Newton raphson method
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The document describes the Newton-Raphson method to find an approximate root of an equation. It explains that given an approximate root x0, the method iteratively calculates better approximations x1, x2, etc. until converging on the exact root. Each new approximation xn+1 is calculated as xn - f(xn)/f'(xn), where f(x) is the original equation, and f'(x) is its derivative. An example applies the method to find the real root of x4 - x - 10 = 0, obtaining the value 1.855 after 4 iterations.
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![Let f(x)=0 be the given equation and x0 be an approximate root of the
equation.
If x1 = x0 + h be the exact root then f(x1)=0.
i.e., f(x0 + h)=0
f(x0) + hf′ x0 +
ℎ2
2!
f′′ x0 + ⋯ = 0 [By Taylor
Series]
Since h is small, neglecting ℎ2
and higher powers of h,
f x0 + hf′
x0 = 0
h = −
f(x0)
f′(x0)
∴ x1 = x0 + h = x0 −
f(x0)
f′(x0)
Similarly, starting with x1, a still better approximation x2 is obtained.
x2 = x1 −
f(x1)
f′(x1)
In general, x = x −
f(xn)](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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Numericals Reasoning.pdf
The secant method is a root-finding algorithm that uses the secant line between two points on a function to approximate a root. It requires choosing two initial points and then calculates the x-value where the secant line through those points crosses the x-axis. This new x-value becomes the next approximation in an iterative process repeated until a root is found within a specified accuracy. The method converges faster than linear methods and does not require evaluating derivatives or bracketing roots. It was presented with an example problem solved over 5 iterations to demonstrate the algorithm.
0.5.derivatives
The document outlines key calculus concepts including:
- Functions, derivatives, differentiation rules, and the definition of a derivative as an infinitesimal change in a function with respect to a variable.
- Concepts related to derivatives such as local/absolute extrema, critical points, increasing/decreasing functions, concavity, asymptotes, and inflection points.
- How to use the first and second derivative tests to determine local extrema, concavity, and increasing/decreasing behavior.
differentiate free
This document discusses the concept of the derivative and differentiation. It begins by explaining how the slope of a curve changes at different points, unlike the constant slope of a line. It then defines the derivative of a function f at a point x0 as the limit of the difference quotient as h approaches 0. If this limit exists, then f is said to be differentiable at x0. The derivative f'(x) then represents the slope of the curve y=f(x) at each point x and is a measure of how steeply the curve is rising or falling at that point. Several examples are provided to illustrate how to compute derivatives using this limit definition.
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This document provides Newton's formula for forward difference interpolation and an example of using it to find the value of tan(0.12).
- Newton's formula uses forward difference interpolation to find the value of a polynomial of degree n that fits a set of (n+1) equally spaced (x,y) points.
- The coefficients of the polynomial are determined using forward differences of the y-values.
- In the example, the value of tan(0.12) is found by applying Newton's formula to a table of tan(x) values from 0.10 to 0.30 using forward differences up to degree 4.
19 min max-saddle-points
The document defines relative and absolute maxima and minima for functions z=f(x,y). A relative maximum occurs when f(a,b) is greater than f(x,y) within a neighborhood circle of (a,b). An absolute maximum occurs when f(a,b) is greater than f(x,y) over the entire domain. Similarly for minima with the inequalities reversed. Extrema refer to maxima and minima. For a continuous function over a closed and bounded domain, absolute extrema exist and occur either in the interior or on the boundary. Examples find and classify extrema of functions.
Differntiation
1) The document contains solutions to 10 calculus problems involving finding derivatives and evaluating functions. The problems include finding derivatives of trigonometric, logarithmic, and exponential functions, as well as evaluating functions at given values.
2) The solutions show the steps taken to solve each problem through substitution and application of derivative rules.
3) The techniques demonstrated include finding derivatives using basic rules, evaluating functions by substitution, and simplifying expressions involving trigonometric identities.
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Newton raphson method
- 1. MADE BY : MOHIT AGARWAL – 161080107026 COMPUTER – 4TH SEM
- 2. Let f(x)=0 be the given equation and x0 be an approximate root of the equation. If x1 = x0 + h be the exact root then f(x1)=0. i.e., f(x0 + h)=0 f(x0) + hf′ x0 + ℎ2 2! f′′ x0 + ⋯ = 0 [By Taylor Series] Since h is small, neglecting ℎ2 and higher powers of h, f x0 + hf′ x0 = 0 h = − f(x0) f′(x0) ∴ x1 = x0 + h = x0 − f(x0) f′(x0) Similarly, starting with x1, a still better approximation x2 is obtained. x2 = x1 − f(x1) f′(x1) In general, x = x − f(xn)
- 3. STEP 1 : Find f 0 , f 1 STEP 2 : Check if f 0 . f 1 < 0 STEP 3 : If STEP 2 is true then x0 = 1, and x1 = x0 − f(x0) f′(x0) Continue the iteration xn+1 = xn − f(xn) f′(xn) till you find the root of the equation.
- 4. EXAMPLE : Find the real root of 𝑥4 − 𝑥 − 10 = 0, Correct up to 3 decimal places. SOLUTION : f 0 = −10 f 1 = −10 (x0) f 2 = 4 (x1) f(x) = x4 − x − 10 f′ x = 4x3 − 1 Now, x2 = x1 − f(x1) f′(x1) x2 = 2 − 4 31 = 1.871 x3 = x2 − f(x2) f′(x2) x3 = 1.871 − 0.3835 25.1988 = 1.8558 Similarly, x4 = x3 − f x3 f′ x3 = 1.8556 ∴ x = 1.855