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a double factorial products up 2 at a time not 1
2x!!=product k=1->x 2x
so x+2n!!=x!!*product k=1->n 2x
for something like a gamma function,we can take the log
L(x)=ln(x!!)
N is a huge number
so ln(N+k)-ln(N)=0 for a little k compared to N
L(N+2n)=L(N)+sum k=1->n ln(N+2k)
=L(N)+sum k=1->n ln(N)
=L(N)+nlnN
now do some substitution
L(2n)+sum k=1->N/2 ln(2n+2k)=sum k=1->N/2 ln(2k)+nlnN
move it to right and merge sums
L(2n)=sum k=1->N/2 -ln(2n+2k)+ ln(2k)+nlnN
log rule
L(2n)=sum k=1->N/2 ln(k/k+n)+nlnN
sub n for 2n and exponentiate
n!!=product k=1->N/2 k/k+n/2 +N^(n/2)
minor changes
n!!=product k=1->N/2 2k/k+n +sqrt(N^n)
done!
change the limits a bit N->2N^2 x->n
x!!=lim N->inf 4*N^n*product k=1->N^2 k/(x+2k)
idk if it actually works but i think it would cuz it based on the factorial video from lines that connect
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Hi iccute,
Here is a relevant link: Double Factorial.
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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