Mixing
Simplifying the summation $ (3^isqrt{n/3^i})/log_2(n/3^i)$ from $i=0 text{ to }k-1$
$begingroup$
I have been trying to simplify the following summation with the intention of breaking it into less complex summations, but I keep getting stuck no matter what I try:
$$sum_{i=0}^{k-1} 3^{i} cdot frac{sqrt{frac{n}{3^i}}}{log_{2}frac{n}{3^i}}$$
Among the things I tried was raising to the power of $1/2$* the upper part of the fraction to cancel out the $3^i$, but after that I'm left with $sqrt{n,3^i}$ anyways without any clear steps to simplify further. Also tried putting the lower part of the fraction as a difference of logarithms, but got stuck in the same way. I even tried to use the change of base identity to put the lower part of the fraction in terms of $log_3(2)$ and see if I could have canceled something but no luck.
Just simplifying the sum would be good enough since I can try to take it from there by substituting $k; (k = log_3(n)$ in case that is useful).
summation
edited Jan 28 at 10:16
Jneven
951322
asked Jan 28 at 5:18
MikeKatz45MikeKatz45
399
$endgroup$
add a comment |
$begingroup$
I have been trying to simplify the following summation with the intention of breaking it into less complex summations, but I keep getting stuck no matter what I try:
$$sum_{i=0}^{k-1} 3^{i} cdot frac{sqrt{frac{n}{3^i}}}{log_{2}frac{n}{3^i}}$$
Among the things I tried was raising to the power of $1/2$* the upper part of the fraction to cancel out the $3^i$, but after that I'm left with $sqrt{n,3^i}$ anyways without any clear steps to simplify further. Also tried putting the lower part of the fraction as a difference of logarithms, but got stuck in the same way. I even tried to use the change of base identity to put the lower part of the fraction in terms of $log_3(2)$ and see if I could have canceled something but no luck.
Just simplifying the sum would be good enough since I can try to take it from there by substituting $k; (k = log_3(n)$ in case that is useful).
summation
edited Jan 28 at 10:16
Jneven
951322
asked Jan 28 at 5:18
MikeKatz45MikeKatz45
399
$endgroup$
$begingroup$
May be this is simpler (assuming the Algebra is correct :)): $$sqrt{n} cdot sum_{i=0}^{k-1} frac{sqrt{3^{i}} }{p-i cdot k}$$ where $k=log_{2}{3}, p=log_{2}{n}.$ Note that both p, k are constants.
$endgroup$
– NoChance
Jan 28 at 11:25
$begingroup$
it is kinda simplier yeah, but the idea is to basically put the 2 terms that depend on the index in separate sums somehow. That way they could be solved using identities.
$endgroup$
– MikeKatz45
Jan 28 at 19:36
2
$begingroup$
I don't see any reason to think it's possible to do better than what @NoChance has done. But NoChance has been careless in using $k$ to stand for $log_23$, when $k$ is already being used in the upper limit of the summation.
$endgroup$
– Gerry Myerson
Mar 14 at 1:04
1
$begingroup$
@GerryMyerson thank you for your comment. You are correct.
$endgroup$
– NoChance
Mar 14 at 8:33
add a comment |
$begingroup$
I have been trying to simplify the following summation with the intention of breaking it into less complex summations, but I keep getting stuck no matter what I try:
$$sum_{i=0}^{k-1} 3^{i} cdot frac{sqrt{frac{n}{3^i}}}{log_{2}frac{n}{3^i}}$$
Among the things I tried was raising to the power of $1/2$* the upper part of the fraction to cancel out the $3^i$, but after that I'm left with $sqrt{n,3^i}$ anyways without any clear steps to simplify further. Also tried putting the lower part of the fraction as a difference of logarithms, but got stuck in the same way. I even tried to use the change of base identity to put the lower part of the fraction in terms of $log_3(2)$ and see if I could have canceled something but no luck.
Just simplifying the sum would be good enough since I can try to take it from there by substituting $k; (k = log_3(n)$ in case that is useful).
summation
edited Jan 28 at 10:16
Jneven
951322
asked Jan 28 at 5:18
MikeKatz45MikeKatz45
399
$endgroup$
I have been trying to simplify the following summation with the intention of breaking it into less complex summations, but I keep getting stuck no matter what I try:
$$sum_{i=0}^{k-1} 3^{i} cdot frac{sqrt{frac{n}{3^i}}}{log_{2}frac{n}{3^i}}$$
Among the things I tried was raising to the power of $1/2$* the upper part of the fraction to cancel out the $3^i$, but after that I'm left with $sqrt{n,3^i}$ anyways without any clear steps to simplify further. Also tried putting the lower part of the fraction as a difference of logarithms, but got stuck in the same way. I even tried to use the change of base identity to put the lower part of the fraction in terms of $log_3(2)$ and see if I could have canceled something but no luck.
Just simplifying the sum would be good enough since I can try to take it from there by substituting $k; (k = log_3(n)$ in case that is useful).
summation
summation
edited Jan 28 at 10:16
Jneven
951322
asked Jan 28 at 5:18
MikeKatz45MikeKatz45
399
edited Jan 28 at 10:16
Jneven
951322
asked Jan 28 at 5:18
MikeKatz45MikeKatz45
399
edited Jan 28 at 10:16
Jneven
951322
edited Jan 28 at 10:16
Jneven
951322
edited Jan 28 at 10:16
Jneven
951322
951322
asked Jan 28 at 5:18
MikeKatz45MikeKatz45
399
asked Jan 28 at 5:18
MikeKatz45MikeKatz45
399
asked Jan 28 at 5:18
MikeKatz45MikeKatz45
399
399
$begingroup$
May be this is simpler (assuming the Algebra is correct :)): $$sqrt{n} cdot sum_{i=0}^{k-1} frac{sqrt{3^{i}} }{p-i cdot k}$$ where $k=log_{2}{3}, p=log_{2}{n}.$ Note that both p, k are constants.
$endgroup$
– NoChance
Jan 28 at 11:25
$begingroup$
it is kinda simplier yeah, but the idea is to basically put the 2 terms that depend on the index in separate sums somehow. That way they could be solved using identities.
$endgroup$
– MikeKatz45
Jan 28 at 19:36
2
$begingroup$
I don't see any reason to think it's possible to do better than what @NoChance has done. But NoChance has been careless in using $k$ to stand for $log_23$, when $k$ is already being used in the upper limit of the summation.
$endgroup$
– Gerry Myerson
Mar 14 at 1:04
1
$begingroup$
@GerryMyerson thank you for your comment. You are correct.
$endgroup$
– NoChance
Mar 14 at 8:33
add a comment |
$begingroup$
May be this is simpler (assuming the Algebra is correct :)): $$sqrt{n} cdot sum_{i=0}^{k-1} frac{sqrt{3^{i}} }{p-i cdot k}$$ where $k=log_{2}{3}, p=log_{2}{n}.$ Note that both p, k are constants.
$endgroup$
– NoChance
Jan 28 at 11:25
$begingroup$
it is kinda simplier yeah, but the idea is to basically put the 2 terms that depend on the index in separate sums somehow. That way they could be solved using identities.
$endgroup$
– MikeKatz45
Jan 28 at 19:36
2
$begingroup$
I don't see any reason to think it's possible to do better than what @NoChance has done. But NoChance has been careless in using $k$ to stand for $log_23$, when $k$ is already being used in the upper limit of the summation.
$endgroup$
– Gerry Myerson
Mar 14 at 1:04
1
$begingroup$
@GerryMyerson thank you for your comment. You are correct.
$endgroup$
– NoChance
Mar 14 at 8:33
$begingroup$
May be this is simpler (assuming the Algebra is correct :)): $$sqrt{n} cdot sum_{i=0}^{k-1} frac{sqrt{3^{i}} }{p-i cdot k}$$ where $k=log_{2}{3}, p=log_{2}{n}.$ Note that both p, k are constants.
$endgroup$
– NoChance
Jan 28 at 11:25
$begingroup$
May be this is simpler (assuming the Algebra is correct :)): $$sqrt{n} cdot sum_{i=0}^{k-1} frac{sqrt{3^{i}} }{p-i cdot k}$$ where $k=log_{2}{3}, p=log_{2}{n}.$ Note that both p, k are constants.
$endgroup$
– NoChance
Jan 28 at 11:25
$begingroup$
it is kinda simplier yeah, but the idea is to basically put the 2 terms that depend on the index in separate sums somehow. That way they could be solved using identities.
$endgroup$
– MikeKatz45
Jan 28 at 19:36
$begingroup$
it is kinda simplier yeah, but the idea is to basically put the 2 terms that depend on the index in separate sums somehow. That way they could be solved using identities.
$endgroup$
– MikeKatz45
Jan 28 at 19:36
2
2
$begingroup$
I don't see any reason to think it's possible to do better than what @NoChance has done. But NoChance has been careless in using $k$ to stand for $log_23$, when $k$ is already being used in the upper limit of the summation.
$endgroup$
– Gerry Myerson
Mar 14 at 1:04
$begingroup$
I don't see any reason to think it's possible to do better than what @NoChance has done. But NoChance has been careless in using $k$ to stand for $log_23$, when $k$ is already being used in the upper limit of the summation.
$endgroup$
– Gerry Myerson
Mar 14 at 1:04
1
1
$begingroup$
@GerryMyerson thank you for your comment. You are correct.
$endgroup$
– NoChance
Mar 14 at 8:33
$begingroup$
@GerryMyerson thank you for your comment. You are correct.
$endgroup$
– NoChance
Mar 14 at 8:33
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If I plug
$$
sum_{i=0}^{log_3(n)-1}3^i frac{sqrt{frac{n}{3^i}}}{log_2(frac{n}{3^i})}
$$
into Wolfram Mathematica, then I get no simplification, so an nice simplification probably does not exist. The best we can probably do is, with thanks to @NoChance,
$$
sum_{i=0}^{log_3(n)-1}3^i frac{sqrt{frac{n}{3^i}}}{log_2(frac{n}{3^i})} = sum_{i=0}^{log_3(n)-1}3^i frac{sqrt{frac{n}{3^i}}}{log_2(n)-log_2(3^i))} \
=sum_{i=0}^{log_3(n)-1} 3^ifrac{sqrt{frac{n}{3^i}}}{log_2(n)-ilog_2(3))} \
=sum_{i=0}^{log_3(n)-1} sqrt{n}frac{3^isqrt{frac{1}{3^i}}}{log_2(n)-ilog_2(3))} \
= sqrt{n}sum_{i=0}^{log_3(n)-1}frac{sqrt{3^i}}{log_2(n)-ilog_2(3))} \
$$
If I plug this into Wolfram Mathematica, then we see that we get something
$$
sqrt{n}sum_{i=0}^{log_3(n)-1}frac{sqrt{3^i}}{log_2(n)-ilog_2(3))} = -frac{nbigg(log(3-sqrt{3})-log(3)bigg)}{log_2(3)}-frac{sqrt{n}HyperGeometric2F1bigg(1,log_3(3n),2+log_3(n),frac{1}{sqrt{3}}bigg)}{sqrt{3}log_2{3n}}
$$
If we look at Wikipedia, https://en.wikipedia.org/wiki/Hypergeometric_function, for some more info on the Hypergeometric2F1 we see that there are no further simplifications listed.
answered Mar 16 at 0:39
ChocolateRainChocolateRain
1364
$endgroup$
add a comment |
$begingroup$
The issue expression can be presented in the form of
$$S={largesumlimits_{i=0}^{k-1}}3^idfrac{sqrt{dfrac n{3^i}}}{log_2dfrac{n}{3^i}}
= sqrt n log_32{largesumlimits_{i=0}^{k-1}}dfrac{(sqrt3)^i}{log_3{n}-i},$$
so
$$S=sqrt n log_32left(3^{k/2}Phi(sqrt3,1,k-log_3n)-Phi(sqrt3,1,-log_3n)right)$$
(see also Wolfram Alpha), where
$$Phi(z,s,a) = {largesumlimits_{m=0}^infty,dfrac
{z^m}{(m+a)^s}}$$
is Lerch transcendent.
Assuming $k=lceillog_3n -1rceil,$ can be built the plots for the issue sums
and via Lerch transcendent
Easy to see that summation via Lerch transcendent is correct.
answered Mar 16 at 16:09
Yuri NegometyanovYuri Negometyanov
12.2k1729
$endgroup$
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%2f3090509%2fsimplifying-the-summation-3i-sqrtn-3i-log-2n-3i-from-i-0-text-to%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
$begingroup$
If I plug
$$
sum_{i=0}^{log_3(n)-1}3^i frac{sqrt{frac{n}{3^i}}}{log_2(frac{n}{3^i})}
$$
into Wolfram Mathematica, then I get no simplification, so an nice simplification probably does not exist. The best we can probably do is, with thanks to @NoChance,
$$
sum_{i=0}^{log_3(n)-1}3^i frac{sqrt{frac{n}{3^i}}}{log_2(frac{n}{3^i})} = sum_{i=0}^{log_3(n)-1}3^i frac{sqrt{frac{n}{3^i}}}{log_2(n)-log_2(3^i))} \
=sum_{i=0}^{log_3(n)-1} 3^ifrac{sqrt{frac{n}{3^i}}}{log_2(n)-ilog_2(3))} \
=sum_{i=0}^{log_3(n)-1} sqrt{n}frac{3^isqrt{frac{1}{3^i}}}{log_2(n)-ilog_2(3))} \
= sqrt{n}sum_{i=0}^{log_3(n)-1}frac{sqrt{3^i}}{log_2(n)-ilog_2(3))} \
$$
If I plug this into Wolfram Mathematica, then we see that we get something
$$
sqrt{n}sum_{i=0}^{log_3(n)-1}frac{sqrt{3^i}}{log_2(n)-ilog_2(3))} = -frac{nbigg(log(3-sqrt{3})-log(3)bigg)}{log_2(3)}-frac{sqrt{n}HyperGeometric2F1bigg(1,log_3(3n),2+log_3(n),frac{1}{sqrt{3}}bigg)}{sqrt{3}log_2{3n}}
$$
If we look at Wikipedia, https://en.wikipedia.org/wiki/Hypergeometric_function, for some more info on the Hypergeometric2F1 we see that there are no further simplifications listed.
answered Mar 16 at 0:39
ChocolateRainChocolateRain
1364
$endgroup$
add a comment |
$begingroup$
If I plug
$$
sum_{i=0}^{log_3(n)-1}3^i frac{sqrt{frac{n}{3^i}}}{log_2(frac{n}{3^i})}
$$
into Wolfram Mathematica, then I get no simplification, so an nice simplification probably does not exist. The best we can probably do is, with thanks to @NoChance,
$$
sum_{i=0}^{log_3(n)-1}3^i frac{sqrt{frac{n}{3^i}}}{log_2(frac{n}{3^i})} = sum_{i=0}^{log_3(n)-1}3^i frac{sqrt{frac{n}{3^i}}}{log_2(n)-log_2(3^i))} \
=sum_{i=0}^{log_3(n)-1} 3^ifrac{sqrt{frac{n}{3^i}}}{log_2(n)-ilog_2(3))} \
=sum_{i=0}^{log_3(n)-1} sqrt{n}frac{3^isqrt{frac{1}{3^i}}}{log_2(n)-ilog_2(3))} \
= sqrt{n}sum_{i=0}^{log_3(n)-1}frac{sqrt{3^i}}{log_2(n)-ilog_2(3))} \
$$
If I plug this into Wolfram Mathematica, then we see that we get something
$$
sqrt{n}sum_{i=0}^{log_3(n)-1}frac{sqrt{3^i}}{log_2(n)-ilog_2(3))} = -frac{nbigg(log(3-sqrt{3})-log(3)bigg)}{log_2(3)}-frac{sqrt{n}HyperGeometric2F1bigg(1,log_3(3n),2+log_3(n),frac{1}{sqrt{3}}bigg)}{sqrt{3}log_2{3n}}
$$
If we look at Wikipedia, https://en.wikipedia.org/wiki/Hypergeometric_function, for some more info on the Hypergeometric2F1 we see that there are no further simplifications listed.
answered Mar 16 at 0:39
ChocolateRainChocolateRain
1364
$endgroup$
add a comment |
$begingroup$
If I plug
$$
sum_{i=0}^{log_3(n)-1}3^i frac{sqrt{frac{n}{3^i}}}{log_2(frac{n}{3^i})}
$$
into Wolfram Mathematica, then I get no simplification, so an nice simplification probably does not exist. The best we can probably do is, with thanks to @NoChance,
$$
sum_{i=0}^{log_3(n)-1}3^i frac{sqrt{frac{n}{3^i}}}{log_2(frac{n}{3^i})} = sum_{i=0}^{log_3(n)-1}3^i frac{sqrt{frac{n}{3^i}}}{log_2(n)-log_2(3^i))} \
=sum_{i=0}^{log_3(n)-1} 3^ifrac{sqrt{frac{n}{3^i}}}{log_2(n)-ilog_2(3))} \
=sum_{i=0}^{log_3(n)-1} sqrt{n}frac{3^isqrt{frac{1}{3^i}}}{log_2(n)-ilog_2(3))} \
= sqrt{n}sum_{i=0}^{log_3(n)-1}frac{sqrt{3^i}}{log_2(n)-ilog_2(3))} \
$$
If I plug this into Wolfram Mathematica, then we see that we get something
$$
sqrt{n}sum_{i=0}^{log_3(n)-1}frac{sqrt{3^i}}{log_2(n)-ilog_2(3))} = -frac{nbigg(log(3-sqrt{3})-log(3)bigg)}{log_2(3)}-frac{sqrt{n}HyperGeometric2F1bigg(1,log_3(3n),2+log_3(n),frac{1}{sqrt{3}}bigg)}{sqrt{3}log_2{3n}}
$$
If we look at Wikipedia, https://en.wikipedia.org/wiki/Hypergeometric_function, for some more info on the Hypergeometric2F1 we see that there are no further simplifications listed.
answered Mar 16 at 0:39
ChocolateRainChocolateRain
1364
$endgroup$
If I plug
$$
sum_{i=0}^{log_3(n)-1}3^i frac{sqrt{frac{n}{3^i}}}{log_2(frac{n}{3^i})}
$$
into Wolfram Mathematica, then I get no simplification, so an nice simplification probably does not exist. The best we can probably do is, with thanks to @NoChance,
$$
sum_{i=0}^{log_3(n)-1}3^i frac{sqrt{frac{n}{3^i}}}{log_2(frac{n}{3^i})} = sum_{i=0}^{log_3(n)-1}3^i frac{sqrt{frac{n}{3^i}}}{log_2(n)-log_2(3^i))} \
=sum_{i=0}^{log_3(n)-1} 3^ifrac{sqrt{frac{n}{3^i}}}{log_2(n)-ilog_2(3))} \
=sum_{i=0}^{log_3(n)-1} sqrt{n}frac{3^isqrt{frac{1}{3^i}}}{log_2(n)-ilog_2(3))} \
= sqrt{n}sum_{i=0}^{log_3(n)-1}frac{sqrt{3^i}}{log_2(n)-ilog_2(3))} \
$$
If I plug this into Wolfram Mathematica, then we see that we get something
$$
sqrt{n}sum_{i=0}^{log_3(n)-1}frac{sqrt{3^i}}{log_2(n)-ilog_2(3))} = -frac{nbigg(log(3-sqrt{3})-log(3)bigg)}{log_2(3)}-frac{sqrt{n}HyperGeometric2F1bigg(1,log_3(3n),2+log_3(n),frac{1}{sqrt{3}}bigg)}{sqrt{3}log_2{3n}}
$$
If we look at Wikipedia, https://en.wikipedia.org/wiki/Hypergeometric_function, for some more info on the Hypergeometric2F1 we see that there are no further simplifications listed.
answered Mar 16 at 0:39
ChocolateRainChocolateRain
1364
answered Mar 16 at 0:39
ChocolateRainChocolateRain
1364
answered Mar 16 at 0:39
ChocolateRainChocolateRain
1364
answered Mar 16 at 0:39
ChocolateRainChocolateRain
1364
1364
add a comment |
add a comment |
$begingroup$
The issue expression can be presented in the form of
$$S={largesumlimits_{i=0}^{k-1}}3^idfrac{sqrt{dfrac n{3^i}}}{log_2dfrac{n}{3^i}}
= sqrt n log_32{largesumlimits_{i=0}^{k-1}}dfrac{(sqrt3)^i}{log_3{n}-i},$$
so
$$S=sqrt n log_32left(3^{k/2}Phi(sqrt3,1,k-log_3n)-Phi(sqrt3,1,-log_3n)right)$$
(see also Wolfram Alpha), where
$$Phi(z,s,a) = {largesumlimits_{m=0}^infty,dfrac
{z^m}{(m+a)^s}}$$
is Lerch transcendent.
Assuming $k=lceillog_3n -1rceil,$ can be built the plots for the issue sums
and via Lerch transcendent
Easy to see that summation via Lerch transcendent is correct.
answered Mar 16 at 16:09
Yuri NegometyanovYuri Negometyanov
12.2k1729
$endgroup$
add a comment |
$begingroup$
The issue expression can be presented in the form of
$$S={largesumlimits_{i=0}^{k-1}}3^idfrac{sqrt{dfrac n{3^i}}}{log_2dfrac{n}{3^i}}
= sqrt n log_32{largesumlimits_{i=0}^{k-1}}dfrac{(sqrt3)^i}{log_3{n}-i},$$
so
$$S=sqrt n log_32left(3^{k/2}Phi(sqrt3,1,k-log_3n)-Phi(sqrt3,1,-log_3n)right)$$
(see also Wolfram Alpha), where
$$Phi(z,s,a) = {largesumlimits_{m=0}^infty,dfrac
{z^m}{(m+a)^s}}$$
is Lerch transcendent.
Assuming $k=lceillog_3n -1rceil,$ can be built the plots for the issue sums
and via Lerch transcendent
Easy to see that summation via Lerch transcendent is correct.
answered Mar 16 at 16:09
Yuri NegometyanovYuri Negometyanov
12.2k1729
$endgroup$
add a comment |
$begingroup$
The issue expression can be presented in the form of
$$S={largesumlimits_{i=0}^{k-1}}3^idfrac{sqrt{dfrac n{3^i}}}{log_2dfrac{n}{3^i}}
= sqrt n log_32{largesumlimits_{i=0}^{k-1}}dfrac{(sqrt3)^i}{log_3{n}-i},$$
so
$$S=sqrt n log_32left(3^{k/2}Phi(sqrt3,1,k-log_3n)-Phi(sqrt3,1,-log_3n)right)$$
(see also Wolfram Alpha), where
$$Phi(z,s,a) = {largesumlimits_{m=0}^infty,dfrac
{z^m}{(m+a)^s}}$$
is Lerch transcendent.
Assuming $k=lceillog_3n -1rceil,$ can be built the plots for the issue sums
and via Lerch transcendent
Easy to see that summation via Lerch transcendent is correct.
answered Mar 16 at 16:09
Yuri NegometyanovYuri Negometyanov
12.2k1729
$endgroup$
The issue expression can be presented in the form of
$$S={largesumlimits_{i=0}^{k-1}}3^idfrac{sqrt{dfrac n{3^i}}}{log_2dfrac{n}{3^i}}
= sqrt n log_32{largesumlimits_{i=0}^{k-1}}dfrac{(sqrt3)^i}{log_3{n}-i},$$
so
$$S=sqrt n log_32left(3^{k/2}Phi(sqrt3,1,k-log_3n)-Phi(sqrt3,1,-log_3n)right)$$
(see also Wolfram Alpha), where
$$Phi(z,s,a) = {largesumlimits_{m=0}^infty,dfrac
{z^m}{(m+a)^s}}$$
is Lerch transcendent.
Assuming $k=lceillog_3n -1rceil,$ can be built the plots for the issue sums
and via Lerch transcendent
Easy to see that summation via Lerch transcendent is correct.
answered Mar 16 at 16:09
Yuri NegometyanovYuri Negometyanov
12.2k1729
edited Mar 16 at 18:07
answered Mar 16 at 16:09
Yuri NegometyanovYuri Negometyanov
12.2k1729
answered Mar 16 at 16:09
Yuri NegometyanovYuri Negometyanov
12.2k1729
answered Mar 16 at 16:09
Yuri NegometyanovYuri Negometyanov
12.2k1729
12.2k1729
add a comment |
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.
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%2f3090509%2fsimplifying-the-summation-3i-sqrtn-3i-log-2n-3i-from-i-0-text-to%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
$begingroup$
May be this is simpler (assuming the Algebra is correct :)): $$sqrt{n} cdot sum_{i=0}^{k-1} frac{sqrt{3^{i}} }{p-i cdot k}$$ where $k=log_{2}{3}, p=log_{2}{n}.$ Note that both p, k are constants.
$endgroup$
– NoChance
Jan 28 at 11:25
$begingroup$
it is kinda simplier yeah, but the idea is to basically put the 2 terms that depend on the index in separate sums somehow. That way they could be solved using identities.
$endgroup$
– MikeKatz45
Jan 28 at 19:36
2
$begingroup$
I don't see any reason to think it's possible to do better than what @NoChance has done. But NoChance has been careless in using $k$ to stand for $log_23$, when $k$ is already being used in the upper limit of the summation.
$endgroup$
– Gerry Myerson
Mar 14 at 1:04
1
$begingroup$
@GerryMyerson thank you for your comment. You are correct.
$endgroup$
– NoChance
Mar 14 at 8:33