高校数学の質問スレ（医者・東大卒専用） Part438

# 1-(1-1/n)^x >= cl
cl=0.95
n=10000
x=log(1-cl)/log(1-1/n) ; x # log(alclha)/log(cl[no adverse effect])

"log(1-cl)=x*log(1-1/n)
log((1-cl)^(1/x))=log(1-1/n)
1-1/n=(1-cl)^(1/x)
1/n=1-(1-cl)^(1/x)
"
x=40000
n=1/(1-(1-cl)^(1/x)) ; n

x=n*y=log(1-0.95)/log(1-1/n)
# solve n*y=log(1-0.95)/log(1-1/n) for y
cl=0.95
n=1000
log(1/(1-cl))/(n*log(n/(n-1)))

f=\(n,cl=0.95) log(1/(1-cl))/(n*log(n/(n-1)))
n=10:10000
plot(n,f(n),bty='l',type='l')
abline(h=3,lty=3)