Solve for $k$ if $beta^kk!ge (1-alpha)/alpha$.
Let $beta$ be a constant and $alphain (0,1]$. I want to show that for any $alphain (0,1]$ (no matter how small) there exists $kinmathbb N$ such that $$beta^kk!ge (1-alpha)/alpha$$
I used that $beta^kk!=(beta^{-k}/k!)^{-1}$ and hence, since $x^k/k!to 0$ as $ktoinfty$ for any $xin mathbb R$ (from the summation property of the exponential series), I obtained that such a $k$ always exists.
My question is whether I can solve the above inequality for $k$ and derive a statement using for instance the big $mathcal O$ notation, e.g., something like $kin mathcal O(1/alpha)$. Any ideas? Thank you.
inequality asymptotics
asked Nov 29 at 5:06
Jimmy R.
33k42157
add a comment |
Let $beta$ be a constant and $alphain (0,1]$. I want to show that for any $alphain (0,1]$ (no matter how small) there exists $kinmathbb N$ such that $$beta^kk!ge (1-alpha)/alpha$$
I used that $beta^kk!=(beta^{-k}/k!)^{-1}$ and hence, since $x^k/k!to 0$ as $ktoinfty$ for any $xin mathbb R$ (from the summation property of the exponential series), I obtained that such a $k$ always exists.
My question is whether I can solve the above inequality for $k$ and derive a statement using for instance the big $mathcal O$ notation, e.g., something like $kin mathcal O(1/alpha)$. Any ideas? Thank you.
inequality asymptotics
asked Nov 29 at 5:06
Jimmy R.
33k42157
add a comment |
Let $beta$ be a constant and $alphain (0,1]$. I want to show that for any $alphain (0,1]$ (no matter how small) there exists $kinmathbb N$ such that $$beta^kk!ge (1-alpha)/alpha$$
I used that $beta^kk!=(beta^{-k}/k!)^{-1}$ and hence, since $x^k/k!to 0$ as $ktoinfty$ for any $xin mathbb R$ (from the summation property of the exponential series), I obtained that such a $k$ always exists.
My question is whether I can solve the above inequality for $k$ and derive a statement using for instance the big $mathcal O$ notation, e.g., something like $kin mathcal O(1/alpha)$. Any ideas? Thank you.
inequality asymptotics
asked Nov 29 at 5:06
Jimmy R.
33k42157
Let $beta$ be a constant and $alphain (0,1]$. I want to show that for any $alphain (0,1]$ (no matter how small) there exists $kinmathbb N$ such that $$beta^kk!ge (1-alpha)/alpha$$
I used that $beta^kk!=(beta^{-k}/k!)^{-1}$ and hence, since $x^k/k!to 0$ as $ktoinfty$ for any $xin mathbb R$ (from the summation property of the exponential series), I obtained that such a $k$ always exists.
My question is whether I can solve the above inequality for $k$ and derive a statement using for instance the big $mathcal O$ notation, e.g., something like $kin mathcal O(1/alpha)$. Any ideas? Thank you.
inequality asymptotics
inequality asymptotics
asked Nov 29 at 5:06
Jimmy R.
33k42157
asked Nov 29 at 5:06
Jimmy R.
33k42157
edited Nov 29 at 8:32
asked Nov 29 at 5:06
Jimmy R.
33k42157
asked Nov 29 at 5:06
Jimmy R.
33k42157
asked Nov 29 at 5:06
Jimmy R.
33k42157
33k42157
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
From
$b^kk!ge (1-a)/a
$
we get
$kln b +ln(k!) ge c$
where
$c = ln((1-a)/a)
$.
Since
$0 < a < 1$,
$0 < (1-a)/a
=1/a-1$
so
$c$ can be any real.
To get an
approximate case of equality,
for a first step use
$ln(k!) > kln k - k$.
Then if
$kln b+kln k - k
ge c$,
$k$ is ok.
Write this as
$c
le k(ln(b)-1) + kln(k)
=k(ln(k)+ln(b)-1)
=k(ln(kb/e))
$
or
$cb/e
le (kb/e)(ln(kb/e))
$.
Letting
$r = cb/e$
and
$x = kb/e$,
this becomes
$r le xln(x)$.
The problem of inverting
this equation has been well studied.
As a first approximation,
$x = r/ln(r)$.
This becomes
$kb/e
approx dfrac{cb/e}{ln(cb/e)}
$
or
$k
approx dfrac{c}{ln(cb/e)}
$.
That's all.
answered Nov 29 at 5:23
marty cohen
72.3k549127
+1 Thank you very much!
– Jimmy R.
Nov 29 at 5:45
So, from the last equation is it correct to infer that $k=mathcal O(c(ln c)^{-1})$?
– Jimmy R.
Nov 29 at 5:51
1
If $c$ is large, which occurs when $a$ is small, then yes.
– marty cohen
Nov 29 at 5:58
add a comment |
If you have a look at this question of mine, you will see a magnificent approximation proposed by @robjohn.
Adapted to the problem $beta^k,k!=A$ with $ A>0$, the approximation of $k$ would be given by
$$ksim e beta, e^{W(t)}-frac 12=frac{t e beta}{W(t) }-frac 12 qquad text{where}qquad t=frac{log left(frac{A^2}{2 pi beta }right)}{2 e beta }$$ and, in the real domain, Lambert function $W(t)$ exists as long as $t ge-frac 1e$.
answered Nov 29 at 6:07
Claude Leibovici
118k1157132
It would be interesting to compare our two answers. I'd have to look up the asymptotics of $W(t)$.
– marty cohen
Nov 29 at 6:34
@martycohen. I would like to say that, from the time robjohn provided this approximation (almost two years ago), I used it a lot of times (including in my answers here on MSE) and it is incredibly very good. I suppose that you noticed that, using $beta=1$ we get the solution of $n!=K$.
– Claude Leibovici
Nov 29 at 6:45
+1 looks great! Thank you very much!
– Jimmy R.
Nov 29 at 8:11
1
@JimmyR. Trust me : it is not great ! It is super great. I use it very often in my research and a lot of times in my answers here.
– Claude Leibovici
Nov 29 at 8:56
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
From
$b^kk!ge (1-a)/a
$
we get
$kln b +ln(k!) ge c$
where
$c = ln((1-a)/a)
$.
Since
$0 < a < 1$,
$0 < (1-a)/a
=1/a-1$
so
$c$ can be any real.
To get an
approximate case of equality,
for a first step use
$ln(k!) > kln k - k$.
Then if
$kln b+kln k - k
ge c$,
$k$ is ok.
Write this as
$c
le k(ln(b)-1) + kln(k)
=k(ln(k)+ln(b)-1)
=k(ln(kb/e))
$
or
$cb/e
le (kb/e)(ln(kb/e))
$.
Letting
$r = cb/e$
and
$x = kb/e$,
this becomes
$r le xln(x)$.
The problem of inverting
this equation has been well studied.
As a first approximation,
$x = r/ln(r)$.
This becomes
$kb/e
approx dfrac{cb/e}{ln(cb/e)}
$
or
$k
approx dfrac{c}{ln(cb/e)}
$.
That's all.
answered Nov 29 at 5:23
marty cohen
72.3k549127
+1 Thank you very much!
– Jimmy R.
Nov 29 at 5:45
So, from the last equation is it correct to infer that $k=mathcal O(c(ln c)^{-1})$?
– Jimmy R.
Nov 29 at 5:51
1
If $c$ is large, which occurs when $a$ is small, then yes.
– marty cohen
Nov 29 at 5:58
add a comment |
From
$b^kk!ge (1-a)/a
$
we get
$kln b +ln(k!) ge c$
where
$c = ln((1-a)/a)
$.
Since
$0 < a < 1$,
$0 < (1-a)/a
=1/a-1$
so
$c$ can be any real.
To get an
approximate case of equality,
for a first step use
$ln(k!) > kln k - k$.
Then if
$kln b+kln k - k
ge c$,
$k$ is ok.
Write this as
$c
le k(ln(b)-1) + kln(k)
=k(ln(k)+ln(b)-1)
=k(ln(kb/e))
$
or
$cb/e
le (kb/e)(ln(kb/e))
$.
Letting
$r = cb/e$
and
$x = kb/e$,
this becomes
$r le xln(x)$.
The problem of inverting
this equation has been well studied.
As a first approximation,
$x = r/ln(r)$.
This becomes
$kb/e
approx dfrac{cb/e}{ln(cb/e)}
$
or
$k
approx dfrac{c}{ln(cb/e)}
$.
That's all.
answered Nov 29 at 5:23
marty cohen
72.3k549127
+1 Thank you very much!
– Jimmy R.
Nov 29 at 5:45
So, from the last equation is it correct to infer that $k=mathcal O(c(ln c)^{-1})$?
– Jimmy R.
Nov 29 at 5:51
1
If $c$ is large, which occurs when $a$ is small, then yes.
– marty cohen
Nov 29 at 5:58
add a comment |
From
$b^kk!ge (1-a)/a
$
we get
$kln b +ln(k!) ge c$
where
$c = ln((1-a)/a)
$.
Since
$0 < a < 1$,
$0 < (1-a)/a
=1/a-1$
so
$c$ can be any real.
To get an
approximate case of equality,
for a first step use
$ln(k!) > kln k - k$.
Then if
$kln b+kln k - k
ge c$,
$k$ is ok.
Write this as
$c
le k(ln(b)-1) + kln(k)
=k(ln(k)+ln(b)-1)
=k(ln(kb/e))
$
or
$cb/e
le (kb/e)(ln(kb/e))
$.
Letting
$r = cb/e$
and
$x = kb/e$,
this becomes
$r le xln(x)$.
The problem of inverting
this equation has been well studied.
As a first approximation,
$x = r/ln(r)$.
This becomes
$kb/e
approx dfrac{cb/e}{ln(cb/e)}
$
or
$k
approx dfrac{c}{ln(cb/e)}
$.
That's all.
answered Nov 29 at 5:23
marty cohen
72.3k549127
From
$b^kk!ge (1-a)/a
$
we get
$kln b +ln(k!) ge c$
where
$c = ln((1-a)/a)
$.
Since
$0 < a < 1$,
$0 < (1-a)/a
=1/a-1$
so
$c$ can be any real.
To get an
approximate case of equality,
for a first step use
$ln(k!) > kln k - k$.
Then if
$kln b+kln k - k
ge c$,
$k$ is ok.
Write this as
$c
le k(ln(b)-1) + kln(k)
=k(ln(k)+ln(b)-1)
=k(ln(kb/e))
$
or
$cb/e
le (kb/e)(ln(kb/e))
$.
Letting
$r = cb/e$
and
$x = kb/e$,
this becomes
$r le xln(x)$.
The problem of inverting
this equation has been well studied.
As a first approximation,
$x = r/ln(r)$.
This becomes
$kb/e
approx dfrac{cb/e}{ln(cb/e)}
$
or
$k
approx dfrac{c}{ln(cb/e)}
$.
That's all.
answered Nov 29 at 5:23
marty cohen
72.3k549127
answered Nov 29 at 5:23
marty cohen
72.3k549127
answered Nov 29 at 5:23
marty cohen
72.3k549127
answered Nov 29 at 5:23
marty cohen
72.3k549127
72.3k549127
+1 Thank you very much!
– Jimmy R.
Nov 29 at 5:45
So, from the last equation is it correct to infer that $k=mathcal O(c(ln c)^{-1})$?
– Jimmy R.
Nov 29 at 5:51
1
If $c$ is large, which occurs when $a$ is small, then yes.
– marty cohen
Nov 29 at 5:58
add a comment |
+1 Thank you very much!
– Jimmy R.
Nov 29 at 5:45
So, from the last equation is it correct to infer that $k=mathcal O(c(ln c)^{-1})$?
– Jimmy R.
Nov 29 at 5:51
1
If $c$ is large, which occurs when $a$ is small, then yes.
– marty cohen
Nov 29 at 5:58
+1 Thank you very much!
– Jimmy R.
Nov 29 at 5:45
+1 Thank you very much!
– Jimmy R.
Nov 29 at 5:45
So, from the last equation is it correct to infer that $k=mathcal O(c(ln c)^{-1})$?
– Jimmy R.
Nov 29 at 5:51
So, from the last equation is it correct to infer that $k=mathcal O(c(ln c)^{-1})$?
– Jimmy R.
Nov 29 at 5:51
1
1
If $c$ is large, which occurs when $a$ is small, then yes.
– marty cohen
Nov 29 at 5:58
If $c$ is large, which occurs when $a$ is small, then yes.
– marty cohen
Nov 29 at 5:58
add a comment |
If you have a look at this question of mine, you will see a magnificent approximation proposed by @robjohn.
Adapted to the problem $beta^k,k!=A$ with $ A>0$, the approximation of $k$ would be given by
$$ksim e beta, e^{W(t)}-frac 12=frac{t e beta}{W(t) }-frac 12 qquad text{where}qquad t=frac{log left(frac{A^2}{2 pi beta }right)}{2 e beta }$$ and, in the real domain, Lambert function $W(t)$ exists as long as $t ge-frac 1e$.
answered Nov 29 at 6:07
Claude Leibovici
118k1157132
It would be interesting to compare our two answers. I'd have to look up the asymptotics of $W(t)$.
– marty cohen
Nov 29 at 6:34
@martycohen. I would like to say that, from the time robjohn provided this approximation (almost two years ago), I used it a lot of times (including in my answers here on MSE) and it is incredibly very good. I suppose that you noticed that, using $beta=1$ we get the solution of $n!=K$.
– Claude Leibovici
Nov 29 at 6:45
+1 looks great! Thank you very much!
– Jimmy R.
Nov 29 at 8:11
1
@JimmyR. Trust me : it is not great ! It is super great. I use it very often in my research and a lot of times in my answers here.
– Claude Leibovici
Nov 29 at 8:56
add a comment |
If you have a look at this question of mine, you will see a magnificent approximation proposed by @robjohn.
Adapted to the problem $beta^k,k!=A$ with $ A>0$, the approximation of $k$ would be given by
$$ksim e beta, e^{W(t)}-frac 12=frac{t e beta}{W(t) }-frac 12 qquad text{where}qquad t=frac{log left(frac{A^2}{2 pi beta }right)}{2 e beta }$$ and, in the real domain, Lambert function $W(t)$ exists as long as $t ge-frac 1e$.
answered Nov 29 at 6:07
Claude Leibovici
118k1157132
It would be interesting to compare our two answers. I'd have to look up the asymptotics of $W(t)$.
– marty cohen
Nov 29 at 6:34
@martycohen. I would like to say that, from the time robjohn provided this approximation (almost two years ago), I used it a lot of times (including in my answers here on MSE) and it is incredibly very good. I suppose that you noticed that, using $beta=1$ we get the solution of $n!=K$.
– Claude Leibovici
Nov 29 at 6:45
+1 looks great! Thank you very much!
– Jimmy R.
Nov 29 at 8:11
1
@JimmyR. Trust me : it is not great ! It is super great. I use it very often in my research and a lot of times in my answers here.
– Claude Leibovici
Nov 29 at 8:56
add a comment |
If you have a look at this question of mine, you will see a magnificent approximation proposed by @robjohn.
Adapted to the problem $beta^k,k!=A$ with $ A>0$, the approximation of $k$ would be given by
$$ksim e beta, e^{W(t)}-frac 12=frac{t e beta}{W(t) }-frac 12 qquad text{where}qquad t=frac{log left(frac{A^2}{2 pi beta }right)}{2 e beta }$$ and, in the real domain, Lambert function $W(t)$ exists as long as $t ge-frac 1e$.
answered Nov 29 at 6:07
Claude Leibovici
118k1157132
If you have a look at this question of mine, you will see a magnificent approximation proposed by @robjohn.
Adapted to the problem $beta^k,k!=A$ with $ A>0$, the approximation of $k$ would be given by
$$ksim e beta, e^{W(t)}-frac 12=frac{t e beta}{W(t) }-frac 12 qquad text{where}qquad t=frac{log left(frac{A^2}{2 pi beta }right)}{2 e beta }$$ and, in the real domain, Lambert function $W(t)$ exists as long as $t ge-frac 1e$.
answered Nov 29 at 6:07
Claude Leibovici
118k1157132
edited Nov 29 at 7:32
answered Nov 29 at 6:07
Claude Leibovici
118k1157132
answered Nov 29 at 6:07
Claude Leibovici
118k1157132
answered Nov 29 at 6:07
Claude Leibovici
118k1157132
118k1157132
It would be interesting to compare our two answers. I'd have to look up the asymptotics of $W(t)$.
– marty cohen
Nov 29 at 6:34
@martycohen. I would like to say that, from the time robjohn provided this approximation (almost two years ago), I used it a lot of times (including in my answers here on MSE) and it is incredibly very good. I suppose that you noticed that, using $beta=1$ we get the solution of $n!=K$.
– Claude Leibovici
Nov 29 at 6:45
+1 looks great! Thank you very much!
– Jimmy R.
Nov 29 at 8:11
1
@JimmyR. Trust me : it is not great ! It is super great. I use it very often in my research and a lot of times in my answers here.
– Claude Leibovici
Nov 29 at 8:56
add a comment |
It would be interesting to compare our two answers. I'd have to look up the asymptotics of $W(t)$.
– marty cohen
Nov 29 at 6:34
@martycohen. I would like to say that, from the time robjohn provided this approximation (almost two years ago), I used it a lot of times (including in my answers here on MSE) and it is incredibly very good. I suppose that you noticed that, using $beta=1$ we get the solution of $n!=K$.
– Claude Leibovici
Nov 29 at 6:45
+1 looks great! Thank you very much!
– Jimmy R.
Nov 29 at 8:11
1
@JimmyR. Trust me : it is not great ! It is super great. I use it very often in my research and a lot of times in my answers here.
– Claude Leibovici
Nov 29 at 8:56
It would be interesting to compare our two answers. I'd have to look up the asymptotics of $W(t)$.
– marty cohen
Nov 29 at 6:34
It would be interesting to compare our two answers. I'd have to look up the asymptotics of $W(t)$.
– marty cohen
Nov 29 at 6:34
@martycohen. I would like to say that, from the time robjohn provided this approximation (almost two years ago), I used it a lot of times (including in my answers here on MSE) and it is incredibly very good. I suppose that you noticed that, using $beta=1$ we get the solution of $n!=K$.
– Claude Leibovici
Nov 29 at 6:45
@martycohen. I would like to say that, from the time robjohn provided this approximation (almost two years ago), I used it a lot of times (including in my answers here on MSE) and it is incredibly very good. I suppose that you noticed that, using $beta=1$ we get the solution of $n!=K$.
– Claude Leibovici
Nov 29 at 6:45
+1 looks great! Thank you very much!
– Jimmy R.
Nov 29 at 8:11
+1 looks great! Thank you very much!
– Jimmy R.
Nov 29 at 8:11
1
1
@JimmyR. Trust me : it is not great ! It is super great. I use it very often in my research and a lot of times in my answers here.
– Claude Leibovici
Nov 29 at 8:56
@JimmyR. Trust me : it is not great ! It is super great. I use it very often in my research and a lot of times in my answers here.
– Claude Leibovici
Nov 29 at 8:56
add a comment |
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