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Bisection and Newton Raphson Methods Questions and Answers

1.

The root of the equation \(xe^{x} = 1\) between 0 and 1, obtained by using two iterations of bisection method, is

Exam Year(s)

GATE Mathematics - 2012,

Answer: C

Two iterations of bisection give 0.75.

Let \(f(x) = xe^{x} - 1\). Then \(f(0) = -1 \lt 0\) and \(f(1) = e - 1 \gt 0\), so a root lies in [0, 1].

\[ \begin{aligned} x_{1} &= 0.5:\ f(0.5) = 0.5e^{0.5} - 1 = -0.176 \lt 0 \\ x_{2} &= \tfrac{0.5 + 1}{2} = 0.75 \end{aligned} \]

Since \(f(0.5) \lt 0\), the root lies in [0.5, 1], and the second midpoint is 0.75. (The true root is about 0.567.) 0.25 would be the second midpoint only if \(f(0.5)\) were positive; 0.50 is the first iterate, and 0.65 is not a bisection midpoint of [0, 1].

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Answer: A

The iteration is \(x_{n+1} = \dfrac{(1 + x_{n})\, e^{-x_{n}}}{1 + e^{-x_{n}}}\).

Write the equation as \(f(x) = x - e^{-x} = 0\), so \(f'(x) = 1 + e^{-x}\). Newton-Raphson gives

\[ \begin{aligned} x_{n+1} &= x_{n} - \frac{x_{n} - e^{-x_{n}}}{1 + e^{-x_{n}}} \\ &= \frac{x_{n} + x_{n}e^{-x_{n}} - x_{n} + e^{-x_{n}}}{1 + e^{-x_{n}}} \\ &= \frac{(1 + x_{n})\, e^{-x_{n}}}{1 + e^{-x_{n}}} \end{aligned} \]

The options with \(1 - e^{-x_{n}}\) in the denominator use the wrong sign for \(f'\) (as if \(f = x + e^{-x}\)), and the last option is not a rearrangement of Newton's step. Starting from \(x_{0} = 0.5\) the iteration converges quickly to the root 0.5671.

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