On 15/07/11 19:55, achrzesz wrote:
sage: assume(t,'real')
sage: solve(diff(myHH,t),t)
[t == 1/10*log(31)]
Thanks a lot!. This work for the special case of myHH, where:
myHH(t) =(1-e^(-t/0.1))^3*(e^(-t/10.)
However, for a more general function:
sage: t=var('t')
sage: taum, tauh, m = var('taum, tauh, m')
sage: sage: f(t) = (1-e^(-t/taum))^m*(e^(-t/tauh))
sage: solve(diff(f,t)==0,t)
Gives
[(e^(t/taum) - 1)^m*e^(-m*t/taum) == m*tauh*(e^(t/taum) - 1)^(m -
1)*e^(-(m - 1)*t/taum - t/taum)/taum]
No idea how to handle that.
According to Mathematica, the solution should be
>>> t->taum*log(1+(m*tauh/taum))
My idea would be substitute my current taum, tauh and m in this
expresion to study different parameters in the peak of that function.
m is an integer number, but t, taum and tauh are real numbers.
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