Question

6 The equation \cot{\frac{1}{2}x} = 3x has one root in the interval $0 < x < \pi$, denoted by $\alpha$. (a) Show by calculation that $\alpha$ lies between 0.5 and 1. (b) Show that, if a sequence of positive values given by the iterative formula $x_{n+1} = \frac{1}{3}(x_n + 4 \tan^{-1}(\frac{1}{3x_n}))$ converges, then it converges to $\alpha$.

          6 The equation \cot{\frac{1}{2}x} = 3x has one root in the interval $0 < x < \pi$, denoted by $\alpha$.
(a) Show by calculation that $\alpha$ lies between 0.5 and 1.
(b) Show that, if a sequence of positive values given by the iterative formula
$x_{n+1} = \frac{1}{3}(x_n + 4 \tan^{-1}(\frac{1}{3x_n}))$ converges, then it converges to $\alpha$.

6 The equation cot(1)/(2)x = 3x has one root in the interval 0 < x < π, denoted by α.
(a) Show by calculation that α lies between 0.5 and 1.
(b) Show that, if a sequence of positive values given by the iterative formula
xn+1 = (1)/(3)(xn + 4 tan^-1((1)/(3xn))) converges, then it converges to α.

Added by Ryan C.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Kyle Sower

Step 1

** Show more…

Show all steps

Thanks for your feedback!

(b) Show that, if a sequence of positive values given by the iterative formula x_(n+1)=(1)/(3)(x_(n)+4tan^(-1)((1)/(3x_(n)))) converges, then it converges to alpha . The equation cot x = 3x has one root in the interval 0 < x < x, denoted by (a) Show by calculation that lies between 0.5 and 1. [2] (b) Show that, if a sequence of positive values given by the iterative formula 1= (+4tan~() converges, then it converges to . [2]