時間序列︰生成函數‧漸近展開︰白努利多項式之根《四》

若說『科學發現』依賴『觀察』或許毋庸置疑︰

Observation

Observation is the active acquisition of information from a primary source. In living beings, observation employs the senses. In science, observation can also involve the recording of data via the use of instruments. The term may also refer to any data collected during the scientific activity. Observations can be qualitative, that is, only the absence or presence of a property is noted, or quantitative if a numerical value is attached to the observed phenomenon by counting or measuring.

Observer is someone who gathers information about observed phenomenon, but does not intervene. Observing the air traffic in Rõuge, Estonia.

Observation in science

The scientific method requires observations of nature to formulate and test hypotheses.[1] It consists of these steps:[2][3]

  1. Asking a question about a natural phenomenon
  2. Making observations of the phenomenon
  3. Hypothesizing an explanation for the phenomenon
  4. Predicting logical, observable consequences of the hypothesis that have not yet been investigated
  5. Testing the hypothesis’ predictions by an experiment, observational study, field study, or simulation
  6. Forming a conclusion from data gathered in the experiment, or making a revised/new hypothesis and repeating the process
  7. Writing out a description of the method of observation and the results or conclusions reached
  8. Review of the results by peers with experience researching the same phenomenon

Observations play a role in the second and fifth steps of the scientific method. However the need for reproducibility requires that observations by different observers can be comparable. Human sense impressions are subjective and qualitative, making them difficult to record or compare. The use of measurement developed to allow recording and comparison of observations made at different times and places, by different people. Measurement consists of using observation to compare the phenomenon being observed to a standard. The standard of comparison can be an artifact, process, or definition which can be duplicated or shared by all observers, if not by direct measurement then by counting the number of aspects or properties of the object that are comparable to the standard. Measurement reduces an observation to a number which can be recorded, and two observations which result in the same number are equal within the resolution of the process.

Senses are limited, and are subject to errors in perception such as optical illusions. Scientific instruments were developed to magnify human powers of observation, such as weighing scales, clocks, telescopes, microscopes, thermometers, cameras, and tape recorders, and also translate into perceptible form events that are unobservable by human senses, such as indicator dyes, voltmeters, spectrometers, infrared cameras, oscilloscopes, interferometers, geiger counters, x-ray machines, and radio receivers.

One problem encountered throughout scientific fields is that the observation may affect the process being observed, resulting in a different outcome than if the process was unobserved. This is called the observer effect. For example, it is not normally possible to check the air pressure in an automobile tire without letting out some of the air, thereby changing the pressure. However, in most fields of science it is possible to reduce the effects of observation to insignificance by using better instruments.

Considered as a physical process itself, all forms of observation (human or instrumental) involve amplification and are thus thermodynamically irreversible processes, increasing entropy.

 

要講歷史上許多『數學發現』常因『觀察』,恐怕甚為可疑!簡單的講︰宇宙時空裡物質之性質,物質間作用的數理表達 …… ,如何『發現』呢?因為東西已在那裡,現象早就浮現 …… ,『好奇心』驅使『觀察者』研究『量測數據』,尋找『現象關係』,然後形成了『假說』,藉著『理化實驗』持續向大自然『發問』…… 乎!!抽象言之,什麼科學沒有數學居其中耶??那麼『科學方法』── 觀察 ── 焉不能用於數學咦☆

更何況數學還有好處哩,不必作『實驗』,直接就提出『命題』,一般可以『證明』真假勒☆

難到『觀察』不能發現關係嗎??!!

pi@raspberrypi:~ ipython3 Python 3.4.2 (default, Oct 19 2014, 13:31:11)  Type "copyright", "credits" or "license" for more information.  IPython 2.3.0 -- An enhanced Interactive Python. ?         -> Introduction and overview of IPython's features. %quickref -> Quick reference. help      -> Python's own help system. object?   -> Details about 'object', use 'object??' for extra details.  In [1]: from sympy import *  In [2]: from sympy.plotting import plot  In [3]: init_printing()  In [4]: x = symbols('x')  In [5]: B_1_X = bernoulli(1,x)  In [6]: B_2_X = bernoulli(2,x)  In [7]: B_3_X = bernoulli(3,x)  In [8]: B_4_X = bernoulli(4,x)  In [9]: B_5_X = bernoulli(5,x)  In [10]: plot(B_1_X, B_2_X , (x, -1,2), ylim = (-0.5, 0.5)) Out[10]: <sympy.plotting.plot.Plot at 0x74d7bb90>  In [11]: plot(B_2_X, B_3_X , (x, -1,2), ylim = (-0.5, 0.5)) Out[11]: <sympy.plotting.plot.Plot at 0x72a8a3d0>  In [12]: plot(B_3_X, B_4_X , (x, -1,2), ylim = (-0.5, 0.5)) Out[12]: <sympy.plotting.plot.Plot at 0x71544cb0>  In [13]: plot(B_4_X, B_5_X , (x, -1,2), ylim = (-0.5, 0.5)) Out[13]: <sympy.plotting.plot.Plot at 0x7277ceb0>  In [14]:  </pre>     \frac{d \ B_2 (x) }{dx} = \frac{d}{dx} \left( x^2 - x + \frac{1}{6} \right)= 2 \cdot B_1 (x) = 2 \cdot \left( x - \frac{1}{2} \right) <span style="color: #003300;">B_2 (x)︰在閉區間[0, \frac{1}{2}]嚴格遞減↘。從B_2 (0)正→零→負→負極小值B_2 (\frac{1}{2})。</span>  <span style="color: #003300;">在閉區間[\frac{1}{2}, 1]嚴格遞增↗。從負極小值B_2 (\frac{1}{2})→負→零→正→B_2(1)。</span>  <span style="color: #003300;">B_1 (x) ︰在開區間(0, \frac{1}{2})恆負B_1 (x) < 0 。從B_1 (0) = - \frac{1}{2}→B_1 (\frac{1}{2}) = 0。</span>  <span style="color: #003300;">在開區間(\frac{1}{2}, 1)恆正B_1 (x) > 0。從B_1 (\frac{1}{2}) = 0→B_1 (1) = \frac{1}{2}。</span>  <img class="alignnone size-full wp-image-68422" src="http://www.freesandal.org/wp-content/uploads/Figure-1X_2X.png" alt="" width="652" height="553" />   \frac{d \ B_3 (x) }{dx} = \frac{d}{dx} \left( x^3 - \frac{3}{2} x^2 + \frac{1}{2} x \right)= 3 \cdot B_2 (x) = 3 \cdot \left( x^2 - x + \frac{1}{6} \right) <span style="color: #003300;">B_3 (x) ︰在開區間(0, \frac{1}{2})恆正B_3 (x) > 0 。從B_3 (0) = 0→B_3 (\frac{1}{2}) = 0。</span>  <span style="color: #003300;">在開區間(\frac{1}{2}, 1)恆負B_3 (x) < 0。從B_3 (\frac{1}{2}) = 0→B_3 (1) = 0。</span>  <img class="alignnone size-full wp-image-68423" src="http://www.freesandal.org/wp-content/uploads/Figure-2X_3X.png" alt="" width="652" height="553" />   \frac{d \ B_4 (x) }{dx} = \frac{d}{dx} \left( x^4 - 2 x^3 + x^2 -\frac{1}{30} \right)= 4 \cdot B_3 (x) = 4 \cdot \left( x^3 - \frac{3}{2} x^2 + \frac{1}{2} x \right) <span style="color: #003300;">B_4 (x)︰在閉區間[0, \frac{1}{2}]嚴格遞增↗。從B_4 (0)負→零→正→正極大值B_4 (\frac{1}{2})。</span>  <span style="color: #003300;">在閉區間[\frac{1}{2}, 1]嚴格遞減↘。從正極大值B_4 (\frac{1}{2})→正→零→負→B_4(1)。</span>  <img class="alignnone size-full wp-image-68424" src="http://www.freesandal.org/wp-content/uploads/Figure-3X_4X.png" alt="" width="652" height="553" />   \frac{d \ B_5 (x) }{dx} = \frac{d}{dx} \left( x^5 - \frac{5}{2} x^4 + \frac{5}{3} x^3 - \frac{1}{6} x \right)= 5 \cdot B_4 (x) = 5 \cdot \left( x^4 - 2 x^3 + x^2 -\frac{1}{30} \right) <span style="color: #003300;">B_5 (x) ︰在開區間(0, \frac{1}{2})恆負B_5 (x) < 0 。從B_5 (0) = 0→B_5 (\frac{1}{2}) = 0。</span>  <span style="color: #003300;">在開區間(\frac{1}{2}, 1)恆正B_5 (x) > 0。從B_5 (\frac{1}{2}) = 0→B_5 (1) = 0。</span>  <img class="alignnone size-full wp-image-68425" src="http://www.freesandal.org/wp-content/uploads/Figure-4X_5X.png" alt="" width="652" height="553" />     <span style="color: #003300;">因此除了B_1 (x) = x - \frac{1}{2}之外,我們可將白努力多項式B_m (x)歸納成四種</span>  <span style="color: #003300;">B_{4k}、B_{4k+1}、B_{4k+2}、B_{4k+3,兩類</span>  <span style="color: #003300;">◎ 偶次2n類︰{(-1)}^n \cdot B_{2n} (x)</span>  <span style="color: #003300;">在閉區間[0, \frac{1}{2}]嚴格遞增↗。從{(-1)}^n \cdot B_{2n} (0)負→零→正→正極大值{(-1)}^n \cdot B_{2n} (\frac{1}{2})。</span>  <span style="color: #003300;">在閉區間[\frac{1}{2}, 1]嚴格遞減↘。從正極大值{(-1)}^n \cdot B_{2n} (\frac{1}{2})→正→零→負→{(-1)}^n B_{2n}(1)。</span>  <span style="color: #003300;">◎ 奇次2n+1類︰{(-1)}^n \cdot B_{2n+1} (x)</span>  <span style="color: #003300;">在開區間(0, \frac{1}{2})恆負{(-1)}^n \cdot B_{2n+1} (x) < 0 。從B_{2n+1} (0) = 0→B_{2n+1} (\frac{1}{2}) = 0。</span>  <span style="color: #003300;">在開區間(\frac{1}{2}, 1)恆正{(-1)}^n \cdot B_{2n+1} (x) > 0。從B_{2n+1} (\frac{1}{2}) = 0→B_{2n+1} (1) = 0。</span>  <span style="color: #003300;">且以『<a style="color: #003300;" href="http://www.freesandal.org/?p=3297">數學歸納法</a>』簡略證明如下︰</span>  <span style="color: #808080;">‧已知(-1) \cdot B_2 (x)為真。</span>  <span style="color: #808080;">‧偶次2n類\longrightarrow奇次2n+1類為真。</span>  <span style="color: #808080;">假設B (x) = {(-1)}^n \cdot B_{2n+1} (x),由B{'} (x) = (2n +1) \left( {(-1)}^n B_{2n} (x) \right)與歸納假設{(-1)}^n B_{2n} 性質為真可知矣。</span>  <span style="color: #808080;">‧奇次2n+1類\longrightarrow偶次2(n+1)類為真。</span>  <span style="color: #808080;">假設B (x) = {(-1)}^{n+1} \cdot B_{2n+2} (x),從B{'} (x) = - (2n +2) \left( {(-1)}^n B_{2n+1} (x) \right)及前述證明{(-1)}^n B_{2n+1} $ 性質為真可得也。

證明完畢☆