ANSWERS: 1
  • That's an evil one. S=(1+x)^(1/x) can be found using the general binomial expansion: (1+x)^n = 1 + nx + n(n-1)x^2/2 + n(n-1)(n-2)x^3/3! + ... putting in n=(1/x) gets: S=1 + 1 + (1-x)/2 + (1-x)(1-2x)/3! + (1-x)(1-2x)(1-3x)/4! +... Which can be seen to tend to the series for e when x tends to zero. But what is the error? Looking at the coefficient of x gets: C = 0 + 0 - 1/2 - 3/3! - 6/4! - 10/5! - 15/6! + ... double that: 2C = 0 + 0 - 1*2/2 - 2*3/3! - 3*4/4! - 4*5/5! - 5*6/6! + ... The first two terms of the factorial cancel: 2C = -1 - 1 - 1/2! - 1/3! - 1/4! + ... which is the series for -e So S = e - xe/2 + O(x^2) S/e is 1 - x/2 + O(x^2) So now the formula is to a first approximation: (1 - x/2)^(1/x) Using the binomial expansion AGAIN... (1-x/2)^n = 1 - nx/2 + n(n-1)x^2/2/2^2 - n(n-1)(n-2)x^3/3!/2^3 + ... put n = (1/x) (1+x/2)^(1/x) = 1 - 1/2 + (1-x)/2/2^2 - (1-x)(2-x)/3!/2^3 + ... when x is zero this is 1 - 1/2 + 1/2/2^2 - 1/3!/2^3 + 1/4!/2^4 OR 2^0/1 - 2^-1/2 + 2^-2/2! - 2^-3/3! + 2^-4/4! +... which I think is e^(-1/2) = 1/sqrt(e) Check: evaluate the original equation with x=0.0001 answer 0.60655 Close enough. Phew. There's an easier way if you can use differentiation. Just differentiate the equation on the right and use a single taylor expansion. f'(x) = f(0) + xf'(0) + O(x^2) But differentiation is defined in terms of limits, so that's kind of circular reasoning so I did it all from first principles.

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