TenaliRaman
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 Apr7 comment Prove that if $0\leq a,b$ and $a+b=1$ then $x^ay^b\leq ax+by$ for $x, y >0$ This is known as Weighted AM-GM Inequality. Check out proofs here and here. Mar24 comment Why is the conditional probability treated as a definition in Kolmogorov's probability theory? Interesting. So, you are basically asking why should we not define conditional probability as the most basic form of probability, so that the current definition of conditional probability comes out as a some lemma or theorem. This will probably require some notion of conditional measures. Most likely what might happen is we will end up pushing the current conditional probability definition and generalize it to general measures (if that is possible). So, we won't be able to get rid of the intuitive definition. Also, I can't see any particular advantage of a general conditional measure. Mar20 comment $T$ is a linear operator You seem to have the right idea, but your notation is messed up. From what is given to you, $(Tx)_i = \frac{x_i}{i}$, use it to write your earlier post clearly. Mar20 comment $T$ is a linear operator Look at $(Tx)_i$, $(Ty)_i$ and $(T(x+y))_i$, what can you conclude? Look at $(T(kx))_i$ and $k(Tx)_i$, what can you conclude? What does the above tell you about T? For the norm part, what is norm $\|Tx\|$ (remember $Tx \in l^2$)? Does it remind you of some well known inequality? Mar18 comment How to use CVX to solve this problem? CVX can accept the form as it is, although I am not quite convinced that your objective function is convex. Mar18 comment Prove Trigonometric Identitiy @ArpanBanerjee You are correct. Mar18 comment Prove Trigonometric Identitiy Indeed, you are correct. Mar18 comment Prove Trigonometric Identitiy It is not a good idea to multiply and divide by a term that can be zero for certain values. In your case, your term is zero for a = 0. Mar13 comment calculating $\mathbb E\left(\exp\left(\frac{1}{2}\sum_{i=1}^n X_i^2\right)\right)$ If $X_i$'s are independent, $$\mathbb E\left(\exp\left(\frac{1}{2}\sum_{i=1}^n X_i^2\right)\right) = \prod_{i = 1}^n \mathbb E\left(\exp\left(\frac{X_i^2}{2}\right)\right)$$ Mar11 comment Positive numbers inequality The last step has a bug. Mar11 comment Positive numbers inequality @pharmine, $y_1 \leq y_2 \leq \cdots \leq y_n$ need not necessarily be used (it is a mere convenience). Finally, note that $$\left(\frac{2}{n(n+1)}\sum_{i = 1}^n y_i\right)^{\log_2 3}$$ Since sum y_i = 1, $$\geq \left(\frac{2}{n(n+1)}\right)^{\log_2 3}$$ Since (n+1)/2 <= n $$\geq \left(\frac{1}{n^2}\right)^{\log_2 3}$$ Since log_2 3 >= 1 $$\geq \frac{1}{n^2}$$ Mar11 comment Positive numbers inequality What the hell? Please explain the downvote, whoever did it, you owe me at least that much! X( Mar11 comment how to show $\frac{1+\cos x+\sin x}{1+\cos x-\sin x} = \frac{1+\sin x}{\cos x}$ t = -2 is completely justified, essentially you go back in time by 2 units and then throw the ball. It is just time travel 101 :P ;-) Mar11 comment how to show $\frac{1+\cos x+\sin x}{1+\cos x-\sin x} = \frac{1+\sin x}{\cos x}$ If only indeed! I tried to put something on top, but nothing I tried made much sense :D Mar11 comment how to show $\frac{1+\cos x+\sin x}{1+\cos x-\sin x} = \frac{1+\sin x}{\cos x}$ I left it as such because of the diamond actually :-) Mar11 comment Prove $(1 +\frac{ 1}{n}) ^ {n} \ge 2$ Nope. This is a direct proof using binomial expansion. Mar6 comment Show that if $n$ is composite, then $\phi(n) \leq n-\sqrt{n}$ math.stackexchange.com/questions/896920/… Jan21 comment show that $4 = \sum_{n=1}^\infty (-2)^{n+1}\frac{n+2}{n!}$ @learnmore 2*(-1)^n is not (-2)^n Jan20 comment Does $\int_0^\infty \sin(x^{2/3}) dx$ converges? $x = y^{3/2}$ then $dx = \frac{3}{2}y^{1/2}dy$ Jan14 comment Showing $d(x,y) = \frac{|x-y|}{1+|x-y|}$ is a distance. math.stackexchange.com/questions/686792/…