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Numerical solutions of algebraic equations

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Avneet Singh Lal

The document discusses numerical methods for solving algebraic and transcendental equations. It describes direct and iterative methods. Bisection, regula falsi, and Newton Raphson are iterative root-finding algorithms explained in detail with examples. The order of convergence of iterative methods is defined as the rate at which error decreases between successive approximations. The document serves as seminar material on engineering mathematics covering numerical solutions of equations.

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SEMINAR On engineering mathematics TOPIC: NUMERICAL SOLUTION OF ALGEBRAIC EQUATIONS AVNEET SINGH LAL EC-2
SOLUTION OF ALGEBRAIC AND TRANSCENDENTAL EQUATIONS Consider the equation of the form f(x)=0. If f(x) is a quadratic, cubic or biquadratic expression then algebraic formulae are available for expressing the roots. But when f(x) is a polynomial of higher degree or an expression involving transcendental functions e.g., 1 + cos x 5x, x tan x cosh x, e-x sin x etc., algebraic methods are not available.
METHODS In order to solve these type of equations following methods exist: DIRECTIVE METHODS: The methods which are used to find solutions of given equations in the direct process is called as directive methods. Example: remainder theorem, Factorization method etc Note: By using Directive Methods, it is possible to find exact solutions of the given equation. ITERATIVE METHODS (INDIRECT METHODS): The methods which are used to find solutions of the given equation in some indirect process is called as Iterative Methods Note: By using Iterative methods, it is possible to find approximate solution of the given equation and also it is possible to find single solution of the given equation at the same time.
DESCARTES RULE OF SIGNS Number of positive roots of f(x) = 0 is equal to the number of sign changes of the coefficients or is less than this number by an even integer. The number of negative roots of f(x) = 0 is obtained by considering the number of sign changes in f(x). EXAMPLE: f (x) = x5 + 2x4 5x3 7x2 + 12x + 2 No. of sign change = 2 = Positive Roots f (-x) = x5 + 2x4 + 5x3 7x2 12x + 2 No. of sign change = 3 = Negative Roots
ITERATIVE METHODS BISECTION METHOD REGULA FALSI NEWTON RAPHSON
BISECTION METHOD This method is based on the repeated application of intermediate value property. Locate the interval (a, b) in which root lies. Bisect the interval (a, b). Choose the half interval in which the root lies. Bisect the half interval. Repeat the process until the root converges.
EXAMPLE: Find the real positive root of the equation- f (x) = x log10x 1.2 Correct upto 3 decimal places. SOLUTION: S.No. a f (a) b f (b) xi f (xi) 1 2.4 -ve 2.8 +ve 2.6 -ve 2 2.6 -ve 2.8 +ve 2.7 -ve 3 2.7 -ve 2.8 +ve 2.75 +ve 4 2.7 -ve 2.75 +ve 2.725 -ve 5 2.725 -ve 2.75 +ve 2.7375 -ve 6 2.7375 -ve 2.75 +ve 2.74375 +ve 7 2.7375 -ve 2.74375 +ve 2.743625 -ve
REGULA FALSI METHOD This is the oldest method for finding the real roots of a numerical equation. It is sometimes known as method of linear interpolation. Locate the interval (a, b) in which the root lies. First Approximation to the root Locate the next interval (a, X0) or ( X0, b ) in which root lies. Repeat the process until the root converges.
EXAMPLE: Find the real positive root of the equation- f (x) = 3x cos x 1 Correct upto 4 decimal places. Use Regula falsi method. SOLUTION: S.No. a f (a) b f (b) xi f (xi) 1 0 -2 1 1.459698 0.578085 -0.103255 2 0.578085 -0.103256 1 1.459698 0.605959 -0.004080 3 0.605959 -0.004079 1 1.459698 0.607157 -0.000159 4 0.607057 -0.000159 1 1.459698 0.607196 0.000006
NEWTON RAPHSON METHOD This method is generally used to improve the result obtained by one of the previous methods. Locate the interval (a, b). Choose a or b which is nearer to the root as the first approximation x0 to the root. Next approximation: Repeat the process until the root converges.
EXAMPLE: Find a positive value of (17)1/3 correct to six decimal places by Newton Raphson method. SOLUTION: The root lies between 2.5 and 2.6. Since, |f (2.6)| < |f (2.5)| Therefore, the root is nearer to 2.6. Let us take x0 = 2.58. Putting n = 0, 1, 2, ..., successively in Newton Raphsons formula, we get- x1 = 2.571311 x2 = 2.571281 x3 = 2.5712815 Since x2 and x3 are same up to 6 decimal places, hence, the required Positive root is 2.571281.
ORDER OF CONVERGENCE OF ITERATIVE METHODS The order of convergence is the order at which the error between two successive approximation to the root decreases. Where: A is a non-zero finite number called asymptotic error constant. ei and ei+1 are the errors in successive approximations. k is the largest positive real number
REFERENCES ENGINEERING MATHEMATICS ( N.P. BALI) http://www.sakshieducation.com/Engg/Engg Academia/CommonSubjects/MM- Algebraic&TransdentialEquations.pdf
!!!THE END!!! THANK YOU FOR YOUR ATTENTION

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Numerical solutions of algebraic equations

  • 1. SEMINAR On engineering mathematics TOPIC: NUMERICAL SOLUTION OF ALGEBRAIC EQUATIONS AVNEET SINGH LAL EC-2
  • 2. SOLUTION OF ALGEBRAIC AND TRANSCENDENTAL EQUATIONS Consider the equation of the form f(x)=0. If f(x) is a quadratic, cubic or biquadratic expression then algebraic formulae are available for expressing the roots. But when f(x) is a polynomial of higher degree or an expression involving transcendental functions e.g., 1 + cos x 5x, x tan x cosh x, e-x sin x etc., algebraic methods are not available.
  • 3. METHODS In order to solve these type of equations following methods exist: DIRECTIVE METHODS: The methods which are used to find solutions of given equations in the direct process is called as directive methods. Example: remainder theorem, Factorization method etc Note: By using Directive Methods, it is possible to find exact solutions of the given equation. ITERATIVE METHODS (INDIRECT METHODS): The methods which are used to find solutions of the given equation in some indirect process is called as Iterative Methods Note: By using Iterative methods, it is possible to find approximate solution of the given equation and also it is possible to find single solution of the given equation at the same time.
  • 4. DESCARTES RULE OF SIGNS Number of positive roots of f(x) = 0 is equal to the number of sign changes of the coefficients or is less than this number by an even integer. The number of negative roots of f(x) = 0 is obtained by considering the number of sign changes in f(x). EXAMPLE: f (x) = x5 + 2x4 5x3 7x2 + 12x + 2 No. of sign change = 2 = Positive Roots f (-x) = x5 + 2x4 + 5x3 7x2 12x + 2 No. of sign change = 3 = Negative Roots
  • 5. ITERATIVE METHODS BISECTION METHOD REGULA FALSI NEWTON RAPHSON
  • 6. BISECTION METHOD This method is based on the repeated application of intermediate value property. Locate the interval (a, b) in which root lies. Bisect the interval (a, b). Choose the half interval in which the root lies. Bisect the half interval. Repeat the process until the root converges.
  • 7. EXAMPLE: Find the real positive root of the equation- f (x) = x log10x 1.2 Correct upto 3 decimal places. SOLUTION: S.No. a f (a) b f (b) xi f (xi) 1 2.4 -ve 2.8 +ve 2.6 -ve 2 2.6 -ve 2.8 +ve 2.7 -ve 3 2.7 -ve 2.8 +ve 2.75 +ve 4 2.7 -ve 2.75 +ve 2.725 -ve 5 2.725 -ve 2.75 +ve 2.7375 -ve 6 2.7375 -ve 2.75 +ve 2.74375 +ve 7 2.7375 -ve 2.74375 +ve 2.743625 -ve
  • 8. REGULA FALSI METHOD This is the oldest method for finding the real roots of a numerical equation. It is sometimes known as method of linear interpolation. Locate the interval (a, b) in which the root lies. First Approximation to the root Locate the next interval (a, X0) or ( X0, b ) in which root lies. Repeat the process until the root converges.
  • 9. EXAMPLE: Find the real positive root of the equation- f (x) = 3x cos x 1 Correct upto 4 decimal places. Use Regula falsi method. SOLUTION: S.No. a f (a) b f (b) xi f (xi) 1 0 -2 1 1.459698 0.578085 -0.103255 2 0.578085 -0.103256 1 1.459698 0.605959 -0.004080 3 0.605959 -0.004079 1 1.459698 0.607157 -0.000159 4 0.607057 -0.000159 1 1.459698 0.607196 0.000006
  • 10. NEWTON RAPHSON METHOD This method is generally used to improve the result obtained by one of the previous methods. Locate the interval (a, b). Choose a or b which is nearer to the root as the first approximation x0 to the root. Next approximation: Repeat the process until the root converges.
  • 11. EXAMPLE: Find a positive value of (17)1/3 correct to six decimal places by Newton Raphson method. SOLUTION: The root lies between 2.5 and 2.6. Since, |f (2.6)| < |f (2.5)| Therefore, the root is nearer to 2.6. Let us take x0 = 2.58. Putting n = 0, 1, 2, ..., successively in Newton Raphsons formula, we get- x1 = 2.571311 x2 = 2.571281 x3 = 2.5712815 Since x2 and x3 are same up to 6 decimal places, hence, the required Positive root is 2.571281.
  • 12. ORDER OF CONVERGENCE OF ITERATIVE METHODS The order of convergence is the order at which the error between two successive approximation to the root decreases. Where: A is a non-zero finite number called asymptotic error constant. ei and ei+1 are the errors in successive approximations. k is the largest positive real number
  • 13. REFERENCES ENGINEERING MATHEMATICS ( N.P. BALI) http://www.sakshieducation.com/Engg/Engg Academia/CommonSubjects/MM- Algebraic&TransdentialEquations.pdf
  • 14. !!!THE END!!! THANK YOU FOR YOUR ATTENTION