Why $ lim_{nrightarrow infty} frac{n!}{n^{k}(n-k)! } =1 $?
$begingroup$
I was on brilliant.org learning probability. There was a process explaining how the distribution of a Poisson Random Variable can be obtained from a Binomial Random Variable.
Consider the binomial distribution:
$$
begin{equation}begin{aligned}
P(X=k) &={binom n k} p^k (1-p)^{n-k}\
&=frac{n!}{k!(n-k)!} p^k (1-p)^{n-k}
end{aligned}end{equation}
$$
Substitute $m=np$ , or $p=frac{m}{n}$ :
$$
begin{equation}begin{aligned}
P(X=k) &=frac{n!}{k!(n-k)!} left(frac{m}{n}right)^k left(1-frac{m}{n}right)^{n-k}\
&=frac{n!}{k!(n-k)!} frac{m^k}{n^k} left(1-frac{m}{n}right)^{n}left(1-frac{m}{n}right)^{-k}
end{aligned}end{equation}
$$
Slightly rearrange
$$
begin{equation}begin{aligned}
&=frac{n!}{n^k(n-k)!} left(1-frac{m}{n}right)^{-k}frac{m^k}{k!}left(1-frac{m}{n}right)^{n}
end{aligned}end{equation}
$$
Note that
$$
begin{equation}begin{aligned}
& lim_{nrightarrow infty} frac{n!}{n^{k}(n-k)! } =1,quadlim_{nrightarrow infty} left(1-frac{m}{n}right)^{-k} =1,quad lim_{nrightarrow infty} left(1-frac{m}{n}right)^{n} =e^{-m}
end{aligned}end{equation}
$$
Thus, we have the final result which is equal to the formula for the Poisson distribution.
$$
=frac{m^k e^{-m}}{k!}
$$
In all these steps, what I don't understand is the following limit:
$$
lim_{nrightarrow infty} frac{n!}{n^{k}(n-k)! } =1
$$
limits factorial
asked yesterday
billyandrbillyandr
237
$endgroup$
add a comment |
$begingroup$
I was on brilliant.org learning probability. There was a process explaining how the distribution of a Poisson Random Variable can be obtained from a Binomial Random Variable.
Consider the binomial distribution:
$$
begin{equation}begin{aligned}
P(X=k) &={binom n k} p^k (1-p)^{n-k}\
&=frac{n!}{k!(n-k)!} p^k (1-p)^{n-k}
end{aligned}end{equation}
$$
Substitute $m=np$ , or $p=frac{m}{n}$ :
$$
begin{equation}begin{aligned}
P(X=k) &=frac{n!}{k!(n-k)!} left(frac{m}{n}right)^k left(1-frac{m}{n}right)^{n-k}\
&=frac{n!}{k!(n-k)!} frac{m^k}{n^k} left(1-frac{m}{n}right)^{n}left(1-frac{m}{n}right)^{-k}
end{aligned}end{equation}
$$
Slightly rearrange
$$
begin{equation}begin{aligned}
&=frac{n!}{n^k(n-k)!} left(1-frac{m}{n}right)^{-k}frac{m^k}{k!}left(1-frac{m}{n}right)^{n}
end{aligned}end{equation}
$$
Note that
$$
begin{equation}begin{aligned}
& lim_{nrightarrow infty} frac{n!}{n^{k}(n-k)! } =1,quadlim_{nrightarrow infty} left(1-frac{m}{n}right)^{-k} =1,quad lim_{nrightarrow infty} left(1-frac{m}{n}right)^{n} =e^{-m}
end{aligned}end{equation}
$$
Thus, we have the final result which is equal to the formula for the Poisson distribution.
$$
=frac{m^k e^{-m}}{k!}
$$
In all these steps, what I don't understand is the following limit:
$$
lim_{nrightarrow infty} frac{n!}{n^{k}(n-k)! } =1
$$
limits factorial
asked yesterday
billyandrbillyandr
237
$endgroup$
$begingroup$
There are several posts about this: Why does $lim_{ntoinfty} frac{n!}{(n-k)!n^k}$ equal 1, Finding limit of sequence: $lim _{n to infty} {frac{n!}{n^k(n-k)!}}=1$, Proof that $limlimits_{h to infty} frac{h!}{h^k(h-k)!}=1$ for any $k$, Limits involing Factorials $lim_{Ntoinfty} frac{N!}{(N-k)!N^{k}}$
$endgroup$
– Martin Sleziak
yesterday
$begingroup$
I found the posts in the above comment using Approach0. For some useful tips on searching here see: How to search on this site?
$endgroup$
– Martin Sleziak
yesterday
$begingroup$
Some of the other posts treating the same question painfully lack details and context. Maybe you'd want to put them on hold or close them.
$endgroup$
– billyandr
yesterday
$begingroup$
billyandr: If you actually have a look at those links, you can see that two of those posts are closed (as duplicates) now. Let me also say that the fact that you have added some more context to your question is certainly appreciated. (After all, that's what lead to reopening.)
$endgroup$
– Martin Sleziak
yesterday
add a comment |
$begingroup$
I was on brilliant.org learning probability. There was a process explaining how the distribution of a Poisson Random Variable can be obtained from a Binomial Random Variable.
Consider the binomial distribution:
$$
begin{equation}begin{aligned}
P(X=k) &={binom n k} p^k (1-p)^{n-k}\
&=frac{n!}{k!(n-k)!} p^k (1-p)^{n-k}
end{aligned}end{equation}
$$
Substitute $m=np$ , or $p=frac{m}{n}$ :
$$
begin{equation}begin{aligned}
P(X=k) &=frac{n!}{k!(n-k)!} left(frac{m}{n}right)^k left(1-frac{m}{n}right)^{n-k}\
&=frac{n!}{k!(n-k)!} frac{m^k}{n^k} left(1-frac{m}{n}right)^{n}left(1-frac{m}{n}right)^{-k}
end{aligned}end{equation}
$$
Slightly rearrange
$$
begin{equation}begin{aligned}
&=frac{n!}{n^k(n-k)!} left(1-frac{m}{n}right)^{-k}frac{m^k}{k!}left(1-frac{m}{n}right)^{n}
end{aligned}end{equation}
$$
Note that
$$
begin{equation}begin{aligned}
& lim_{nrightarrow infty} frac{n!}{n^{k}(n-k)! } =1,quadlim_{nrightarrow infty} left(1-frac{m}{n}right)^{-k} =1,quad lim_{nrightarrow infty} left(1-frac{m}{n}right)^{n} =e^{-m}
end{aligned}end{equation}
$$
Thus, we have the final result which is equal to the formula for the Poisson distribution.
$$
=frac{m^k e^{-m}}{k!}
$$
In all these steps, what I don't understand is the following limit:
$$
lim_{nrightarrow infty} frac{n!}{n^{k}(n-k)! } =1
$$
limits factorial
asked yesterday
billyandrbillyandr
237
$endgroup$
I was on brilliant.org learning probability. There was a process explaining how the distribution of a Poisson Random Variable can be obtained from a Binomial Random Variable.
Consider the binomial distribution:
$$
begin{equation}begin{aligned}
P(X=k) &={binom n k} p^k (1-p)^{n-k}\
&=frac{n!}{k!(n-k)!} p^k (1-p)^{n-k}
end{aligned}end{equation}
$$
Substitute $m=np$ , or $p=frac{m}{n}$ :
$$
begin{equation}begin{aligned}
P(X=k) &=frac{n!}{k!(n-k)!} left(frac{m}{n}right)^k left(1-frac{m}{n}right)^{n-k}\
&=frac{n!}{k!(n-k)!} frac{m^k}{n^k} left(1-frac{m}{n}right)^{n}left(1-frac{m}{n}right)^{-k}
end{aligned}end{equation}
$$
Slightly rearrange
$$
begin{equation}begin{aligned}
&=frac{n!}{n^k(n-k)!} left(1-frac{m}{n}right)^{-k}frac{m^k}{k!}left(1-frac{m}{n}right)^{n}
end{aligned}end{equation}
$$
Note that
$$
begin{equation}begin{aligned}
& lim_{nrightarrow infty} frac{n!}{n^{k}(n-k)! } =1,quadlim_{nrightarrow infty} left(1-frac{m}{n}right)^{-k} =1,quad lim_{nrightarrow infty} left(1-frac{m}{n}right)^{n} =e^{-m}
end{aligned}end{equation}
$$
Thus, we have the final result which is equal to the formula for the Poisson distribution.
$$
=frac{m^k e^{-m}}{k!}
$$
In all these steps, what I don't understand is the following limit:
$$
lim_{nrightarrow infty} frac{n!}{n^{k}(n-k)! } =1
$$
limits factorial
limits factorial
asked yesterday
billyandrbillyandr
237
asked yesterday
billyandrbillyandr
237
edited yesterday
billyandr
asked yesterday
billyandrbillyandr
237
asked yesterday
billyandrbillyandr
237
asked yesterday
billyandrbillyandr
237
237
$begingroup$
There are several posts about this: Why does $lim_{ntoinfty} frac{n!}{(n-k)!n^k}$ equal 1, Finding limit of sequence: $lim _{n to infty} {frac{n!}{n^k(n-k)!}}=1$, Proof that $limlimits_{h to infty} frac{h!}{h^k(h-k)!}=1$ for any $k$, Limits involing Factorials $lim_{Ntoinfty} frac{N!}{(N-k)!N^{k}}$
$endgroup$
– Martin Sleziak
yesterday
$begingroup$
I found the posts in the above comment using Approach0. For some useful tips on searching here see: How to search on this site?
$endgroup$
– Martin Sleziak
yesterday
$begingroup$
Some of the other posts treating the same question painfully lack details and context. Maybe you'd want to put them on hold or close them.
$endgroup$
– billyandr
yesterday
$begingroup$
billyandr: If you actually have a look at those links, you can see that two of those posts are closed (as duplicates) now. Let me also say that the fact that you have added some more context to your question is certainly appreciated. (After all, that's what lead to reopening.)
$endgroup$
– Martin Sleziak
yesterday
add a comment |
$begingroup$
There are several posts about this: Why does $lim_{ntoinfty} frac{n!}{(n-k)!n^k}$ equal 1, Finding limit of sequence: $lim _{n to infty} {frac{n!}{n^k(n-k)!}}=1$, Proof that $limlimits_{h to infty} frac{h!}{h^k(h-k)!}=1$ for any $k$, Limits involing Factorials $lim_{Ntoinfty} frac{N!}{(N-k)!N^{k}}$
$endgroup$
– Martin Sleziak
yesterday
$begingroup$
I found the posts in the above comment using Approach0. For some useful tips on searching here see: How to search on this site?
$endgroup$
– Martin Sleziak
yesterday
$begingroup$
Some of the other posts treating the same question painfully lack details and context. Maybe you'd want to put them on hold or close them.
$endgroup$
– billyandr
yesterday
$begingroup$
billyandr: If you actually have a look at those links, you can see that two of those posts are closed (as duplicates) now. Let me also say that the fact that you have added some more context to your question is certainly appreciated. (After all, that's what lead to reopening.)
$endgroup$
– Martin Sleziak
yesterday
$begingroup$
There are several posts about this: Why does $lim_{ntoinfty} frac{n!}{(n-k)!n^k}$ equal 1, Finding limit of sequence: $lim _{n to infty} {frac{n!}{n^k(n-k)!}}=1$, Proof that $limlimits_{h to infty} frac{h!}{h^k(h-k)!}=1$ for any $k$, Limits involing Factorials $lim_{Ntoinfty} frac{N!}{(N-k)!N^{k}}$
$endgroup$
– Martin Sleziak
yesterday
$begingroup$
There are several posts about this: Why does $lim_{ntoinfty} frac{n!}{(n-k)!n^k}$ equal 1, Finding limit of sequence: $lim _{n to infty} {frac{n!}{n^k(n-k)!}}=1$, Proof that $limlimits_{h to infty} frac{h!}{h^k(h-k)!}=1$ for any $k$, Limits involing Factorials $lim_{Ntoinfty} frac{N!}{(N-k)!N^{k}}$
$endgroup$
– Martin Sleziak
yesterday
$begingroup$
I found the posts in the above comment using Approach0. For some useful tips on searching here see: How to search on this site?
$endgroup$
– Martin Sleziak
yesterday
$begingroup$
I found the posts in the above comment using Approach0. For some useful tips on searching here see: How to search on this site?
$endgroup$
– Martin Sleziak
yesterday
$begingroup$
Some of the other posts treating the same question painfully lack details and context. Maybe you'd want to put them on hold or close them.
$endgroup$
– billyandr
yesterday
$begingroup$
Some of the other posts treating the same question painfully lack details and context. Maybe you'd want to put them on hold or close them.
$endgroup$
– billyandr
yesterday
$begingroup$
billyandr: If you actually have a look at those links, you can see that two of those posts are closed (as duplicates) now. Let me also say that the fact that you have added some more context to your question is certainly appreciated. (After all, that's what lead to reopening.)
$endgroup$
– Martin Sleziak
yesterday
$begingroup$
billyandr: If you actually have a look at those links, you can see that two of those posts are closed (as duplicates) now. Let me also say that the fact that you have added some more context to your question is certainly appreciated. (After all, that's what lead to reopening.)
$endgroup$
– Martin Sleziak
yesterday
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is rather obvious if you cancel the factorials:
$$frac{n!}{n^{k}(n-k)! } =frac{overbrace{n(n-1)cdots (n-k+1)}^{k; factors}}{n^k}= 1cdot left(1-frac{1}{n}right)cdots left(1-frac{k-1}{n}right)stackrel{n to infty}{longrightarrow} 1$$
answered yesterday
trancelocationtrancelocation
14.1k1829
$endgroup$
$begingroup$
Thank you so much. I didn't know it was right there under my eyes.
$endgroup$
– billyandr
yesterday
$begingroup$
You are welcome. This "not seeing the obvious" just happens once in a while, I think, to all who do maths. So, it is good to have a math platform like this one. :-)
$endgroup$
– trancelocation
yesterday
add a comment |
$begingroup$
$$a_n=frac{n!}{n^{k}(n-k)! }implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$
Use Stirling approximation and continue with Taylor series to get
$$log(a_n)=frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$ Continue with Taylor
$$a_n=e^{log(a_n)}=1+frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$
answered yesterday
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
1
$begingroup$
This has already a slight touch of overkill, hasn't it? :-)
$endgroup$
– trancelocation
yesterday
$begingroup$
@trancelocation. You are totally right for the limit. One of my manias is to always look at the approach to the limit. Have a look at matheducators.stackexchange.com/questions/8339/… . Cheers :-)
$endgroup$
– Claude Leibovici
yesterday
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
It is rather obvious if you cancel the factorials:
$$frac{n!}{n^{k}(n-k)! } =frac{overbrace{n(n-1)cdots (n-k+1)}^{k; factors}}{n^k}= 1cdot left(1-frac{1}{n}right)cdots left(1-frac{k-1}{n}right)stackrel{n to infty}{longrightarrow} 1$$
answered yesterday
trancelocationtrancelocation
14.1k1829
$endgroup$
$begingroup$
Thank you so much. I didn't know it was right there under my eyes.
$endgroup$
– billyandr
yesterday
$begingroup$
You are welcome. This "not seeing the obvious" just happens once in a while, I think, to all who do maths. So, it is good to have a math platform like this one. :-)
$endgroup$
– trancelocation
yesterday
add a comment |
$begingroup$
It is rather obvious if you cancel the factorials:
$$frac{n!}{n^{k}(n-k)! } =frac{overbrace{n(n-1)cdots (n-k+1)}^{k; factors}}{n^k}= 1cdot left(1-frac{1}{n}right)cdots left(1-frac{k-1}{n}right)stackrel{n to infty}{longrightarrow} 1$$
answered yesterday
trancelocationtrancelocation
14.1k1829
$endgroup$
$begingroup$
Thank you so much. I didn't know it was right there under my eyes.
$endgroup$
– billyandr
yesterday
$begingroup$
You are welcome. This "not seeing the obvious" just happens once in a while, I think, to all who do maths. So, it is good to have a math platform like this one. :-)
$endgroup$
– trancelocation
yesterday
add a comment |
$begingroup$
It is rather obvious if you cancel the factorials:
$$frac{n!}{n^{k}(n-k)! } =frac{overbrace{n(n-1)cdots (n-k+1)}^{k; factors}}{n^k}= 1cdot left(1-frac{1}{n}right)cdots left(1-frac{k-1}{n}right)stackrel{n to infty}{longrightarrow} 1$$
answered yesterday
trancelocationtrancelocation
14.1k1829
$endgroup$
It is rather obvious if you cancel the factorials:
$$frac{n!}{n^{k}(n-k)! } =frac{overbrace{n(n-1)cdots (n-k+1)}^{k; factors}}{n^k}= 1cdot left(1-frac{1}{n}right)cdots left(1-frac{k-1}{n}right)stackrel{n to infty}{longrightarrow} 1$$
answered yesterday
trancelocationtrancelocation
14.1k1829
answered yesterday
trancelocationtrancelocation
14.1k1829
answered yesterday
trancelocationtrancelocation
14.1k1829
answered yesterday
trancelocationtrancelocation
14.1k1829
14.1k1829
$begingroup$
Thank you so much. I didn't know it was right there under my eyes.
$endgroup$
– billyandr
yesterday
$begingroup$
You are welcome. This "not seeing the obvious" just happens once in a while, I think, to all who do maths. So, it is good to have a math platform like this one. :-)
$endgroup$
– trancelocation
yesterday
add a comment |
$begingroup$
Thank you so much. I didn't know it was right there under my eyes.
$endgroup$
– billyandr
yesterday
$begingroup$
You are welcome. This "not seeing the obvious" just happens once in a while, I think, to all who do maths. So, it is good to have a math platform like this one. :-)
$endgroup$
– trancelocation
yesterday
$begingroup$
Thank you so much. I didn't know it was right there under my eyes.
$endgroup$
– billyandr
yesterday
$begingroup$
Thank you so much. I didn't know it was right there under my eyes.
$endgroup$
– billyandr
yesterday
$begingroup$
You are welcome. This "not seeing the obvious" just happens once in a while, I think, to all who do maths. So, it is good to have a math platform like this one. :-)
$endgroup$
– trancelocation
yesterday
$begingroup$
You are welcome. This "not seeing the obvious" just happens once in a while, I think, to all who do maths. So, it is good to have a math platform like this one. :-)
$endgroup$
– trancelocation
yesterday
add a comment |
$begingroup$
$$a_n=frac{n!}{n^{k}(n-k)! }implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$
Use Stirling approximation and continue with Taylor series to get
$$log(a_n)=frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$ Continue with Taylor
$$a_n=e^{log(a_n)}=1+frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$
answered yesterday
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
1
$begingroup$
This has already a slight touch of overkill, hasn't it? :-)
$endgroup$
– trancelocation
yesterday
$begingroup$
@trancelocation. You are totally right for the limit. One of my manias is to always look at the approach to the limit. Have a look at matheducators.stackexchange.com/questions/8339/… . Cheers :-)
$endgroup$
– Claude Leibovici
yesterday
add a comment |
$begingroup$
$$a_n=frac{n!}{n^{k}(n-k)! }implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$
Use Stirling approximation and continue with Taylor series to get
$$log(a_n)=frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$ Continue with Taylor
$$a_n=e^{log(a_n)}=1+frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$
answered yesterday
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
1
$begingroup$
This has already a slight touch of overkill, hasn't it? :-)
$endgroup$
– trancelocation
yesterday
$begingroup$
@trancelocation. You are totally right for the limit. One of my manias is to always look at the approach to the limit. Have a look at matheducators.stackexchange.com/questions/8339/… . Cheers :-)
$endgroup$
– Claude Leibovici
yesterday
add a comment |
$begingroup$
$$a_n=frac{n!}{n^{k}(n-k)! }implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$
Use Stirling approximation and continue with Taylor series to get
$$log(a_n)=frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$ Continue with Taylor
$$a_n=e^{log(a_n)}=1+frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$
answered yesterday
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
$$a_n=frac{n!}{n^{k}(n-k)! }implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$
Use Stirling approximation and continue with Taylor series to get
$$log(a_n)=frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$ Continue with Taylor
$$a_n=e^{log(a_n)}=1+frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$
answered yesterday
Claude LeiboviciClaude Leibovici
126k1158135
answered yesterday
Claude LeiboviciClaude Leibovici
126k1158135
answered yesterday
Claude LeiboviciClaude Leibovici
126k1158135
answered yesterday
Claude LeiboviciClaude Leibovici
126k1158135
126k1158135
1
$begingroup$
This has already a slight touch of overkill, hasn't it? :-)
$endgroup$
– trancelocation
yesterday
$begingroup$
@trancelocation. You are totally right for the limit. One of my manias is to always look at the approach to the limit. Have a look at matheducators.stackexchange.com/questions/8339/… . Cheers :-)
$endgroup$
– Claude Leibovici
yesterday
add a comment |
1
$begingroup$
This has already a slight touch of overkill, hasn't it? :-)
$endgroup$
– trancelocation
yesterday
$begingroup$
@trancelocation. You are totally right for the limit. One of my manias is to always look at the approach to the limit. Have a look at matheducators.stackexchange.com/questions/8339/… . Cheers :-)
$endgroup$
– Claude Leibovici
yesterday
1
1
$begingroup$
This has already a slight touch of overkill, hasn't it? :-)
$endgroup$
– trancelocation
yesterday
$begingroup$
This has already a slight touch of overkill, hasn't it? :-)
$endgroup$
– trancelocation
yesterday
$begingroup$
@trancelocation. You are totally right for the limit. One of my manias is to always look at the approach to the limit. Have a look at matheducators.stackexchange.com/questions/8339/… . Cheers :-)
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– Claude Leibovici
yesterday
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@trancelocation. You are totally right for the limit. One of my manias is to always look at the approach to the limit. Have a look at matheducators.stackexchange.com/questions/8339/… . Cheers :-)
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– Claude Leibovici
yesterday
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There are several posts about this: Why does $lim_{ntoinfty} frac{n!}{(n-k)!n^k}$ equal 1, Finding limit of sequence: $lim _{n to infty} {frac{n!}{n^k(n-k)!}}=1$, Proof that $limlimits_{h to infty} frac{h!}{h^k(h-k)!}=1$ for any $k$, Limits involing Factorials $lim_{Ntoinfty} frac{N!}{(N-k)!N^{k}}$
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– Martin Sleziak
yesterday
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I found the posts in the above comment using Approach0. For some useful tips on searching here see: How to search on this site?
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– Martin Sleziak
yesterday
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Some of the other posts treating the same question painfully lack details and context. Maybe you'd want to put them on hold or close them.
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– billyandr
yesterday
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billyandr: If you actually have a look at those links, you can see that two of those posts are closed (as duplicates) now. Let me also say that the fact that you have added some more context to your question is certainly appreciated. (After all, that's what lead to reopening.)
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– Martin Sleziak
yesterday