bo198214 Wrote:Then I dont understand how you compute \( \log_a^{\circ n} AB=\log_a^{\circ n}(A)+\epsilon_1 \).
This is one of the logarithmic laws:
\( \log(AB) = \log(A)+\log(B) \)
Then, we know that \( \log(1)=0 \), and that \( \log(1+\epsilon) \approx \log_b(e)\ \times\ \epsilon \). As epsilon goes to 0, this is in fact exact.
Therefore, if B is (1+epsilon), then we can state:
\( \log(AB) = \log(A) + \log(B) = \log(A) + \epsilon \)
Quote:It is always unclear what you mean by "exact" solution. Isnt everything exact in mathematics except if we say its approximate? A limit for example is exact if it exists regardless what difficulties we have with numeric computation.
I mean exact only when the limits I provide are taken to infinity. Otherwise, they are definitely approximate, just like I say the following value is "approximately" equal to e for a finite n:
\( e = \left(1+\frac{1}{n}\right)^n \)
Do you see the distinction? For numerical computation purposes, the solution will always be approximate. However, I still claim exactness when the specified limits are taken to infinity.
