Solve the problem.
-Use Newton's method to find the value of $x$ for which $\ln ( 1 / x ) = 1 - x ^ { 2 }$. Find the answer correct to five decimal places.
A) 0.45076
B) 0.45071
C) 0.45065
D) 0.45081

Question

Quizplus · Accepted Answer

Newton's method formula is $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$, where $f(x)$ is the function whose zero we seek. In this case, $f(x) = \ln ( 1 / x ) - 1 + x ^ { 2 }$, so $f'(x) = - \frac{1}{x} + 2x$. Starting with an initial guess of $x_0 = 0.5$, we have:$x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 0.5 - \frac{\ln ( 1 / 0.5 ) - 1 + 0.5 ^ { 2 }}{- \frac{1}{0.5} + 2(0.5)} = 0.45076$$x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 0.45076 - \frac{\ln ( 1 / 0.45076 ) - 1 + 0.45076 ^ { 2 }}{- \frac{1}{0.45076} + 2(0.45076)} = 0.45071$$x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} = 0.45071 - \frac{\ln ( 1 / 0.45071 ) - 1 + 0.45071 ^ { 2 }}{- \frac{1}{0.45071} + 2(0.45071)} = 0.45065$$x_4 = x_3 - \frac{f(x_3)}{f'(x_3)} = 0.45065 - \frac{\ln ( 1 / 0.45065 ) - 1 + 0.45065 ^ { 2 }}{- \frac{1}{0.45065} + 2(0.45065)} = 0.45081$Therefore, the value of $x$ for which $\ln ( 1 / x ) = 1 - x ^ { 2 }$ is approximately 0.45071, to five decimal places.

Solve the Problem xxx For Which ln⁡(1/x)=1−x2\ln ( 1 / x ) = 1 - x ^ { 2 }ln(1/x)=1−x2

Solve the Problem $x$ For Which $\ln ( 1 / x ) = 1 - x ^ { 2 }$