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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 )
𝑙𝑒𝑡𝑛 = 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
𝑙𝑒𝑡𝑛 = 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
0.000573𝑖𝑠𝑎𝑐𝑐𝑢𝑟𝑎𝑡𝑒𝑡𝑜𝑡𝑤𝑜𝑑𝑒𝑐𝑖𝑚𝑎𝑙𝑝𝑙𝑎𝑐𝑒𝑠
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 )
𝑙𝑒𝑡𝑛 = 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
𝑙𝑒𝑡𝑛 = 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
−0.000351𝑖𝑠𝑎𝑐𝑐𝑢𝑟𝑎𝑡𝑒𝑡𝑜𝑡𝑤𝑜𝑑𝑒...
