Mixing
What is the limit of $(1-x)^frac{1}{x}$ as x approaches 0?
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Basically, if something has a probability of 1/5, doing it 5 times gives a probability of it not happening that's close to that formula. Doing something with a probability of 1/100 100 times makes it even closer. I'm wondering if this is a well-known named constant, the way these probabilities converge is interesting to me.
limits
asked Jan 31 at 20:56
VyvianVyvian
82
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add a comment |
$begingroup$
Basically, if something has a probability of 1/5, doing it 5 times gives a probability of it not happening that's close to that formula. Doing something with a probability of 1/100 100 times makes it even closer. I'm wondering if this is a well-known named constant, the way these probabilities converge is interesting to me.
limits
asked Jan 31 at 20:56
VyvianVyvian
82
$endgroup$
2
$begingroup$
If $x=1/n$ then this becomes $$left(1-frac1nright)^n$$ which has a ver well-known limit as $ntoinfty$.
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– Lord Shark the Unknown
Jan 31 at 20:57
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Seee (sorry), See e en.wikipedia.org/wiki/E_(mathematical_constant)#History
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– Tito Eliatron
Jan 31 at 20:58
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This is about the second most famous mathematical constant.
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– Yves Daoust
Jan 31 at 21:01
$begingroup$
wolframalpha.com/input/?i=series+expansion+of+(1-x)%5E(1%2Fx)
$endgroup$
– Yves Daoust
Jan 31 at 21:04
add a comment |
$begingroup$
Basically, if something has a probability of 1/5, doing it 5 times gives a probability of it not happening that's close to that formula. Doing something with a probability of 1/100 100 times makes it even closer. I'm wondering if this is a well-known named constant, the way these probabilities converge is interesting to me.
limits
asked Jan 31 at 20:56
VyvianVyvian
82
$endgroup$
Basically, if something has a probability of 1/5, doing it 5 times gives a probability of it not happening that's close to that formula. Doing something with a probability of 1/100 100 times makes it even closer. I'm wondering if this is a well-known named constant, the way these probabilities converge is interesting to me.
limits
limits
asked Jan 31 at 20:56
VyvianVyvian
82
asked Jan 31 at 20:56
VyvianVyvian
82
asked Jan 31 at 20:56
VyvianVyvian
82
asked Jan 31 at 20:56
VyvianVyvian
82
asked Jan 31 at 20:56
VyvianVyvian
82
82
2
$begingroup$
If $x=1/n$ then this becomes $$left(1-frac1nright)^n$$ which has a ver well-known limit as $ntoinfty$.
$endgroup$
– Lord Shark the Unknown
Jan 31 at 20:57
$begingroup$
Seee (sorry), See e en.wikipedia.org/wiki/E_(mathematical_constant)#History
$endgroup$
– Tito Eliatron
Jan 31 at 20:58
$begingroup$
This is about the second most famous mathematical constant.
$endgroup$
– Yves Daoust
Jan 31 at 21:01
$begingroup$
wolframalpha.com/input/?i=series+expansion+of+(1-x)%5E(1%2Fx)
$endgroup$
– Yves Daoust
Jan 31 at 21:04
add a comment |
2
$begingroup$
If $x=1/n$ then this becomes $$left(1-frac1nright)^n$$ which has a ver well-known limit as $ntoinfty$.
$endgroup$
– Lord Shark the Unknown
Jan 31 at 20:57
$begingroup$
Seee (sorry), See e en.wikipedia.org/wiki/E_(mathematical_constant)#History
$endgroup$
– Tito Eliatron
Jan 31 at 20:58
$begingroup$
This is about the second most famous mathematical constant.
$endgroup$
– Yves Daoust
Jan 31 at 21:01
$begingroup$
wolframalpha.com/input/?i=series+expansion+of+(1-x)%5E(1%2Fx)
$endgroup$
– Yves Daoust
Jan 31 at 21:04
2
2
$begingroup$
If $x=1/n$ then this becomes $$left(1-frac1nright)^n$$ which has a ver well-known limit as $ntoinfty$.
$endgroup$
– Lord Shark the Unknown
Jan 31 at 20:57
$begingroup$
If $x=1/n$ then this becomes $$left(1-frac1nright)^n$$ which has a ver well-known limit as $ntoinfty$.
$endgroup$
– Lord Shark the Unknown
Jan 31 at 20:57
$begingroup$
Seee (sorry), See e en.wikipedia.org/wiki/E_(mathematical_constant)#History
$endgroup$
– Tito Eliatron
Jan 31 at 20:58
$begingroup$
Seee (sorry), See e en.wikipedia.org/wiki/E_(mathematical_constant)#History
$endgroup$
– Tito Eliatron
Jan 31 at 20:58
$begingroup$
This is about the second most famous mathematical constant.
$endgroup$
– Yves Daoust
Jan 31 at 21:01
$begingroup$
This is about the second most famous mathematical constant.
$endgroup$
– Yves Daoust
Jan 31 at 21:01
$begingroup$
wolframalpha.com/input/?i=series+expansion+of+(1-x)%5E(1%2Fx)
$endgroup$
– Yves Daoust
Jan 31 at 21:04
$begingroup$
wolframalpha.com/input/?i=series+expansion+of+(1-x)%5E(1%2Fx)
$endgroup$
– Yves Daoust
Jan 31 at 21:04
add a comment |
3 Answers
3
active
oldest
votes
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Let $n=frac1x$. You get $lim_{ntoinfty}(1-frac1n)^n=frac1e$.
answered Jan 31 at 21:02
Chris CusterChris Custer
14.3k3827
$endgroup$
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I didn't identify that it was just the reciprocal of e, thanks
$endgroup$
– Vyvian
Jan 31 at 21:30
add a comment |
$begingroup$
Hint:
Determine first the limit of the logarithm: $;dfrac1x,ln (1-x)=dfrac{ln(1-x)-ln 1}x;$ is a rate of variation.
answered Jan 31 at 21:00
BernardBernard
124k741117
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add a comment |
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By the Binomial theorem,
$$left(1-frac1nright)^n=sum_{k=0}^nbinom nkleft(-frac1nright)^k
\=1-frac nn+frac{n(n-1)}{2n^2}-frac{n(n-1)(n-2)}{3!n^3}+frac{n(n-1)(n-2)(n-3)}{4!n^4}-cdotsleft(-frac1nright)^n
\=frac12left(1-frac1nright)-frac1{3!}left(1-frac1nright)left(1-frac2nright)+frac1{4!}left(1-frac1nright)left(1-frac2nright)left(1-frac3nright)-cdotsleft(-frac1nright)^n$$
The constant term of this expression is, for unbounded $n$,
$$frac12-frac1{3!}+frac1{4!}-cdotsto e^{-1}$$
Then coefficient of $dfrac1n$ is
$$-frac12+frac{1+2}{3!}-frac{1+2+3}{4!}-cdots frac{(n-1)n}{2n!}cdotsto -frac{e^{-1}}2$$
So the first order approximation is
$$left(1-xright)^x=frac1e-frac x{2e}.$$
answered Jan 31 at 21:21
Yves DaoustYves Daoust
132k676230
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Undue downvote.
$endgroup$
– Yves Daoust
Jan 31 at 21:22
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $n=frac1x$. You get $lim_{ntoinfty}(1-frac1n)^n=frac1e$.
answered Jan 31 at 21:02
Chris CusterChris Custer
14.3k3827
$endgroup$
$begingroup$
I didn't identify that it was just the reciprocal of e, thanks
$endgroup$
– Vyvian
Jan 31 at 21:30
add a comment |
$begingroup$
Let $n=frac1x$. You get $lim_{ntoinfty}(1-frac1n)^n=frac1e$.
answered Jan 31 at 21:02
Chris CusterChris Custer
14.3k3827
$endgroup$
$begingroup$
I didn't identify that it was just the reciprocal of e, thanks
$endgroup$
– Vyvian
Jan 31 at 21:30
add a comment |
$begingroup$
Let $n=frac1x$. You get $lim_{ntoinfty}(1-frac1n)^n=frac1e$.
answered Jan 31 at 21:02
Chris CusterChris Custer
14.3k3827
$endgroup$
Let $n=frac1x$. You get $lim_{ntoinfty}(1-frac1n)^n=frac1e$.
answered Jan 31 at 21:02
Chris CusterChris Custer
14.3k3827
answered Jan 31 at 21:02
Chris CusterChris Custer
14.3k3827
answered Jan 31 at 21:02
Chris CusterChris Custer
14.3k3827
answered Jan 31 at 21:02
Chris CusterChris Custer
14.3k3827
14.3k3827
$begingroup$
I didn't identify that it was just the reciprocal of e, thanks
$endgroup$
– Vyvian
Jan 31 at 21:30
add a comment |
$begingroup$
I didn't identify that it was just the reciprocal of e, thanks
$endgroup$
– Vyvian
Jan 31 at 21:30
$begingroup$
I didn't identify that it was just the reciprocal of e, thanks
$endgroup$
– Vyvian
Jan 31 at 21:30
$begingroup$
I didn't identify that it was just the reciprocal of e, thanks
$endgroup$
– Vyvian
Jan 31 at 21:30
add a comment |
$begingroup$
Hint:
Determine first the limit of the logarithm: $;dfrac1x,ln (1-x)=dfrac{ln(1-x)-ln 1}x;$ is a rate of variation.
answered Jan 31 at 21:00
BernardBernard
124k741117
$endgroup$
add a comment |
$begingroup$
Hint:
Determine first the limit of the logarithm: $;dfrac1x,ln (1-x)=dfrac{ln(1-x)-ln 1}x;$ is a rate of variation.
answered Jan 31 at 21:00
BernardBernard
124k741117
$endgroup$
add a comment |
$begingroup$
Hint:
Determine first the limit of the logarithm: $;dfrac1x,ln (1-x)=dfrac{ln(1-x)-ln 1}x;$ is a rate of variation.
answered Jan 31 at 21:00
BernardBernard
124k741117
$endgroup$
Hint:
Determine first the limit of the logarithm: $;dfrac1x,ln (1-x)=dfrac{ln(1-x)-ln 1}x;$ is a rate of variation.
answered Jan 31 at 21:00
BernardBernard
124k741117
answered Jan 31 at 21:00
BernardBernard
124k741117
answered Jan 31 at 21:00
BernardBernard
124k741117
answered Jan 31 at 21:00
BernardBernard
124k741117
124k741117
add a comment |
add a comment |
$begingroup$
By the Binomial theorem,
$$left(1-frac1nright)^n=sum_{k=0}^nbinom nkleft(-frac1nright)^k
\=1-frac nn+frac{n(n-1)}{2n^2}-frac{n(n-1)(n-2)}{3!n^3}+frac{n(n-1)(n-2)(n-3)}{4!n^4}-cdotsleft(-frac1nright)^n
\=frac12left(1-frac1nright)-frac1{3!}left(1-frac1nright)left(1-frac2nright)+frac1{4!}left(1-frac1nright)left(1-frac2nright)left(1-frac3nright)-cdotsleft(-frac1nright)^n$$
The constant term of this expression is, for unbounded $n$,
$$frac12-frac1{3!}+frac1{4!}-cdotsto e^{-1}$$
Then coefficient of $dfrac1n$ is
$$-frac12+frac{1+2}{3!}-frac{1+2+3}{4!}-cdots frac{(n-1)n}{2n!}cdotsto -frac{e^{-1}}2$$
So the first order approximation is
$$left(1-xright)^x=frac1e-frac x{2e}.$$
answered Jan 31 at 21:21
Yves DaoustYves Daoust
132k676230
$endgroup$
$begingroup$
Undue downvote.
$endgroup$
– Yves Daoust
Jan 31 at 21:22
add a comment |
$begingroup$
By the Binomial theorem,
$$left(1-frac1nright)^n=sum_{k=0}^nbinom nkleft(-frac1nright)^k
\=1-frac nn+frac{n(n-1)}{2n^2}-frac{n(n-1)(n-2)}{3!n^3}+frac{n(n-1)(n-2)(n-3)}{4!n^4}-cdotsleft(-frac1nright)^n
\=frac12left(1-frac1nright)-frac1{3!}left(1-frac1nright)left(1-frac2nright)+frac1{4!}left(1-frac1nright)left(1-frac2nright)left(1-frac3nright)-cdotsleft(-frac1nright)^n$$
The constant term of this expression is, for unbounded $n$,
$$frac12-frac1{3!}+frac1{4!}-cdotsto e^{-1}$$
Then coefficient of $dfrac1n$ is
$$-frac12+frac{1+2}{3!}-frac{1+2+3}{4!}-cdots frac{(n-1)n}{2n!}cdotsto -frac{e^{-1}}2$$
So the first order approximation is
$$left(1-xright)^x=frac1e-frac x{2e}.$$
answered Jan 31 at 21:21
Yves DaoustYves Daoust
132k676230
$endgroup$
$begingroup$
Undue downvote.
$endgroup$
– Yves Daoust
Jan 31 at 21:22
add a comment |
$begingroup$
By the Binomial theorem,
$$left(1-frac1nright)^n=sum_{k=0}^nbinom nkleft(-frac1nright)^k
\=1-frac nn+frac{n(n-1)}{2n^2}-frac{n(n-1)(n-2)}{3!n^3}+frac{n(n-1)(n-2)(n-3)}{4!n^4}-cdotsleft(-frac1nright)^n
\=frac12left(1-frac1nright)-frac1{3!}left(1-frac1nright)left(1-frac2nright)+frac1{4!}left(1-frac1nright)left(1-frac2nright)left(1-frac3nright)-cdotsleft(-frac1nright)^n$$
The constant term of this expression is, for unbounded $n$,
$$frac12-frac1{3!}+frac1{4!}-cdotsto e^{-1}$$
Then coefficient of $dfrac1n$ is
$$-frac12+frac{1+2}{3!}-frac{1+2+3}{4!}-cdots frac{(n-1)n}{2n!}cdotsto -frac{e^{-1}}2$$
So the first order approximation is
$$left(1-xright)^x=frac1e-frac x{2e}.$$
answered Jan 31 at 21:21
Yves DaoustYves Daoust
132k676230
$endgroup$
By the Binomial theorem,
$$left(1-frac1nright)^n=sum_{k=0}^nbinom nkleft(-frac1nright)^k
\=1-frac nn+frac{n(n-1)}{2n^2}-frac{n(n-1)(n-2)}{3!n^3}+frac{n(n-1)(n-2)(n-3)}{4!n^4}-cdotsleft(-frac1nright)^n
\=frac12left(1-frac1nright)-frac1{3!}left(1-frac1nright)left(1-frac2nright)+frac1{4!}left(1-frac1nright)left(1-frac2nright)left(1-frac3nright)-cdotsleft(-frac1nright)^n$$
The constant term of this expression is, for unbounded $n$,
$$frac12-frac1{3!}+frac1{4!}-cdotsto e^{-1}$$
Then coefficient of $dfrac1n$ is
$$-frac12+frac{1+2}{3!}-frac{1+2+3}{4!}-cdots frac{(n-1)n}{2n!}cdotsto -frac{e^{-1}}2$$
So the first order approximation is
$$left(1-xright)^x=frac1e-frac x{2e}.$$
answered Jan 31 at 21:21
Yves DaoustYves Daoust
132k676230
answered Jan 31 at 21:21
Yves DaoustYves Daoust
132k676230
answered Jan 31 at 21:21
Yves DaoustYves Daoust
132k676230
answered Jan 31 at 21:21
Yves DaoustYves Daoust
132k676230
132k676230
$begingroup$
Undue downvote.
$endgroup$
– Yves Daoust
Jan 31 at 21:22
add a comment |
$begingroup$
Undue downvote.
$endgroup$
– Yves Daoust
Jan 31 at 21:22
$begingroup$
Undue downvote.
$endgroup$
– Yves Daoust
Jan 31 at 21:22
$begingroup$
Undue downvote.
$endgroup$
– Yves Daoust
Jan 31 at 21:22
add a comment |
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2
$begingroup$
If $x=1/n$ then this becomes $$left(1-frac1nright)^n$$ which has a ver well-known limit as $ntoinfty$.
$endgroup$
– Lord Shark the Unknown
Jan 31 at 20:57
$begingroup$
Seee (sorry), See e en.wikipedia.org/wiki/E_(mathematical_constant)#History
$endgroup$
– Tito Eliatron
Jan 31 at 20:58
$begingroup$
This is about the second most famous mathematical constant.
$endgroup$
– Yves Daoust
Jan 31 at 21:01
$begingroup$
wolframalpha.com/input/?i=series+expansion+of+(1-x)%5E(1%2Fx)
$endgroup$
– Yves Daoust
Jan 31 at 21:04