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I'd like to solve the following problem: Given

MI[o_, n_] := Log2[1 + 4 o (Sqrt@n + Sqrt[1 + n])^2]

Find a function f[n_] that tells which o to use to optimise (maximise) MI.

My first pen and paper approach was to calculate the gradient of the function. Then, starting from the origin, the idea is to step "a little bit" into this direction to get to the next, best, point. However, for a closed solution, this means solving an iteration problem. So I went to Mathematica. Here, the approach looks like

Maximize[{MI[o, n], 0 <= Sqrt[o^2 + n^2] <= b, b > 0}, {o, n}]

This is out of Mathematica's (10.2) power. What works at least is

a = 3;
Simplify@Maximize[{MI[o, n], 0 <= Sqrt[o^2 + n^2] <= a}, {o, n}]

If someone has an idea of an approach that leads to a symbolic solution, I'd be happy!

PS: As @march, @rhermans and @Jack LaVigne pointed out, the question only makes sense when using a bound to o and n, which I did not mention (besides implicitly the last code).

I'd like to solve the following problem: Given

MI[o_, n_] := Log2[1 + 4 o (Sqrt@n + Sqrt[1 + n])^2]

Find a function f[n_] that tells which o to use to optimise (maximise) MI.

My first pen and paper approach was to calculate the gradient of the function. Then, starting from the origin, the idea is to step "a little bit" into this direction to get to the next, best, point. However, for a closed solution, this means solving an iteration problem. So I went to Mathematica. Here, the approach looks like

Maximize[{MI[o, n], 0 <= Sqrt[o^2 + n^2] <= b, b > 0}, {o, n}]

This is out of Mathematica's (10.2) power. What works at least is

a = 3;
Simplify@Maximize[{MI[o, n], 0 <= Sqrt[o^2 + n^2] <= a}, {o, n}]

If someone has an idea of an approach that leads to a symbolic solution, I'd be happy!

PS: As @march, @rhermans and @Jack LaVigne pointed out, the question only makes sense when using a bound to o and n, which I did not mention (besides implicitly in the last code).

I'd like to solve the following problem: Given

MI[o_, n_] := Log2[1 + 4 o (Sqrt@n + Sqrt[1 + n])^2]

Find a function f[n_] that tells which o to use to optimise (maximise) MI.

My first pen and paper approach was to calculate the gradient of the function. Then, starting from the origin, the idea is to step "a little bit" into this direction to get to the next, best, point. However, for a closed solution, this means solving an iteration problem. So I went to Mathematica. Here, the approach looks like

Maximize[{MI[o, n], 0 <= Sqrt[o^2 + n^2] <= b, b > 0}, {o, n}]

This is out of Mathematica's (10.2) power. What works at least is

a = 3;
Simplify@Maximize[{MI[o, n], 0 <= Sqrt[o^2 + n^2] <= a}, {o, n}]

If someone has an idea of an approach that leads to a symbolic solution, I'd be happy!

I'd like to solve the following problem: Given

MI[o_, n_] := Log2[1 + 4 o (Sqrt@n + Sqrt[1 + n])^2]

Find a function f[n_] that tells which o to use to optimise (maximise) MI.

My first pen and paper approach was to calculate the gradient of the function. Then, starting from the origin, the idea is to step "a little bit" into this direction to get to the next, best, point. However, for a closed solution, this means solving an iteration problem. So I went to Mathematica. Here, the approach looks like

Maximize[{MI[o, n], 0 <= Sqrt[o^2 + n^2] <= b, b > 0}, {o, n}]

This is out of Mathematica's (10.2) power. What works at least is

a = 3;
Simplify@Maximize[{MI[o, n], 0 <= Sqrt[o^2 + n^2] <= a}, {o, n}]

If someone has an idea of an approach that leads to a symbolic solution, I'd be happy!

PS: As @march, @rhermans and @Jack LaVigne pointed out, the question only makes sense when using a bound to o and n, which I did not mention (besides implicitly the last code).