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Problem solving for a variable in Prime
12-28-2014, 10:48 PM (This post was last modified: 12-29-2014 12:39 AM by Snorre.)
Post: #8
RE: Problem solving for a variable in Prime
Hello compsystems,

Your expression
Quote:solve( ∫( (1/(√(2*π)*s))*e^(-(t-u)^2/(2*s^2)),t, t,∞), t=0.1 )
looks odd because you're using the same name "t" for two different variables. You should rename one of them (either the red or blue), e.g. to τ (tau).

(1) t within the integrand refers to the integration variable t.
(2) t to be solved refers to the integration lower limit t.
(3) t and t are two distinct variables (each having their own context/scope/lifetime).

So you are trying to find a lower integration limit t (somewhere near 0.1) so that the integral becomes zero (the to-solve-equation is implicitly set to =0), which would only hold if both limits are equal (since your integrand has everywhere the same sign).

But your expression differs from that of factor. Look at his screenshots:
Quote:solve( ∫( (1/(√(2*π)*s))*e^(-(t-u)^2/(2*s^2))=0.1, t, t, ∞), s )
Here we have two distinct t again. Now we're searching for s so that the integral becomes 0.1. Since the lower limit t is a free variable the result depends on whatever t is set to.

Hello factor,

I do not know the original problem, but I think the lower limit should not be t but something like either 0 or -∞. (On your calc you seem to have set t to 40, which becomes the lower integration limit). Maybe you're not searching for an s but for t as compsystems suggested (if so, don't name your integration variable also t since it's not forbidden but a bit confusing).
Furthermore I doubt the Prime is able to solve that integral. You should paraphrase it in terms of "erf"/"erfc"/"NORMALD_CDF"/"NORMALD_ICDF" functions and set your lower integration limit and u to concrete values before solving numerically.
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RE: Problem solving for a variable in Prime - Snorre - 12-28-2014 10:48 PM



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