Newton Iteration Method Project

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Enxhmna

Mathematics

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Use Newton's iteration method to approximate the solution of the nonlinear algebraic equation f(x) = 2x – cos(x). Let 10 = 1 be your initial guess and note that your domain is the set of real numbers (radian) and your estimate is correct to two decimal places. 2 Use Newton's iteration method to approximate the intersection point (,y) between the two curves y = 2-r and y = 24 where x > 0. Let to = 0.5 be your initial guess and your estimate must be correct to two decimal places. 3 Find the general solution of the finite difference equation A+yn + Apyn = 0. 4 Find the general solution of yn+3 – 2yn+2 + 4yn+1 - Syn = 0. 5 The forward velocity of a new plane is given in the following table: Table 1: Forward velocity of new air plane Time (seconds) 0 25 45 55 75 Velocity ( meter per seconds) 0 220 340 490 600 (a) Determine the velocity at t = 51 seconds using Newton forward divided-difference method. (b) Determine the velocity at t = 100 seconds by using Lagrange polynomial of order four. (c) Using Least squares method to find the velocity at t = 200 seconds. 6 Use Range-Kutta method of fourth order to find the approximate value of y(0.4) correct to four decimal dy places where y(1) is the solution of the initial value problem +r? - y = = 1 with y(0) = 0.5 and h = 0.4. dc = 7 Maximize f(x,y) = 2x + 5y subject to the constraints: 2y - 158 1-y
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Explanation & Answer

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1)
Use Newton’s iteration method to approximate the solution of the nonlinear algebraic equation
𝑓(𝑥) = 2𝑥 − cos(𝑥)
Let 𝑥0 = 1 be your initial guess and note that your domain is the set of real numbers (radian) and
your estimate is correct to two decimal places.

First, we need to take the first derivative of the given function.
𝑓 ′ (𝑥) = 2 + sin⁡(𝑥)
Using the following equation for Newton’s method:
𝑥𝑛 = 𝑥𝑛−1 −

𝑓(𝑥𝑛−1 )
𝑓′(𝑥𝑛−1 )

𝑙𝑒𝑡⁡𝑛 = 1
𝑓(1) = 2 ∗ 1 − cos(1) = 1.4597
𝑓 ′ (1) = 2 + sin(1) = 2.84147
𝑥1 = 𝑥0 −
𝑥1 = 1 −

𝑓(𝑥0 )
𝑓′(𝑥0 )

1.4597
2.84147

𝑥1 = 0.486287
𝑙𝑒𝑡⁡𝑛 = 2
𝑓(0.486287) = 2 ∗ 0.486287 − cos(0.486287) = 0.0884998
𝑓 ′ (0.486287) = 2 + sin(0.486287) = 2.46735
𝑥2 = 𝑥1 −

𝑓(𝑥1 )
𝑓′(𝑥1 )

𝑥2 = 0.486287 −

0.0884998
2.46735

𝑥2 = 0.450419
𝑓(0.450419) = 2 ∗ 0.450419 − cos(0.450419) = 0.000573
0.000573⁡𝑖𝑠⁡𝑎𝑐𝑐𝑢𝑟𝑎𝑡𝑒⁡𝑡𝑜⁡𝑡𝑤𝑜⁡𝑑𝑒𝑐𝑖𝑚𝑎𝑙⁡𝑝𝑙𝑎𝑐𝑒𝑠
𝑥2 = 0.450419

2)
Use Newton’s iteration method to approximate the intersection point (x,y) between the two
curves 𝑦 = 2 − 𝑥 2 and 𝑦 = 2𝑥 where 𝑥 > 0. Let 𝑥0 = 0.5 be your initial guess and your
estimate must be correct to two decimal places.
First, we need to set the two equations equal to one another so that we can determine the
intersection point between the two curves.
2 − 𝑥 2 = 2𝑥
𝑓(𝑥) = 2 − 𝑥 2 − 2𝑥 = 0
Next, we need to take the first derivative of the combined function.
𝑓 ′ (𝑥) = − ln(2) ∗ 2𝑥 − 2𝑥
Using the following equation for Newton’s method:
𝑥𝑛 = 𝑥𝑛−1 −

𝑓(𝑥𝑛−1 )
𝑓′(𝑥𝑛−1 )

𝑙𝑒𝑡⁡𝑛 = 1
𝑓(0.5) = 2 − 0.52 − 20.5 = 0.335786
𝑓 ′ (0.5) = − ln(2) ∗ 20.5 − 2 ∗ 0.5 = −1.98026
𝑥1 = 𝑥0 −
𝑥1 = 0.5 −

𝑓(𝑥0 )
𝑓′(𝑥0 )

0.335786
−1.98026

𝑥1 = 0.669567
𝑙𝑒𝑡⁡𝑛 = 2
𝑓(0.669567) = 2 − 0.6695672 − 20.669567 = −0.038915
𝑓 ′ (0.669567) = − ln(2) ∗ 20.669567 − 2 ∗ 0.669567 = −2.44165
𝑥2 = 𝑥1 −

𝑓(𝑥1 )
𝑓′(𝑥1 )

𝑥2 = 0.669567 −

0.038915
2.44165

𝑥2 = 0.653629

𝑓(0.653629) = 2 − 0.6536292 − 20.653629 = −0.000351
−0.000351⁡𝑖𝑠⁡𝑎𝑐𝑐𝑢𝑟𝑎𝑡𝑒⁡𝑡𝑜⁡𝑡𝑤𝑜⁡𝑑𝑒...

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