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 Aug 7 asked Minimizing a quadratic form with orthogonality constraints May 15 awarded Curious May 15 comment Computing a limit similar to the exponential function @Lucian -- I was unaware of your question, but that is a very nice result! I encountered the limit in this question while trying to find an asymptotic formula for the variance of the number of distinct items present when bootstrapping a data set of size $n$. The answer ended up being $n (e^{-1} - 2 e^{-2})$. May 14 asked Computing a limit similar to the exponential function Aug 6 awarded Tumbleweed Jul 9 answered Dimensions of symmetric and skew-symmetric matrices Sep 4 accepted Monotonicity of $\ell_{p}$ norm and Holder's inequality Sep 4 comment Monotonicity of $\ell_{p}$ norm and Holder's inequality Nice. I would only add that the second inequality is equivalent to $\sum |z_{i}|^{k} \leq (\sum |z_{i}|)^{k}$, which is obvious because the difference between the right and left sides of the inequality is the sum of all the cross terms in $(\sum |z_{i}|)^{k}$, all of which are nonnegative. Sep 4 asked Monotonicity of $\ell_{p}$ norm and Holder's inequality Jul 12 accepted Logarithm inequality for vectors Jul 12 awarded Yearling Jul 10 revised Logarithm inequality for vectors corrected missing absolute value Jul 10 comment Logarithm inequality for vectors That works, but I don't think it's that easy to show the starting inequality. However, I was able to do it by differentiating $f(x) = \log(1+x^{2}) - x^{2} + \frac{x^{4}}{2 (1-x^{2})}$ to show that $f(x)$ is increasing on $(0,1)$. The desired inequality follows. ZachL's comment inspired my answer below, which I think is easier. Jul 10 answered Logarithm inequality for vectors Jul 10 answered Monty Hall/Bayes' Theorem Jul 10 awarded Critic Jul 10 comment Logarithm inequality for vectors I don't think that this works. I want my inequality to hold for all $d \in \mathbf{R}^{n}$ with $||d||_{\infty} < 1$, but the inequality $\log(1+x) \geq x - \frac{1}{2} x^{2}$ only holds for $x \geq 0$. For example, suppose $n=1$ and $d = -0.9$. Then, we have that $\sum_{i=1}^{n} \log(1+d_{i}) = \log(0.1) \approx -2.3 < \mathbf{1}^{T} d - \frac{1}{2} ||d||_{2}^{2} = -0.9 - 0.5 * (-0.9)^{2} \approx -1.3$. Jul 10 asked Logarithm inequality for vectors Apr 23 accepted Maximizing a convex function Apr 22 asked Maximizing a convex function