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visits member for 2 years, 2 months
seen Aug 5 at 20:44

Aug
6
awarded  Tumbleweed
Jul
30
asked A different proof of the distributive law for gcd and lcm
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
Dec
11
awarded  Scholar
Dec
11
accepted Regularity Conditions for Constrained Optimization
Dec
10
comment Variance Formula
Looks correct to me.