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To prove: $$x-\frac{x^2}{2}<\ln(1+x)<x-\frac{x^2}{2(1+x)},\quad\forall x>0$$ I have used Taylor series expansion at 0 for both the inequalites. The greater than by expanding $\ln(1+x)$ and the less than by expanding $\int \ln(1+x)\,dx$ at 0. Is there a cleaner / more elegant way of achieving the same? |
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Another way to do the lower bound, for example, would be to consider $f(x) = x - \dfrac{x^2}{2} - \ln(1+x)$. $f(0) = 0$. But also $f'(x) = 1 - x - \dfrac{1}{1+x} = \dfrac{(1-x)(1+x) - 1}{1+x} = \dfrac{1 - x^2 - 1}{1+x} = \dfrac{-x^2}{1+x} < 0$. So since $f(0) = 0$ and $f' < 0$, $f$ is monotonically and strictly decreasing. Thus $f(x) \leq 0$, and if $x > 0$ we have that $f(x) < 0$. And this says exactly that $x - x^2/2 < \ln(1 + x)$. |
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The upper bound also yields to standard calculus methods. Equivalently, we want to prove that $(1+x)\ln(1+x) \gt x(1+x)-\frac{x^2}{2}$, or more simply that $x+\frac{x^2}{2} \gt (1+x)\ln(1+x)$. Let $f(x)=x+\frac{x^2}{2}-(1+x)\ln(1+x)$. We have $f(0)=0$. We will show that $f$ is increasing, and hence posiive, for positive $x$. We have $$f'(x)=1+x-(1+\ln(1+x))=x-\ln(1+x).$$ To show that $f'(x)$ is positive for $x \gt 0$, note that $f'(0)=0$. But $f'(x)$ is increasing, since $f''(x)=1-\frac{1}{1+x} \gt 0$ for $x\gt 0$. |
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