[Tutor] Math: integers to a fractional power

[Evert Rol]

> 	OK, so I have a question for you math people: I am trying to solve a quartic equation (Ax^4 + Bx^3 + Cx^2 + Dx + E) using Ferrari's Method, which I found on Wikipedia at this location.
> 
> 	I'm using Python 3.1.2 (on Mac OS X 10.6, in case that matters).
> 
> 	First, since I don't know Python very well, I was wondering if there is an easier way to do this, without having to install any modules. (I can't figure out how to put anything on my PYTHONPATH).
> 
> 	If not there is not an easier way without installing a module, could someone recommend a module AND give me a link to detailed instructions for how to install it? 
> 
> 	I would PERFER to continue using Ferrari's Method, but am not opposed to trying something else.
> 
> 	My only problem when using Ferrari's Method was this error:
> 
> Traceback (most recent call last):
>   File "Problem309.py", line 135, in <module>
>     print(main())
>   File "Problem309.py", line 127, in main
>     math.pow(L1, 2)-math.pow(L2, 2))
>   File "Problem309.py", line 73, in sQuartic
>     W = math.pow(alpha + 2 * y, 1/2)
> TypeError: can't convert complex to float

This may be more a result from the underlying C math library (also the exception you got below); I think the functions in that library are called when using functions from the math module, and the C library only works for non-complex numbers.
You could have a look at the cmath module. However, you won't find a pow() function there.

Instead, try using the built-in power operator: **, and use complex():
>>> complex(-27)**1/3.
(-9+0j)
>>> complex(-4)**(1/2.)
(1.2246467991473532e-16+2j)

Note the (classic) inaccuracy problem of floating points: the real part should simply be 0.
(Also, create complex numbers with an imaginary part != 0 as:
>>> complex(5, -2)
(5-2j)
)

Would that help you?

Cheers,

  Evert

> 	Now, your first thought is probably that I have a negative number in the square root, or that I should be using math.sqrt. According to this method, I need to use cube roots as well, and I wanted to keep my code consistent. I also have checked, and the only variable that could be ending up as a complex number is the variable y. So, here is the code that assigns y:
> 
> u = math.pow(Q, 2)/4 + math.pow(P, 3)/27
> if u < 0:
>     return None
> R = -(Q / 2) + math.pow(u, 1/2)
> U = R ** (1/3)
> y = -(5/6) * alpha + U
> if U == 0:
>     y = y - Q ** (1/3)
> elif U != 0:
>     y = y - P/(3*U)
> W = math.pow(alpha + 2 * y, (1/2))
> 
> 
> 	The problem I am having is that, as far as I can tell,  is that U is turning into a complex number, which is impossible! You can take the cube root of any number, positive or negative!
> 
> 	So I tried in the Python Interpreter, and got this:
> 
> >>> -27**(1/3)
> -3.0
> >>> math.pow(-27, 1/3)
> Traceback (most recent call last):
>   File "<pyshell#23>", line 1, in <module>
>     math.pow(-27, 1/3)
> ValueError: math domain error
> 
> >>>
> 
> 	Is there something going on here that I am unaware of?
> 
> 	Please let me know!! Thanks so much!
> 
> ---Matthew Denaburg
> 
> 
> 
> 
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