:( help This problem is our first introduction to Newton\'s Method for the solut

Question

:( help

This problem is our first introduction to Newton's Method for the solution of nonlinear equations. The positive solution of f(x) = X2 - 2 = 0 is obviously How might one calculate a numerical value of ? The idea of Newton's method is to pick a guess x0, compute the tangent to the graph of f at (x0, f(x0)), and then replacc the guess x0 with X1 which is the T intercept of the tangent. (You should draw a picture of this idea.) Suppose x0 = 1.5. The tangent to the graph of f at the point (x0,f(x0)) can be written in slope intercept form as y = x+ . The tangent intercepts the x-axis at x1 = . Now let's repeat the process. The tangent to the graph of f at the point (x1 , f(x1)) can be written in slope intercept form as y = x+ . The tangent intercepts the x-axis at X2 = . Note the accuracy of this approximation: Note: To obtain the same result in your last answer as ww you need to compute all intermediate answers to the full accuracy of your calculator. To accomplish this store all intermediate results in your calculator. Do not copy them on paper and then rccntcr them by hand later. Doing so introduces rounding errors and compromises the accuracy of your calculations and solutions. In fact, you should make a habit of this every time you use your calculator: store intermediate results and avoid having to rccntcr them. Most calculators keep more digits internally than they can display, so losing accuracy by copying and reentering is inevitable.

Explanation / Answer

f(x) = x^2 - 2
x^2 -2=0 => x= sqrt(2)

as newton method fourmula => Xn+1 = Xn - f(x) / f'(x)

f ' (x) = 2.x
x0 = 1.5
tangent to the graph of f at the point ( 1.5, 0.25) => y = f'(x0) .x + c
f'(x0) = 2. 1.5 = 3
y= 3x + c => at x= 1.5 , we get y= 0.25

0.25 = 3. 1.5 +c
c = -4.25
y = 3x - 4.25

X1 = 1.5 - 0.25/3
= 1.4166669

[(x1, f(x1)] = [ 1.4166669 , 0.006944]

0.006944 = f'(x1) . x + c1
  
= 2.8333333 . (1.416669) + c1

c1 = -4.00695

y = 2.83333 x - 4.00695

x2 = x1 - f(x1)/ f'(x1)

= 1.4166669 - f(1.4166669) / f'(1.4166669)

= 1.4166669 - 0.0069444/2.8333333

= 1.4166669 - 0.00245098

= 1.4142166

[(x2, f(x2)] = [ 1.4142166, 0.0000085]
f'(x2) = 2.8284332

0.0000085 = f'(x2) .x +c
  
= 2.8284332 . ( 1.4142166) + c

c = -4.00000086

y = 2.8284338 x - 4.00000086

the accuracy of this approx.

sqrt(2) - x2
=> 1.4142135- 1.4142166
=> 0.00000303

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