Computer Science\/Numerical Methods\/ Bisection method example<\/p>\n
[fvplayer id=”110″]<\/p>\n
The Bisection Method<\/strong> is a root-finding technique that applies to continuous functions<\/strong> where the root lies between two given points. It is a simple and reliable method based on the Intermediate Value Theorem<\/strong>.<\/p>\n \u00a0Select two points a<\/strong> and b<\/strong> such that f(a)f(a)<\/span>f<\/span>(<\/span>a<\/span>)<\/span><\/span><\/span><\/span> and f(b)f(b)<\/span>f<\/span>(<\/span>b<\/span>)<\/span><\/span><\/span><\/span> have opposite signs<\/strong> (i.e., f(a)\u00d7f(b)<0f(a) \\times f(b) < 0<\/span>f<\/span>(<\/span>a<\/span>)<\/span>\u00d7<\/span><\/span>f<\/span>(<\/span>b<\/span>)<\/span><<\/span><\/span>0<\/span><\/span><\/span><\/span>), ensuring a root exists in between. c=a+b2c = \\frac{a + b}{2}<\/span>c<\/span>=<\/span><\/span>2a<\/span>+<\/span>b<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n \u00a0Check f(c)f(c)<\/span>f<\/span>(<\/span>c<\/span>)<\/span><\/span><\/span><\/span>:<\/p>\n If f(c)=0f(c) = 0<\/span>f<\/span>(<\/span>c<\/span>)<\/span>=<\/span><\/span>0<\/span><\/span><\/span><\/span>, then cc<\/span>c<\/span><\/span><\/span><\/span> is the root.<\/p>\n<\/li>\n If f(a)\u00d7f(c)<0f(a) \\times f(c) < 0<\/span>f<\/span>(<\/span>a<\/span>)<\/span>\u00d7<\/span><\/span>f<\/span>(<\/span>c<\/span>)<\/span><<\/span><\/span>0<\/span><\/span><\/span><\/span>, then the root lies in [a, c]<\/strong>, set b=cb = c<\/span>b<\/span>=<\/span><\/span>c<\/span><\/span><\/span><\/span>.<\/p>\n<\/li>\n Else, the root lies in [c, b]<\/strong>, set a=ca = c<\/span>a<\/span>=<\/span><\/span>c<\/span><\/span><\/span><\/span>. \u2223b\u2212a\u2223<tolerance|b – a| < \\text{tolerance}<\/span>\u2223<\/span>b<\/span>\u2212<\/span><\/span>a<\/span>\u2223<\/span><<\/span><\/span>tolerance<\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n f(1)=13\u22121\u22122=\u22122f(1) = 1^3 – 1 – 2 = -2<\/span>f<\/span>(<\/span>1<\/span>)<\/span>=<\/span><\/span>13<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u2212<\/span><\/span>1<\/span>\u2212<\/span><\/span>2<\/span>=<\/span><\/span>\u2212<\/span>2<\/span><\/span><\/span><\/span>\u00a0Algorithm of Bisection Method<\/strong><\/h3>\n
\u00a0Compute the midpoint cc<\/span>c<\/span><\/span><\/span><\/span>:<\/p>\n\n
\u00a0Repeat the process until the error tolerance is met:<\/p>\n<\/li>\n<\/ul>\n\u00a0Example: Find the root of f(x)=x3\u2212x\u22122f(x) = x^3 – x – 2<\/span>f<\/span>(<\/span>x<\/span>)<\/span>=<\/span><\/span>x<\/span>3<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u2212<\/span><\/span>x<\/span>\u2212<\/span><\/span>2<\/span><\/span><\/span><\/span> in [1, 2]<\/strong><\/h3>\n
Step 1: Check function values<\/strong><\/h3>\n
f(2)=23\u22122\u22122=4f(2) = 2^3 – 2 – 2 = 4<\/span>f<\/span>(<\/span>2<\/span>)<\/span>=<\/span><\/span>23<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u2212<\/span><\/span>2<\/span>\u2212<\/span><\/span>2<\/span>=<\/span><\/span>4<\/span><\/span><\/span><\/span>
Since f(1)\u00d7f(2)<0f(1) \\times f(2) < 0<\/span>f<\/span>(<\/span>1<\/span>)<\/span>\u00d7<\/span><\/span>f<\/span>(<\/span>2<\/span>)<\/span><<\/span><\/span>0<\/span><\/span><\/span><\/span>, a root exists between 1 and 2<\/strong>.<\/p>\nStep 2: Iterations<\/strong><\/h3>\n