Hi guys:
I have an function whose numerical test shows it is independent with parameter "s". So I tried Simplify and FullSimplify to see whether parameter "s" could be got rid of. However, these two commands do not work. (Take quite long time) Wondering is there any command I can use to simplify the expression until "s" is gone. Also, I need to mention "s" is greater or equal to 0.
You may find nb file in the attachment.
Many thanks in advance
Please read the rules: http://wolfr.am/READ-1ST
Please EDIT your posts and make your titles more meaningful, less generic.
The derivative of f with respect to s is not zero.
f
s
D[f[delta, t, s, d, lambda], s] // Simplify (* (((-1 + d) E^(-(delta + d lambda) (s + t)) - d E^(-(delta + lambda + d lambda) (s + t))) (delta + lambda - d E^(-(delta + lambda + d lambda) t) (delta + d lambda) + (-1 + d) E^(-(delta + d lambda) t) (delta + lambda + d lambda)))/((delta + d lambda) (delta + lambda + d lambda)) + (d delta E^(-(delta + lambda + d lambda) t) (delta + lambda) (delta - d lambda) + d delta E^(-(delta + lambda + d lambda) (s + 2 t)) (E^((delta + lambda + d lambda) t) + E^((delta + lambda + d lambda) (s + t))) (delta + lambda) (-delta + d lambda) - (-1 + d) E^(-(delta + d lambda) t) (-delta + (-1 + d) lambda) (lambda - \[Delta]) \[Delta] + (-1 + d) E^(-(delta + d lambda) (s + 2 t)) (E^((delta + d lambda) t) + E^((delta + d lambda) (s + t))) (-delta + (-1 + d) lambda) (lambda - \[Delta]) \[Delta] + E^(-(2 delta + lambda + d lambda) t) (d delta (delta + lambda) ((-1 + d) delta E^(lambda t) + d (-delta + lambda)) - d delta E^(-(delta + lambda + d lambda) s) (1 + E^((delta + lambda + d lambda) s)) (delta + lambda) ((-1 + d) delta E^(lambda t) + d (-delta + lambda)) + (1 - d) (d delta - (-1 + d) E^( lambda t) (delta + lambda)) (lambda - \[Delta]) \[Delta] - (1 - d) E^(-(delta + d lambda) (s + t)) (E^((delta + d lambda) t) + E^((delta + d lambda) (s + t))) (d delta - (-1 + d) E^( lambda t) (delta + lambda)) (lambda - \[Delta]) \[Delta]))/(delta (delta + lambda) (lambda - \[Delta]) \[Delta]) *)
Hey guys:
I figure our one way to deal with it now.
FIrst, differentiate w.r.t s in f, you will see result is 0, which implies that there is no dependency between s and f.
Then, letting s=0 in f gives the final result.
The derivative $\frac{\partial}{\partial s} f[\delta,t,s,d,\lambda]$ is not $0$. You may verify this by using the attached file. More the more, the dependence of f with respect to s is not linear. See the following graph:
Sorry, I forgot to mention "s" is greater or equal to 0. It turns out "s" does not have affect from graph when "s" comes into its domain. However, it is true the derivative with respect to "s" is not zero. I am still working on it now.
Appreciate any ideas.
Let us define the partial derivative function with respect to $s$ and let us plot its graph:
derfs[\[Delta]_, t_, s_, d_, \[Lambda]_] := Evaluate@D[f[\[Delta], t, s, d, \[Lambda]], s] Plot[derfs[1, 2, s, 1.5, 2], {s, 0, 4}]
Hope, this would help you.
Thanks.
My only concern with this plot is even when partial derivative is not zero, the figure is pretty small. Would it be a calculation accuracy problem? How can I check on that?
We can compute the Limit[] when s tends to infinity:
In[1]:= Limit[f[1/5, 7, s, 11/100, 1/10], s -> \[Infinity]] // N Out[1]= 0.548733
Together with the fact that the derivative is non-negative, i.e. the function f is non-decreasing with respect to s, we can conclude that the function f is dependent with respect to s.
Thanks for your explanation. Good idea.
To see dependency on s look at Series expansion in s
Series
Series[f[?, t, s, d, ?], {s, 0, 1}] // Simplify
I do not understand why a taylor series might be helpful here. Could yo please explain more? Sorry for my stupidness.
The series expansion includes terms with non-zero powers of s, i.e., there is a dependency on s.
Thank you. I am still thinking the maths behind that. If my understanding is correct, your way is only checking s on a particular point as you expand function f at s=0. Is that correct? On the other hand, we can only check first finite terms from the result, there is no way to guarantee that s would not appear in the later terms. Is that right? If that is the case, how can we fix that?
If your conjecture that the function is independent of s were true, then the series expansion about any value of s would not include s. Only one counterexample is needed to disprove the conjecture. You can stop the search with the first counterexample. You can use as many terms of the series expansion as you want but additional terms just pile on additional evidence that the conjecture is false.
You are absolutely correct. Thank you.
It can't, there is an s dependency:
LogPlot[f[1, 2, s, 1.5, 2], {s, -5, 5}]
With more digits you can see the dependency:
f[1, 2, 1.0`30, 3/2, 2] f[1, 2, 2.0`30, 3/2, 2]
