Is $|\ln(1+t) - t| \leq t^2$ for $|t| \leq \frac{1}{2}$ obvious? In a textbook of probability and statistics there is a use of approximation by Taylor series that $\ln(1 + t) \doteq t$. It is stated that that the accuracy of the approximation is due to $|\ln(1+t) - t| \leq t^2$ for $|t| \leq \frac{1}{2}$.
While I am aware that $x$ is the first term of the Taylor series of $\ln(1+t)$, I am not sure, however, where $|\ln(1+t) - t| \leq t^2$ for $|t| \leq \frac{1}{2}$ comes from and whether it should be obvious that it holds.
Edit: Changed $t$ to $t^2$
 A: With the edit, consider the function $f(t) = \ln(1+t)- t + t^2$ for $ t\in [-1/2,1/2]$. The minimum of this function must occur at a a boundary point or a point where $f'(t) = 0$. Note that $$f'(t) = \frac 1{1+t} - 1 + 2t = 0 \,\,\,\, \implies \,\,\,\, 2t^2 + t = 0  \,\,\,\, \implies \,\,\,\, t = 0,-1/2.$$ Checking the points $-1/2, 0, 1/2$ we find that $f$ has a minimum of $0$ at $t = 0$. Thus for all $t \in [-1/2,1/2]$, we have $$\ln(1+t) - t + t^2\ge 0 \,\,\, \implies \,\,\,\, t - \ln(1+t) \le t^2.$$ But $t - \ln(1+t) = \big \lvert \ln(1+t) - t \big \rvert$ so this proves the inequality.
A: Since $\log(1+t)<t$ for all $|t|<1/2$, we can consider the function
$$g(t)=t-\log(1+t)-t^2,$$
and observe that its derivative is
$$g'(t)=-\frac{t(1+2t)}{1+t}$$
and its second derivative is
$$g''(t)=-2+\frac{1}{(1+t)^2}$$
so the point $t=0$ is the only maximum in the interval $(-1/2,1/2)$, where the value is zero. The points $\pm 1/2$ can be checked separately, if desired. The conclusion is that $g(t)<0$ for all $|t|\leq 1/2$.
A: The second term of the Taylor expansion, which will dominate when you subtract away the $x$, is $-\frac12t^2$. When you're close enough to zero, the terms of higher degree will not be able to add up to another half of $t^2$. So it indeed "clear" that your inequality will hold in some neighborhood of $0$.
More computation is necessary to see that this neighborhood is at least $\frac12$ wide, though. (I suspect the error-bound versions of Taylor's theorem may be helpful there).
