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

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