Lieber Eckhard,
Thanks a lot for your . This is exactly what I was looking for! thanks a
bunch!!!
On 08/02/2010 05:24 PM, Eckhard Kosin wrote:
Hi Jose,
Am Montag, den 02.08.2010, 08:09 +0200 schrieb Jose Guzman:
I am trying to solve at which x-values the function takes a given y
value. In its most simplistic case, this is easy to get with Solve:
sage: var('t')
sage: y=1-e**(-t/5)
sage: sol = solve(y==0.632,t)
sage: n(sol[0].rhs())
Which gives me 4.9983, the value I was looking for is 5, so it is OK.
However, in more complicated expressions (a biexponential of the from:
sage: y=-17.6666666666666*(e^(-1/5*t) - 1)*e^(-3.33333333333333*t
+0.971148468732930)
which varies between 0 an 1 in the interval between x=0 and x=5, the
task becomes complicated:
Trying to get the value at which the functions is 20% of its maximum I get
sage: solve(y==.2,t)
sage: [e^(1/5*t) == 265*e^(-10/3*t + 3467/3570)/(265*e^(-10/3*t +
3467/3570) - 3)]
which is indeed another equation which depends on t?
sage didn't find a better algebraic solution. Try to find solutions
numerically.
sage: plot(y, (t,0,5))
shows that y==0.2 will have one solution in [0,0.5] and another solution
in [1,1.5]. You'll get them numerically with
sage: t1 = find_root(y==0.2, 0,0.5); t1
0.023210775112474515
sage: t2 = find_root(y==0.2, 1,1.5); t2
1.1642130321591619
Verify the result:
sage: y.subs(t=t1)
0.200000000000621
sage: y.subs(t=t2)
0.200000000000000
Regards,
Eckhard
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