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I have the following equation.

f[a,b,Q]=1 - Q*Exp[b] + Exp[a]*(Exp[b]^3 - Exp[b]^4 + 2*Exp[b]^5 - 
2*Q*Exp[b]^5 - Q*Exp[b]^6 + Q^2*Exp[b]^7) + Exp[a]^2*(-Exp[b]^9 + Exp[b]^(10))

and I want to solve for its first derivative at a specific point, dictated by the logarithm of a solution/rule from a previous calculation, i.e. s1b.

fUV[a[b], b, 2] == 0 /. b -> Log[b /. s1b[[4]][[2]]]

As you can see I am solving for "a" as a series in "b", and the fourth substitution rule, s1b[[4]], has a "b"-value for its second argument. (The first argument would be an "a"-value). Anyway, I get the following;

(1.588455019 - 1.285970719 I) + 
(0.510032785 + 0.330500867 I) E^a[-0.3465735903 + 1.9999490798 I] - 
(0.0163998482 - 0.0617256728 I) E^(2 a[-0.3465735903 + 1.9999490798 I])
 == 0

which is quadratic in what I want to solve for.

But when I do;

Solve[%, a[Log[b /. s1b[[4]][[2]]]]]

I only get an empty list. Due to the relative simplicity I am thinking that, since this "index" of "a" is a solved-for real number, possibly Mathematica may not have the same "index" each time.... An accuracy thing?

Thanks for your suggestions and help in advance!

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  • $\begingroup$ Mathematica needs to recognise the second argument of Solve as a variable. It can be a bit picky in what it will accept. Try replacing a[-0.3465735903 + 1.9999490798 I] with a fixed symbol e.g. %/.a[_]->x and solve for that. $\endgroup$
    – mikado
    Commented Jul 4, 2016 at 19:31
  • $\begingroup$ Oh I get it... Yes, it worked. Thanks! $\endgroup$
    – nate
    Commented Jul 4, 2016 at 20:14

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