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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Equality of field extensions given Splitting field is $S_n$.

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Clash Royale CLAN TAG #URR8PPP up vote 4 down vote favorite 1 Let $fin mathbbQ[x]$ be an irreducible polynomial of degree $ngeq 5$. Let $L$ be the splitting of $f$ and let $alphain L$ be a zero of $f$. Claim If $[L:mathbbQ]=n!$, then $mathbbQ[alpha] = mathbbQ[alpha^4]$. Attempt $[L:mathbbQ]= n!$, and $L' leq S_n$ implies $L'= S_n$. $[mathbbQ[alpha]:mathbbQ]=n$ since $f$ is irreducible. $mathbbQ[alpha^4] subset mathbbQ[alpha]implies [mathbbQ[alpha^4]:mathbbQ] leq n$. Now, there exists no subgroup of $S_n$ with index $2<t<n$, hence it is either $2$ or $n$. If it is $2$, $mathbbQ[alpha^4]'=A_n$, which is a normal subgroup hence $mathbbQ[alpha^4]$ is a splitting field over $mathbbQ$. How can I claim it is not possible? Notation If $mathbbQ subset K subset L$, $mathbbQ' = S_n$, $L'=e$ and $K'=sin S_n: sk=k text for all kin K$ galois-theory splitting-field share | cite | improve this question asked Aug 6 at 23:40 Jo' 14...
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