simplification using substitution

[Dave M]

Hi
I've tried subs() for this but it's not working for me. I have the following expression:

j*(3*x**4 + 6*x**2*y**2 - 24*x**2*z**2 + 3*y**4 - 24*y**2*z**2 + 8*z**4)/(2*(x**2 + y**2 + z**2)**(9/2))

I want to simplify it by making the substitutions r = sympy.sqrt(x*x + y*y + z*z) and s = sympy.sqrt(x*x + y*y) so that it becomes something like

j/r**9*(4*z**2*(z**2 - 3*s**2) + 3/2*s**4)

can this be done with sympy, and if so how?
Thanks

      

[Aaron Meurer]

The subs for s = sqrt(x**2 + y**2 + z**2) should work.

To make the other one work, try subs(x**2, s**2 - y**2), followed by a call to factor().

Aaron Meurer

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[Dave M]

Thanks for responding Aaron. Here is the relevant part of my session for the attempted r substitution. Can you tell me what I'm doing incorrectly and how to do it right?

In [113]: f = j*(3*x**4 + 6*x**2*y**2 - 24*x**2*z**2 + 3*y**4 - 24*y**2*z**2 + 8*z**4)/(2*(x**2 + y**2 + z**2)**(9/2))

In [114]: f
Out[114]: j*(x**2 + y**2 + z**2)**(-4.5)*(3*x**4 + 6*x**2*y**2 - 24*x**2*z**2 + 3*y**4 - 24*y**2*z**2 + 8*z**4)/2

In [115]: f.subs(sympy.sqrt(x**2+y**2+z**2),r)
Out[115]: j*(x**2 + y**2 + z**2)**(-4.5)*(3*x**4 + 6*x**2*y**2 - 24*x**2*z**2 + 3*y**4 - 24*y**2*z**2 + 8*z**4)/2

[Aaron Meurer]

First off, make sure not to have 9/2, as Python converts this into a Float. Use Rational(9, 2) or S(9)/2.

What is your r? It works for me:

In [10]: r = Symbol('r')

In [11]: f = j*(3*x**4 + 6*x**2*y**2 - 24*x**2*z**2 + 3*y**4 - 24*y**2*z**2 + 8*z**4)/(2*(x**2 + y**2 + z**2)**(9/2))

In [12]: print(f.subs(sqrt(x**2+y**2+z**2),r))

j*(3*x**4 + 6*x**2*y**2 - 24*x**2*z**2 + 3*y**4 - 24*y**2*z**2 + 8*z**4)/(2*r**9)

I suspect you need to read http://docs.sympy.org/latest/tutorial/gotchas.html.

Aaron Meurer

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[Dave M]

Thanks for the tips. My problem seems to have been I had r already defined (to sympy.sqrt(x**2 + y**2 + z**2). I guess it cannot be pre-defined for subs to work correctly. Works now. Thanks!

BTW I really like sympy. I just stumbled across it last week by accident and had to try it out. And very cool that it has special functions. The predefined Legendre polynomials are saving me lots of time from having to key them in.

[Aaron Meurer]

Yes, this is what I figured. Read the page I referred you to. It discusses this. You need to be clear about the difference between Python variables and SymPy symbols.

Aaron Meurer

On Sun, Apr 6, 2014 at 2:23 PM, Dave M <daveme...@gmail.com> wrote:

Thanks for the tips. My problem seems to have been I had r already defined (to sympy.sqrt(x**2 + y**2 + z**2). I guess it cannot be pre-defined for subs to work correctly. Works now. Thanks!

BTW I really like sympy. I just stumbled across it last week by accident and had to try it out. And very cool that it has special functions. The predefined Legendre polynomials are saving me lots of time from having to key them in.

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