For any real number x, show that x - 0.5*x^2 < ln(1+x)?
I suggest to form a new function `f(x) = x - x^2/2 - ln(1+x)` and to use derivative to check is the function is negative or positive.
Differentiating the function yields:
`f'(x) = 1 - 2x/2 - 1/(1+x) =gt f'(x) = (1 + x - x - x^2 - 1)/((1+x)^2)`
Reducing like terms in brackets yields:
`f'(x) = (- x^2)/((1+x)^2)`
`` Notice that the derivative is negative for x `in` R.
Since the derivative is negative =>`f(x) lt 0 =gt x - x^2/2 - ln(1+x) lt 0 =gt x - x^2/2lt ln(1+x).`
The Maclaurin series for ln(1 + x) where |x| < 1 is given by
ln(1 + x) = `x - x^2/2 + x^3/3 - x^4/4` ...
From the series for ln(1+x) it can be inferred that `x - x^2/2 < x - x^2/2 + x^3/3 - x^4/4`
=> `x - x^2/2 < ln(1 + x)`
It has to be kept in mind that the series is valid only for |x|< 1.
Using the Maclaurin expansion for ln(1 + x), the given inequality can be proved