TenaliRaman
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 Apr 23 awarded Yearling Apr 20 reviewed Approve Proof of Equivalence of Sets Apr 19 reviewed Approve Prove that for all positive integers $n$, $2^1+2^2+2^3+…+2^n=2^{n+1}-2$ Apr 18 reviewed Approve $X_1, \dots, X_n$ are independent random variables. Suppose $M = \min(X_1, X_2, \dots, X_n)$ Apr 18 reviewed Approve coordinate geometry : intersection of a curve with a line Apr 7 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. Mar 24 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. Mar 23 answered Prove that if $\alpha, \beta, \gamma$ are angles in triangle, then $(tan(\frac{\alpha}{2}))^2+(tan(\frac{\beta}{2}))^2+(tan(\frac{\gamma}{2}))^2\geq1$ Mar 23 answered Simple integral calculation: $\int \sqrt{x^2-4}\,dx$ Mar 20 reviewed Approve Sum of a power series $n x^n$ Mar 20 reviewed Approve analysis problem of continity Mar 20 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. Mar 20 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? Mar 20 revised $T$ is a linear operator Latex Corrections Mar 19 awarded Citizen Patrol Mar 18 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. Mar 18 comment Prove Trigonometric Identitiy @ArpanBanerjee You are correct. Mar 18 comment Prove Trigonometric Identitiy Indeed, you are correct. Mar 18 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. Mar 18 answered Prove Trigonometric Identitiy