CODEWITHSUNDEEP
python
sadesoda
sadesoda

Reputation: 75

python epsilon wrong calculation

epsilon is set to 0.1 but it gives me results until 1.2. I dont know what causes that. Can anyone help? 

def evaluate_poly(poly, x):
    total = 0.0
    for i in range(len(poly)):
        total += poly[i] * (x ** i)

    return total

def compute_deriv(poly):
    deriv = []
    if len(poly) < 2:
        return  [0.0]
    else:
        for i in range(1, len(poly)):
            deriv.append(float(poly[i] * i))
        return deriv

def compute_root(poly, x_0, epsilon):
    num = 0
    root = x_0
    while abs(evaluate_poly(poly, root)) >= epsilon:
        root = root - evaluate_poly(poly, root) / evaluate_poly(compute_deriv(poly), root)
        num += 1
    return [root, num]

print(compute_root((1.0,-2.0,1.0), 14, 0.1))

Upvotes: 1

Views: 122

Answers (2)

Serge Ballesta
Serge Ballesta

Reputation: 149155

You are trying to solve numerically x2 - 2 * x + 1 = 0. We know that the equation has a double root at x=1. But at that point (x=1), the derivative (2 * x - 2) is 0. That mean that the error in y will always be one magnitude order below the error in x, so the result of x = 1.2, y ~ 0.04 < 0.1 is not at all a surprise.

The difference xn - xn-1 would be a much better guess of the error in x:

def compute_root1(poly, x_0, epsilon):
    num = 0
    root = x_0
    while True:
        root1 = root - evaluate_poly(poly, root) / evaluate_poly(compute_deriv(poly), root)
        if abs(root1 - root) < epsilon: break
        num += 1
        root = root1
    return [root1, num+1]

It gives:

>>> print(compute_root1((1.0,-2.0,1.0), 14, 0.1))
[1.05078125, 8]

Upvotes: 3

snow
snow

Reputation: 438

epsilon represents y error (evaluate poly if you will), but your result of 1.2... is x value, where y is 0. in this case your y is 0.0412 and is lower than 0.1 so code is ok.

change your return in compute_root to:

return [root, num, abs(evaluate_poly(poly, root))]

Upvotes: 2

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