On 2:59 PM, Matthew Denaburg wrote:
Hi,
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
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
You don't have to guess. Once you come to that conclusion, check it.
Put a print( y), or perhaps print (repr(y)) and check the value.
<snip>
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!
<snip>
I don't know in particular, but if you were to raise a negative number
to the .3333 power, the answer would be complex. Perhaps you're getting
hit with some roundoff/quantization error. 1/3 cannot be exactlly
represented, of course. I'd be curious if that's the situation here.
If so, you could finesse the problem with the expression:
sign * (abs(x) ** (1.0/3))
where sign is +1 or -1 representing the sign of x.
DaveA
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