Would it be possible to define a tree of equations as:
root: f(x)=f1(x)*f3(x)
leaf 1: f1(x)=...
leaf 0: f3(x)=...+f2(x)
leaf 00: f2(x)=...
or equivalently
leaf 1
/
root
\
leaf 0
\
leaf 00
and then recurse the tree by applying the diff operator to the current
equation?
Thanks
V
On Apr 5, 2:31 am, Chris Smith <[email protected]> wrote:
> The following might get you started:
>
> def difcse(cse, x, n=1):
> r, e = cse
> nreps = len(r)
> diffs = []
> for i in range(nreps):
> v, ri = r[i]
> dri = simplify(ri.diff(x, n))
> if dri:
> func = Symbol('d_'+v.name+'_d'+x.name+
> str(n if n>1 else ''))
> r.append((v, func(x)))
> diffs.append((r[-1][1].diff(x, n), func, dri))
> if len(r) > nreps:
> derivs = r[nreps:]
> for j in range(len(e)):
> e[j] = e[j].subs(derivs).diff(x, n).subs(
> [d[:2] for d in diffs]).subs(
> [(b, a) for a, b in r[nreps:]])
> for j in range(nreps, len(r)):
> i = j - nreps
> r[j] = diffs[i][-2:]
> return r, e
>
> >>> re = cse(cos(x+y)*(x**2+1)+exp((x+y)/(x**2+1)))
> >>> print difcse(re, x, 1)
>
> ([(x0, x + y), (x1, x**2 + 1), (d_x0_dx, 1), (d_x1_dx, 2*x)],
> [-d_x0_dx*x1*sin(x
> 0) + d_x1_dx*cos(x0) + (d_x0_dx/x1 - d_x1_dx*x0/x1**2)*exp(x0/x1)])>>> re =
> cse(cos(x+y)*(x**2+1)+exp((x+y)/(x**2+1)))
> >>> print difcse(re, x, 2)
>
> ([(x0, x + y), (x1, x**2 + 1), (d_x1_dx2, 2)],
> [-d_x1_dx2*x0*exp(x0/x1)/x1**2 +
> d_x1_dx2*cos(x0) + x0**2*exp(x0/x1)*Derivative(x1, x)**2/x1**4 +
> 2*x0*exp(x0/x1)
> *Derivative(x1, x)**2/x1**3])
>
> Notice that when taking the second derivative, some first derivatives
> remain, so perhaps the difcse (which modifies the cse result in place)
> should also always include the first derivative if n > 1.
>
> /c
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