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Newton's Method
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We wanted to solve f(x) = x + tan(x) = 0 using the Newton's method. First, we compute the derivative f'(x) = 1 + sec2(x). Then we start the Newton's iteration by setting x0=2. Recall the Newton's iteration formula is xn=xn-1-f(x)/f(x0). We have x0=2 x1=2.027314579 x2=2.028754298 x3=2.028757838 Therefore the first root, up to three place accuracy, is 2.029 For the second root, we do exactly the same thing starting with x0=5, so we haveWe have x0=5 x1=4.879393859 x2=4.907699753 x3=4.91303811 x4=4.913180344 Therefore, the second root, up to three place accuracy, is 4.913.
其他解答:
因為 Andrew 答得好。|||||Masterijk,點解贊助2點俾我既?^^
Newton's Method
發問:
此文章來自奇摩知識+如有不便請留言告知
The equation x + tanx = 0 is important in a variety of applications--- for example, in the study of the diffusion of heat. It has a sequence a1, a2, a3, ... of positive roots, with the nth root slightly larger than (n-0.5)π. Use Newton's method to compute a1 and a2 to three-place accuracy.最佳解答:
We wanted to solve f(x) = x + tan(x) = 0 using the Newton's method. First, we compute the derivative f'(x) = 1 + sec2(x). Then we start the Newton's iteration by setting x0=2. Recall the Newton's iteration formula is xn=xn-1-f(x)/f(x0). We have x0=2 x1=2.027314579 x2=2.028754298 x3=2.028757838 Therefore the first root, up to three place accuracy, is 2.029 For the second root, we do exactly the same thing starting with x0=5, so we haveWe have x0=5 x1=4.879393859 x2=4.907699753 x3=4.91303811 x4=4.913180344 Therefore, the second root, up to three place accuracy, is 4.913.
其他解答:
因為 Andrew 答得好。|||||Masterijk,點解贊助2點俾我既?^^
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