Friday, January 1, 2016

calculus - Bounding ln(1+1/x) for x>0




I am hoping to somewhat rigorously establish the bound
1x+1/2<ln(1+1/x)<1/x
for any x>0. The upper bound is clear since
$$
1+1/x$$
The lower bound seems dicier. A taylor expansion argument wouldn't seem to work near zero, since 1/x blows up.




Geometrically at least it seems clear near zero. Since ln(1+1/x) is decreasing and convex, and lim and at 0
\frac{1}{1/2}=2
I can establish the lower bound near the origin. Should I then take over with taylor expansion for larger x? From the graph, it seems like a pretty fine lower bound.



I still feel like I should be able to make a neater argument using the convexity and decreasing properties of \ln(1+1/x) and any pointers would be appreciated!



edit: Not a duplicate, the term in the denominator on the lhs has a 1/2.



Answer



Hint: Define
f_1\colon \mathbf R_{>0} \to \mathbf R, \qquad f_1(t):=\ln \left(1+ \frac{1}{t} \right) - \frac{1}{t}
and
f_2\colon \mathbf R_{>0} \to \mathbf R, \qquad f_2(t):=\ln \left(1+ \frac{1}{t} \right) - \frac{1}{t+\frac{1}{2}}
and use the intermediate value theorem twice.


No comments:

Post a Comment

analysis - Injection, making bijection

I have injection f \colon A \rightarrow B and I want to get bijection. Can I just resting codomain to f(A)? I know that every function i...