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Simplification

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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

Attachments:
POSTED BY: 喆昊 张
Answer
16 days ago

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]
POSTED BY: Sander Huisman
Answer
15 days ago

To see dependency on s look at Series expansion in s

Series[f[δ, t, s, d, λ], {s, 0, 1}] // Simplify
POSTED BY: Robert Hanlon
Answer
15 days ago

I do not understand why a taylor series might be helpful here. Could yo please explain more? Sorry for my stupidness.

POSTED BY: 喆昊 张
Answer
14 days ago

The series expansion includes terms with non-zero powers of s, i.e., there is a dependency on s.

POSTED BY: Robert Hanlon
Answer
13 days ago

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?

POSTED BY: 喆昊 张
Answer
13 days ago

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.

POSTED BY: Robert Hanlon
Answer
12 days ago

You are absolutely correct. Thank you.

POSTED BY: 喆昊 张
Answer
12 days ago

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.

POSTED BY: 喆昊 张
Answer
15 days ago

The derivative of $\frac{\partial}{\partial s} f[\delta,t,s,d,\lambda]$ 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:

enter image description here

Attachments:
POSTED BY: Valeriu Ungureanu
Answer
14 days ago

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.

POSTED BY: 喆昊 张
Answer
14 days ago

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}]

enter image description here

Hope, this would help you.

POSTED BY: Valeriu Ungureanu
Answer
13 days ago

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?

POSTED BY: 喆昊 张
Answer
13 days ago

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.

POSTED BY: Valeriu Ungureanu
Answer
12 days ago

Thanks for your explanation. Good idea.

POSTED BY: 喆昊 张
Answer
12 days ago

The derivative of f with respect to s is not zero.

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])  *)
POSTED BY: Robert Hanlon
Answer
14 days ago

Please read the rules: http://wolfr.am/READ-1ST

Please EDIT your posts and make your titles more meaningful, less generic.

POSTED BY: Moderation Team
Answer
6 days ago

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