Use Newtons method to find the roots of the equation 1/x=1+x^3
f(x)=x^3-1/x + 1
guess, f(x)
1, 1 positive
2, 8.5 postive go the other way.
1/2, 1/8-2+1=-7/8 so root is between 1, and 1/2
try .75 as root.
Now is the time to use the google search window
.75^3-1/.75+1= (put that in google window), .089 postive, so go lower toward 1/2
.65^3-1/.65+1=-.26 go postive now
.70^3-1/.70+1=-.08 negative, but notice it is close to zero....
.73^3-1/.73+1=+.019, go back
.72^2-1/.72+1=-.0156 go back
.726^2-1/.726 + 1=.005 go back
.724^2-1/.724+1= you do it. You can get the roots as accurate as you wish. Calculator required.
multiply by x
1 = x + x^4
x^4 + x - 1 = 0
let f(x) = x^4 + x - 1
f'(x) = 4x^3 + 1
newton's formula for his method says
f(newx) = x - f(x)/f'(x) , where x is some intial guess
= x - (x^4 + x -1)/(4x^2 + 1)
= (3x^4 + 1)/(4x^3 + 1)
let's start with bobpursleys first guess of x = .75
x ---- newx
.75------ .72529
.72529 -- .724492755
.72449275-- .7244919
x = .72449 correct to 5 decimals in 3 iterations
there appears to be another root around x = -1.25
x = -1.25
x ---- newx
-1.25 ----- +1.2219..
-1.2219.. -- -1.2207459..
-1.220749.. -- -1.2207440..
-1.220744 --- -1.2207440..
x = -1.220744 correct to 6 decimals in 4 iterations
my rough sketch shows only 2 real roots,
since it is a 4th degree, there have to be 2 complex roots.
you could do two consecutive synthetic divisions using
x=.72449 followed by x = -1.220744
to reduce it down to a quadratic, then using the formula to find the imaginary roots.
To use Newton's method to find the roots of an equation, you will need an initial guess and a function for which you want to find the root. In this case, let's solve the equation 1/x = 1 + x^3.
Step 1: Let's rearrange the equation to the form f(x) = 0, where f(x) = 1/x - (1 + x^3).
So, f(x) = 1/x - 1 - x^3 = 0.
Step 2: Find the derivative of f(x) with respect to x. Let's call it f'(x).
f(x) = 1/x - 1 - x^3
f'(x) = -1/x^2 - 3x^2
Step 3: Start with an initial guess, let's say x0 = 1.
Step 4: Use the formula for Newton's method to update the guess x until f(x) is close to zero.
xn+1 = xn - (f(xn) / f'(xn))
Let's perform a few iterations to find the root:
Iteration 1:
x1 = x0 - (f(x0) / f'(x0))
f(x0) = 1/x0 - 1 - x0^3
f'(x0) = -1/x0^2 - 3x0^2
Substituting the values:
x1 = 1 - ((1/1 - 1 - 1^3) / (-1/1^2 - 3(1)^2))
Iteration 2:
x2 = x1 - (f(x1) / f'(x1))
Repeat this process until the value of f(x) is close to zero or you achieve the desired accuracy.
By using this iterative process, you can find the roots of the equation 1/x = 1 + x^3 using Newton's method.