Numerical Methods Vedamurthy Solution 203
Numerical Methods Vedamurthy Solution 203: A Step-by-Step Guide and Explanation
Numerical methods are techniques that use numerical calculations to solve mathematical problems that are difficult or impossible to solve analytically. Numerical methods are widely used in science, engineering, and other fields that require accurate and efficient solutions to complex problems.
One of the numerical methods that is commonly used is the Newton-Raphson method, which is a root-finding algorithm that iteratively approximates the solution of a nonlinear equation. The Newton-Raphson method is based on the idea that a function can be approximated by a tangent line at a given point, and the intersection of the tangent line with the x-axis gives a better approximation of the root than the original point.
In this article, we will show you how to use the Newton-Raphson method to solve numerical methods vedamurthy solution 203, which is a nonlinear equation that arises in the study of heat transfer. We will also explain the steps and the logic behind the method, and provide some examples and tips to help you understand and apply it.
What is Numerical Methods Vedamurthy Solution 203?
Numerical methods vedamurthy solution 203 is a nonlinear equation that is given by:
$$f(x) = x^3 – 3x + 1 = 0$$
This equation represents the steady-state temperature distribution in a rod that is heated at one end and insulated at the other end. The equation has one real root and two complex roots, but we are only interested in finding the real root, which represents the temperature at the heated end of the rod.
To find the real root of this equation using the Newton-Raphson method, we need to follow these steps:
- Choose an initial guess for the root, denoted by $x_0$.
- Find the value of the function and its derivative at $x_0$, denoted by $f(x_0)$ and $f'(x_0)$.
- Use the formula $$x_1 = x_0 – \frac{f(x_0)}{f'(x_0)}$$ to find a better approximation of the root, denoted by $x_1$.
- Repeat steps 2 and 3 until the difference between two successive approximations is smaller than a desired tolerance, or until a maximum number of iterations is reached.
- The final approximation is taken as the root of the equation.
How to Apply the Newton-Raphson Method to Numerical Methods Vedamurthy Solution 203?
To apply the Newton-Raphson method to numerical methods vedamurthy solution 203, we need to choose an initial guess for the root, and then use the formula $$x_{n+1} = x_n – \frac{f(x_n)}{f'(x_n)}$$ to find better approximations of the root, until we reach a desired accuracy or a maximum number of iterations.
Let us illustrate this process with an example. Suppose we choose $x_0 = 1$ as our initial guess for the root. Then we can find the value of the function and its derivative at $x_0$ as follows:
$$f(x_0) = f(1) = 1^3 – 3(1) + 1 = -1$$
$$f'(x_0) = f'(1) = 3(1)^2 – 3 = 0$$
Using the formula, we can find the next approximation of the root as follows:
$$x_1 = x_0 – \frac{f(x_0)}{f'(x_0)} = 1 – \frac{-1}{0}$$
However, we encounter a problem here. The denominator of the formula is zero, which means we cannot divide by it. This indicates that the Newton-Raphson method fails to converge for this initial guess, and we need to choose a different initial guess.
Let us try another initial guess, say $x_0 = 2$. Then we can find the value of the function and its derivative at $x_0$ as follows:
$$f(x_0) = f(2) = 2^3 – 3(2) + 1 = 3$$
$$f'(x_0) = f'(2) = 3(2)^2 – 3 = 9$$
Using the formula, we can find the next approximation of the root as follows:
$$x_1 = x_0 – \frac{f(x_0)}{f'(x_0)} = 2 – \frac{3}{9} = 1.6667$$
This time, we do not encounter any problem with the formula, and we can continue to find better approximations of the root by repeating the process. Here are some more iterations of the method:
| n | $x_n$ | $f(x_n)$ | $f'(x_n)$ | $x_{n+1}$ |
|---|---|---|---|---|
| 0 | 2 | 3 | 9 | 1.6667 |
| 1 | 1.6667 | -0.1852 | 6.6667 | 1.6926 |
| 2 | 1.6926 | -0.0065 | 7.1852 | 1.6932 |
| 3 | 1.6932 | -0.000001 | 7.1935 | 1.6932 |
| 4 | 1.6932 | -0.00000000000001 | 7.1935 | 1.6932 |
| … | ||||
| The method converges to $x \approx 1.6932$ as the root of the equation. | ||||