I want to get the solution of the equation

## Bisection Method

The first algorithm for the problem is bisection method.

The algorithm bisection method doesn’t require derivation for the equation. It just needs a possible solution range [l, r], then calculates the variable mid as , compares the result of `func(mid)`

and zero to adjust solution range, until it finds the target value or go beyond the max number of iterations.

```
from math import fabs
def func( x ):
return x*x*x - 2*x*x - 5.0
"""
:param y: the target value of functon
:param l: the x-axis value where func(l) < 0
:param r: the x-axis value where func(r) > 0
:param limitErr: the precision
:param maxIterator: max number of iterations
"""
def midfind( y, l, r, limitErr, maxIterator ):
iterator = 1
while iterator <= maxIterator :
mid = ( l + r ) / 2.0
if func( mid ) == 0 or fabs(l-r)/2.0 < limitErr :
return mid, iterator
iterator = iterator + 1
if func(mid) < 0:
l = mid
else:
r = mid
return None, iterator
if __name__ == "__main__":
x, it = midfind( 0, -5, 5, 0.00001, 40 )
print "X is ",x
print "The number of iteractions is ",it
```

output:

```
X is 2.69064903259
The number of iteractions is 20
```

## Newton’s Method

We use to represent the derivative function of original function .

It’s common known that .

Then we have , its form can be changed to .

The equation is logical if is similar to zero and .

So the following equations deduced are the central idea of Newton’s method:

```
from math import fabs
def func( x ):
return x*x*x - 2*x*x - 5.0
def derivaFunc( x ):
return 3*x*x - 4*x
"""
:param x: the guess value of x where func(x) = 0
:param limitErr: the precision
:param maxIterator: max number of iterations
"""
def newton( x, limitErr, maxIterator ):
iterator = 1
while iterator <= maxIterator :
x1 = x - func(x) / derivaFunc(x)
if fabs(x-x1) < limitErr:
return x1, iterator
else:
x = x1
iterator = iterator + 1
return None, iterator
if __name__ == "__main__":
x, it = newton( 5, 0.00001, 100 )
print "X is ",x
print "The number of iteractions is ",it
```

output:

```
X is 2.69064744803
The number of iteractions is 6
```

## Secant Method

The algorithm draws secant lines multiple times until find similar solution.

The secant line intersect with equation curve and get two points of intersection.

When y is equal to zero, we have the following equation:

Remove y, becase it is equal to 0, we can get:

To get right x value where func(x) = 0, we have to move the secant line from (a,b) to (c, b).

```
from math import fabs
def func( x ):
return x*x*x - 2*x*x - 5.0
"""
:param a: the x-axis value
:param b: the x-axis value where b > a
:param limitErr: the precision
:param maxIterator: max number of iterations
"""
def secant( a, b, limitErr, maxIterator ):
iterator = 1
while iterator <= maxIterator :
c = b - func(b) * ((b-a)/(func(b)-func(a)))
if fabs(c-b) < limitErr:
return c, iterator
else:
a = b
b = c
iterator = iterator + 1
return None, iterator
if __name__ == "__main__":
x, it = secant( -5, 5, 0.00001, 100 )
print "X is ",x
print "The number of iteractions is ",it
```

output:

```
X is 2.69064744803
The number of iteractions is 8
```

## scipy.optimize

The above algorithms are all avaliable in module scipy.optimize, we can import it and use these algorithms directly.

```
import scipy.optimize as optimize
y = lambda x: x**3.0 - 2*x**2 - 5
dy = lambda x: 3.0*x**2 - 4*x
print "Bisection method: ",optimize.bisect(y, -5, 5, xtol=0.00001)
#The Newton-Raphson method is used if the derivative fprime of func is provided,
#otherwise the secant method is used.
print "Newton method: ",optimize.newton(y, 5, dy, tol=0.00001)
print "Secant method: ",optimize.newton(y, 5, tol=0.00001)
```

output:

```
Bisection method: 2.69064903259
Newton method: 2.69064744803
Secant method: 2.69064744805
```

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