न्यूटन-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?

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  1. 2.687
  2. 2.678
  3. 3.607
  4. 3.597

Answer (Detailed Solution Below)

Option 1 : 2.687

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संकल्पना:

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) = x- 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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