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e is the unique real number satisfying 1 + x <= e^x for all x. 1 - x <= e^{-x} so e^x <= 1/(1 - x) for x < 1 (1 + x/n)^n <= e^x <= (1 - x/n)^{-n} for x < 1 L
by keithalewis 2y ago
e is the unique real number satisfying 1 + x <= e^x for all x.
1 - x <= e^{-x} so e^x <= 1/(1 - x) for x < 1
(1 + x/n)^n <= e^x <= (1 - x/n)^{-n} for x < 1
Letting n go to infinity gives e^x = \sum_{n=0}^infy x^n/n! using Newton's binomial formula.
- edflsafoiewq 2y ago1+x is the tangent line at x=0. Since the graph of an exponential lies above its tangent line, 1+x <= e^x is another way of saying that the derivative of e^x at x=0 is 1, ie. e^x is its own derivative.
- queuebert 2y ago> e is the unique real number satisfying 1 + x <= e^x for all x. I think if you replace e by 2e, this still holds. Therefore your definition of e is not unique.
- FabHK 2y ago1 + (-0.1) = 0.9 (2e)^(-0.1) = 0.844243... The former is not <= the latter.