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anonymous

  • 5 years ago

Solve sinx=x^2 correct to four decimals using Newton's method with x sub zero=1.

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  1. anonymous
    • 5 years ago
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    Newton's method was developed to find the roots of a function (i.e. what f(x)=0). As such, when you have something like you have, you need to put it in that form first. So, sin(x) =x^2 then f(x):=sinx-x2=0. You can now proceed. \[x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)} \rightarrow x_{n+1}=x_n-\frac{\sin x_n - x_n^2}{\cos x_n -2x_n}\]

  2. anonymous
    • 5 years ago
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    So,\[x_1=1-\frac{\sin(1)-(1)^2}{\cos(1)-2(1)} \approx 0.891395995\]and this number becomes your second input, and you keep going (technically, forever).

  3. anonymous
    • 5 years ago
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    \[x_2=0.891395995-\frac{\sin (0.891395995)-(0.891395995)^2}{\cos(0.891395995)-2(0.891395995)}\]\[\approx 0.876984844\]

  4. anonymous
    • 5 years ago
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    Mathematica spits a terminating output of about 0.8767262153950624.

  5. anonymous
    • 5 years ago
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    The other solution is x=0.

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