Newtons Method - Assignment Example

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