I have a rather nasty-looking function Ratio, given by,

f[n_, t_] = 
  Assuming[Element[n, Integers] && n > 0 && 0 < t < Pi/2, 
   Integrate[Cos[x]^n, {x, Pi/2 - t, Pi/2}]];
cn[n_] = Pi^(n/2)/Gamma[n/2 + 1]; 
g[n_, t_] = cn[n - 1]*(Sin[(t)]^(n - 1)*Cos[(t)]/n + f[n, t]);
Ratio[n_, t_] = FullSimplify[g[n^2, t]/cn[n^2]]

Given values p_n that fall in the range from $0$ to $1/4$, I wish to solve Ratio[n,t]==p_n for values of t in the range from $0$ to $\pi/2$.

But when I run

Assuming[0 < t < Pi/2, NSolve[Ratio[2, t] == 0.250277, t, Reals]]

Mathematica doesn't seem to ever arrive at an answer. The Ratio-function looks normal enough and there should exist solutions in the given intervals:

enter image description here

What goes wrong? Thanks.


closed as off-topic by march, MarcoB, m_goldberg, corey979, bbgodfrey Jan 4 '17 at 1:55

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  • 3
    $\begingroup$ Try FindRoot instead. NSolve is mainly for polynomial equations and equations involving some small number of elementary functions. $\endgroup$ – march Jan 3 '17 at 21:10
  • $\begingroup$ @march Thanks, that worked beautifully! If you post this as an answer, I will accept it. Or should I delete this question (as it is probably a duplicate)? $\endgroup$ – Bobson Dugnutt Jan 3 '17 at 21:20

You need to tell NSolve about the bounds explicitly:

NSolve[Ratio[2, t] == 0.250277 && 0 < t < Pi/2, t, Reals]
(* {{t -> 1.15546}} *)
  • $\begingroup$ Can I, in general, simply add my list of assumptions after the equality, or is this only valid for the range of the variable which is being solved for? $\endgroup$ – Bobson Dugnutt Jan 3 '17 at 23:05
  • $\begingroup$ @Lovsovs: See under "Details" in the NSolve documentation for the kinds of things you can put there; there are a lot of possibilities. $\endgroup$ – Rahul Jan 3 '17 at 23:26

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