e^x → 1 + x (as x → 0).
Because log e = 1, this could be written as.
e^x → 1 + x(log e) as x → 0.
Interestingly if we now generalise for n^x,
n^x → 1 + x(log n) as x → 0.
This therefore provides a ready means of approximating the value of log n (for any value of n).
So log n → (n^x - 1)/x as x → 0.
So for example if we set x = .000001,
log 2 → (2^.000001 - 1)/.000001 = (1.000000693147... - 1)/.000001 = .693147...
So this approximation is already correct to 6 decimal places with respect to its true value and the relative accuracy of the approximation can be continually increased through taking a smaller value of x.
Monday, June 18, 2012
Interesting Relationship
As n becomes larger Cos (2π/n) → 1.
So for example when n = 1000, Cos (2π/n) = .99998026...
Also when n becomes larger Sin (2π/n) → 2π/n.
Thus again when n = 1000, Sin (2π/n) = .0062831439... = (6.2831439.../1000)
(and 2π = 6.2831853...),
and with respect to Euler's Identity,
e^(iπ) = - 1;
therefore
e^(2iπ) = 1 (i.e. 1^1)
Also,
e^(2iπ) = Cos (2π) + i Sin (2π).
Therefore e^{(2iπ)/n} = Cos {(2π)/n} + i Sin {(2π)/n}.
So as n → ∞,
e^{(2iπ)/n} = 1^(1/n) → 1 + (2iπ)/n
So for example when n = 1000, Cos (2π/n) = .99998026...
Also when n becomes larger Sin (2π/n) → 2π/n.
Thus again when n = 1000, Sin (2π/n) = .0062831439... = (6.2831439.../1000)
(and 2π = 6.2831853...),
and with respect to Euler's Identity,
e^(iπ) = - 1;
therefore
e^(2iπ) = 1 (i.e. 1^1)
Also,
e^(2iπ) = Cos (2π) + i Sin (2π).
Therefore e^{(2iπ)/n} = Cos {(2π)/n} + i Sin {(2π)/n}.
So as n → ∞,
e^{(2iπ)/n} = 1^(1/n) → 1 + (2iπ)/n
Subscribe to:
Posts (Atom)