Regula Falsi Method (False Position)| Numerical Methods -1| MAT202 | MAT204 | Module 4 | Part 3

This video demonstrates solving equations using numerical methods like the Regula Falsi method and Newton-Raphson method. Examples include finding roots for equations involving trigonometric functions, logarithms, and exponentials, iteratively refining approximations to achieve solutions correct to several decimal places. point eight zero one two into f of two, point five, the whole by f of b minus f of a that is f of two, point eight zero, one, two minus f of a nano f of two, point five in the value in the next number two, point five into another number f of two, point eight, zero, one, two table minus two, point eight zero one two into f of two, point five, number nine, and the minus point eight, six, seven five, I know the whole by point zero, zero, seven, nine minus of minus point eight, six, seven five, which is equal to and greatest book, not the answer. Another two point seven, nine, eight, four, one, two, point seven, nine, eight, four, nine, and two taken, you know, dividends f of two point seven, nine, eight, four, five, you know, that is equal to our formula. Another two point seven, nine, eight, four into ten of two point seven, nine, eight, four plus one numbers. And then therefore fixing the number of x tan x plus one other substitute for the answer calculated equal to okay, x is equal to two point seven, nine, eight, four and one another positive value. So root lies between two point five, and you put the opposite e value. Suppose the negative I need two point seven, nine, eight, four, nine, two, point eight, zero, one, two, nine, okay, then the root lies between road lies between the negative value two point five 2.7984 2.5 and 2.7984 again 2.5 number in the area 2.7984 number in the billionaire so this is f4 2.7984 minus 2.7984 into f of 2.5 the whole by f of b inner f of 2.7984 minus f of a f of n and the f of 2.5 are formulaely 2.5 into three x plus sine x minus e raise to x is equal to zero using regular falsie method find the approximate root between zero and one another between zero and one above nothing is um 2.5 into f of 3 minus 3 into f of 2.5 either a and f of b minus b into f of a by f of b minus f of the real root of x log to the base 10 x is equal to 1.2 by method of force position same problem number Newton-Raphson method is minus b into f of a by f of b minus f of e. Okay i'm kind of making a function for fixed numbers for fixed is equal to x log to the base 10 x Calculate logic logarithm low to the base 10 to minus 1.2 which is equal to the characteristics. This is equal to minus 0.59 7. 9. 4 Okay correct to 4 decimal place 0.8675 Another but three times three plus one on the line the answer point five seven two four four decimal in the background I'm going to remember minus 3 into f of a f of 2.5.06:52the whole by f of b naught f of 2.5 minus f of a normal f of 3. that is equal to the you can the idea minus point eight six seven five the whole by f of b minus f of nine point five seven two four minus I'm gonna get them for my fourth three and zero one two plus one calculated to simplify making the answer. Newton-Raphson Method: An iterative method for finding successively better approximations to the roots (or zeroes) of a real-valued function. It uses the derivative of the function to find the tangent line at a given point, and then finds the x-intercept of that tangent line as the next approximation. Secant Method: A root-finding algorithm that uses a succession of roots of secant lines to better approximate a root of a function. It's similar to the Newton-Raphson method but doesn't require calculating the derivative. Root (of a function): A value of x for which f(x) = 0, where f(x) is a given function. Iteration: A repetitive procedure to obtain successively closer approximations to a solution. Approximation: An estimate of a number or quantity obtained by a process that is not exact. Function: A mathematical relationship where each input value (x) corresponds to exactly one output value (f(x)). Derivative: A measure of how a function changes as its input changes. In simpler terms, it represents the instantaneous rate of change of a function. Logarithm (log): The exponent or power to which a base must be raised to yield a given number. For example, the logarithm of 100 to base 10 is 2, because 10 raised to the power of 2 is 100. The transcript mentions log base 10. Okay, let's outline the steps to find the root of the equation f(x) = x³ - 25 = 0 using the Newton-Raphson method. While the provided context mentions the Newton-Raphson method , it applies it to a different function (x log₁₀(x) - 1.2 = 0). Here are the general steps for your equation: Define the function: f(x) Perform iterations: Iteration 1: x₁ = 3 - (3³ - 25) / (3 * 3²) = 3 - 2/27 ≈ 2.9259 Iteration 2: x₂ = 2.9259 - (2.9259³ - 25) / (3 * 2.9259²) ≈ 2.9240 Iteration 3: x₃ = 2.9240 - (2.9240³ - 25) / (3 * 2.9240²) ≈ 2.9240 Conclusion: The iterations quickly converge. The root of the equation x³ - 25 = 0 is approximately 2.9240 .