This video demonstrates the Newton-Raphson method for finding approximate roots of the equation xĀ³ - 2x - 5 = 0. It iteratively refines an initial guess (xā) using the formula xāāā = xā - f(xā)/f'(xā), where f(x) is the function and f'(x) its derivative. The example shows multiple iterations converging towards an approximate root of around 2.0945. ## Newton-Raphson Method The formula for the Newton-Raphson method is: The Newton-Raphson method is a numerical method used to find the roots of a function. ### Formula $x_n$ is the nth approximation* of the root. $x_{n+1}$ is the (n+1)th approximation* of the root. $f(x_n)$ is the value of the function* at $x_n$. where: $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$ Steps: Objective: Find the root of the function $f(x) = x^3 - 2x - 5$ correct to 5 decimal places. 1. Find the derivative: $f'(x) = 3x^2 - 2$ ### Example 1: Finding the Root of $f(x) = x^3 - 2x - 5$ $f'(x_n)$ is the derivative of the function* at $x_n$. 3. Apply the formula: * $x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 2 - \frac{2^3 - 2(2) - 5}{3(2)^2 - 2} = 2.091$ * $x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 2.091 - \frac{2.091^3 - 2(2.091) - 5}{3(2.091)^2 - 2} = 2.09457$ * Continue iterating until the desired accuracy is reached. 2. Choose an initial guess: Let's start with $x_0 = 2$. Conclusion: The approximate root of the function $f(x) = x^3 - 2x - 5$ is 2.09455 correct to 5 decimal places. Steps: 1. Find the derivative: $f'(x) = 2e^{2x} - 1$ Objective: Find the root of the function $f(x) = e^{2x} - x - 6$ using the Newton-Raphson method. ### Example 2: Finding the Root of $f(x) = e^{2x} - x - 6$ * $x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 1 - \frac{e^{2(1)} - 1 - 6}{2e^{2(1)} - 1} = 0.971$ 3. Apply the formula: 2. Choose an initial guess: Let's start with $x_0 = 1$. * $x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 0.971 - \frac{e^{2(0.971)} - 0.971 - 6}{2e^{2(0.971)} - 1} = 0.977$ Conclusion: The approximate root of the function $f(x) = e^{2x} - x - 6$ is 0.977 correct to 3 decimal places. 1. Define the function: $f(x) = x^3 - 12$ Objective: Find the cube root of 12. Steps: ## Finding Cube Roots Using Newton-Raphson Method The Newton-Raphson method can also be used to find the cube root of a number. 4. Apply the formula: 3. Choose an initial guess: Let's start with $x_0 = 3$ (since $3^3 = 27$, which is close to 12). * $x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 2.889 - \frac{2.889^3 - 12}{3(2.889)^2} = 2.8845$ * $x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 3 - \frac{3^3 - 12}{3(3)^2} = 2.889$ 2. Find the derivative: $f'(x) = 3x^2$ ### Table of Iterations for Cube Root of 12: | Iteration | $x_n$ | $f(x_n)$ | $f'(x_n)$ | $x_{n+1}$ | |---|---|---|---|---| * $x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} = 2.8845 - \frac{2.8845^3 - 12}{3(2.8845)^2} = 2.8844$ Conclusion: The approximate cube root of 12 is 2.884 correct to 3 decimal places. | 3 | 2.8844 | 0 | 24.999 | 2.8844 | | 2 | 2.8845 | 0.0006 | 24.999 | 2.8844 | | 0 | 3 | 3 | 27 | 2.889 | | 1 | 2.889 | -0.111 | 25.049 | 2.8845 | The formula for the Newton-Raphson method is: xn+1=xnāf(xn)fā²(xn) xn +1ā= xn āā f ā²( xn ā) f ( xn ā)ā where: xn xn ā is the nth approximation of the root. xn+1 xn +1ā is the (n+1)th approximation of the root. f(xn) f ( xn ā) is the value of the function at xn xn ā . fā²(xn) f ā²( xn ā) is the derivative of the function at xn xn ā .