Understanding a very elementary property of factorials
I've seen this stated in a few places.
If $$vartheta(x) = sum_{ple{x}} log (p) qquad psi(x) = sum_{m=1}^{infty}varthetaleft(sqrt[m]{x}right)$$ Then $$log(x!) = sum_{m=1}^{infty} psileft(frac{x}{m}right).$$
It is used by Ramanujan here. It is used by Jitsuro Nagura here.
Can anyone provide a proof for why it is true or provide a link to a proof?
Thanks very much.
number-theory analytic-number-theory factorial
edited Mar 22 '13 at 21:31
anon
71.8k5108211
asked Mar 22 '13 at 20:57
Larry Freeman
3,23421239
add a comment |
I've seen this stated in a few places.
If $$vartheta(x) = sum_{ple{x}} log (p) qquad psi(x) = sum_{m=1}^{infty}varthetaleft(sqrt[m]{x}right)$$ Then $$log(x!) = sum_{m=1}^{infty} psileft(frac{x}{m}right).$$
It is used by Ramanujan here. It is used by Jitsuro Nagura here.
Can anyone provide a proof for why it is true or provide a link to a proof?
Thanks very much.
number-theory analytic-number-theory factorial
edited Mar 22 '13 at 21:31
anon
71.8k5108211
asked Mar 22 '13 at 20:57
Larry Freeman
3,23421239
add a comment |
I've seen this stated in a few places.
If $$vartheta(x) = sum_{ple{x}} log (p) qquad psi(x) = sum_{m=1}^{infty}varthetaleft(sqrt[m]{x}right)$$ Then $$log(x!) = sum_{m=1}^{infty} psileft(frac{x}{m}right).$$
It is used by Ramanujan here. It is used by Jitsuro Nagura here.
Can anyone provide a proof for why it is true or provide a link to a proof?
Thanks very much.
number-theory analytic-number-theory factorial
edited Mar 22 '13 at 21:31
anon
71.8k5108211
asked Mar 22 '13 at 20:57
Larry Freeman
3,23421239
I've seen this stated in a few places.
If $$vartheta(x) = sum_{ple{x}} log (p) qquad psi(x) = sum_{m=1}^{infty}varthetaleft(sqrt[m]{x}right)$$ Then $$log(x!) = sum_{m=1}^{infty} psileft(frac{x}{m}right).$$
It is used by Ramanujan here. It is used by Jitsuro Nagura here.
Can anyone provide a proof for why it is true or provide a link to a proof?
Thanks very much.
number-theory analytic-number-theory factorial
number-theory analytic-number-theory factorial
edited Mar 22 '13 at 21:31
anon
71.8k5108211
asked Mar 22 '13 at 20:57
Larry Freeman
3,23421239
edited Mar 22 '13 at 21:31
anon
71.8k5108211
asked Mar 22 '13 at 20:57
Larry Freeman
3,23421239
edited Mar 22 '13 at 21:31
anon
71.8k5108211
edited Mar 22 '13 at 21:31
anon
71.8k5108211
edited Mar 22 '13 at 21:31
anon
71.8k5108211
71.8k5108211
asked Mar 22 '13 at 20:57
Larry Freeman
3,23421239
asked Mar 22 '13 at 20:57
Larry Freeman
3,23421239
asked Mar 22 '13 at 20:57
Larry Freeman
3,23421239
3,23421239
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
We find
$$sum_{mge1}psileft(frac{x}{m}right)=sum_{mge1}sum_{kge1}varthetaleft(sqrt[k]{frac{x}{m}}right)=sum_{mge1}sum_{kge1}sum_{ple sqrt[k]{x/m}}log p$$
$$=sum_{mge1}sum_{kge1}sum_{mp^kle x}log p=log prod_{ple x}p^{#{(m,k):mp^kle x}}=log x!$$
since when counting $#{(m,k):mp^kle x}$, one sees for every $1le nle x$ there are $t=v_p(n)$ different tuples $(m,t),(mp,t-1),cdots,(mp^{t-1},1)$ counted in the set (note $kge1$).
answered Mar 22 '13 at 21:08
anon
71.8k5108211
Could you please expand a bit on your last paragraph? It looks to me like you're counting only the pairs $(m,v_p(n))$ with $m = 1$.
– A.P.
Apr 28 '15 at 9:13
@A.P. Sorry, it's been two years so I don't quite remember what went wrong, but I suspect my brain might have went dark when typing out what I saw in my head. I've edited to give the correct list of tuples corresponding to each $1le nle x$. Basically, compute the sizes of the fibers of the obvious map ${(m,k):mp^kle x}to{n:1le nle x}$ given by $(m,k)mapsto mp^k$ then sum the fibers' sizes.
– anon
Apr 28 '15 at 10:56
1
Thank you, now I understand. It still took me a while to parse your wording, so for the benefit of other here is how I see it: for every positive integer $n$ there are exactly $v_p(n)$ ways of writing $n$ as $m p^t$ with $m in Bbb{Z}$ and $t geq 1$.
– A.P.
Apr 28 '15 at 11:06
add a comment |
A slightly different variation based upon the prime factorisation of $x!$.
We obtain for integers $x>0$
begin{align*}
color{blue}{log x!}&=logprod_{pleq x}p^{leftlfloorfrac{x}{p}rightrfloor+leftlfloorfrac{x}{p^2}rightrfloor+cdots}tag{1}\
&=logprod_{pleq x}p^{sum_{m=1}^inftyleftlfloorfrac{x}{p^m}rightrfloor}tag{2}\
&=sum_{pleq x}sum_{m=1}^inftylog pleftlfloorfrac{x}{p^m}rightrfloortag{3}\
&=sum_{m=1}^inftysum_{pleq sqrt[m]{x}}log pleftlfloorfrac{x}{p^m}rightrfloortag{4}\
&=sum_{m=1}^inftysum_{pleq sqrt[m]{x}}log psum_{j=1}^{x/p^m}1tag{5}\
&=sum_{m=1}^inftysum_{j=1}^xsum_{pleq sqrt[m]{x}}log ptag{6}\
&,,color{blue}{=sum_{j=1}^xpsileft(frac{x}{j}right)}
end{align*}
and the claim follows.
Comment:
In (1) we do the prime factorisation of $x!$ and observe that $leftlfloorfrac{x}{p}rightrfloor$ counts the numbers $leq x$ which are a multiple of $p$. Since we also have to add $1$ for each number which is a multiple of $p^2$ we add $leftlfloorfrac{x}{p^2}rightrfloor$, etc.
In (2) we use the series notation. Note the series is finite.
In (3) we use properties of the logarithm.
In (4) we exchange the order of the series.
In (5) we write the factor $leftlfloorfrac{x}{p^m}rightrfloor$ as sum.
In (6) we exchange the order of the series again.
answered Dec 1 at 22:10
Markus Scheuer
60k455143
+1. It's a nice job.
– Felix Marin
Dec 11 at 4:07
1
@FelixMarin: Thanks Felix. This answer follows a proof which was given by E. Landau in 1909.
– Markus Scheuer
Dec 11 at 5:11
I didn't know that. Thanks because everyday we learn something new.
– Felix Marin
Dec 11 at 16:22
1
@FelixMarin: You're welcome. The same holds true for me. :-)
– Markus Scheuer
Dec 11 at 16:29
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f338295%2funderstanding-a-very-elementary-property-of-factorials%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
We find
$$sum_{mge1}psileft(frac{x}{m}right)=sum_{mge1}sum_{kge1}varthetaleft(sqrt[k]{frac{x}{m}}right)=sum_{mge1}sum_{kge1}sum_{ple sqrt[k]{x/m}}log p$$
$$=sum_{mge1}sum_{kge1}sum_{mp^kle x}log p=log prod_{ple x}p^{#{(m,k):mp^kle x}}=log x!$$
since when counting $#{(m,k):mp^kle x}$, one sees for every $1le nle x$ there are $t=v_p(n)$ different tuples $(m,t),(mp,t-1),cdots,(mp^{t-1},1)$ counted in the set (note $kge1$).
answered Mar 22 '13 at 21:08
anon
71.8k5108211
Could you please expand a bit on your last paragraph? It looks to me like you're counting only the pairs $(m,v_p(n))$ with $m = 1$.
– A.P.
Apr 28 '15 at 9:13
@A.P. Sorry, it's been two years so I don't quite remember what went wrong, but I suspect my brain might have went dark when typing out what I saw in my head. I've edited to give the correct list of tuples corresponding to each $1le nle x$. Basically, compute the sizes of the fibers of the obvious map ${(m,k):mp^kle x}to{n:1le nle x}$ given by $(m,k)mapsto mp^k$ then sum the fibers' sizes.
– anon
Apr 28 '15 at 10:56
1
Thank you, now I understand. It still took me a while to parse your wording, so for the benefit of other here is how I see it: for every positive integer $n$ there are exactly $v_p(n)$ ways of writing $n$ as $m p^t$ with $m in Bbb{Z}$ and $t geq 1$.
– A.P.
Apr 28 '15 at 11:06
add a comment |
We find
$$sum_{mge1}psileft(frac{x}{m}right)=sum_{mge1}sum_{kge1}varthetaleft(sqrt[k]{frac{x}{m}}right)=sum_{mge1}sum_{kge1}sum_{ple sqrt[k]{x/m}}log p$$
$$=sum_{mge1}sum_{kge1}sum_{mp^kle x}log p=log prod_{ple x}p^{#{(m,k):mp^kle x}}=log x!$$
since when counting $#{(m,k):mp^kle x}$, one sees for every $1le nle x$ there are $t=v_p(n)$ different tuples $(m,t),(mp,t-1),cdots,(mp^{t-1},1)$ counted in the set (note $kge1$).
answered Mar 22 '13 at 21:08
anon
71.8k5108211
Could you please expand a bit on your last paragraph? It looks to me like you're counting only the pairs $(m,v_p(n))$ with $m = 1$.
– A.P.
Apr 28 '15 at 9:13
@A.P. Sorry, it's been two years so I don't quite remember what went wrong, but I suspect my brain might have went dark when typing out what I saw in my head. I've edited to give the correct list of tuples corresponding to each $1le nle x$. Basically, compute the sizes of the fibers of the obvious map ${(m,k):mp^kle x}to{n:1le nle x}$ given by $(m,k)mapsto mp^k$ then sum the fibers' sizes.
– anon
Apr 28 '15 at 10:56
1
Thank you, now I understand. It still took me a while to parse your wording, so for the benefit of other here is how I see it: for every positive integer $n$ there are exactly $v_p(n)$ ways of writing $n$ as $m p^t$ with $m in Bbb{Z}$ and $t geq 1$.
– A.P.
Apr 28 '15 at 11:06
add a comment |
We find
$$sum_{mge1}psileft(frac{x}{m}right)=sum_{mge1}sum_{kge1}varthetaleft(sqrt[k]{frac{x}{m}}right)=sum_{mge1}sum_{kge1}sum_{ple sqrt[k]{x/m}}log p$$
$$=sum_{mge1}sum_{kge1}sum_{mp^kle x}log p=log prod_{ple x}p^{#{(m,k):mp^kle x}}=log x!$$
since when counting $#{(m,k):mp^kle x}$, one sees for every $1le nle x$ there are $t=v_p(n)$ different tuples $(m,t),(mp,t-1),cdots,(mp^{t-1},1)$ counted in the set (note $kge1$).
answered Mar 22 '13 at 21:08
anon
71.8k5108211
We find
$$sum_{mge1}psileft(frac{x}{m}right)=sum_{mge1}sum_{kge1}varthetaleft(sqrt[k]{frac{x}{m}}right)=sum_{mge1}sum_{kge1}sum_{ple sqrt[k]{x/m}}log p$$
$$=sum_{mge1}sum_{kge1}sum_{mp^kle x}log p=log prod_{ple x}p^{#{(m,k):mp^kle x}}=log x!$$
since when counting $#{(m,k):mp^kle x}$, one sees for every $1le nle x$ there are $t=v_p(n)$ different tuples $(m,t),(mp,t-1),cdots,(mp^{t-1},1)$ counted in the set (note $kge1$).
answered Mar 22 '13 at 21:08
anon
71.8k5108211
edited Apr 28 '15 at 10:55
answered Mar 22 '13 at 21:08
anon
71.8k5108211
answered Mar 22 '13 at 21:08
anon
71.8k5108211
answered Mar 22 '13 at 21:08
anon
71.8k5108211
71.8k5108211
Could you please expand a bit on your last paragraph? It looks to me like you're counting only the pairs $(m,v_p(n))$ with $m = 1$.
– A.P.
Apr 28 '15 at 9:13
@A.P. Sorry, it's been two years so I don't quite remember what went wrong, but I suspect my brain might have went dark when typing out what I saw in my head. I've edited to give the correct list of tuples corresponding to each $1le nle x$. Basically, compute the sizes of the fibers of the obvious map ${(m,k):mp^kle x}to{n:1le nle x}$ given by $(m,k)mapsto mp^k$ then sum the fibers' sizes.
– anon
Apr 28 '15 at 10:56
1
Thank you, now I understand. It still took me a while to parse your wording, so for the benefit of other here is how I see it: for every positive integer $n$ there are exactly $v_p(n)$ ways of writing $n$ as $m p^t$ with $m in Bbb{Z}$ and $t geq 1$.
– A.P.
Apr 28 '15 at 11:06
add a comment |
Could you please expand a bit on your last paragraph? It looks to me like you're counting only the pairs $(m,v_p(n))$ with $m = 1$.
– A.P.
Apr 28 '15 at 9:13
@A.P. Sorry, it's been two years so I don't quite remember what went wrong, but I suspect my brain might have went dark when typing out what I saw in my head. I've edited to give the correct list of tuples corresponding to each $1le nle x$. Basically, compute the sizes of the fibers of the obvious map ${(m,k):mp^kle x}to{n:1le nle x}$ given by $(m,k)mapsto mp^k$ then sum the fibers' sizes.
– anon
Apr 28 '15 at 10:56
1
Thank you, now I understand. It still took me a while to parse your wording, so for the benefit of other here is how I see it: for every positive integer $n$ there are exactly $v_p(n)$ ways of writing $n$ as $m p^t$ with $m in Bbb{Z}$ and $t geq 1$.
– A.P.
Apr 28 '15 at 11:06
Could you please expand a bit on your last paragraph? It looks to me like you're counting only the pairs $(m,v_p(n))$ with $m = 1$.
– A.P.
Apr 28 '15 at 9:13
Could you please expand a bit on your last paragraph? It looks to me like you're counting only the pairs $(m,v_p(n))$ with $m = 1$.
– A.P.
Apr 28 '15 at 9:13
@A.P. Sorry, it's been two years so I don't quite remember what went wrong, but I suspect my brain might have went dark when typing out what I saw in my head. I've edited to give the correct list of tuples corresponding to each $1le nle x$. Basically, compute the sizes of the fibers of the obvious map ${(m,k):mp^kle x}to{n:1le nle x}$ given by $(m,k)mapsto mp^k$ then sum the fibers' sizes.
– anon
Apr 28 '15 at 10:56
@A.P. Sorry, it's been two years so I don't quite remember what went wrong, but I suspect my brain might have went dark when typing out what I saw in my head. I've edited to give the correct list of tuples corresponding to each $1le nle x$. Basically, compute the sizes of the fibers of the obvious map ${(m,k):mp^kle x}to{n:1le nle x}$ given by $(m,k)mapsto mp^k$ then sum the fibers' sizes.
– anon
Apr 28 '15 at 10:56
1
1
Thank you, now I understand. It still took me a while to parse your wording, so for the benefit of other here is how I see it: for every positive integer $n$ there are exactly $v_p(n)$ ways of writing $n$ as $m p^t$ with $m in Bbb{Z}$ and $t geq 1$.
– A.P.
Apr 28 '15 at 11:06
Thank you, now I understand. It still took me a while to parse your wording, so for the benefit of other here is how I see it: for every positive integer $n$ there are exactly $v_p(n)$ ways of writing $n$ as $m p^t$ with $m in Bbb{Z}$ and $t geq 1$.
– A.P.
Apr 28 '15 at 11:06
add a comment |
A slightly different variation based upon the prime factorisation of $x!$.
We obtain for integers $x>0$
begin{align*}
color{blue}{log x!}&=logprod_{pleq x}p^{leftlfloorfrac{x}{p}rightrfloor+leftlfloorfrac{x}{p^2}rightrfloor+cdots}tag{1}\
&=logprod_{pleq x}p^{sum_{m=1}^inftyleftlfloorfrac{x}{p^m}rightrfloor}tag{2}\
&=sum_{pleq x}sum_{m=1}^inftylog pleftlfloorfrac{x}{p^m}rightrfloortag{3}\
&=sum_{m=1}^inftysum_{pleq sqrt[m]{x}}log pleftlfloorfrac{x}{p^m}rightrfloortag{4}\
&=sum_{m=1}^inftysum_{pleq sqrt[m]{x}}log psum_{j=1}^{x/p^m}1tag{5}\
&=sum_{m=1}^inftysum_{j=1}^xsum_{pleq sqrt[m]{x}}log ptag{6}\
&,,color{blue}{=sum_{j=1}^xpsileft(frac{x}{j}right)}
end{align*}
and the claim follows.
Comment:
In (1) we do the prime factorisation of $x!$ and observe that $leftlfloorfrac{x}{p}rightrfloor$ counts the numbers $leq x$ which are a multiple of $p$. Since we also have to add $1$ for each number which is a multiple of $p^2$ we add $leftlfloorfrac{x}{p^2}rightrfloor$, etc.
In (2) we use the series notation. Note the series is finite.
In (3) we use properties of the logarithm.
In (4) we exchange the order of the series.
In (5) we write the factor $leftlfloorfrac{x}{p^m}rightrfloor$ as sum.
In (6) we exchange the order of the series again.
answered Dec 1 at 22:10
Markus Scheuer
60k455143
+1. It's a nice job.
– Felix Marin
Dec 11 at 4:07
1
@FelixMarin: Thanks Felix. This answer follows a proof which was given by E. Landau in 1909.
– Markus Scheuer
Dec 11 at 5:11
I didn't know that. Thanks because everyday we learn something new.
– Felix Marin
Dec 11 at 16:22
1
@FelixMarin: You're welcome. The same holds true for me. :-)
– Markus Scheuer
Dec 11 at 16:29
add a comment |
A slightly different variation based upon the prime factorisation of $x!$.
We obtain for integers $x>0$
begin{align*}
color{blue}{log x!}&=logprod_{pleq x}p^{leftlfloorfrac{x}{p}rightrfloor+leftlfloorfrac{x}{p^2}rightrfloor+cdots}tag{1}\
&=logprod_{pleq x}p^{sum_{m=1}^inftyleftlfloorfrac{x}{p^m}rightrfloor}tag{2}\
&=sum_{pleq x}sum_{m=1}^inftylog pleftlfloorfrac{x}{p^m}rightrfloortag{3}\
&=sum_{m=1}^inftysum_{pleq sqrt[m]{x}}log pleftlfloorfrac{x}{p^m}rightrfloortag{4}\
&=sum_{m=1}^inftysum_{pleq sqrt[m]{x}}log psum_{j=1}^{x/p^m}1tag{5}\
&=sum_{m=1}^inftysum_{j=1}^xsum_{pleq sqrt[m]{x}}log ptag{6}\
&,,color{blue}{=sum_{j=1}^xpsileft(frac{x}{j}right)}
end{align*}
and the claim follows.
Comment:
In (1) we do the prime factorisation of $x!$ and observe that $leftlfloorfrac{x}{p}rightrfloor$ counts the numbers $leq x$ which are a multiple of $p$. Since we also have to add $1$ for each number which is a multiple of $p^2$ we add $leftlfloorfrac{x}{p^2}rightrfloor$, etc.
In (2) we use the series notation. Note the series is finite.
In (3) we use properties of the logarithm.
In (4) we exchange the order of the series.
In (5) we write the factor $leftlfloorfrac{x}{p^m}rightrfloor$ as sum.
In (6) we exchange the order of the series again.
answered Dec 1 at 22:10
Markus Scheuer
60k455143
+1. It's a nice job.
– Felix Marin
Dec 11 at 4:07
1
@FelixMarin: Thanks Felix. This answer follows a proof which was given by E. Landau in 1909.
– Markus Scheuer
Dec 11 at 5:11
I didn't know that. Thanks because everyday we learn something new.
– Felix Marin
Dec 11 at 16:22
1
@FelixMarin: You're welcome. The same holds true for me. :-)
– Markus Scheuer
Dec 11 at 16:29
add a comment |
A slightly different variation based upon the prime factorisation of $x!$.
We obtain for integers $x>0$
begin{align*}
color{blue}{log x!}&=logprod_{pleq x}p^{leftlfloorfrac{x}{p}rightrfloor+leftlfloorfrac{x}{p^2}rightrfloor+cdots}tag{1}\
&=logprod_{pleq x}p^{sum_{m=1}^inftyleftlfloorfrac{x}{p^m}rightrfloor}tag{2}\
&=sum_{pleq x}sum_{m=1}^inftylog pleftlfloorfrac{x}{p^m}rightrfloortag{3}\
&=sum_{m=1}^inftysum_{pleq sqrt[m]{x}}log pleftlfloorfrac{x}{p^m}rightrfloortag{4}\
&=sum_{m=1}^inftysum_{pleq sqrt[m]{x}}log psum_{j=1}^{x/p^m}1tag{5}\
&=sum_{m=1}^inftysum_{j=1}^xsum_{pleq sqrt[m]{x}}log ptag{6}\
&,,color{blue}{=sum_{j=1}^xpsileft(frac{x}{j}right)}
end{align*}
and the claim follows.
Comment:
In (1) we do the prime factorisation of $x!$ and observe that $leftlfloorfrac{x}{p}rightrfloor$ counts the numbers $leq x$ which are a multiple of $p$. Since we also have to add $1$ for each number which is a multiple of $p^2$ we add $leftlfloorfrac{x}{p^2}rightrfloor$, etc.
In (2) we use the series notation. Note the series is finite.
In (3) we use properties of the logarithm.
In (4) we exchange the order of the series.
In (5) we write the factor $leftlfloorfrac{x}{p^m}rightrfloor$ as sum.
In (6) we exchange the order of the series again.
answered Dec 1 at 22:10
Markus Scheuer
60k455143
A slightly different variation based upon the prime factorisation of $x!$.
We obtain for integers $x>0$
begin{align*}
color{blue}{log x!}&=logprod_{pleq x}p^{leftlfloorfrac{x}{p}rightrfloor+leftlfloorfrac{x}{p^2}rightrfloor+cdots}tag{1}\
&=logprod_{pleq x}p^{sum_{m=1}^inftyleftlfloorfrac{x}{p^m}rightrfloor}tag{2}\
&=sum_{pleq x}sum_{m=1}^inftylog pleftlfloorfrac{x}{p^m}rightrfloortag{3}\
&=sum_{m=1}^inftysum_{pleq sqrt[m]{x}}log pleftlfloorfrac{x}{p^m}rightrfloortag{4}\
&=sum_{m=1}^inftysum_{pleq sqrt[m]{x}}log psum_{j=1}^{x/p^m}1tag{5}\
&=sum_{m=1}^inftysum_{j=1}^xsum_{pleq sqrt[m]{x}}log ptag{6}\
&,,color{blue}{=sum_{j=1}^xpsileft(frac{x}{j}right)}
end{align*}
and the claim follows.
Comment:
In (1) we do the prime factorisation of $x!$ and observe that $leftlfloorfrac{x}{p}rightrfloor$ counts the numbers $leq x$ which are a multiple of $p$. Since we also have to add $1$ for each number which is a multiple of $p^2$ we add $leftlfloorfrac{x}{p^2}rightrfloor$, etc.
In (2) we use the series notation. Note the series is finite.
In (3) we use properties of the logarithm.
In (4) we exchange the order of the series.
In (5) we write the factor $leftlfloorfrac{x}{p^m}rightrfloor$ as sum.
In (6) we exchange the order of the series again.
answered Dec 1 at 22:10
Markus Scheuer
60k455143
answered Dec 1 at 22:10
Markus Scheuer
60k455143
answered Dec 1 at 22:10
Markus Scheuer
60k455143
answered Dec 1 at 22:10
Markus Scheuer
60k455143
60k455143
+1. It's a nice job.
– Felix Marin
Dec 11 at 4:07
1
@FelixMarin: Thanks Felix. This answer follows a proof which was given by E. Landau in 1909.
– Markus Scheuer
Dec 11 at 5:11
I didn't know that. Thanks because everyday we learn something new.
– Felix Marin
Dec 11 at 16:22
1
@FelixMarin: You're welcome. The same holds true for me. :-)
– Markus Scheuer
Dec 11 at 16:29
add a comment |
+1. It's a nice job.
– Felix Marin
Dec 11 at 4:07
1
@FelixMarin: Thanks Felix. This answer follows a proof which was given by E. Landau in 1909.
– Markus Scheuer
Dec 11 at 5:11
I didn't know that. Thanks because everyday we learn something new.
– Felix Marin
Dec 11 at 16:22
1
@FelixMarin: You're welcome. The same holds true for me. :-)
– Markus Scheuer
Dec 11 at 16:29
+1. It's a nice job.
– Felix Marin
Dec 11 at 4:07
+1. It's a nice job.
– Felix Marin
Dec 11 at 4:07
1
1
@FelixMarin: Thanks Felix. This answer follows a proof which was given by E. Landau in 1909.
– Markus Scheuer
Dec 11 at 5:11
@FelixMarin: Thanks Felix. This answer follows a proof which was given by E. Landau in 1909.
– Markus Scheuer
Dec 11 at 5:11
I didn't know that. Thanks because everyday we learn something new.
– Felix Marin
Dec 11 at 16:22
I didn't know that. Thanks because everyday we learn something new.
– Felix Marin
Dec 11 at 16:22
1
1
@FelixMarin: You're welcome. The same holds true for me. :-)
– Markus Scheuer
Dec 11 at 16:29
@FelixMarin: You're welcome. The same holds true for me. :-)
– Markus Scheuer
Dec 11 at 16:29
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f338295%2funderstanding-a-very-elementary-property-of-factorials%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
