Math Problem Statement

  1. Find the root of x2-2=0 by using

a. Newton Raphson method( ) where Xo=1.5

b. Secant formula () Xo=1.45 X1=1.5

Solution

Let's solve the problem step by step for each method.

(a) Newton-Raphson Method

The Newton-Raphson formula is:

xn+1=xnf(xn)f(xn)x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}

Here, f(x)=x22f(x) = x^2 - 2 and f(x)=2xf'(x) = 2x.

Iteration Steps:

  1. Start with x0=1.5x_0 = 1.5:

    f(1.5)=(1.5)22=0.25f(1.5) = (1.5)^2 - 2 = 0.25 f(1.5)=21.5=3f'(1.5) = 2 \cdot 1.5 = 3 x1=1.50.253=1.50.0833=1.4167x_1 = 1.5 - \frac{0.25}{3} = 1.5 - 0.0833 = 1.4167

  2. Next iteration (x1=1.4167x_1 = 1.4167):

    f(1.4167)=(1.4167)220.00694f(1.4167) = (1.4167)^2 - 2 \approx 0.00694 f(1.4167)=21.4167=2.8334f'(1.4167) = 2 \cdot 1.4167 = 2.8334 x2=1.41670.006942.83341.41670.00245=1.4143x_2 = 1.4167 - \frac{0.00694}{2.8334} \approx 1.4167 - 0.00245 = 1.4143

  3. Next iteration (x2=1.4143x_2 = 1.4143):

    f(1.4143)=(1.4143)220.000007f(1.4143) = (1.4143)^2 - 2 \approx 0.000007 f(1.4143)=21.4143=2.8286f'(1.4143) = 2 \cdot 1.4143 = 2.8286 x3=1.41430.0000072.82861.4142x_3 = 1.4143 - \frac{0.000007}{2.8286} \approx 1.4142

The root converges to x1.4142x \approx 1.4142 after sufficient iterations.


(b) Secant Method

The Secant formula is:

xn+1=xnf(xn)xnxn1f(xn)f(xn1)x_{n+1} = x_n - f(x_n) \cdot \frac{x_n - x_{n-1}}{f(x_n) - f(x_{n-1})}

Given x0=1.45x_0 = 1.45 and x1=1.5x_1 = 1.5:

Iteration Steps:

  1. Compute f(x0)f(x_0) and f(x1)f(x_1): f(1.45)=(1.45)22=0.1025f(1.45) = (1.45)^2 - 2 = 0.1025 f(1.5)=(1.5)22=0.25f(1.5) = (1.5)^2 - 2 = 0.25

  2. Find x2x_2: x2=1.50.251.51.450.250.1025=1.50.250.050.1475x_2 = 1.5 - 0.25 \cdot \frac{1.5 - 1.45}{0.25 - 0.1025} = 1.5 - 0.25 \cdot \frac{0.05}{0.1475} x2=1.50.08471.4153x_2 = 1.5 - 0.0847 \approx 1.4153

  3. Compute f(x2)f(x_2): f(1.4153)=(1.4153)220.00156f(1.4153) = (1.4153)^2 - 2 \approx 0.00156

  4. Find x3x_3: x3=1.41530.001561.41531.50.001560.251.4142x_3 = 1.4153 - 0.00156 \cdot \frac{1.4153 - 1.5}{0.00156 - 0.25} \approx 1.4142

The root converges to x1.4142x \approx 1.4142 after sufficient iterations.


Final Results:

  1. Newton-Raphson root: 1.4142\approx 1.4142
  2. Secant method root: 1.4142\approx 1.4142

Would you like a detailed explanation of any part or need further assistance?

Here are 5 related questions to explore:

  1. Why does the Newton-Raphson method converge faster than the Secant method?
  2. What are the conditions for convergence of these iterative methods?
  3. How can the initial guesses influence the convergence of the Secant method?
  4. What happens if the derivative in the Newton-Raphson method is zero at some point?
  5. Can you apply these methods to find the root of a cubic polynomial?

Tip: Always check if the function and its derivative behave well around the initial guesses to avoid divergence in iterative methods.

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Math Problem Analysis

Mathematical Concepts

Numerical Methods
Root Finding
Iteration Methods

Formulas

Newton-Raphson formula: x_(n+1) = x_n - f(x_n)/f'(x_n)
Secant formula: x_(n+1) = x_n - f(x_n) * (x_n - x_(n-1)) / (f(x_n) - f(x_(n-1)))

Theorems

Convergence of Newton-Raphson method
Convergence of Secant method

Suitable Grade Level

Undergraduate - First Year