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"The quieter you become, the more you are able to hear..."


22h
asked conditional probability about sum and product rule
Jul
13
accepted Terminology on algebra.
Jul
12
asked Terminology on algebra.
Jul
9
awarded  Civic Duty
Jul
2
awarded  Curious
Jul
2
awarded  Inquisitive
Jul
2
asked Independent of random variables.
Jun
9
asked The maximum term in Binomial distribution
Jun
7
accepted Asymptotic in hypergeometric distribution.
Jun
6
asked Asymptotic in hypergeometric distribution.
May
31
accepted Does the inequality $\frac{1-x_1x_2}{1-y_1y_2} \leq \frac{1-x_1}{1-y_1} + \frac{1-x_2}{1-y_2}$ hold?
May
31
asked Does the inequality $\frac{1-x_1x_2}{1-y_1y_2} \leq \frac{1-x_1}{1-y_1} + \frac{1-x_2}{1-y_2}$ hold?
May
28
asked $\sigma$-algebra and measurable set.
May
27
awarded  Popular Question
May
26
comment Derivative of $f(x)=\int_1^{\frac{1}{x}}\frac{\text{d}t}{\sqrt{(t^2-1)(1-t^2x^2)}}$
@ReneSchipperus because $\varphi(x)=\frac{1}{x}$. then $F(x,\varphi(x))=\frac{1}{\sqrt{(1-\frac{1}{x^2})(1-1)}}$ so it is meaningless
May
26
comment Derivative of $f(x)=\int_1^{\frac{1}{x}}\frac{\text{d}t}{\sqrt{(t^2-1)(1-t^2x^2)}}$
@DavidH I have edited the post. the integral lower bound is $1$, and $x \in (0,1)$. then the integral is real-valued.
May
26
comment Derivative of $f(x)=\int_1^{\frac{1}{x}}\frac{\text{d}t}{\sqrt{(t^2-1)(1-t^2x^2)}}$
@JPi, sorry .I made a mistake. the integral is from $1$,not $0$
May
26
revised Derivative of $f(x)=\int_1^{\frac{1}{x}}\frac{\text{d}t}{\sqrt{(t^2-1)(1-t^2x^2)}}$
added 19 characters in body
May
26
asked Derivative of $f(x)=\int_1^{\frac{1}{x}}\frac{\text{d}t}{\sqrt{(t^2-1)(1-t^2x^2)}}$
May
24
comment Something about $\frac{\log x}{x}$
@SanathDevalapurkar I am not tried this method. My way is using Cauchy Inequality: Because $x_1x_2 > e^2$ equivalent to $\log (x_1x_2) > 2$. then I have $\log(x_1x_2) = \int_1^{x_1x_2} \frac{1}{t}\text{d}t \geq \left(\int_1^{x_1x_2}\frac{1}{\sqrt{t}}\text{d}t\right)^2 = (2\sqrt{x_1x_2}-2)^2 = 4(\sqrt{x_1x_2}-1)^2>2$. in order to satisfy the last inequality. we must have $x_1x_2 > \left(1+\frac{\sqrt{2}}{2}\right)^2$. the right hand side of the last inequality is less than $e^2$. I hope this way would solve this question.But up to now I can't go any further.