不能發售的OX瑪莉歐……好像還滿好玩的XDD 


不能發售的OX瑪莉歐……好像還滿好玩的XDD


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A curious connection : What's the function $f(x)$?

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Clash Royale CLAN TAG #URR8PPP up vote 16 down vote favorite 4 Hello I was wondering what was the function $f$ defines like this : Let $f(x)$ a continuous and differentiable function such that : $$f(x)=sum_k=0^inftyfracf'(k)k!(-x)^k$$ In fact can't solve it but it makes a connection between Ramanujan's Master theorem and Frullani's integral via the Fundamental theorem of calculus I explain : We have : $$int_0^inftyx^-s-1f(x)dx=Gamma(-s)f'(s)$$ Or : $$int_0^inftyfracx^-s-1f(x)Gamma(-s)dx=f'(s)$$ Now we use the Fundamental theorem of calculus to get : $$int_0^sint_0^inftyfracx^-s-1f(x)Gamma(-s)dxds=f(s)-f(0)$$ Now we take the limit to get : $$lim_stoinftyint_0^sint_0^inftyfracx^-s-1f(x)Gamma(-s)dxds=f(infty)-f(0)$$ Wich is equal to: $$int_0^inftyfracf(ax)-f(bx)ln(fracab)x$$ So my question is what is the function $f(x)$ , there exists a closed form to this ,is it trivial or not ? Thanks Ps:I know it's not very rigorous but I think it's intere...
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Issue with $int_0^inftyfraccos(x^2)x^2dx$

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Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 2 I have stumbled into a issue with gamma function, I will show the approach first: $$Gamma(x)=int_0^inftyt^x-1e^-tdt$$ subtituting $t=iu^2$ gives: $$Gamma(x)=2int_0^infty(iu^2)^x-1e^-iu^2iudurightarrowfracGamma(x)2i^x=int_0^inftyu^2x-1e^-iu^2du$$ Doing the same thing using $t=-iu^2,$results in$$fracGamma(x)-2i^x=int_0^inftyu^2x-1e^iu^2du$$ Now, summing those two and using that $i^x=e^fracipi2x ,$gives $$fracGamma(x)2(e^fracipi2x+e^frac-ipi2x)=int_0^inftyu^2x-1(e^iu^2+e^-iu^2)du$$ which is just $$fracGamma(x)2cos(fracpi2x)=int_0^inftyu^2x-1cos(u^2)du$$ plugging $x=-frac12$ we get that$$int_0^inftyfraccos(x^2)x^2dx=-sqrtfracpi2$$ Well, obviously this integral diverges... But if Instead of summing we subtract we get that $$fracGamma(x)2sin(fracpi2x)=int_0^inftyu^2x-1sin(u^2)du$$ simmilarly with $$x=-frac12 rightarrow int_0^inftyfracsin(x^2)x^2dx=sqrtfracpi2$$ So its not that completely garbage. Now my question is, w...
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Let $U$ and $V$ be independent random variables both having Bernoulli , Compute $E(W)$

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Clash Royale CLAN TAG #URR8PPP up vote 0 down vote favorite Let $U$ and $V$ be independent random variables both having Bernoulli distribution, i.e. $U$ ~ Bernoulli($0.2$) and $V$ ~ Bernoulli($0.2$) Let $W = U cdot V$ (a) Compute $E(W)$ (b) Compute $E(UW)$ My attempt: $(a)$ $$E(W) = E(UV) = E(U)E(V)$$ Because they're independent $E(U) = E(V) = 0.2$, so $E(W) = 0.04$ $(b)$ $E(UW) = E(UUW) = E(U^2V) = E(U^2V) = E(U^2)E(V)$ $E(U^2) = 0.2^2 = 0.04$ [Not sure about this] Thus, $E(UW) = 0.04 cdot 0.2 = 0.008$ probability expectation share | cite | improve this question edited Jul 26 at 2:55 Graham Kemp 80.1k 4 32 75 asked Jul 25 at 23:45 Tinler 404 3 11 1 Note $U^2=U$ here so $E[U^2]=E[U]$ not $left(E[U]right)^2$ – Henry Jul 25 at 23:53 add a comment  |  up vote 0 down vote favorite Let $U$ and $V$ be independent random variables both having Bernoulli dist...
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