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Use Newton's method to approximate a root of the equation    \cos(x^2 + 3) = x^3  as follows. Let  x_1 = 1  be the initial approximation. The second approximation  x_2 is?

Sagot :

crisforp
x_2 = x_1 -  f(x_1) / f ' (x_1) => x_2 = 1 + (cos4 -1 )/ ( 2sin4 + 3) ≈ 1 - 7.64 ≈ - 6.64;
yoodyannapolis

The second approximation [tex]x_2[/tex] is -0.112

Newton iteration formula is:

[tex]x_{n+1}=x_n - \frac{f(x_n)}{f'(x_{n+1})}[/tex]

We have the initial root is [tex]x_1 = 1[/tex] function:

[tex]\cos(x^2 + 3) = x^3 \\ \cos(x^2 + 3) - x^3 =0\\f(x)=\cos(x^2 + 3) - x^3\\f'(x)=-2x \ sinx(x^2 + 3) - 3x^2\\f(1)=\cos(1^2 + 3) - 1^3\\f(1)=\cos(4) - 1\\f(1)=-1.6534\\f'(1)=-2.1 \ sinx(1^2 + 3) - 3.1^2\\f'(1)=-1.486[/tex]

Now, substitute the values in the newton iteration formula, we get

[tex]x_{2}=x_1- \frac{f(x_1)}{f'(x_{1})}\\x_{2}=1-\frac{-1.6534}{-1.486} \\x_{2}=-0.112[/tex]

Therefore, the second root is -0.112

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