Math Problem Statement
Come si poassa dalla prima espressione alla seconda? Prima: [(1 + i1) ⋅ (1 + i2) ⋅ … ⋅ (1 + iN)] 1/N - 1= i*. Seconda: [( i1) ⋅ (i2) ⋅ (iN)] 1/N = i*
Solution
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Math Problem Analysis
Mathematical Concepts
Geometric Mean
Exponential Approximation
Taylor Series
Formulas
(1 + i_1)(1 + i_2)...(1 + i_N)^{1/N} - 1 = i^*
(i_1 ⋅ i_2 ⋅ ... ⋅ i_N)^{1/N} = i^*
1 + i_k ≈ e^{i_k} for small i_k
e^x ≈ 1 + x for small x
Theorems
Exponential Approximation for Small Values
Geometric Mean
Suitable Grade Level
University Level