Above 0, the function is monotonic but is ill-behaved in that it is very
flat; this can cause troubles for different solver routines. A slow but
robust method is bisection
>>> nsolve(eq,c2,(1,1000),solver='bisect')
mpf('22.964256014441664')
Using that, you don't need a very precise range for the root.
Also, once you have the root, continuation to a new root at a new set of
parameters can be attempted by slowly varying the old parameters until they
become the new parameters. This is called "the continuation method" and
should be able to easily find information on that. Here is an example.
Your equation now is 3/10 - 1/log(c2 + 1) + 17/(50*c2). Let's say you want
to vary the 3/10 to 5/10, the numerator of 1 on the log term to 2 and the
17/50 to 2, too. Let's make a function to give a linear variation from a to
b in n steps:
>>> def vary(a,b,n=10):
... return a + i*(b-a)/S(n)
...
>>> n = 10
>>> ceq=vary(S(3)/10, s(5)/10, n)-vary(1, 2, n)/log(c2+1)+vary(S(17)/50, 2,
n)/c2
>>> ceq
i/50 - (i/10 + 1)/log(c2 + 1) + 3/10 + (83*i/500 + 17/50)/c2
We already know that a solution for the original equation is about 22 so
make that your initial guess
>>> g=22
Then increment the i value to move you toward your desired equation,
keeping the new solution as your new guess.
>>> for iv in range(n+1):
... g=nsolve(ceq.subs(i,iv),c2,g)
...
So what is the final solution? The guess that we got when using i=n
>>> g
mpf('35.36155261017116')
And this is the desired equation that we got to by slowly varying the
coefficients:
>>> ceq.subs(i, iv)
1/2 - 2/log(c2 + 1) + 2/c2
You can't use continuation blindly; graphing is your friend. Don't forget
about plot in SymPy, too: e.g. plot(x+1,(x,-1,1)).
/c
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