davriy funksiyalarni integrallash

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1576504035.doc p [ ] ) 1 . 4 ( ) 2 , 0 ( ) ( ) ( ) ( ) ( 2 0 1 ) ( p p î » ò å = n k n k n k n k x x f a dx x f ( ) 6816 , 1 1419 , 1 0691 , 1 0009 , 1 6 = + + = p õ = - n k n k x x 1 ) ( 2 2 sin ) ( ) ( 2 ) ( 1 ,..., , n n n n x x x [ ] x x x x x x n k n k n k sin sin cos cos 1 2 1 2 sin ) ( ) ( ) ( 2 - - = - 0 2 sin ) ( 2 0 2 0 1 2 > - = ò …

= n j n j ik ik n j n j ik n j j e e e x f 1 2 ) 1 ( 1 2 ) 1 ( 1 ) ( p a p a 0 1 1 1 1 2 2 2 = - - = - - · n ik ki ik n ik n n ik ik e e e e e e p p a p a p a ú û ù ê ë é î n t , 0 a ) 2 sin 2 cos ( ) ( 1 1 0 1 kx t b kx t a a x t k n k k n p p + + = å - = - ò - = 2 0 2 2 sin 1 ) ( p j j k d k k ò - - = 2 2 2 sin 25 , …

x) = 1 bo’lib, (4.2) formulaning aniq ekani ravshandir.endi 0<k<n-1 bo’lsin. u holda shu bilan bir vaqtda kvadratur yig’indi ham nolga teng: shunday qilib, (4.2) formulaning trigonometrik aniqlik tartibiga teng ekan ixtiyoriy uchun (4.3) kvadratur formulaning (n - 1) - tartibli ixtiyoriy trigonometrik ko’phad uchun aniq tenglikka aylanishini ko’rsatish qiyin emas. misol sifatida ushbu to’liq elliptik integralning k= 0,5 dagi qiymatini to’rt xona aniqligida hisoblaylik. integral ostidagi funksiya juft va л davrli bo’lgani sababli k(0,5) ni ko’rinishda yozish mumkin. bu integralni hisoblash uchun (4.3) da т = л deb olamiz. endi п = 6 deb olib, tugunlarni = 0 nuqtaga nisbatan simmetrik ravishda joylashtiramiz: u holda k(0,5) ning jadvaldagi qiymati 1,6858. demak xato 0,0042 ga teng. foydalanilgan adabiyotlar: 1. соболев с.л. введение в теорию кубатурных формул. –м.: «наука». -2008г. 2. никольский с.м. квадратурные формулы. 2-е изд. –м.: «наука». -1992г. 3. крылов в.н. приближённые вычисления интегралов. –м.: «наука». -2007 г. 4. …

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1576504035.doc p [ ] ) 1 . 4 ( ) 2 , 0 ( ) ( ) ( ) ( ) ( 2 0 1 ) ( p p î » ò å = n k n k n k n k x x f a dx x f ( ) 6816 , 1 1419 , 1 0691 , 1 0009 , 1 6 = + + = p õ = - n k n k x x 1 ) ( 2 2 sin ) ( ) ( 2 ) ( 1 ,..., , n n n n x x x [ ] x x x x x x n k n k n k sin sin cos cos 1 …

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