User:Chela5808/My sandbox Sam2

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Before Styling Sample 1 Sample 2 Sample 3 Sample 4 Sample 5 Sample 6
After Styling Sample 1 Sample 2 Sample 3 Sample 4 Sample 5 Sample 6


CONVERTING SAMPLE 2 CONTENT TO WE FORMAT

Semi-automated procedure using apps: AdobeAcrobat, OpenOFFice and NotePad, avoiding usage of sam2.tex file:

STEP 1. Using app. Adobe Acrobat 8 Pro Version 8.1.3

  • Open file "sam2.pdf"
  • Export file to HTML3.2
  • File generated is "sam2.htm"

STEP 2. Using app. OppenOffice.org Writer/Web Version 3.0.0

  • Open file "sam2.htm"
  • Export file to MediaWiki(.txt)
  • File generated is "sam2.txt"

STEP 3. Using app. Notepad -Windox XP SP2

  • Open file "sam2.txt"
  • Selected all, copy to clipboard

STEP 4. In WikiEducator

  • Generate new page User:Chela5808/My sandbox Sam2
  • Paste-Save
  • Output below


Comments:

  • Conversion time from Step 1 to Step 4: 20 min.
  • Further text styling by hand is required
  • Further [math]\mathrm{L\!\!^{{}_{\scriptstyle A}} \!\!\!\!\!\;\; T\!_{\displaystyle E} \! X} \,[/math] styling by hand is required ( tags <math></math> )




Sam 2 Output (Before any styling by hand)

Created from PDF via Acrobat SaveAsXMLMapping table version: 28-February-2003The function defined in (??) has amazing properties. Let’s multiply two such series together, using two different values of x (call them x = p in one series and x = q in the other):

� 2 �� 2 �

pq

f (p)f (q)= 1+ p + + ... 1+ q +

+ ...

2! 2!

� 22 �

pq

= 1+(p + q)+ + pq +

+ ...

2! 2! (p + q)2

= 1+(p + q)+ + ... , (1)

2!

– including terms only up to the ‘second degree’ (i.e. those with not more than two variables multiplied together). The result seems to be just the same function (??), but with the new variable x = p + q. And if you go on, always putting together products of the same degree, you’ll find the next terms are

32 2

(p + q)3 /3! = (p +3pq +3pq + q 3 )/3! (third degree)

and

43 22 3

(p + q)4 /4! = (p +4pq +6pq +4pq + q 4 )/4! (degree 4.) As you can guess, if we take more terms we’re going to get the result

(p + q)2 (p + q)3

f (p)f (q)=1+(p + q)+ + + ... = f (p + q). (2)

2! 3! To get a proof of this result is much harder: you have to

11
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