4.13-bo'lim : Nyuton usuli
2. Agar \({x_{\,0}} = 1\) bo'lsa, \(f\left( x \right) = x\cos \left( x \right) - {x^2}\) uchun \({x_{\,2}}\) ni aniqlash uchun Nyuton usulidan foydalaning.
Yechim
Haqiqatan ham, bu muammoni hal qilish uchun unchalik ko'p narsa yo'q. Biz bilamizki, Nyuton usulining asosiy formulasi:
\[{x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\]shuning uchun biz qilishimiz kerak bo'lgan narsa buni ikki marta bosib o'tishdir.
Mana funktsiyaning hosilasi, chunki bizga bu kerak bo'ladi.
\[f'\left( x \right) = \cos \left( x \right) - x\sin \left( x \right) - 2x\]Endi biz yuqoridagi formulani ikki marta bajarishimiz kerak.
2-qadam
\({x_{\,1}}\) formulasi orqali birinchi iteratsiya:
\[{x_{\,1}} = {x_{\,0}} - \frac{{f\left( {{x_{\,0}}} \right)}}{{f'\left( {{x_{\,0}}} \right)}} = 1 - \frac{{f\left( 1 \right)}}{{f'\left( 1 \right)}} = 1 - \frac{{ - 0.4596976941}}{{ - 2.301168679}} = 0.8002329432\]Shuni unutmangki, biz uchun burchaklar har doim radyanlarda bo'ladi, shuning uchun kalkulyatoringiz radyanlarda hisoblash uchun o'rnatilganligiga ishonch hosil qiling.
3-qadam
\({x_{\,2}}\) formulasi orqali ikkinchi iteratsiya:
\[\begin{align*}{x_{\,2}} = {x_{\,1}} - \frac{{f\left( {{x_{\,1}}} \right)}}{{f'\left( {{x_{\,1}}} \right)}} & = 0.8002329432 - \frac{{f\left( {0.8002329432} \right)}}{{f'\left( {0.8002329432} \right)}}\\ & = 0.8002329432 - \frac{{ - 0.08297883948}}{{ - 1.478108132}} = 0.7440943985\end{align*}\] Shunday qilib, bu muammoning javobi \(\require{bbox} \bbox[2pt,border:1px solid black]{{{x_{\,2}} = 0.7440943985}}\).Muammo bayonotida so'ralmagan bo'lsa-da, haqiqiy ildiz 0,739085133215161. Shuni ham yodda tutingki, bu olish uchun ba'zi bir hisoblash yordami talab qilinadi va bu umuman qo'lda ololmaydigan narsa emas.