Newton Raphson method Algorithm & Example-1 f(x)=x^3 | newton method algorithm
AlgorithmBisectionmethodSteps(Rule)Step-1:Findpoints`a`and`b`suchthat`aStep-2:Taketheinterval`[a,b]`andfindnextvalue`x_0=(a+b)/2`Step-3:If`f(x_0)=0`then`x_0`isanexactroot,elseif`f(a)*f(x_0)elseif`f(x_0)*f(b)Step-4:Repeatsteps2&3until`f(x_i)=0`or`|f(x_i)|Example-1Findarootofanequation`f(x)=x3-x-1`usingBisectionmethodSolution:Here`x3-x-1=0`Let`f(x)=x3-x-1`Here`1(st)`iteration:Here`f(1)=-1<0`and`f(2)=5>0``:.`Now,Rootliesbetween`1`and`2``x_0=(1+2)/2=1.5``f(x_0)=f(1.5)=0.875>0``2(nd)...
Algorithm
Bisection method Steps (Rule) Step-1: Find points `a` and `b` such that `a Step-2: Take the interval `[a, b]` and find next value `x_0 = (a+b)/2` Step-3: If `f(x_0) = 0` then `x_0` is an exact root, else if `f(a) * f(x_0) else if `f(x_0) * f(b) Step-4: Repeat steps 2 & 3 until `f(x_i) = 0` or `|f(x_i)| Example-1Find a root of an equation `f(x)=x3-x-1` using Bisection methodSolution:Here `x3-x-1=0`
Let `f(x) = x3-x-1`
Here
`1(st)` iteration :
Here `f(1) = -1 < 0` and `f(2) = 5 > 0`
`:.` Now, Root lies between `1` and `2`
`x_0 = (1 + 2)/2 = 1.5`
`f(x_0) = f(1.5) = 0.875 > 0`
`2(nd)` iteration :
Here `f(1) = -1 < 0` and `f(1.5) = 0.875 > 0`
`:.` Now, Root lies between `1` and `1.5`
`x_1 = (1 + 1.5)/2 = 1.25`
`f(x_1) = f(1.25) = -0.29688 < 0`
`3(rd)` iteration :
Here `f(1.25) = -0.29688 < 0` and `f(1.5) = 0.875 > 0`
`:.` Now, Roo...