Renter - page 21
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It's a pretty good "solution." Only it's already obtained on page 2 of the thread.
This is not a solution, but just a function whose zero needs to be found. We have been trying to find this zero based on this function for a long time.
I thought you said, that conversion of diphuples to class of algebraic equations can help to solve this problem. But so far it only got to the point where you got the same function whose zero we are looking for. If you think you got the same result "more reasonably" than Neutron, justify why.
Take a close look at this function...
are you going to keep insisting it's the same thing?
.
I gave you the solution to the problem at hand. Check it out. If you find a discrepancy, point it out.
There might be an error here. OK, I have another formula, take a look at it. I entered it into MS XL, and the graphs turn out quite plausible.
f' = -q/(k-q)^2 * (1-(1+q-k)^t) + (1 + q/(k-q)) * t * (1+q-k)^(t-1) = 0
And second: you haven't given a solution, neither have I. So far we only have a function whose zero needs to be found. We arrived at this function in different ways, and hopefully it should coincide.
However, Sergey set the task to find the zero of this function in analytical form. This will be the solution.
another picture... Is it really the same?
Oleg, you got it wrong. You should differentiate not by t, but by k.
And the derivative is already there - see my post above. That's exactly this function, without any fantasies.
Or, if you don't believe me, differentiate by k this one:
Take X0 to be 1.
There might be a mistake here. OK, I have another formula, take a look at it.
f' = -q/(k-q)^2 * (1-(1+q-k)^t) + (1 + q/(k-q)) * t * (1+q-k)^(t-1) = 0
And second: you haven't given the solution, and neither have I. So far we only have a function whose zero needs to be found. We arrived at this function in different ways, and hopefully it should coincide.
However, Sergey set the task to find the zero of this function in analytical form. This will be the solution.
if it exists!
Only a small fraction of real world problems can be represented analytically.
But it is possible to linearize the problem, and thereby simplify the formula, and perhaps it will have an "analytical solution" in such a form -- but it will no longer be the solution of the original problem, but of the linearized one. And you won't be satisfied with that again.
I see, you've given up. Have you? I haven't given up yet.
Under the last conditions announced by Sergei(t=50 or more, q=0.1...0.3), the solution exists. I intend to obtain it by a single iteration of the tangent method. It will be approximate, but the accuracy, I hope, should suit the author of the branch.
Oleg, you have made a mess of things. You should differentiate not by t, but by k.
correct...
I see, you've given up. Have you? I haven't given up yet.
Under the last conditions voiced by Sergei(t=50 or more, q=0.1...0.3), a solution exists.
Oleg, you are differentiating some function of yours, and something is not converging there. This is the wrong function, because the right one should have a denominator(k-q). This is a key feature of the function, you can't get rid of it.
I have already offered you the correct function of accumulated withdrawals and its derivative.
Oleg, you got it wrong. It is not necessary to differentiate by t, but by k.
And the derivative is already there - see my post above. That's exactly this function, without any fantasies.
Or, if you don't believe me, differentiate by k this one:
Take X0 as equal to 1.
.
what else do we differentiate?