樹莓派 0W 狂想曲︰薄片電腦《♪》

莫非是勞侖次吸子之致命吸引力!

勞侖次吸子

勞侖次吸子勞侖次振子(Lorenz oscillator)的長期行為對應的碎形結構,以愛德華·諾頓·勞侖次的姓氏命名。勞侖次振子是能產生混沌流的三維動力系統,是一種吸子,以其雙紐線形狀而著稱。映射展示出動力系統(三維系統的三個變量)的狀態是如何以一種複雜且不重複的模式,隨時間的推移而演變的。

A sample solution in the Lorenz attractor when ρ = 28, σ = 10, and β = 8/3

簡述

 勞侖次方程式的一條軌跡被描繪成金屬線,以展現方向以及三維結構

勞侖次吸子及其導出的方程組是由愛德華·諾頓·勞侖次於1963年發表,最初是發表在《大氣科學雜誌》(Journal of the Atmospheric Sciences)雜誌的論文《Deterministic Nonperiodic Flow》中提出的,是由大氣方程式中出現的對流卷方程式簡化得到的。

這一勞侖次模型不只對非線性數學有重要性,對於氣候和天氣預報來說也有著重要的含義。行星和恆星大氣可能會表現出多種不同的准周期狀態,這些准周期狀態雖然是完全確定的,但卻容易發生突變,看起來似乎是隨機變化的,而模型對此現象有明確的表述。

從技術角度看來,勞侖次振子具有非線性、三維性和確定性。2001年,沃里克·塔克爾(Warwick Tucker)證明出在一組確定的參數下 ,系統會表現出混沌行為,顯示出人們今天所知的奇異吸子。這樣的奇異吸子是豪斯多夫維數在2與3之間的碎形彼得·格拉斯伯格(Peter Grassberger)已於1983年估算出豪斯多夫維數為2.06 ± 0.01,而關聯維數為2.05 ± 0.01。

此系統也會出現在單模雷射[1]發電機[2]的簡化模型中。除此之外 ,閉環對流、水輪轉動等物理模型也有此系統的應用。

 

鄉村人口持續流向城市,派生城鄉差距逐年擴大的問題乎?因此連教育資源也都分高山平地耶??樹莓派物美價廉,理念宗旨為的是教育學子而來。這個新生的樹莓派 0W 貌似口香糖,能嚼出電腦味吧!故而想打造薄片電腦也!也算縮小城鄉差距的棉薄之力矣!!

熟知三日不見江山改, PIXEL 跑得像烏龜,動則負載百分百?難到樹莓派 0W 並非為此生??論壇查訪探分明︰

[GUIDE] Raspbian Lite with PIXEL/LXDE/XFCE/MATE/Openbox GUI

by GhostRaider » Sat Jan 23, 2016 3:19 am
Requirements:
1. Any Raspberry Pi microcomputer
2. Raspbian Lite image
3. SD / microSD card (At least 4GB or higher, but it is possible to use smaller storage space depending on your configuration)
4. Keyboard and Mouse (For using the GUI, but can vary depending on how the Raspberry Pi will be used)
5. TV / Monitor (For seeing the GUI)
6. A normal computer with Linux, macOS, or Windows (For burning Rasbian Lite image to SD / microSD)
7. Internet connectionTable of Contents
1. Introduction
2. Memory Usage
3. Part 1 – Build the Foundation
4. Part 2 – Bring in the Furniture
5. Sample Screenshots
6. GUI Customization
7. Virtual Machine Playground
8. Advanced – Custom Desktop Environment using i3 WM
9. Extras

 

看來又得萬丈高樓平地起的呦??!!

依法炮制想在先,昔日 LXDE 安裝需網路,那個系統韌體

bcm2708-rpi-0-w.dtb

卻又需新版!!??

一時想起過牆梯,不必強求張良計,假借樹莓派 3B 為媒介,只待嘛搞定後再移植☆

【眼見為憑】

 

 

 

 

 

 

 

 

 

樹莓派 0W 狂想曲《𝄞》

韶光飛逝,轉眼樹莓派已經五歲了,如今維基百科的詞條

Raspberry Pi

The Raspberry Pi is a series of small single-board computers developed in the United Kingdom by the Raspberry Pi Foundation to promote the teaching of basic computer science in schools and in developing countries.[4][5][6] The original model became far more popular than anticipated,[7] selling outside of its target market for uses such as robotics. Peripherals (including keyboards, mice and cases) are not included with the Raspberry Pi. Some accessories however have been included in several official and unofficial bundles.[7]

According to the Raspberry Pi Foundation, over 5 million Raspberry Pis have been sold before February 2015, making it the best-selling British computer.[8] By November 2016 they had sold 11 million units.[9][10]

………

 

內容早已十分豐富。今年生日時發表『樹莓派 0W』

New product! Raspberry Pi Zero W joins the family

Today is Raspberry Pi’s fifth birthday: it’s five years since we launched the original Raspberry Pi, selling a hundred thousand units in the first day, and setting us on the road to a lifetime total (so far) of over twelve million units. To celebrate, we’re announcing a new product: meet Raspberry Pi Zero W, a new variant of Raspberry Pi Zero with wireless LAN and Bluetooth, priced at only $10.

Multum in parvo

So what’s the story?

 

述說著什麼樣的故事呢?

In November 2015, we launched Raspberry Pi Zero, the diminutive $5 entry-level Raspberry Pi. This represented a fivefold reduction in cost over the original Model A: it was cheap enough that we could even stick it on the front cover of The MagPi, risking civil insurrection in newsagents throughout the land.

Over the ensuing fifteen months, Zero grew a camera connector and found its way into everything from miniature arcade cabinets to electric skateboards. Many of these use cases need wireless connectivity. The homebrew “People in Space” indicator in the lobby at Pi Towers is a typical example, with an official wireless dongle hanging off the single USB port: users often end up adding a USB hub to allow them to connect a keyboard, a mouse and a network adapter, and this hub can easily cost more than the Zero itself.

Zero W fixes this problem by integrating more functionality into the core product. It uses the same Cypress CYW43438 wireless chip as Raspberry Pi 3 Model B to provide 802.11n wireless LAN and Bluetooth 4.0 connectivity.

 

這麼個美金十元小小板子

Raspberry Pi Zero W

Raspberry Pi Zero W

The Raspberry Pi Zero W extends the Pi Zero family. Launched at the end of February 2017, the Pi Zero W has all the functionality of the original Pi Zero but with added connectivity, consisting of:

  • 802.11 b/g/n wireless LAN
  • Bluetooth 4.1
  • Bluetooth Low Energy (BLE)

Like the Pi Zero, it also has:

  • 1GHz, single-core CPU
  • 512MB RAM
  • Mini HDMI and USB On-The-Go ports
  • Micro USB power
  • HAT-compatible 40-pin header
  • Composite video and reset headers
  • CSI camera connector

 

據聞有無限之應用潛力,只需谷歌

raspberry pi zero w projects

便曉。作者好奇也得了一片,把玩之餘,狂想可如何用於教具教材耶?故而才奏此曲的乎!

 

 

 

 

 

 

 

 

 

時間序列︰生成函數‧漸近展開︰歐拉的天空《奇門》

若已熟悉白努利周期函數之性質︰

{\tilde{B}}_{n}^{'} (x) = n \cdot {\tilde{B}}_{n-1},  \ n \ge1

\int_{k}^{k+1} {\tilde{B}}_n (x) dx = 0, \ n \ge 1

{\tilde{B}}_{2n} (k+1) = {\tilde{B}}_{2n} (1) = {\tilde{B}}_{2n} (0) = B_{2n}, \ n \ge 1

{\tilde{B}}_{2n+1} (k+1) = {\tilde{B}}_{2n+1} (1) = - {\tilde{B}}_{2n+1} (0) = B_{2n+1} = 0, \ n \ge 1

果真知 B_0 (x) \equiv 1 ,白努利數唯一奇 B_1 \neq 0

既得白努利數之生成函數 G(t) = \frac{t}{e^t -1} 且先探其奇偶性乎?

……

關係一現機鋒出 G(-t) - G(t) = e^t G(t) - G(t) = t 。原來這個白努利數唯一奇 B_1- B_1 - B_1 = 1 ,不假它求數自知 B_1 = -\frac{1}{2} 。遞迴關係無覓處

(m+1) B_m = - \sum \limits_{k=0}^{m-1} \left( \begin{array}{ccc}   m+1 \\ k \end{array} \right) B_k

。恰恰此中得

t = \sum \limits_{m=0}^{\infty} B_m \frac{t^m}{m !} \cdot (e^t -1)

── 摘自《時間序列︰生成函數‧漸近展開︰白努利 □○《六》

 

那麼一招一式之推導

\int_{k}^{k+1} f(x) dx = \int_{k}^{k+1}  B_0 (x)  f(x) dx = \int_{k}^{k+1} \frac{d {\tilde{B}}_1 (x)}{dx} f(x) dx

= {\tilde{B}}_1 (x) f(x) |_{k^{+}}^{k+1{-}} - \int_{k}^{k+1} {\tilde{B}}_1 (x) f^{'} (x) dx

= {\tilde{B}}_1 (k+1^{-}) f(k+1) - {\tilde{B}}_1 (k^{+}) f(k) - \int_{k}^{k+1} {\tilde{B}}_1 (x) f^{'} (x) dx

= \frac{1}{2} \left[ f(k) + f(k+1) \right] - \int_{k}^{k+1} {\tilde{B}}_1 (x) f^{'} (x) dx

能入『求和』 vs. 『求積』關聯之門也。

\therefore \frac{1}{2} \left[ f(k) + f(k+1) \right] = \int_{k}^{k+1} f(x) dx + \int_{k}^{k+1} {\tilde{B}}_1 (x) f^{'} (x) dx

亦能曉

\frac{1}{2} \left[ f(0) + f(1) \right] = \int_{0}^{1} f(x) dx + \int_{0}^{1} {\tilde{B}}_1 (x) f^{'} (x) dx

\frac{1}{2} \left[ f(1) + f(2) \right] = \int_{1}^{2} f(x) dx + \int_{1}^{2} {\tilde{B}}_1 (x) f^{'} (x) dx

……

\frac{1}{2} \left[ f(k) + f(k+1) \right] = \int_{k}^{k+1} f(x) dx + \int_{k}^{k+1} {\tilde{B}}_1 (x) f^{'} (x) dx

……

\frac{1}{2} \left[ f(n-1) + f(n) \right] = \int_{n-1}^{n} f(x) dx + \int_{n-1}^{n} {\tilde{B}}_1 (x) f^{'} (x) dx

加之得

\sum \limits_{k=0}^{n} f(k) - \frac{1}{2} \left[ f(n)+f(0) \right] = \int_{0}^{n} f(x) dx + \int_{0}^{n} {\tilde{B}}_1 (x) f^{'} (x) dx

\sum \limits_{k=1}^{n} f(k) - \frac{1}{2} \left[ f(n)-f(0) \right] = \int_{0}^{n} f(x) dx + \int_{0}^{n} {\tilde{B}}_1 (x) f^{'} (x) dx

『取其便』之理矣。

如是者豈不會閱讀

證明

證明使用數學歸納法以及黎曼-斯蒂爾傑斯積分,下文中假設 {\begin{smallmatrix}f(x)\end{smallmatrix}}的可微次數足夠大,  {\begin{smallmatrix}a,b\in {\mathbb {Z}}\end{smallmatrix}}
為了方便,將原式的各項用不同顏色表示:
\sum _{{a<n\leq b}}f(n)={\color {red}\int _{{a}}^{{b}}f(t)\,{\mathrm {d}}t}+{\color {OliveGreen}\sum _{{r=0}}^{{k}}{\frac {(-1)^{{r+1}}B_{{r+1}}}{(r+1)!}}\cdot (f^{{(r)}}(b)-f^{{(r)}}(a))}+{\color {blue}{\frac {(-1)^{k}}{(k+1)!}}\int _{{a}}^{{b}}{\bar {B}}_{{k+1}}(t)f^{{(k+1)}}(t)}

  {\begin{smallmatrix}k=0\end{smallmatrix}}的情形

容易算出
{\bar {B}}_{1}(t)={\color {Purple}\left\langle t\right\rangle -{\frac {1}{2}}}
{\begin{aligned}\sum _{{a<n\leq b}}f(n)&=\int _{{a}}^{{b}}f(t)\,{\mathrm {d}}\left\lfloor t\right\rfloor \\&={\color {red}\int _{{a}}^{{b}}f(t)\,{\mathrm {d}}t}-\int _{{a}}^{{b}}f(t)\,{\mathrm {d}}\left\langle t\right\rangle \\&={\color {red}\int _{{a}}^{{b}}f(t)\,{\mathrm {d}}t}-\int _{{a}}^{{b}}f(t)\,{\mathrm {d}}({\color {Purple}\left\langle t\right\rangle -{\frac {1}{2}}})\\&={\color {red}\int _{{a}}^{{b}}f(t)\,{\mathrm {d}}t}-{\color {BurntOrange}\int _{{a}}^{{b}}f(t)\,{\mathrm {d}}{\bar {B_{1}}}(t)}\\\end{aligned}}
其中橙色的項通過分部積分可化為
{\begin{aligned}{\color {BurntOrange}\int _{{a}}^{{b}}f(t)\,{\mathrm {d}}{\bar {B_{1}}}(t)}&=(f(t){\bar {B_{1}}}(t))|_{{t=a}}^{{t=b}}-\int _{{a}}^{{b}}{\bar {B_{1}}}(t)\,{\mathrm {d}}f(t)\\&=f(b)B_{1}(\left\langle b\right\rangle )-f(a)B_{1}(\left\langle a\right\rangle )-{\color {blue}\int _{{a}}^{{b}}{\bar {B_{1}}}(t)f'(t)\,{\mathrm {d}}t}\\&={\color {OliveGreen}B_{1}\cdot (f(b)-f(a))}-{\color {blue}\int _{{a}}^{{b}}{\bar {B_{1}}}(t)f'(t)\,{\mathrm {d}}t}\\\end{aligned}}

假設  {\begin{smallmatrix}k=n-1\end{smallmatrix}}時原式成立

  \sum _{{a<n\leq b}}f(n)={\color {red}\int _{{a}}^{{b}}f(t)\,{\mathrm {d}}t}+{\color {OliveGreen}\sum _{{r=0}}^{{n-1}}{\frac {(-1)^{{r+1}}B_{{r+1}}}{(r+1)!}}\cdot (f^{{(r)}}(b)-f^{{(r)}}(a))}+{\color {blue}{\frac {(-1)^{{n-1}}}{n!}}\int _{{a}}^{{b}}{\bar {B}}_{{n}}(t)f^{{(n)}}(t)\,{\mathrm {d}}t}

處理積分(藍色項)

  {\begin{aligned}{\color {blue}{\frac {(-1)^{{n-1}}}{n!}}\int _{{a}}^{{b}}{\bar {B}}_{{n}}(t)f^{{(n)}}(t)\,{\mathrm {d}}t}&={\frac {(-1)^{{n-1}}}{n!}}\int _{{a}}^{{b}}{\frac {{\bar {B'}}_{{n+1}}(t)}{n+1}}f^{{(n)}}(t)\,{\mathrm {d}}t\\&={\frac {(-1)^{{n-1}}}{(n+1)!}}\int _{{a}}^{{b}}{\bar {B'}}_{{n+1}}(t)f^{{(n)}}(t)\,{\mathrm {d}}t\\&={\frac {(-1)^{{n-1}}}{(n+1)!}}\int _{{a}}^{{b}}f^{{(n)}}(t)\,{\mathrm {d}}{\bar {B}}_{{n+1}}(t)\\&={\frac {(-1)^{{n-1}}}{(n+1)!}}((f^{{(n)}}(t){\bar {B_{{n+1}}}}(t))|_{{t=a}}^{{t=b}}-\int _{{a}}^{{b}}{\bar {B}}_{{n+1}}(t)\,{\mathrm {d}}f^{{(n)}}(t))\\&={\frac {(-1)^{{n-1}}}{(n+1)!}}(f^{{(n)}}(b)B_{{n+1}}(\left\langle b\right\rangle )-f^{{(n)}}(a)B_{{n+1}}(\left\langle a\right\rangle )-\int _{{a}}^{{b}}{\bar {B}}_{{n+1}}(t)f^{{(n+1)}}(t)\,{\mathrm {d}}t)\\&={\frac {(-1)^{{n-1}}B_{{n+1}}}{(n+1)!}}\cdot (f^{{(n)}}(b)-f^{{(n)}}(a))-{\frac {(-1)^{{n-1}}}{(n+1)!}}\int _{{a}}^{{b}}{\bar {B}}_{{n+1}}(t)f^{{(n+1)}}(t)\,{\mathrm {d}}t)\\&={\color {OliveGreen}{\frac {(-1)^{{n+1}}B_{{n+1}}}{(n+1)!}}\cdot (f^{{(n)}}(b)-f^{{(n)}}(a))}+{\color {blue}{\frac {(-1)^{{n}}}{(n+1)!}}\int _{{a}}^{{b}}{\bar {B}}_{{n+1}}(t)f^{{(n+1)}}(t)\,{\mathrm {d}}t)}\\\end{aligned}}

將處理後的積分代入

  {\begin{aligned}\sum _{{a<n\leq b}}f(n)&={\color {red}\int _{{a}}^{{b}}f(t)\,{\mathrm {d}}t}+{\color {OliveGreen}\sum _{{r=0}}^{{n-1}}{\frac {(-1)^{{r+1}}B_{{r+1}}}{(r+1)!}}\cdot (f^{{(r)}}(b)-f^{{(r)}}(a))}+{\color {blue}{\frac {(-1)^{{n-1}}}{n!}}\int _{{a}}^{{b}}{\bar {B}}_{{n}}(t)f^{{(n)}}(t)\,{\mathrm {d}}t}\\&={\color {red}\int _{{a}}^{{b}}f(t)\,{\mathrm {d}}t}+{\color {OliveGreen}\sum _{{r=0}}^{{n-1}}{\frac {(-1)^{{r+1}}B_{{r+1}}}{(r+1)!}}\cdot (f^{{(r)}}(b)-f^{{(r)}}(a))}+{\color {OliveGreen}{\frac {(-1)^{{n+1}}B_{{n+1}}}{(n+1)!}}\cdot (f^{{(n)}}(b)-f^{{(n)}}(a))}+{\color {blue}{\frac {(-1)^{{n}}}{(n+1)!}}\int _{{a}}^{{b}}{\bar {B}}_{{n+1}}(t)f^{{(n+1)}}(t)\,{\mathrm {d}}t)}\\&={\color {red}\int _{{a}}^{{b}}f(t)\,{\mathrm {d}}t}+{\color {OliveGreen}\sum _{{r=0}}^{{n}}{\frac {(-1)^{{r+1}}B_{{r+1}}}{(r+1)!}}\cdot (f^{{(r)}}(b)-f^{{(r)}}(a))}+{\color {blue}{\frac {(-1)^{{(n)}}}{(n+1)!}}\int _{{a}}^{{b}}{\bar {B}}_{{n+1}}(t)f^{{(n+1)}}(t)\,{\mathrm {d}}t}\\\end{aligned}}

得到想要的結果。

 

,焉不能得出公式

The formula

If  m and  n are natural numbers and  f(x) is a complex or real valued continuous function for real numbers  x in the interval {\displaystyle [m,n],} then the integral

  I = \int_m^n f(x)\,dx

can be approximated by the sum (or vice versa)

  {\displaystyle S=f(m+1)+\cdots +f(n-1)+f(n)}

(see rectangle method). The Euler–Maclaurin formula provides expressions for the difference between the sum and the integral in terms of the higher derivatives  {\displaystyle f^{(k)}(x)} evaluated at the end points of the interval, that is to say when  {\displaystyle x=m} and {\displaystyle x=n.}

Explicitly, for a natural number  p and a function f(x) that is  p times continuously differentiable in the interval  {\displaystyle [m,n],} we have

{\displaystyle S-I=\sum _{k=1}^{p}{{\frac {B_{k}}{k!}}(f^{(k-1)}(n)-f^{(k-1)}(m))}+R}

where  B_{k} is the kth Bernoulli number (with  {\displaystyle B_{1}=1/2}) and  R is an error term which is normally small for suitable values of p and depends on  {\displaystyle n,m,p,} and  f.

The formula is often written with the subscript taking only even values, since the odd Bernoulli numbers are zero except for  {\displaystyle B_{1},} in which case we have[1][2]

{\displaystyle \sum _{i=m}^{n}f(i)=\int _{m}^{n}f(x)\,dx+{\frac {f(n)+f(m)}{2}}+\sum _{k=1}^{\lfloor p/2\rfloor }{\frac {B_{2k}}{(2k)!}}(f^{(2k-1)}(n)-f^{(2k-1)}(m))+R}

or alternatively

{\displaystyle \sum _{i=m+1}^{n}f(i)=\int _{m}^{n}f(x)\,dx+{\frac {f(n)-f(m)}{2}}+\sum _{k=1}^{\lfloor p/2\rfloor }{\frac {B_{2k}}{(2k)!}}(f^{(2k-1)}(n)-f^{(2k-1)}(m))+R.}

 

耶?!

 

 

 

 

 

 

 

 

時間序列︰生成函數‧漸近展開︰歐拉的天空《遁甲》

奇門遁甲傳說

據煙波釣叟歌中記載,奇門遁甲起源於傳說時代,黃帝炎帝聯軍和蚩尤在涿鹿展開的一場大戰,蚩尤身高七尺,鐵頭銅身,刀槍不入 ,能呼風喚雨並在戰場上製造迷霧,使得炎黃聯軍陷入不利境地。黃帝於是向天祈禱,終於獲得九天玄女給的河圖洛書和彩鳳銜來的太乙、六壬、遁甲之書,黃帝以此發明了指南車,逆轉了戰局,取得了勝利。黃帝令風后演繹天書,並最終演繹成三式之法:大六壬 、太乙神數、奇門遁甲一千零八十局(陽遁、陰遁各五百四十局) 。後來該術數為姜子牙所習得,由姜子牙刪減為七十二局(陽遁、陰遁各三十六局),再經過姜子牙傳給黃石公,再由黃石公傳給張良,最終由張良將其精簡為現今的一十八局(陽遁、陰遁各九局) 。

天干首起『甲』,易曰︰用九,見群龍無首,吉。故『遁甲』也。在天有三光,日光乙乙萬物生,月光炳炳照大地,星光指向引路灯 ,乙丙丁三奇出矣。紫白飛星九宮八門,太上曰︰禍福無門,惟人自召。雖然,河圖洛書陰陽五行所以言生剋制化沖和之象,所以極其數,蓋揭露天地人三才生殺有時乎??

黃帝陰符經》又稱《陰符經》,全書一卷三篇,傳聞是黃帝所撰,學者大多認為是後人偽托,現有三說:戰國時的蘇秦北魏寇謙之,或唐朝李荃。這部經即使在中國古代的哲學和兵法中都有一定的地位。《陰符經》更是道教的一部重要道經,歷代對它的註解僅次於《道德經》和《南華真經》。《陰符經》有多種版本,在此僅舉一本,略探其自然人生之旨。

黃帝陰符經

天之道,天之行,矣。天有五賊之者五賊在心施行於天,宇宙在乎,萬物生乎天性,人也人性,機也天之道以人也。機,斗轉星移;機,龍蛇起陸;機,天地反覆;天人合德,萬變定基性有巧拙,可以伏藏九竅,在乎三要。可以動靜火生於木,禍發必克,奸生於國 ,時動必潰修練,謂之聖人

也。天地萬物之萬物人之萬物之也 。三盜三才。故曰:食其時,百骸治;動其機,萬化安 。人神而神不知不神所以神。日月有數,大小有定,聖功焉,神明焉。其盜機也,天下莫能見莫能知也。君子得之固躬,小人得之輕命

善聽善視絕利一源,用師倍;三反晝夜,用師倍 。心生於物,死於物,機在於目。天之無恩大恩生,迅雷烈風,莫不蠢然。至樂性餘,至靜性廉。天之至私,用之至公。生者死之根,死者生之根。恩生於害,害生於恩。愚人以天地文理聖,我以時物文理哲。人以,我以不愚。人以,我以不奇。沈水入火,自取滅亡。自然之道靜,故天地萬物生天地之道浸,故陰陽勝陰陽相推,而變化順矣。是故聖人知自然之道不可違,因而制之至靜之道律曆不能契。爰有奇器 ,是生萬象,八卦甲子,神陰陽相勝之術,昭昭乎盡乎象矣。

─── 摘自《天地文理聖,時物文理哲?!

 

古代早有面積術,是微積分發展史上重要之『求積』問題︰

Quadrature (mathematics)

In mathematics, quadrature is a historical term which means determining area. Quadrature problems served as one of the main sources of problems in the development of calculus, and introduce important topics in mathematical analysis.

 
The area of a segment of a parabola is 4/3 that of the area of a certain inscribed triangle.

 

直至今日,數值分析裡仍然常用『拉格朗日插值法』

Lagrange polynomial

In numerical analysis, Lagrange polynomials are used for polynomial interpolation. For a given set of distinct points  x_{j} and numbers  y_{j}, the Lagrange polynomial is the polynomial of lowest degree that assumes at each point  x_{j} the corresponding value  y_{j} (i.e. the functions coincide at each point). The interpolating polynomial of the least degree is unique, however, and since it can be arrived at through multiple methods, referring to “the Lagrange polynomial” is perhaps not as correct as referring to “the Lagrange form” of that unique polynomial.

Although named after Joseph Louis Lagrange, who published it in 1795, the method was first discovered in 1779 by Edward Waring. It is also an easy consequence of a formula published in 1783 by Leonhard Euler.[1]

Uses of Lagrange polynomials include the Newton–Cotes method of numerical integration and Shamir’s secret sharing scheme in cryptography.

Lagrange interpolation is susceptible to Runge’s phenomenon of large oscillation. And changing the points  x_{j} requires recalculating the entire interpolant, so it is often easier to use Newton polynomials instead.

This image shows, for four points ((−9, 5), (−4, 2), (−1, −2), (7, 9)), the (cubic) interpolation polynomial L(x) (dashed, black), which is the sum of the scaled basis polynomials y00(x), y11(x), y22(x) and y33(x). The interpolation polynomial passes through all four control points, and each scaled basis polynomial passes through its respective control point and is 0 where x corresponds to the other three control points.

 

求取『定積分』之值哩!!

Simpson’s rule

In numerical analysis, Simpson’s rule is a method for numerical integration, the numerical approximation of definite integrals. Specifically, it is the following approximation:

{\displaystyle \int _{a}^{b}f(x)\,dx\approx {\tfrac {b-a}{6}}\left[f(a)+4f\left({\tfrac {a+b}{2}}\right)+f(b)\right],}

for points that are equally spaced. For unequally spaced points, see Cartwright.[1]

Simpson’s rule also corresponds to the three-point Newton-Cotes quadrature rule.

The method is credited to the mathematician Thomas Simpson (1710–1761) of Leicestershire, England. Kepler used similar formulas over 100 years prior. For this reason the method is sometimes called Kepler’s rule, or Keplersche Fassregel in German.

Simpson’s rule can be derived by approximating the integrand f (x) (in blue) by the quadratic interpolant P(x) (in red).

Quadratic interpolation

One derivation replaces the integrand  f(x) by the quadratic polynomial (i.e. parabola)  P(x) which takes the same values as  f(x) at the end points a and b and the midpoint m = (a + b) / 2. One can use Lagrange polynomial interpolation to find an expression for this polynomial,

  P(x)=f(a){\tfrac {(x-m)(x-b)}{(a-m)(a-b)}}+f(m){\tfrac {(x-a)(x-b)}{(m-a)(m-b)}}+f(b){\tfrac {(x-a)(x-m)}{(b-a)(b-m)}}.

An easy (albeit tedious) integration by substitution shows that

\int _{a}^{b}P(x)\,dx={\tfrac {b-a}{6}}\left[f(a)+4f\left({\tfrac {a+b}{2}}\right)+f(b)\right].[2]

This calculation can be carried out more easily if one first observes that (by scaling) there is no loss of generality in assuming that  a=-1 and  b=1.

 

若能深知『梯形規則』之來歷,或得見歐拉的天空耶??!!

Trapezoidal rule

In mathematics, and more specifically in numerical analysis, the trapezoidal rule (also known as the trapezoid rule or trapezium rule) is a technique for approximating the definite integral

  \int _{a}^{b}f(x)\,dx.

The trapezoidal rule works by approximating the region under the graph of the function  f(x) as a trapezoid and calculating its area. It follows that

  \int _{a}^{b}f(x)\,dx\approx (b-a)\left[{\frac {f(a)+f(b)}{2}}\right].

The function f(x) (in blue) is approximated by a linear function (in red).

 

An animation showing how the trapezoidal rule approximation improves with more strips.

 

試問什麼條件下 \int_{a}^{b} f(x) dx = (b-a) \left[ \frac{f(a)+f(b)}{2} \right] 呢?

由『微積分基本定理』及『分部積分法』可知

\int_{a}^{b} f(x) dx = \int_{a}^{b} \frac{d (x+c)}{dx} f(x) dx

= (x+c) f(x) |_{a}^{b} - \int_{a}^{b} (x+c) f^{'} (x) dx

假使 f^{'} (x) = k 是常數,如果定『積分常數』為 c = - \frac{a+b}{2} ,那麼原式等於

= (b-a) \left[ \frac{f(a)+f(b)}{2} \right] - k \int_{a}^{b} (x - \frac{a+b}{2}) dx

= (b-a) \left[ \frac{f(a)+f(b)}{2} \right] - k \left[ \frac{x^2}{2} - \frac{a+b}{2} x \right]  |_{a}^{b}

= (b-a) \left[ \frac{f(a)+f(b)}{2} \right] 呦!

將能通以和為貴之道吧☆

 

 

 

 

 

 

 

 

 

 

 

時間序列︰生成函數‧漸近展開︰當白努利遇上傅立葉《V》

『光』的『祕密』是什麼呢?『光』對所有『慣性觀察者』都是『恆定』的『光速』 c !因此『光』將『時間』 t 與『空間』 x 聯繫了起來 x = c \cdot t ,凌波寰宇 \Psi ( 2 \pi  \frac{x}{\lambda} - 2 \pi \frac{t}{T} ), c = \frac{\lambda}{T} = f \cdot \lambda 亙古至今不知幾百億光年耶!!故而『光』之本性可謂『純真』乎??

據『歷史典故』上說,東晉慧遠大師主持東林寺,立下了規矩『影不出山迹不入谷』;一過虎溪,寺後山虎則吼。一日大詩人陶淵明和道士陸修靜來訪,談的投機,送行時不覺過了虎溪橋,待聞得虎嘯後方恍然大悟,相視大笑而別,後世稱作『虎溪三笑』。其後有清朝唐蝸寄題的廬山東林寺三笑庭名聯:

橋跨虎溪,三教三源流,三人三笑語;
蓮開僧舍,一花一世界,一葉一如來。

今天的人或許較熟悉英國詩人布莱克的『一沙一世界,一花一天堂。』名句。這個名句出自一首長詩《純真的徵兆》的起頭︰

snadworld

天堂鳥-花

220px-Blake_jacobsladder

Auguries of Innocence

To see a world in a grain of sand,
一粒沙裡世界
And a heaven in a wild flower,
一朵花中天堂
Hold infinity in the palm of your hand,
掌尺足無限
And eternity in an hour.
時針能永恆

布莱克生於 1757 年,幼年就個性獨特討厭正統學校的教條氣息,因而拒絕入學,博覽眾書自學成家,由於潛心研讀洛克博克經驗主義哲學著作,於是對這個大千世界有了深刻認識早熟的他為減輕家計重擔和考慮弟妹前途,放棄了畫家夢想,十四歲時就選擇了去雕版印刷作坊當個學徒,二十二歲出師,…
是英國浪漫主義詩人的第一人
雅各的天梯,布莱克的版畫,布莱克『自爬』?

博克的名著【壯美優美觀念起源之哲學探究】,布莱克用來觀察飛鳥之姿』── Auguries ──,體驗預示藝術參與,果然恰當!!就像『掌尺』的可成無限,用時針的『循環』以度永恆一樣;也許布莱克的浪漫充滿著理性思辨,其要總在觀察

─── 摘自《一個奇想!!

 

那位詩人布萊克深得『時空』 t, x『周期』 T, \lambda 的『奧義』

掌尺足握無限
時針能持永恆

也。

那麼考察白努利周期函數

{\tilde{B}}_1 (x) = - \frac{1}{\pi} \sum \limits_{k=1}^{\infty}  \frac{\sin(2 \pi k \cdot x)}{k}

{\tilde{B}}_{2n} (x) = {(-1)}^{n+1} \frac{2 (2n)!}{{(2 \pi )}^{2n}} \sum \limits_{k=1}^{\infty}  \frac{\cos(2 \pi k \cdot x)}{k^{2n}}

{\tilde{B}}_{2n+1} (x) = {(-1)}^{n+1} \frac{2 (2n+1)!}{{(2 \pi )}^{2n+1}} \sum \limits_{k=1}^{\infty}  \frac{\sin(2 \pi k \cdot x)}{k^{2n+1}}

\sin(2 \pi k \cdot x) = \sin (2 \pi \left[ \frac{x}{\frac{1}{k}} \right] )\cos(2 \pi k \cdot x) = \cos (2 \pi \left[ \frac{x}{\frac{1}{k}} \right] )

『最大周期』為 \frac{1}{k} = 1 ,所以除了 {\tilde{B}}_1 (x) 以外,

{\tilde{B}}_1 (0^{+}) = -\frac{1}{2}, \ {\tilde{B}}_1 (1^{-}) = \frac{1}{2}

對任一自然數 m

{\tilde{B}}_{2n} (m) = {\tilde{B}}_{2n} (1) = {\tilde{B}}_{2n} (0)

{\tilde{B}}_{2n+1} (m) = {\tilde{B}}_{2n+1} (1) = {\tilde{B}}_{2n+1} (0) = 0

或可發現歐拉之求和原始矣☆

歐拉-麥克勞林求和公式有時也被寫成如下形式:[3]
\sum _{{y<n\leq x}}f(n)=\int _{{y}}^{{x}}f(t)\,{\mathrm {d}}t+\int _{{y}}^{{x}}(t-\left\lfloor t\right\rfloor )f'(t)\,{\mathrm {d}}t+f(x)(\left\lfloor x\right\rfloor -x)-f(y)(\left\lfloor y\right\rfloor -y)

這是歐拉給出的原始形式。