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Jun
3
comment Why do we use gradient descent in the backpropagation algorithm?
@moose I will have to think about your suggestion carefully. That would require me to elaborate on the answer a little more than I would like to do, but let me see.
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
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
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
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
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
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 :-)
Mar
11
comment Prove $(1 +\frac{ 1}{n}) ^ {n} \ge 2$
Nope. This is a direct proof using binomial expansion.
Mar
6
comment Show that if $n$ is composite, then $\phi(n) \leq n-\sqrt{n}$
math.stackexchange.com/questions/896920/…
Jan
21
comment show that $ 4 = \sum_{n=1}^\infty (-2)^{n+1}\frac{n+2}{n!} $
@learnmore 2*(-1)^n is not (-2)^n
Jan
20
comment Does $\int_0^\infty \sin(x^{2/3}) dx$ converges?
$x = y^{3/2}$ then $dx = \frac{3}{2}y^{1/2}dy$