Michael Zhao
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 Feb5 awarded Nice Answer Jan6 awarded Popular Question Sep6 awarded Yearling Feb14 answered please help me understand the lecture note? heat equation and fourier series Sep6 awarded Yearling Apr22 comment how to show that a function$f$ is contained in all natural numbers? For the proof, that's not necessary to know. It's sufficient to show that the expression for $f(a,b)$ is always a positive integer, for any positive integers $a$ and $b$, which is what we did. Apr22 answered how to show that a function$f$ is contained in all natural numbers? Apr18 comment I have some questions for double integrals, please help… I really need help If you had $d\theta$ instead of $dx$ or $dy$ respectively, you could, unless the function inside had any $\theta$'s involved. Apr17 revised I have some questions for double integrals, please help… I really need help type-setted it in latex Apr17 comment I have some questions for double integrals, please help… I really need help No, that is not acceptable since the bounds of the integral tell you that $x$ runs from 0 to $2\pi$ (respectively, that y runs from 0 to $\pi/2$). Just because you see $2\pi$ or $\pi/2$, doesn't mean it has anything to do with the variable $\theta,$ which I think is the mistake you're making. Apr17 suggested approved edit on I have some questions for double integrals, please help… I really need help Mar25 accepted Concurrency of A'L, B'M, C'N. Mar19 asked Concurrency of A'L, B'M, C'N. Jan30 comment Funny thing. Multiplying both the sides by 0? Yes, I know, so I'm saying that the "usual way" doesn't really work in that you supposedly address the zero value of $\cos \theta = 0$ in the same step as you address non-zero values of $\cos\theta$, i.e. the "usual way" that the OP talks about overlooks a case. Sure, it might be trivial, but it still is overlooked. Jan28 answered Funny thing. Multiplying both the sides by 0? Oct31 answered Expected Value of a Binomial distribution? Oct22 comment Expected Number of Successes in a Sample @Austin Mohr: Thanks. At any rate, it depends on whether you're sampling with replacement or without replacement (are the probabilities of being broken independent from one calculator to the next?). If you're sampling with replacement (probabilities are independent), then you still use the fact that if $X \sim B(n,p)$, then $E[X] = np = 10\cdot 20/200 = 1$, as below. Oct21 awarded Commentator Oct21 comment Expected Number of Successes in a Sample You're assuming sampling with replacement when you're calculating the probabilities. I'm not sure you can do that. If you can, then it's standard knowledge that the expected value of a binomial distribution with $n$ trials and probability $p$ of success is $np$, so in this case $E[X] = 200\cdot20/200 = 20$. Oct21 comment Is this question proper? “Solve $\log_{10}x\in\mathbb{R}$.” Hm, first, do you mean "infinitely" many? This is fairly subjective, I suppose, but I wouldn't use the phrase "Solve." What resonates better for me is: "For what $x$ is $x+1 > 0$?" Alternately if you go with what BobaFret says, "Solve $x+1>0$." makes sense because you are simplifying the inequality and isolating $x$ to one side, thereby getting information about $x$ itself.