Scilab/C4/Integration/Khasi

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Time Narration
00:01 Paralok, ngi pdiangsngewbha ia phi sha ka Spoken Tutorial halor ka Composite Numerical Integration.
00:07 Ha kaba kut jong kane ka jinghikai, phin sa nang kumno ban:
00:11 Develop ia ka Scilab code na ka bynta bun jait ki Composite Numerical Integration algorithms
00:17 Divide ia ka integral ha ki equal intervals
00:21 Apply ia ka algorithm sha man kawei ka interval bad
00:24 Khein ia ka composite value of the integral.
00:28 Ban record ia kane ka jinghikai, nga pyndonkam
00:30 Ubuntu 12.04 kum ka operating system
00:34 Ryngkat bad ka Scilab 5.3.3 version.
00:38 Shwa ban pyrshang ia kane ka jinghikai, u ne ka nongpule kidei ban don ia ka jingtip ba donkam jong
00:42 Ka Scilab bad
00:44 Integration using Numerical Methods.
00:47 Na ka bynta ka Scilab, sngewbha peit ia ki jinghikai ba iadei ba don ha ka Spoken Tutorial website.
00:55 Numerical Integration kadei ka
00:58 Jingpule halor kumno ba u numerical value jong ka integral lah ban wad.
01:03 La pyndonkam ia u haba u mathematical integration ba dei thik um don.
01:08 U antad ia u definite integral na ki values jong ka integrand.
01:15 To ngin pule ia ka Composite Trapezoidal Rule.
01:18 Kane ka rule kadei ka tnad baiar jong ka trapezoidal rule.
01:22 Ngi phiah ia ka interval a comma b ha ki n intervals ba iaryngkat.
01:29 Nangta, h equals to b minus a divided by n kadei ka common lengths jong ki intervals.
01:36 Nangta ka composite trapezoidal rule la ai da:
01:41 The integral of the function F of x in the interval a to b is approximately equal to h multiplied by the sum of the values of the function at x zero to x n
01:57 To ngin solve ia kawei ka nuksa da kaba pyndonkam ia ka composite trapezoidal rule.
02:02 Shu shim ia u number jong ki intervals n is equal to 10 (n=10)
02:09 To ngin peit ia u code na ka bynta ka Composite Trapezoidal Rule ha ka Scilab editor
02:16 Nyngkong ngi define ia ka function ryngkat bad ki parameters f , a , b , n.
02:22 f u thew ia ka function ba ngi hap ban solve.
02:25 a udei u lower limit jong ka integral,
02:28 b kadei ka upper limit jong ka integral bad
02:31 n udei u number jong ki intervals.
02:34 linspace function la pyndonkam ban create shiphew tylli ki intervals ba iaryngkat hapdeng zero bad one.
02:42 Ngi wad ia u value jong ka integral bad buh ia u ha I one.
02:49 Nion ha Execute ha ka Scilab editor bad jied Save and execute ia u code.
03:02 Define ia ka example function da kaba type:
03:05 d e f f open parenthesis open single quote open square bracket y close square bracket is equal to f of x close quote comma open quote y is equal to one by open parenthesis two asterisk x plus one close parenthesis close quote close parenthesis
03:30 Nion Enter . Type Trap underscore composite open parenthesis f comma zero comma one comma ten close parenthesis
03:41 Nion Enter .
03:43 Ka jubab la pyni ha ka console .
03:47 Nangta ngin sa pule ia ka Composite Simpson's rule.
03:51 Ha kane ka rule, ngi thep ia ka interval a comma b sha ka n is greater than 1 sub-intervals kaba ka jingjrong ka iaryngkat .
04:03 Apply Simpson's rule sha man kawei ka interval.
04:06 Ngi ioh ia u value jong ka integral u ban dei:
04:10 h by three multiplied by the sum of f zero, four into f one , two into f two to f n.
04:19 To ngin solve ia kawei ka nuksa da kaba pyndonkam ia ka Composite Simpson's rule.
04:24 Ngi ai ia ka function one by one plus x cube d x in the interval one to two.
04:32 Ai ba ka number jong ki intervals kan dei twenty .
04:37 To ngin phai sha u code na ka bynta ka Composite Simpson's rule.
04:42 Nyngkong eh ngi define ia ka function ryngkat ki parameters f , a , b , n.
04:49 f ka thew sha ka function ba ngi hap ban solve,
04:52 a udei u lower limit jong ka integral,
04:56 b udei u upper limit jong ka integral bad
04:58 n udei u number jong ki intervals.
05:02 Ngi wad artylli ki sets jong ki points
05:04 Ngi wad ia u value jong ka function ryngkat bad uwei u set bad multiply ia u da two.
05:10 Bad kawei ka set, ngi wad ia u value bad multiply ia u da four.
05:16 Ngi sum ia kine ki values bad muliply ia u da h by three and store the final value in I .
05:24 To ngin execute ia u code.
05:28 Save bad execute ia ka file Simp underscore composite dot s c i.
05:39 Nyngkong to ngan clear ia ka screen.
05:42 Define ia ka function ba la ai ha ka nuksa da kaba type:
05:45 d e f f open parenthesis open single quote open square bracket y close square bracket is equal to f of x close quote comma open quote y is equal to one divided by open parenthesis one plus x cube close parenthesis close quote close parenthesis
06:12 Nion Enter .
06:14 Type Simp underscore composite open parenthesis f comma one comma two comma twenty close parenthesis
06:24 Nion Enter .
06:26 Ka jubab la pyni ha ka console.
06:31 To mynta ngin peit ia ka Composite Midpoint Rule.
06:35 Ka integrate ia ki polynomials jong ka degree one lane duna ia one.
06:40 Divide ia ka interval a comma b sha ki sub-intervalsba iaryngkat ka width.
06:49 Wad ia u midpoint jong man kawei ka interval ba la kdew da x i .
06:54 Ngi wad ia ka sum jong ki values jong ka integral ha man uwei u midpoint.
07:00 To ngin solve ia kane ka problem da kaba pyndonkam ia ka Composite Midpoint Rule.
07:05 We are given a function one minus x square d x in the interval zero to one point five.
07:15 Ngi shim n is equal to twenty .
07:18 To ngin peit ia u code na ka bynta ka Composite Midpoint rule.
07:24 Nyngkong ngi define ia ka function ryngkat ki parameters f , a , b , n.
07:30 f ka thew sha ka function ba ngi hap ban solve,
07:33 a udei u lower limit jong ka integral,
07:36 b udei u upper limit jong ka integral bad
07:39 n udei u number jong ki intervals.
07:41 Ngi wad ia u midpoint jong man kawei ka interval.
07:45 Wad ia u value jong ka integral ha man uwei u midpoint bad nangta wad ia ka sum bad buh ia ka ha I.
07:53 To mynta ngin ia solve ia ka nuksa.
07:55 Save bad execute ia ka file mid underscore composite dot s c i.
08:04 To ngan clear ia ka screen.
08:08 Ngi define ia ka function ba la ai ha ka nuksa da kaba type:
08:13 d e f f open parenthesis open single quote open square bracket y close square bracket is equal to f of x close quote comma open quote y is equal to one minus x square close quote close parenthesis
08:37 Nion Enter.
08:39 Nangta type mid underscore composite open parenthesis f comma zero comma one point five comma twenty close parenthesis
08:53 Nion Enter. Ka jubab la pyni ha ka console.
08:59 To ngin batai kyllum ia kane ka jinghikai.
09:02 Ha kane ka jinghikai ngi la pule ban:
09:04 Develop ia u Scilab code na ka bynta kanumerical integration
09:08 Wad ia u value jong ka integral.
09:11 Peit ia ka video ba don ha ka link ba la pyni harum.
09:15 Ka kyllum lang ia ka Spoken Tutorial project.
09:18 Lada phim don ia ka bandwidth kaba biang, phi lah ban shu download bad peit ia ka.
09:23 Ka kynhun jong ka Spoken Tutorial
09:25 Ka pynlong ia ki workshops da kaba pyndonkam da ki spoken turorials
09:29 Ka ai certificates sha kito kiba pass ha ka online test.
09:32 Na ka bynta kham bun ki jingtip ba bniah, sngewbha thoh sha ka contact@spoken-tutorial.org.
09:40 Spoken Tutorial Project kadei shibynta jong ka Talk to a Teacher project.
09:45 La kyrshan ia ka da ka National Mission on Eduction lyngba ka ICT, MHRD, Sorkar India
09:52 Kham bun ki jingtip halor kane ka mission kidon ha http://spoken-tutorial.org/NMEICT-Intro.
10:03 Nga i Meboreen na Shillong nga pynkut ia kane. Khublei shibun.

Contributors and Content Editors

Meboreen Mary