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Newtons Method - Assignment Example

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The author of the "Newton’s Method" paper explains Newton’s method the process, how it works when it doesn’t work, and why you would use it in detail. Newton's method begins with identifying the equation and then a guess is made about a figure for xo. …
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Extract of sample "Newtons Method"

Running head: Newton Method Newton Method Author Author Affiliation QUESTION 1 Newton’s method is a method of solving equations. Explain the process, how it works, when it doesn’t work and why you would use it in detail. Answer: Newton method begins with identifying the equation and then a guess is made about a figure for xo. This becomes a stating point. In particular, we can approximate roots ƒ(x) = xk- α by taking an overestimate xn of this roots, and choosing xn +1 = xn – as the next estimate (where f denotes the derivative). In the case f (x) = xk- a we have f(x) = kxk-1, and we have just verified that this procedure does yield a convergent sequence whose limit is the positive root of the equation xk=α. Assumed an equation of x2+2x+3. This equation is recognised as f(x) = x2+2x+3. The next step is to compute first derivative as 2x+2 and the first guess is -3 for xo. Then the next step is to make the iteration formula for first order as X1 = xo - = -3 – = -2.75 x f(x) -3 6 -2.75 5.0625 -2.5 4.25 -2.25 3.5625 -2 3 -1.75 2.5625 -1.5 2.25 -1.25 2.0625 -1 2 -0.75 2.0625 -0.5 2.25 -0.25 2.5625 0 3 0.25 3.5625 0.5 4.25 0.75 5.0625 1 6 1.25 7.0625 1.5 8.25 1.75 9.5625 2 11 Consider the sequence given by x0 = α + 1, xn+1=xn(1 +) for a fixed a>0 and fixed k N. then: xn >0, xn+1 < xn and > a. to see this, we again use induction. All three claims are true with n=0, and we assume them for some fixed n. now since xn > 0, a < means that xn+1 < xn; however xn +1 remains positive, since + a - > 0 for all k 1; and finally we can use the Bernoulli inequality with x = (which is greater than -1 because + a - > 0) to conclude that = (1+ > {1+k() } =a Hence by induction all three claims are true for all n. so (xn) is decreasing and bounded below by 0. It therefore converges to a = infnxn, which is non-negative. The recursion formula can be written: kxn+1 = (k-1) + a, and by the algebra of limits we can let n on both sides to obtain kααk-1 = (k – 1 ) αk +α. Hence αk = α, in other words α=α1/k . we have shown, therefore, that every positive real number has a positive kth root for every fixed k N. this root is unique, since if we have two positive numbers α, β with α Read More
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Newtons Method Assignment Example | Topics and Well Written Essays - 1500 Words. https://studentshare.org/mathematics/2051969-newtons-method.
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