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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
Mar
17
answered Why (x'Ay)^2 <=(x'Ax)(y'Ay), when A is positive definite?
Mar
13
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, $$\mathbb E\left(\exp\left(\frac{1}{2}\sum_{i=1}^n X_i^2\right)\right) = \prod_{i = 1}^n \mathbb E\left(\exp\left(\frac{X_i^2}{2}\right)\right)$$
Mar
11
comment Positive numbers inequality
The last step has a bug.
Mar
11
revised Positive numbers inequality
added 87 characters in body
Mar
11
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 $$\geq \frac{1}{n^2}$$
Mar
11
comment Positive numbers inequality
What the hell? Please explain the downvote, whoever did it, you owe me at least that much! X(
Mar
11
awarded  Custodian
Mar
11
reviewed Approve Wedge Notation in probability
Mar
11
revised Positive numbers inequality
deleted 16 characters in body
Mar
11
revised Positive numbers inequality
edited body
Mar
11
revised Positive numbers inequality
edited body
Mar
11
answered Positive numbers inequality
Mar
11
answered Evaluating $\int_0^{\pi/2} \frac{a}{a^2+\cos^2 \theta} \, d\theta$
Mar
11
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 ;-)
Mar
11
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
Mar
11
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 :-)