R: one-dimensional optimization -

I optimization () , or something similar, at least the maximum value / function , I'm unsure about the exact range on which the function should be optimized, which is the required parameter for the function 'opplease' (such as optimize (f = fun, interval = c (lowerbound, upperbound)) ).

In the problem of this optimization, I am able to estimate a value a , which is "close" of the optimal solution, but depends on the "proximity" situation.

Is there a function in R which can use the initial value a , which is not required to specify the function that the function is optimized?

When you say that you are not sure about the lower limit, I suspect that its This means that the parameters that you are trying to guess are not bound down

If this is the case, then a move is to change the function so that the lower bound on the parameter

This trivial function is minimal on x = 4:

  have fun  

Through which we can find:

  optimization ( F = funny, interval = c (0,8)) # & gt; $ Min # & gt; [1] 4  

But for a moment, do not argue that we are not sure whether there is a lower limit or not, and we know that the upper limit is 8. If we try, then an error:

  optimization (f = funny, interval = c (-Inf, 8))  

Because the boundaries are finite In this situation, we can use the exponential changes ( exp () ), which, like the map with the positive numbers, the positive numbers like:

  optimization (F = function (x) fun (log (x)), interval = exp (c (-inf, 8))) # & gt; $ Min # & gt; [1] 54.59815  

And to get root, you just need to change the solution via the following:

  log (54.59815) ) # & Gt; 4  

If you do not know the upper or lower bound on the underlying parameters, then you can use the log-axis conversion instead of log (): :

  function (x) log (x / (1-x))  

and it is inverted in place of exp () Code>:

  function (y) exp (y) / (1 + exp (y))  

Note that log-os conversion unit interval Modes the actual numbers, so the interval parameter becomes 0: 1 .

There are some numerical limitations in these solutions (for example, if we had defined interval = exp (c (-inf, 16)) in the first code, then we had an error Must have met). Tip, you can scale these changes again in the center of a given point, which can reduce the numerical limitations.