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1h
comment Differentiate the Function: $g(u)=\ln\left(\frac{\ln\ u}{1+\ln\ (2u)}\right)$
-1 Obviously not a hint, but some formula (probably produced by some program) with zero mathematical explanation.
1d
awarded  Revival
Jul
3
awarded  Nice Answer
Jul
3
comment Does $\sum_{n=1}^{\infty}\frac{x}{1+n^4x^2}$ uniform converge?
@Elimination What did you try to prove it?
Jul
3
comment Convolute exponential with a gaussian
@StefanWager Let me suggest that you ponder more carefully your use of the word "incorrect". Please read slowly how this answer is written, how it refers to the question, and please indicate which part is "incorrect", if you still think so.
Jul
3
comment On the equality $p(A) = \int_{x} P(A|X=x)\ dF(x)$ in probability
@HowDoIMath There are two concepts at hand here, related but distinct: conditional probability of an event conditioned by an event with positive probability (the result being a number), and conditional probability of an event conditioned by a random variable (the result being a random variable). Recalling the definition of the latter might prove useful: by definition $P(A\mid X)$ is a random variable $Y$ such that (1) $Y$ is $\sigma(X)$-measurable and (2) $E(Y\mathbf 1_B)=P(A\cap B)$ for every $B$ in $\sigma(X)$.
Jul
1
comment Studying this sum: $\sum_{k=0}^{\infty} (k+1)(x)^k$
Third question by this OP on this exact theme. Third time the OP will insist that $1-2+3-4+\ldots$ is you-know-what because their prof told them so, will be explained the difference between Ramanujan summation of a sequence and convergence of the series, and will not listen. Is this an abuse of the site or what?
Jul
1
comment Infinite summation (sum of natural numbers)
@OP No (deliberate) duplicates, please.
Jul
1
comment Infinite summation (sum of natural numbers)
@BusyAnt Actually one can use the word: if $(a_n)$ does not converge to zero then "obviously" the series $\sum\limits_na_n$ diverges. That some users here repeatedly misapply the symbols $=$ and $\sum$ is another subject (probably worthy of investigations on its own, in view of the concentration of these misapplications about series such as $\sum\limits_nn$ and its like, but another subject).
Jul
1
revised Non linear Differential Equation
deleted 17 characters in body
Jun
30
comment How do we know the joint probability distribution measure is valid?
Stick to $w$ or to $\omega$ but avoid switching between these constantly.
Jun
30
comment Law of Iterated Expectation with Probability?
$P(X\leq z - Y)= E[P(X \leq z - Y | Y=y)]$ is incorrect (what would be this $y$ in the RHS), use instead $P(X\leq z - Y)= E[P(X \leq z - Y | Y)]$.
Jun
30
comment Probability that after 10,000 steps (+-1) you'll end up at the origin. How to use Central Limit Theorem?
Yes, yes, sure, please continue to address NONE of the specific points I made. Sorry but I am out.
Jun
30
comment Prove or disprove that $φ_v:u\mapsto \langle\mathcal A u,v\rangle$ is in $V^*$
Modified the title.
Jun
30
revised Prove or disprove that $φ_v:u\mapsto \langle\mathcal A u,v\rangle$ is in $V^*$
edited title
Jun
30
comment How to calculate this limit as $x\rightarrow 0$?
+1. And even $$\lim \frac{\sin x}{x}\cdot\lim \sin ^2 x\cdot\lim \frac1{\cos^3 x}.$$
Jun
30
comment Probability that after 10,000 steps (+-1) you'll end up at the origin. How to use Central Limit Theorem?
.../... (Yes this comment is merely rephrasing my previous-to-last comment. Yes it is unfortunate that I have to.)
Jun
30
comment Probability that after 10,000 steps (+-1) you'll end up at the origin. How to use Central Limit Theorem?
@BruceZ You see? To justify the statements in your answer, you are naturally led to be much more specific than you were when I posted my comments advising for caution about the applications of CLT. Sure there are results allowing to approximate binomials by normals but these do not follow from the general statement of the CLT. If your answer is using some lattice-type CLT, just mention the fact and be specific (but note that the obfuscations about the size of the interval in your last comments are squarely misleading). .../...
Jun
30
comment Solve this logarithmic equation: $2^{2-\ln x}+2^{2+\ln x}=8$
Sure, in the reals, $2^t=1\to t=0$, the trouble is that you have $2^t(2-2^t)=1(2-1)$, not $2^t=1$. It happens that the former implies the latter but this must be proved for the method to be complete.
Jun
29
comment Solve this logarithmic equation: $2^{2-\ln x}+2^{2+\ln x}=8$
The first method was modified. Now I fail to understand why $2^{\ln\:x}\:(2-2^{\ln x})=2^0\cdot 1$ implies $x=1$ (last step). Since this is the heart of the matter, I consider this first method as not being a proof. The second method is allright.