Maths Question!!! Thx~

If x is sufficiently small, exp x ~ 1 + x by Maclaurin series

So, 1/(1 + exp(x))

~ 1/(1 + 1 + x)

= 1/(x + 2)

Now, as powers of x above the second to be neglected

(x + 2)(2 - x) = 4 - x^2 ~ 4

Then, (2 - x)/4 = 1/(x + 2)

In fact, this sum is fairly simple: { }內的表示刪去!!!

1/[1+e^(x)] = (2-x)/4

LHS=1/[1+e^(x)]

=[1+e^(x)]x16e/1x16e

=[16e+16e^(x)^2]/16e

={2}(2-x){(2e)}/{16}4{e}

=(2-4)/4

RHS=(2-x)/4

Since (2-x)/4=(2-x)/4
Therefore 1/(1+e^(x)) = (2-x)/4 is an identity.

2012-12-01 21:57:06 補充：
揀我!!!