = 0.367879··· NB. 0.065988 NB. 0.001 NB. 3248036 NB. <<<<< NB. 0.36738 NB. expe (1x_1%:1x_1),1e_4 NB. 0.065988 NB. 0.0001 NB. 324804743 NB. 0.367929
• One of S. RAMANUJAN's formulae for \(\pi\) reads \[ \frac{1}{\pi}=\frac{2\sqrt{2}}{9801}\sum_{{k=0}}^{\infty}\frac{(4k)!\;(1103+26390k)}{(k!)^{4}\;396^{{4k}}} \] and converges really fast, as for \(k=0\) it already yields 7 (correct) leading digits, \[\pi\approx \frac{9801}{1103 \cdot 2 \cdot \sqrt{2}}=\color{green}{3}.\color{green}{141592}\color{gray}{73001...}\] and will compute a further 8 decimal places of π with each term of the series as can be seen comparing the results of these lines of code
9801 % 1103 * (* %:) 2 NB. Ramanujan's approximation (k=0; sum=1103) 3.141592730013306 pir 0 NB. piR approx (k=0) 3.141592730013306 pir 1 NB. piR approx (k=1) 3.141592653589794 1p1 NB. value of π (to 16 leading digits) 3.141592653589793
• And then there's \(1\cdot1=1\) at the very base …
The SSI feature depends on being provided by the server platform. These single lines embedded in HTML code may, among other things, be used to call and execute a script file. Example:
!--#exec cgi="/cgi/envvar00.cgi" --Enclosing the expression with
Server environment variables may also be called by PHP of course; here is the result of a reverse DNS lookup for your machine (probably revealing some information on your ISP), managed by a SSI call to a PHP script on the same server:
The last line on this page is also the result of a SSI call.
My first provider is maintaining MET/MEST (CET/CEST) since the server farm used resides within Germany and it serves mainly German customers.
Another
site
is running on a different provider's server, where
they are maintaining GMT.
[Sidetrack: On 2009-03-07 we became aware that this particular site (among others) had been hacked in early 2009 to send out malware (a trojan); the situation was brought back to normal with the provider's help a day later on 2009-03-08.]
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This code snippet (applet) uses your computer clock and requires Java to be enabled. |
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ienvenue auureau International des Poids et Mesures |
|
This is the home of UTC and SI, the Bureau International des Poids et Mesures ( BIPM ). |
|
Have a view at their UTC/TAI Time Server applet. It is presented here as a link only, to avoid excessive loading times. (Note the estimate given for the transmission delay.) |
Just for the pleasure of it there is my
Gopher
page, a remeniscence to earlier times of the internet. Hosted
by a (presently) non-commercial site, it is not available 24/7
(as one might exspect).
Hint: If you are an
Opera
user you will want to configure a Gopher Proxy Server at e.g.
File > Preferences > Network > Proxy servers
to be able to view it.
In case you want to send a message, please use this link.