Question
Download Solution PDFन्यूटन-रफसन विधि का उपयोग करके एक पुनरावृत्ति के बाद x3 - 13 = 0 के मूल का अनुमानित मान क्या है जब 3.5 के रूप में प्रारंभिक मान है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंकल्पना:
न्यूटन-रफसन विधि:
निम्न द्वारा पुनरावृति सूत्र दिया गया है
\({x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\)
जहाँ x0 समीकरण f(x) = 0 का प्रारंभिक मान / मूल है
गणना:
दिया हुआ:
f(x) = x3 - 13, x0 = 3.5
f'(x) = 3x2
f(x0) = f(3.5) = 3.53 - 13 = 29.875
f'(x0) = f'(3.5) = 3 × 3.52 = 36.75
हम जानते हैं कि
\({x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\)
\({x_{1}} = {x_0} - \frac{{f\left( {{x_0}} \right)}}{{f'\left( {{x_0}} \right)}}\)
\(x_1=3.5-\frac {29.875}{36.75}\)
∴ x1 = 2.6871
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