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CTK Exchange
gin&tonic
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Oct-09-05, 06:25 AM (EST) |
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"e as the min or max of z*log(x,z)"
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Using the notation log(x,z) as x^log(x,z) = z,I think that the minimum value of z*log(x,z) is at z = e when x > 1, and the maximum value is also at z = e, when 0 < x < 1. If this is true, could somebody prove it please? |
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gin&tonic
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Oct-09-05, 08:45 AM (EST) |
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2. "RE: e as the min or max of z*log(x,z)"
In response to message #1
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Sorry I meant log(x,z) to mean z^log(x,z)=x, ie z is the base!Maybe I could use N instead of x to make sure that z, the base, is understood to be the variable. Is it known what the derivative or the integral of f(z) = log(N,z) is? |
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mr_homm
Member since May-22-05
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Oct-09-05, 10:21 AM (EST) |
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4. "RE: e as the min or max of z*log(x,z)"
In response to message #2
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Perhaps it might be useful to recall that log(z,N) = ln(N)/ln(z). In that case, the expression you are trying to maximize or minimize becomes z*ln(N)/ln(z), and since you are treating z as the variable, the factor of ln(N) is just a constant that scales up the function z*/ln(z). Your two cases correspond to whether ln(N) is positive or negative. I have checked the derivative, which does have a zero at z = e. This makes this value a CANDIDATE for max or min, but of course you must still check the sign of the second derivative to see whether you really have a local minimum, check the derivative for other zero-crossings at points of discontinuity, and check values of the function at endpoints of the interval where it is defined, or asymptotes if the interval is infinite. In your case, the domain of the function is 0 < z < 1 together with 1 < z, because there is a vertical asymptote at z=1. For z < 1, this asymptote goes to negative infinite, and for z>1 it goes to positive infinity. So there is really NO global maximum or minimum for your function. If you restrict to just z>1, then your function goes to positive infinity at both ends of its domain, has no discontinuities in the function or the derivative, and has a positive second derivative at z=e. Therefore z=e is the minimum of z/ln(z) on the region z>1. Your choice of N then either inverts the graph of this function vertically (converting a min to a max) when N<1, or does not invert it, when N>1. Of course the choice of N also rescales the graph of your function on the vertical axis, but that only affects the value of the minimum, not its location. Hope this helps. --Stuart Anderson |
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gin&tonic
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Oct-09-05, 07:00 PM (EST) |
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5. "RE: e as the min or max of z*log(x,z)"
In response to message #4
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Yes, thank you. Usually these things become quite obvious to me WHEN I receive an explanation like yours. I do appreciate the fact that you spent the time and effort on such a detailed note. |
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