[VA] SRC#005- April, 1st Mean Minichallenge

[Albert Chan]

Hi, Maximilian:

I too were afraid of posting a "stupid" solution (never heard of Generalized Mean)

M_k^k = (a₁^k + ... + a_n^k) / n

The case for k=0 doesn't fit quite right, getting meaningless result, M₀ = 1^∞
I had to convince myself this work for Geometric Mean.

For k=ε (small value, but not exactly 0):

LHS = M_ε^ε = e^(ε ln(M_ε)) ≈ 1 + ε ln(M_ε)

RHS = (a₁^ε + ... + a_n^ε) / n ≈ ((1 + ε ln(a₁)) + ... + (1 + ε ln(a_n))) / n ≈ 1 + ε ln(a₁ ... a_n) / n

⇒ M_ε ≈ ⁿ√(a₁ ... a_n)
⇒ M₀ = ⁿ√(a₁ ... a_n)