Translate pseudocode into python (secant method)

Question

I am working on this assignment:

First, implement the f-function defined by: f(x)= exp(x)-sin(x) closest to zero.

Second, implement the Secant method on page 95 and use it to find the root of the f-function, given the input values x0 = -3.5 and x1 = -2.5

Add the following
- an absolute test: abs(f(x) ) < epsilon
- a relative test: abs(x^k - x^{k-1})/ abs(x^{k}) \leq delta
- a maximum iteration guard: k < iter_max

In each iteration print out the iteration number k, the value of current root and current f-value. Print float numbers with 20 digits.

This is the code I have to complete:

import numpy as np
from math import exp, sin
import matplotlib.pyplot as plt

def f(x: float) -> float:
    return

def secant(x0: float, x1: float, f, epsilon: float, delta: float, iter_max: int) -> float:
    return

This is the pseudocode from page 95:

input: x_0, x_1, delta, epsilon, iter_max
fx_0 <- f(x_0); fx_1 <- f(x_1)
output: 0, x_0, fx_0
output: 1, x_1, fx_1
for k = 2 to iter_max do
    if |fx_0| > |fx_1| then
        x_0 <-> x_1; fx_0 <-> fx_1
    end if
    s <- (x_1 - x_0) / (fx_1 - fx_0)
    x_1 <- x_0
    fx_1 <- fx_0
    x_0 <- x_0 - fx_0 * s
    fx_0 <- f(x_0)

    output: k, x_0, fx_0
    if |fx_0| < epsilon or |x_1 - x_0| < delta then stop
end do

And this is my own attempt:

def f(x: float) -> float:
    return exp(x) - sin(x) == 0

def secant(x0: float, x1: float, f, epsilon: float, delta: float, iter_max: int) -> float:
    fx0 = f(x0)
    fx1 = f(x1)
    return 0, x0, fx0
    return 1, x1, fx1

    for k in range(2, iter_max):
        if abs(fx0) > abs(fx1):
            x0 = x1
            x1 = x0
            fx0 = fx1
            fx1 = fx0

            s = (x1 - x0) / (fx1 - fx0)
            x1 = x0
            fx1 = fx0
            x0 = x0 - fx0 * s
            fx0 = f(x0)

        return k, x0, fx0

        if abs(fx0) < epsilon or abs(x**k - x**(k - 1))/ abs(x**(k))  <= delta:
            break

If I follow my code with

root = secant(-3.5, -2.5, f, 0.00000000001, 0.000001, 10) 
print(root)

I get: (0, -3.5, False). So it doesn't actually do any iterations. How can I fix it?

Edit:
Photo of pseudocode

Here: a=x_0, b=x_1 and M =iter_max.
I would like the output to be something like this:

if you use return you will not execute the following lines within your function — eyllanesc
– eyllanesc, Commented Sep 27, 2017 at 20:32
Your sec secant code is reduced to the following if you use silly return.: def secant(x0: float, x1: float, f, epsilon: float, delta: float, iter_max: int) -> float: fx0 = f(x0) fx1 = f(x1) return 0, x0, fx0 — eyllanesc
– eyllanesc, Commented Sep 27, 2017 at 20:33
Sorry for being vague. My problem is the output. If for example I follow my code with root = secant(-3.5, -2.5, f, 0.00000000001, 0.000001, 10) print(root) I get: (0, -3.5, False). So it doesn't actually do any iterations. — user405381
– user405381, Commented Sep 27, 2017 at 20:44

eyllanesc · Accepted Answer · 2017-09-27 21:09:35Z

1

Your implementation has the followings errors:

The first is that the function f must return the value of the function from which you want to get the root, you should not compare it with zero.
The second error is caused by not reading the algorithm's objective and understanding its procedure, if it says:

output: 0, x_0, fx_0
output: 1, x_1, fx_1

indicates that it is the result when iter_max is 0 or 1 respectively.

Third, in the form of exchanging values is inadequate, you must use a pivot since otherwise erase the relevant information
Another thing I do not understand is because you execute the following: abs(x**k - x**(k - 1))/ abs(x**(k)) instead of abs(x1 - x0).

So correcting both errors you get the following code:

def f(x: float) -> float:
    return exp(x) - sin(x)

def secant(x0: float, x1: float, f, epsilon: float, delta: float, iter_max: int) -> float:
    fx0 = f(x0)
    fx1 = f(x1)
    if iter_max == 0:
        return 0, x0, fx0

    elif iter_max == 1:
        return 1, x1, fx1

    for k in range(2, iter_max):
        if abs(fx0) > abs(fx1):
            tmp =  x0
            x0 = x1
            x1 =tmp
            tmp = fx0
            fx0 = fx1
            fx1 = tmp

        s = (x1 - x0) / (fx1 - fx0)
        x1 = x0
        fx1 = fx0
        x0 = x0 - fx0 * s
        fx0 = f(x0)
        if abs(fx0) < epsilon or abs(x1 - x0)  <= delta:
            break
    return k, x0, fx0


root = secant(-3.5, -2.5, f, 0.00000000001, 0.000001, 10) 

print(root)

Output:

(5, -3.183063011933318, 4.7351012000262926e-14)

answered Sep 27, 2017 at 21:09

eyllanesc

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1 Comment

user405381 Over a year ago

Thank you, this works! Although, I need it to print k, the value of the current root and the current f-value after each iteration. Also how do I specify that float numbers should be printed with 20 digits?

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