Solving Equations Numerically: Newton's Method, Bisection MethodDate: 11/05/2002 at 07:36:30 From: Emily Bardsley Subject: Solving equations numerically Hey, Would you be able to tell me how you can work out equations that can only be solved numeraically and not algebraically. Surely the only way is not to try out the equation with every algebraic technique, and if this is so then do you know how many methods of algebraic solving there are? Thank you very much, Emily Date: 11/05/2002 at 08:11:15 From: Doctor Jerry Subject: Re: Solving equations numerically Hi Emily, There are several methods for solving equations numerically. If you know calculus, then you can use Newton's method. Suppose you want to find a zero of a function f, that is, you want to find a number c for which f(c)=0. For Newton's method, you must first find an approximation x1 to c. Maybe you can do this graphically. You calculate the next approximation, call it x2, with the formula x2 = x1-f(x1)/f'(x1) where f' is the derivative of f. Once you have found x2, you find x3 with a formula having the same form: x3 = x2-f(x2)/f'(x2) In this way you get a sequence x1,x2,x3,... of improving approximations to c. Newton's method is usually quite efficient. Try f(x) = x^2-2, for example. There is a zero, namely c=sqrt(2), near x1 = 1.5. Okay, f(x) = x^2-2 f'(x) = 2x x1 = 1.5 x2 = 1.41666666667 x3 = 1.41421568628 etc. Or you can use the Bisection Method. For this method of a finding a zero c of a function f, you begin by figuring out an interval [a1,b1] containing c and having the property that f(a1)*f(b1) < 0. (This condition means that the graph of f crosses the x-axis somewhere between a1 and b1. Where it crosses is a zero of f.) Calculate the midpoint m1 of this interval. If f(a1)*f(m1)<0, then a zero of f is in [a1,m]. This is the new interval [a2,b2]. Otherwise, a zero of f must be in [m1,b1], which is the new interval [a2,b2]. Continue with this scheme of bisection an interval containing a zero. The Bisection Method is slower than Newton's method, but a bit more certain. Both of these methods can be programmed on a calculator to speed up the calculations. - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ Date: 11/10/2002 at 11:20:14 From: Emily Bardsley Subject: Re: Solving equations numerically Dear Dr Math, Thank you so much for your help. I was also wondering if you would be able to help in suggesting some good equations with which I could try out these methods on using graphs. I am looking for equations that have about 2 or 3 roots but can't be solved algebraically. I've been trying to find some interesting equations such as ones involving square roots or ones that have a curve which crosses the x axis at one point but only touches the curve (i.e. a parabola) at another. I'm unsure of how I could go about deriving these equations that will produce good graphs - any ideas? Thank you for your time so far. Emily Date: 11/10/2002 at 15:44:09 From: Doctor Jerry Subject: Re: Solving equations numerically Hi Emily, 1. Find the zeros of the function f(x) = 2^x -4*x 2. a. Locate between successive tenths the first four positive zeros of the function f(x) = cos(x)*cosh(x) + 1. b. Find the three smallest positive zeros of the equation cos(68.617*sqrt(x))*cosh(68.617*sqrt(x)) = -1. 3. Find a zero of f(x) = cos(ln(x)), x>0. 4. Find to within 0.1deg the smallest positive zero of the equation tan^3(t) - 4*tan^2(t) + tan(t) - 4 =0. 5. The volume V (in cubic meters) of 1 mole of gas is related to its temperature T (in kelvins) and pressure P (in pascals (Pa)) by the ideal gas law PV=RT. A more accurate equation is van der Waal's equation (P+a/V^2)(V-b) = R*T. The constant R is 8.314. For carbon dioxide, a=3.592 and b=0.04267. Find to two decimal places the volume V of 1 mole of carbon dioxide if P=20,000 Pa and T=320K. - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ Date: 11/10/2002 at 16:33:15 From: Emily Bardsley Subject: Thank you (Solving equations numerically) Thanks a lot. Excellent service! Emily |
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