Use Newton's method to approximate the value of
(543)^(1/5) as follows:
Let x1=2 be the initial approximation.
find x2 and x3 =? approximation
The question is not completely defined.
Newton's method is for solving an equation.
(543)^(1/5) is an expression that does not contain unknowns.
Please check for typo.
543^(1/5) is the solution to
x^5 = 543 or
x^5 - 543 = 0
let y = x^5 - 543
y' = 5x^4
newx = x - y/y' = x - (x^5 - 543)/(5x^4)
= (4x^5 + 543)/(5x^4)
x2 = 8.3875
x3 = 6.7319
x4 = 5.4384
x5 = 4.3507
x6 = 3.480576
x7 = 3.524449
x8 = 3.5233845
x9 = 3.523384
To use Newton's method to approximate the value of (543)^(1/5), we start by setting up the iterative formula:
x[n+1] = x[n] - f(x[n])/f'(x[n])
where x[n] represents the nth approximation, f(x) represents the function we want to approximate, and f'(x) represents the derivative of the function.
In this case, our function f(x) is (x^5) - 543, and its derivative f'(x) is 5(x^4).
Given that x1 = 2, we can now calculate x2 and x3:
Step 1: Apply the iterative formula to find x2:
x2 = x1 - f(x1)/f'(x1)
= 2 - (2^5 - 543)/(5*(2^4))
= 2 - (32 - 543)/(5*16)
= 2 - (-511)/80
= 2 + 6.3875
≈ 8.3875
Therefore, x2 ≈ 8.3875
Step 2: Apply the iterative formula again to find x3:
x3 = x2 - f(x2)/f'(x2)
= 8.3875 - (8.3875^5 - 543)/(5*(8.3875^4))
= 8.3875 - (8.3875^5 - 543)/(5*1236.6424)
= 8.3875 - (4572.51894 - 543)/6183.212
= 8.3875 - 4029.51894/6183.212
= 8.3875 - 0.6513
≈ 7.7362
Therefore, x3 ≈ 7.7362
Hence, the approximations x2 and x3 for (543)^(1/5) using Newton's method are approximately 8.3875 and 7.7362, respectively.