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I defined a function with n terms where n is a variable:

LogLH[μ_, σ_] = n Log[a[σ]] - Sum[Log[y[[i]]] +  
(Log[y[[i]]] - μ)^2 / (2 σ^2), {i, 1, n}]

In this function, the y[[i]] are (assumed to represent) observations of a quantity - picked from a list of n elements.

This statement multiply produces the error "The expression i cannot be used as a part specification". It seems however, that M. correctly evaluates the function. M. also computes the partial derivatives for μ and σ. But the above errors repeat and lateron, I can not apply the Solve command to the resulting system of equations. Hence my question: How can I define an open ended list of n elements in M.?

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    $\begingroup$ Use SetDelayed (:=) instead of Set (=). Afterwards, once you provide a proper list y and a value for n, the sum will evaluate. E.g., n = 10; y = Range@10; LogLH[m, s]. The rest of your question is uncear: what is "M."? What inequalities? What system of equations? What about the derivatives of $\mu$ and $\sigma$? $\endgroup$
    – corey979
    Commented Nov 14, 2016 at 11:11

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First, you need to use SetDelayed (:=) in your function definition. Next, if you use a list in the step definition then the Sum will step through its elements (rather than going over an integer range).

LogLH[μ_, σ_] := n Log[a[σ]] - Sum[
    Log[y] +  (Log[y] - μ)^2 / (2 σ^2), 
    {i, y}
 ]

So now when you use LogLH the y will stay as a symbol like this:

LogLH[1, 2]
(* -y (1/8 (-1 + Log[y])^2 + Log[y]) + n Log[a[2]] *)

until you set y to a list, then i will step through the values of y:

y = {1, 2, 3};
LogLH[1, 2]
(* {-(3/8) + n Log[a[2]], -(3/8) (-1 + Log[2])^2 - 3 Log[2] + 
      n Log[a[2]], -(3/8) (-1 + Log[3])^2 - 3 Log[3] + n Log[a[2]]} *)

If Solve does not work for your system of equations, you could also try NSolve or FindRoot

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  • $\begingroup$ Your approach eliminates the sum - which is however essential for later analytical treadment of the problem. Think e. g. of the basic problem of calculating a mean. You have n observations with values y[[i]], but for the formula you do not have these y[[i]] nor the n. $\endgroup$
    – Robert Brusa
    Commented Nov 14, 2016 at 15:56
  • $\begingroup$ Now consider s2 to be a function of yq and you want to find the derivative of it. $\endgroup$
    – Robert Brusa
    Commented Nov 14, 2016 at 16:04
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First, use SetDelayed (:=) instead of Set (=). Next, I'd propose to incorporate the list y as an argument of the function, as "y[[i]] are (...) observations of a quantity - picked from a list of n elements", hence n = Length[y]:

LogLH[μ_, σ_, y_List] := 
 Length[y] Log[a[σ]] - Sum[Log[y[[i]]] + (Log[y[[i]]] - μ)^2/(2 σ^2), 
    {i, 1, Length[y]}]

E.g.,

LogLH[2, 0.5, Range[5]]

-18.8866 + 5 Log[a[0.5]]

The OP didn't specify neither a[σ], nor "the resulting system of equations". About the derivatives:

D[LogLH[μ, σ, Range[5]], μ]

enter image description here

D[LogLH[μ, σ, Range[5]], σ]

enter image description here

Mathematica won't handle a Sum with an unspecified n.

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  • $\begingroup$ You wrote: Mathematica won't handle a Sum with an unspecified n. Yes, that's exactly what bothers me. Why not? A function which is the sum of n terms can be differenciated - even if one does not know n. $\endgroup$
    – Robert Brusa
    Commented Nov 16, 2016 at 9:50
  • $\begingroup$ See this discussion. $\endgroup$
    – corey979
    Commented Nov 16, 2016 at 9:59

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