second part Question 2(a)[1, 2].Use the Theorem from the course to prove that g(x) = 1+ e-® has a unique fixed point onFor po = 1, compute p1, P2 by using Fixed-Poit iteration. (Show details of each iteration. Youare NOT allowed to use your computer code)How many Fixed-Point iterations are necessary to achieve the accuracy 10-3 ?

Answered: Question 2 (a) [1, 2]. Use the Theorem…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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second part

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Step 1

For the given function $g (x) = 1 + e^{- x}$ has a unique fixed point on $[1, 2]$ .

The theorem says that,

$g (x)$ is defined and differentiable on $[1, 2]$
$g (x) \in [1, 2] \forall x \in [1, 2]$
There is a number $l < 1$ such that $|g' (x)| \leq l < 1 \forall x \in [1, 2]$ then there is a unique fixed point.

Since $g (x) = 1 + e^{- x}$ is defined and differentiable on $[1, 2]$

The values for the exponential function $e^{- x}$ at x = 1, 2

$e^{- 1} = 0.3679 e^{- 2} = 0.1353$

$1 + e^{- 1}, 1 + e^{- 2}$ both belongs to $[1, 2]$

Step 2

Now the derivative ,

$g' (x) = - e^{- x} |g' (x)| = e^{- x} < 1 \forall x \in [1, 2]$

Therefore, there exists a unique fixed point.

As $g (x)$ satisfies all the criterion of the theorem therefore, $g (x_{n})$ converges.

Given that $p_{0} = 1$

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