Proof clarification $-int_0^pi/12log(tan t)dt=frac23mathrm G$ The Next CEO of Stack OverflowDifferent methods to compute $sumlimits_k=1^infty frac1k^2$ (Basel problem)Integral: $int_0^pi/12 ln(tan x),dx$How to prove $int_0^infty J_nu(x)^3dxstackrel?=fracGamma(1/6) Gamma(1/6+nu/2)2^5/3 3^1/2 pi^3/2 Gamma(5/6+nu/2)$?How to prove $int_0^pifracln(2+cosphi)sqrt2+cosphidphi=fracln3sqrt3Kleft(sqrtfrac23right)$?Integral computation of $int_0^pi mathrm d t sin(acos t/2) mathrmsinh(bsin t/2)$How do I find the sum of the infinite geometric series?Prove $largeint_0^1fracln(1+8x)x^2/3,(1-x)^2/3,(1+8x)^1/3dx=fracln3pisqrt3Gamma^3!left(tfrac13right)$Integral that arises from the derivation of Kummer's Fourier expansion of $lnGamma(x)$Challenging definite integral of hypergeometric functionsClosed form of $sum^infty_n=1 fracsin nn$On $sum_n=0^infty2n+3choose n+1 left(frac12^ncdotfrac32n+1right)^4$ and Gieseking's constantHypergeometric representation of Fresnel $S(x)$
Why did early computer designers eschew integers?
Purpose of level-shifter with same in and out voltages
What would be the main consequences for a country leaving the WTO?
Why is the US ranked as #45 in Press Freedom ratings, despite its extremely permissive free speech laws?
Calculate the Mean mean of two numbers
Is French Guiana a (hard) EU border?
Cannot shrink btrfs filesystem although there is still data and metadata space left : ERROR: unable to resize '/home': No space left on device
Reference request: Grassmannian and Plucker coordinates in type B, C, D
How to use ReplaceAll on an expression that contains a rule
Does Germany produce more waste than the US?
Is it ok to trim down a tube patch?
What is the process for purifying your home if you believe it may have been previously used for pagan worship?
Strange use of "whether ... than ..." in official text
Can you teleport closer to a creature you are Frightened of?
how one can write a nice vector parser, something that does pgfvecparseA=B-C; D=E x F;
Why am I getting "Static method cannot be referenced from a non static context: String String.valueOf(Object)"?
Help/tips for a first time writer?
Why don't programming languages automatically manage the synchronous/asynchronous problem?
Airplane gently rocking its wings during whole flight
Is fine stranded wire ok for main supply line?
Can I board the first leg of the flight without having final country's visa?
TikZ: How to fill area with a special pattern?
What was Carter Burke's job for "the company" in Aliens?
Which one is the true statement?
Proof clarification $-int_0^pi/12log(tan t)dt=frac23mathrm G$
The Next CEO of Stack OverflowDifferent methods to compute $sumlimits_k=1^infty frac1k^2$ (Basel problem)Integral: $int_0^pi/12 ln(tan x),dx$How to prove $int_0^infty J_nu(x)^3dxstackrel?=fracGamma(1/6) Gamma(1/6+nu/2)2^5/3 3^1/2 pi^3/2 Gamma(5/6+nu/2)$?How to prove $int_0^pifracln(2+cosphi)sqrt2+cosphidphi=fracln3sqrt3Kleft(sqrtfrac23right)$?Integral computation of $int_0^pi mathrm d t sin(acos t/2) mathrmsinh(bsin t/2)$How do I find the sum of the infinite geometric series?Prove $largeint_0^1fracln(1+8x)x^2/3,(1-x)^2/3,(1+8x)^1/3dx=fracln3pisqrt3Gamma^3!left(tfrac13right)$Integral that arises from the derivation of Kummer's Fourier expansion of $lnGamma(x)$Challenging definite integral of hypergeometric functionsClosed form of $sum^infty_n=1 fracsin nn$On $sum_n=0^infty2n+3choose n+1 left(frac12^ncdotfrac32n+1right)^4$ and Gieseking's constantHypergeometric representation of Fresnel $S(x)$
$begingroup$
$mathrm G$ is Catalan's constant.
I would like clarification on the following proof provided by @M.N.C.E.:
$$
beginalign
&int^fracpi12_0ln(tanx) rm dx\
=&-2sum^infty_n=0frac12n+1int^fracpi12_0cosBig[(4n+2)xBig] rm dx\
=&-sum^infty_n=0fracsinBig[(2n+1)tfracpi6Big](2n+1)^2tag1\
=&color#E2062C-frac12sum^infty_n=0frac1(12n+1)^2color#6F00FF-sum^infty_n=0frac1(12n+3)^2-color#E2062Cfrac12sum^infty_n=0frac1(12n+5)^2\
&color#E2062C+frac12sum^infty_n=0frac1(12n+7)^2color#6F00FF+sum^infty_n=0frac1(12n+9)^2color#E2062C+frac12sum^infty_n=0frac1(12n+11)^2tag2\
=&color#6F00FF-frac19underbracesum^infty_n=0left[frac1(4n+1)^2-frac1(4n+3)^2right]_mathrm Gcolor#E2062C-frac12mathrm G-frac12underbracesum^infty_n=0left[frac1(12n+3)^2-frac1(12n+9)^2right]_frac19mathrm Gtag3\
=&left(-frac19-frac12-frac118right)mathrm G=large-frac23mathrm G
endalign$$
I understand how to obtain $(1)$, but I do not know how to go from $(1)$ to $(2)$ and then from $(2)$ to $(3)$.
A log of my attempts to find an alternate proof for the identity in question.
I tried to represent $(1)$ as a hypergeometric series because the $1/(2n+1)^2$ terms are after all what give us the identity
$$_3F_2left(frac12,frac12,1;frac32,frac32;-1right)=mathrm G$$
So I was hoping that $$fracsinleft[fracpi6(2n+1)right]sinleft[fracpi6(2n+3)right]=-1$$
But alas this is not the case. I see now that I would be missing an extra factor of $2/3$.
Another, more realistic attempt of mine comes by noting that
$$sinleft[fracpi6(2n+1)right]=fracsqrt32sinfracpi n3+frac12cosfracpi n3$$
So, by letting $K$ be $-1cdot$(the quantity $(1)$), we have that
$$K=fracsqrt32S+frac12C$$
Where $$S=sum_ngeq0fracsinfracpi n3(2n+1)^2$$
and $$C=sum_ngeq0fraccosfracpi n3(2n+1)^2$$
Although neither of these seem to have a 'nice' $_pF_q$ representation. I have run out of ideas for proving that $K=frac23mathrm G$.
sequences-and-series proof-explanation special-functions
asked Mar 28 at 2:58
clathratusclathratus
5,0361438
$endgroup$
add a comment |
$begingroup$
$mathrm G$ is Catalan's constant.
I would like clarification on the following proof provided by @M.N.C.E.:
$$
beginalign
&int^fracpi12_0ln(tanx) rm dx\
=&-2sum^infty_n=0frac12n+1int^fracpi12_0cosBig[(4n+2)xBig] rm dx\
=&-sum^infty_n=0fracsinBig[(2n+1)tfracpi6Big](2n+1)^2tag1\
=&color#E2062C-frac12sum^infty_n=0frac1(12n+1)^2color#6F00FF-sum^infty_n=0frac1(12n+3)^2-color#E2062Cfrac12sum^infty_n=0frac1(12n+5)^2\
&color#E2062C+frac12sum^infty_n=0frac1(12n+7)^2color#6F00FF+sum^infty_n=0frac1(12n+9)^2color#E2062C+frac12sum^infty_n=0frac1(12n+11)^2tag2\
=&color#6F00FF-frac19underbracesum^infty_n=0left[frac1(4n+1)^2-frac1(4n+3)^2right]_mathrm Gcolor#E2062C-frac12mathrm G-frac12underbracesum^infty_n=0left[frac1(12n+3)^2-frac1(12n+9)^2right]_frac19mathrm Gtag3\
=&left(-frac19-frac12-frac118right)mathrm G=large-frac23mathrm G
endalign$$
I understand how to obtain $(1)$, but I do not know how to go from $(1)$ to $(2)$ and then from $(2)$ to $(3)$.
A log of my attempts to find an alternate proof for the identity in question.
I tried to represent $(1)$ as a hypergeometric series because the $1/(2n+1)^2$ terms are after all what give us the identity
$$_3F_2left(frac12,frac12,1;frac32,frac32;-1right)=mathrm G$$
So I was hoping that $$fracsinleft[fracpi6(2n+1)right]sinleft[fracpi6(2n+3)right]=-1$$
But alas this is not the case. I see now that I would be missing an extra factor of $2/3$.
Another, more realistic attempt of mine comes by noting that
$$sinleft[fracpi6(2n+1)right]=fracsqrt32sinfracpi n3+frac12cosfracpi n3$$
So, by letting $K$ be $-1cdot$(the quantity $(1)$), we have that
$$K=fracsqrt32S+frac12C$$
Where $$S=sum_ngeq0fracsinfracpi n3(2n+1)^2$$
and $$C=sum_ngeq0fraccosfracpi n3(2n+1)^2$$
Although neither of these seem to have a 'nice' $_pF_q$ representation. I have run out of ideas for proving that $K=frac23mathrm G$.
sequences-and-series proof-explanation special-functions
asked Mar 28 at 2:58
clathratusclathratus
5,0361438
$endgroup$
add a comment |
$begingroup$
$mathrm G$ is Catalan's constant.
I would like clarification on the following proof provided by @M.N.C.E.:
$$
beginalign
&int^fracpi12_0ln(tanx) rm dx\
=&-2sum^infty_n=0frac12n+1int^fracpi12_0cosBig[(4n+2)xBig] rm dx\
=&-sum^infty_n=0fracsinBig[(2n+1)tfracpi6Big](2n+1)^2tag1\
=&color#E2062C-frac12sum^infty_n=0frac1(12n+1)^2color#6F00FF-sum^infty_n=0frac1(12n+3)^2-color#E2062Cfrac12sum^infty_n=0frac1(12n+5)^2\
&color#E2062C+frac12sum^infty_n=0frac1(12n+7)^2color#6F00FF+sum^infty_n=0frac1(12n+9)^2color#E2062C+frac12sum^infty_n=0frac1(12n+11)^2tag2\
=&color#6F00FF-frac19underbracesum^infty_n=0left[frac1(4n+1)^2-frac1(4n+3)^2right]_mathrm Gcolor#E2062C-frac12mathrm G-frac12underbracesum^infty_n=0left[frac1(12n+3)^2-frac1(12n+9)^2right]_frac19mathrm Gtag3\
=&left(-frac19-frac12-frac118right)mathrm G=large-frac23mathrm G
endalign$$
I understand how to obtain $(1)$, but I do not know how to go from $(1)$ to $(2)$ and then from $(2)$ to $(3)$.
A log of my attempts to find an alternate proof for the identity in question.
I tried to represent $(1)$ as a hypergeometric series because the $1/(2n+1)^2$ terms are after all what give us the identity
$$_3F_2left(frac12,frac12,1;frac32,frac32;-1right)=mathrm G$$
So I was hoping that $$fracsinleft[fracpi6(2n+1)right]sinleft[fracpi6(2n+3)right]=-1$$
But alas this is not the case. I see now that I would be missing an extra factor of $2/3$.
Another, more realistic attempt of mine comes by noting that
$$sinleft[fracpi6(2n+1)right]=fracsqrt32sinfracpi n3+frac12cosfracpi n3$$
So, by letting $K$ be $-1cdot$(the quantity $(1)$), we have that
$$K=fracsqrt32S+frac12C$$
Where $$S=sum_ngeq0fracsinfracpi n3(2n+1)^2$$
and $$C=sum_ngeq0fraccosfracpi n3(2n+1)^2$$
Although neither of these seem to have a 'nice' $_pF_q$ representation. I have run out of ideas for proving that $K=frac23mathrm G$.
sequences-and-series proof-explanation special-functions
asked Mar 28 at 2:58
clathratusclathratus
5,0361438
$endgroup$
$mathrm G$ is Catalan's constant.
I would like clarification on the following proof provided by @M.N.C.E.:
$$
beginalign
&int^fracpi12_0ln(tanx) rm dx\
=&-2sum^infty_n=0frac12n+1int^fracpi12_0cosBig[(4n+2)xBig] rm dx\
=&-sum^infty_n=0fracsinBig[(2n+1)tfracpi6Big](2n+1)^2tag1\
=&color#E2062C-frac12sum^infty_n=0frac1(12n+1)^2color#6F00FF-sum^infty_n=0frac1(12n+3)^2-color#E2062Cfrac12sum^infty_n=0frac1(12n+5)^2\
&color#E2062C+frac12sum^infty_n=0frac1(12n+7)^2color#6F00FF+sum^infty_n=0frac1(12n+9)^2color#E2062C+frac12sum^infty_n=0frac1(12n+11)^2tag2\
=&color#6F00FF-frac19underbracesum^infty_n=0left[frac1(4n+1)^2-frac1(4n+3)^2right]_mathrm Gcolor#E2062C-frac12mathrm G-frac12underbracesum^infty_n=0left[frac1(12n+3)^2-frac1(12n+9)^2right]_frac19mathrm Gtag3\
=&left(-frac19-frac12-frac118right)mathrm G=large-frac23mathrm G
endalign$$
I understand how to obtain $(1)$, but I do not know how to go from $(1)$ to $(2)$ and then from $(2)$ to $(3)$.
A log of my attempts to find an alternate proof for the identity in question.
I tried to represent $(1)$ as a hypergeometric series because the $1/(2n+1)^2$ terms are after all what give us the identity
$$_3F_2left(frac12,frac12,1;frac32,frac32;-1right)=mathrm G$$
So I was hoping that $$fracsinleft[fracpi6(2n+1)right]sinleft[fracpi6(2n+3)right]=-1$$
But alas this is not the case. I see now that I would be missing an extra factor of $2/3$.
Another, more realistic attempt of mine comes by noting that
$$sinleft[fracpi6(2n+1)right]=fracsqrt32sinfracpi n3+frac12cosfracpi n3$$
So, by letting $K$ be $-1cdot$(the quantity $(1)$), we have that
$$K=fracsqrt32S+frac12C$$
Where $$S=sum_ngeq0fracsinfracpi n3(2n+1)^2$$
and $$C=sum_ngeq0fraccosfracpi n3(2n+1)^2$$
Although neither of these seem to have a 'nice' $_pF_q$ representation. I have run out of ideas for proving that $K=frac23mathrm G$.
sequences-and-series proof-explanation special-functions
sequences-and-series proof-explanation special-functions
asked Mar 28 at 2:58
clathratusclathratus
5,0361438
asked Mar 28 at 2:58
clathratusclathratus
5,0361438
asked Mar 28 at 2:58
clathratusclathratus
5,0361438
asked Mar 28 at 2:58
clathratusclathratus
5,0361438
asked Mar 28 at 2:58
clathratusclathratus
5,0361438
5,0361438
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Firstly, any $n$ can be written as $6k$ or $6k+1$ or... or $6k+5$. So
$$eqalign
sum^infty_n=0fracsinBig[(2n+1)tfracpi6Big](2n+1)^2
&=sum_k=0^inftyfracsin(12k+1)fracpi6(12k+1)^2+hboxfive more sumscr
&=sum_k=0^inftyfracsinfracpi6(12k+1)^2+cdotscr
&=frac12sum_k=0^inftyfrac1(12k+1)^2+cdots .cr$$
Then if we write
$$S_m=sum_k=0^inftyfrac1(12k+m)^2$$
we have
$$G=S_1-S_3+S_5-S_7+S_9-S_11$$
and your sum $(2)$ is
$$eqalign
-frac12S_1&-S_3-frac12S_5+frac12S_7+S_9+frac12S_11cr
&=-frac12(S_1-S_3+S_5-S_7+S_9-S_11)-frac32S_3+frac32S_9cr
&=-frac12G-frac32frac19Gcr
&=-frac23G .cr$$
answered Mar 28 at 3:18
DavidDavid
69.6k668131
$endgroup$
$begingroup$
This works for me. Just for clarification: "$sin(12k+1)fracpi6$" is the same as $sinleft[(12k+1)fracpi6right]$?
$endgroup$
– clathratus
Mar 28 at 3:31
$begingroup$
Also: How do we have $$mathrm G=S_1-S_3+S_5-S_7+S_9-S_11$$
$endgroup$
– clathratus
Mar 28 at 3:34
1
$begingroup$
We're adding all sums $sum 1/k^2$ with $k=12m+1,,12m+5,,12m+9$, that is, all sums with $k=4n+1$. And subtracting similar sums.
$endgroup$
– David
Mar 28 at 3:49
add a comment |
$begingroup$
As shown in this answer,
$$
log(2cos(x))=cos(2x)-fraccos(4x)2+fraccos(6x)3-dotstag1
$$
Substituting $xmapstofracpi2-x$ gives
$$
log(2sin(x))=-cos(2x)-fraccos(4x)2-fraccos(6x)3-dotstag2
$$
Subtracting $(1)$ from $(2)$ yields
$$
log(tan(x))=-2left(cos(2x)+fraccos(6x)3+fraccos(10x)5+dotsright)tag3
$$
Notice that $sinleft((2k+1)fracpi6right)$ has period $6$:
$$
beginarrayc
kpmod6&0&1&2&3&4&5\
color#C00sinleft((2k+1)fracpi6right)&color#C00frac12&color#C001&color#C00frac12&color#C00-frac12&color#C00-1&color#C00-frac12\
color#0902k+1&color#0901&color#0903&color#0905&color#0907&color#0909&color#09011\
color#C00frac(-1)^k2&color#C00frac12&color#C00-frac12&color#C00frac12&color#C00-frac12&color#C00frac12&color#C00-frac12\
color#C00frac32(-1)^frack-13&&color#C00frac32&&&color#C00-frac32&&kequiv1pmod3\
color#0903left(2frack-13+1right)&&color#0903&&&color#0909&&kequiv1pmod3 \
endarraytag4
$$
Note that the upper red row is the sum of the two lower red rows
Thus,
$$
beginalign
int_0^pi/12log(tan(x)),mathrmdx
&=-sum_k=0^inftyfracsinleft((2k+1)fracpi6right)(2k+1)^2tag5\
&=-sum_k=0^inftyleft(frac(-1)^k2frac1(2k+1)^2+frac(-1)^k9frac32frac1(2k+1)^2right)tag6\
&=-frac23sum_k=0^inftyfrac(-1)^k(2k+1)^2tag7\[3pt]
&=-frac23mathrmGtag8
endalign
$$
Explanation:
$(5)$: integrate $(3)$ term by term
$(6)$: apply $(4)$; $k$ in the right term is $frack-13$ from $(4)$
$(7)$: collect terms
$(8)$: definition of $mathrmG$
answered Mar 28 at 15:10
robjohn♦robjohn
270k27312640
$endgroup$
$begingroup$
Interesting...+1 is there a name for this technique, or is there any sort of generalization to it?
$endgroup$
– clathratus
Mar 28 at 16:45
$begingroup$
$(1)$-$(3)$ are Fourier Series, $(4)$ is picking out patterns in periodic functions.
$endgroup$
– robjohn♦
Mar 28 at 19:06
add a comment |
$begingroup$
I understand how to obtain $(1)$, but I do not know how to go from $(1)$ to $(2)$ and then from $(2)$ to $(3)$.
HINT:
In order to arrive at $(2)$, note that $sinleft((2n+1)fracpi6right)$ takes on the values
$$sinleft((2n+1)fracpi6right)=begincasesfrac12 &,n=0,6,12,dots\\
1 &,n=1,7,13,dots\\
1/2&,n=2,8,14,dots\\
-frac12&,n=3,9,15,cdots\\
-1&,n=4,10,16,cdots\\
-frac12&,n=5,11,17,cdots
endcases$$
answered Mar 28 at 3:08
Mark ViolaMark Viola
134k1278176
$endgroup$
$begingroup$
If you hover your cursor above the accept-answer button, it says "Click to accept this answer because it solved your problem or it was most helpful in finding your solution..." The accepted answer was accepted because it was not a mere hint, but a full answer. Don't get me wrong-I appreciate your hint (as of now I am the only up-voter of it), but the accepted answer happens to be more helpful. Also: the first answer isn't always the best.
$endgroup$
– clathratus
Mar 28 at 22:22
$begingroup$
Thank you for the note. Was the "HINT" useful? I really wanted to give you the best answer I could, and thought the hint would have sufficed to show the way forward.
$endgroup$
– Mark Viola
Mar 28 at 22:30
$begingroup$
The hint was somewhat useful. I must admit I can't exactly see any patterns in the sets of values of $n$ which I could use... Perhaps your answer would be more helpful if you showed a little more how to proceed with this information. I really appreciate your dedication to the quality of your answers across the site.
$endgroup$
– clathratus
Mar 29 at 1:27
add a comment |
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%2f3165412%2fproof-clarification-int-0-pi-12-log-tan-tdt-frac23-mathrm-g%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Firstly, any $n$ can be written as $6k$ or $6k+1$ or... or $6k+5$. So
$$eqalign
sum^infty_n=0fracsinBig[(2n+1)tfracpi6Big](2n+1)^2
&=sum_k=0^inftyfracsin(12k+1)fracpi6(12k+1)^2+hboxfive more sumscr
&=sum_k=0^inftyfracsinfracpi6(12k+1)^2+cdotscr
&=frac12sum_k=0^inftyfrac1(12k+1)^2+cdots .cr$$
Then if we write
$$S_m=sum_k=0^inftyfrac1(12k+m)^2$$
we have
$$G=S_1-S_3+S_5-S_7+S_9-S_11$$
and your sum $(2)$ is
$$eqalign
-frac12S_1&-S_3-frac12S_5+frac12S_7+S_9+frac12S_11cr
&=-frac12(S_1-S_3+S_5-S_7+S_9-S_11)-frac32S_3+frac32S_9cr
&=-frac12G-frac32frac19Gcr
&=-frac23G .cr$$
answered Mar 28 at 3:18
DavidDavid
69.6k668131
$endgroup$
$begingroup$
This works for me. Just for clarification: "$sin(12k+1)fracpi6$" is the same as $sinleft[(12k+1)fracpi6right]$?
$endgroup$
– clathratus
Mar 28 at 3:31
$begingroup$
Also: How do we have $$mathrm G=S_1-S_3+S_5-S_7+S_9-S_11$$
$endgroup$
– clathratus
Mar 28 at 3:34
1
$begingroup$
We're adding all sums $sum 1/k^2$ with $k=12m+1,,12m+5,,12m+9$, that is, all sums with $k=4n+1$. And subtracting similar sums.
$endgroup$
– David
Mar 28 at 3:49
add a comment |
$begingroup$
Firstly, any $n$ can be written as $6k$ or $6k+1$ or... or $6k+5$. So
$$eqalign
sum^infty_n=0fracsinBig[(2n+1)tfracpi6Big](2n+1)^2
&=sum_k=0^inftyfracsin(12k+1)fracpi6(12k+1)^2+hboxfive more sumscr
&=sum_k=0^inftyfracsinfracpi6(12k+1)^2+cdotscr
&=frac12sum_k=0^inftyfrac1(12k+1)^2+cdots .cr$$
Then if we write
$$S_m=sum_k=0^inftyfrac1(12k+m)^2$$
we have
$$G=S_1-S_3+S_5-S_7+S_9-S_11$$
and your sum $(2)$ is
$$eqalign
-frac12S_1&-S_3-frac12S_5+frac12S_7+S_9+frac12S_11cr
&=-frac12(S_1-S_3+S_5-S_7+S_9-S_11)-frac32S_3+frac32S_9cr
&=-frac12G-frac32frac19Gcr
&=-frac23G .cr$$
answered Mar 28 at 3:18
DavidDavid
69.6k668131
$endgroup$
$begingroup$
This works for me. Just for clarification: "$sin(12k+1)fracpi6$" is the same as $sinleft[(12k+1)fracpi6right]$?
$endgroup$
– clathratus
Mar 28 at 3:31
$begingroup$
Also: How do we have $$mathrm G=S_1-S_3+S_5-S_7+S_9-S_11$$
$endgroup$
– clathratus
Mar 28 at 3:34
1
$begingroup$
We're adding all sums $sum 1/k^2$ with $k=12m+1,,12m+5,,12m+9$, that is, all sums with $k=4n+1$. And subtracting similar sums.
$endgroup$
– David
Mar 28 at 3:49
add a comment |
$begingroup$
Firstly, any $n$ can be written as $6k$ or $6k+1$ or... or $6k+5$. So
$$eqalign
sum^infty_n=0fracsinBig[(2n+1)tfracpi6Big](2n+1)^2
&=sum_k=0^inftyfracsin(12k+1)fracpi6(12k+1)^2+hboxfive more sumscr
&=sum_k=0^inftyfracsinfracpi6(12k+1)^2+cdotscr
&=frac12sum_k=0^inftyfrac1(12k+1)^2+cdots .cr$$
Then if we write
$$S_m=sum_k=0^inftyfrac1(12k+m)^2$$
we have
$$G=S_1-S_3+S_5-S_7+S_9-S_11$$
and your sum $(2)$ is
$$eqalign
-frac12S_1&-S_3-frac12S_5+frac12S_7+S_9+frac12S_11cr
&=-frac12(S_1-S_3+S_5-S_7+S_9-S_11)-frac32S_3+frac32S_9cr
&=-frac12G-frac32frac19Gcr
&=-frac23G .cr$$
answered Mar 28 at 3:18
DavidDavid
69.6k668131
$endgroup$
Firstly, any $n$ can be written as $6k$ or $6k+1$ or... or $6k+5$. So
$$eqalign
sum^infty_n=0fracsinBig[(2n+1)tfracpi6Big](2n+1)^2
&=sum_k=0^inftyfracsin(12k+1)fracpi6(12k+1)^2+hboxfive more sumscr
&=sum_k=0^inftyfracsinfracpi6(12k+1)^2+cdotscr
&=frac12sum_k=0^inftyfrac1(12k+1)^2+cdots .cr$$
Then if we write
$$S_m=sum_k=0^inftyfrac1(12k+m)^2$$
we have
$$G=S_1-S_3+S_5-S_7+S_9-S_11$$
and your sum $(2)$ is
$$eqalign
-frac12S_1&-S_3-frac12S_5+frac12S_7+S_9+frac12S_11cr
&=-frac12(S_1-S_3+S_5-S_7+S_9-S_11)-frac32S_3+frac32S_9cr
&=-frac12G-frac32frac19Gcr
&=-frac23G .cr$$
answered Mar 28 at 3:18
DavidDavid
69.6k668131
answered Mar 28 at 3:18
DavidDavid
69.6k668131
answered Mar 28 at 3:18
DavidDavid
69.6k668131
answered Mar 28 at 3:18
DavidDavid
69.6k668131
69.6k668131
$begingroup$
This works for me. Just for clarification: "$sin(12k+1)fracpi6$" is the same as $sinleft[(12k+1)fracpi6right]$?
$endgroup$
– clathratus
Mar 28 at 3:31
$begingroup$
Also: How do we have $$mathrm G=S_1-S_3+S_5-S_7+S_9-S_11$$
$endgroup$
– clathratus
Mar 28 at 3:34
1
$begingroup$
We're adding all sums $sum 1/k^2$ with $k=12m+1,,12m+5,,12m+9$, that is, all sums with $k=4n+1$. And subtracting similar sums.
$endgroup$
– David
Mar 28 at 3:49
add a comment |
$begingroup$
This works for me. Just for clarification: "$sin(12k+1)fracpi6$" is the same as $sinleft[(12k+1)fracpi6right]$?
$endgroup$
– clathratus
Mar 28 at 3:31
$begingroup$
Also: How do we have $$mathrm G=S_1-S_3+S_5-S_7+S_9-S_11$$
$endgroup$
– clathratus
Mar 28 at 3:34
1
$begingroup$
We're adding all sums $sum 1/k^2$ with $k=12m+1,,12m+5,,12m+9$, that is, all sums with $k=4n+1$. And subtracting similar sums.
$endgroup$
– David
Mar 28 at 3:49
$begingroup$
This works for me. Just for clarification: "$sin(12k+1)fracpi6$" is the same as $sinleft[(12k+1)fracpi6right]$?
$endgroup$
– clathratus
Mar 28 at 3:31
$begingroup$
This works for me. Just for clarification: "$sin(12k+1)fracpi6$" is the same as $sinleft[(12k+1)fracpi6right]$?
$endgroup$
– clathratus
Mar 28 at 3:31
$begingroup$
Also: How do we have $$mathrm G=S_1-S_3+S_5-S_7+S_9-S_11$$
$endgroup$
– clathratus
Mar 28 at 3:34
$begingroup$
Also: How do we have $$mathrm G=S_1-S_3+S_5-S_7+S_9-S_11$$
$endgroup$
– clathratus
Mar 28 at 3:34
1
1
$begingroup$
We're adding all sums $sum 1/k^2$ with $k=12m+1,,12m+5,,12m+9$, that is, all sums with $k=4n+1$. And subtracting similar sums.
$endgroup$
– David
Mar 28 at 3:49
$begingroup$
We're adding all sums $sum 1/k^2$ with $k=12m+1,,12m+5,,12m+9$, that is, all sums with $k=4n+1$. And subtracting similar sums.
$endgroup$
– David
Mar 28 at 3:49
add a comment |
$begingroup$
As shown in this answer,
$$
log(2cos(x))=cos(2x)-fraccos(4x)2+fraccos(6x)3-dotstag1
$$
Substituting $xmapstofracpi2-x$ gives
$$
log(2sin(x))=-cos(2x)-fraccos(4x)2-fraccos(6x)3-dotstag2
$$
Subtracting $(1)$ from $(2)$ yields
$$
log(tan(x))=-2left(cos(2x)+fraccos(6x)3+fraccos(10x)5+dotsright)tag3
$$
Notice that $sinleft((2k+1)fracpi6right)$ has period $6$:
$$
beginarrayc
kpmod6&0&1&2&3&4&5\
color#C00sinleft((2k+1)fracpi6right)&color#C00frac12&color#C001&color#C00frac12&color#C00-frac12&color#C00-1&color#C00-frac12\
color#0902k+1&color#0901&color#0903&color#0905&color#0907&color#0909&color#09011\
color#C00frac(-1)^k2&color#C00frac12&color#C00-frac12&color#C00frac12&color#C00-frac12&color#C00frac12&color#C00-frac12\
color#C00frac32(-1)^frack-13&&color#C00frac32&&&color#C00-frac32&&kequiv1pmod3\
color#0903left(2frack-13+1right)&&color#0903&&&color#0909&&kequiv1pmod3 \
endarraytag4
$$
Note that the upper red row is the sum of the two lower red rows
Thus,
$$
beginalign
int_0^pi/12log(tan(x)),mathrmdx
&=-sum_k=0^inftyfracsinleft((2k+1)fracpi6right)(2k+1)^2tag5\
&=-sum_k=0^inftyleft(frac(-1)^k2frac1(2k+1)^2+frac(-1)^k9frac32frac1(2k+1)^2right)tag6\
&=-frac23sum_k=0^inftyfrac(-1)^k(2k+1)^2tag7\[3pt]
&=-frac23mathrmGtag8
endalign
$$
Explanation:
$(5)$: integrate $(3)$ term by term
$(6)$: apply $(4)$; $k$ in the right term is $frack-13$ from $(4)$
$(7)$: collect terms
$(8)$: definition of $mathrmG$
answered Mar 28 at 15:10
robjohn♦robjohn
270k27312640
$endgroup$
$begingroup$
Interesting...+1 is there a name for this technique, or is there any sort of generalization to it?
$endgroup$
– clathratus
Mar 28 at 16:45
$begingroup$
$(1)$-$(3)$ are Fourier Series, $(4)$ is picking out patterns in periodic functions.
$endgroup$
– robjohn♦
Mar 28 at 19:06
add a comment |
$begingroup$
As shown in this answer,
$$
log(2cos(x))=cos(2x)-fraccos(4x)2+fraccos(6x)3-dotstag1
$$
Substituting $xmapstofracpi2-x$ gives
$$
log(2sin(x))=-cos(2x)-fraccos(4x)2-fraccos(6x)3-dotstag2
$$
Subtracting $(1)$ from $(2)$ yields
$$
log(tan(x))=-2left(cos(2x)+fraccos(6x)3+fraccos(10x)5+dotsright)tag3
$$
Notice that $sinleft((2k+1)fracpi6right)$ has period $6$:
$$
beginarrayc
kpmod6&0&1&2&3&4&5\
color#C00sinleft((2k+1)fracpi6right)&color#C00frac12&color#C001&color#C00frac12&color#C00-frac12&color#C00-1&color#C00-frac12\
color#0902k+1&color#0901&color#0903&color#0905&color#0907&color#0909&color#09011\
color#C00frac(-1)^k2&color#C00frac12&color#C00-frac12&color#C00frac12&color#C00-frac12&color#C00frac12&color#C00-frac12\
color#C00frac32(-1)^frack-13&&color#C00frac32&&&color#C00-frac32&&kequiv1pmod3\
color#0903left(2frack-13+1right)&&color#0903&&&color#0909&&kequiv1pmod3 \
endarraytag4
$$
Note that the upper red row is the sum of the two lower red rows
Thus,
$$
beginalign
int_0^pi/12log(tan(x)),mathrmdx
&=-sum_k=0^inftyfracsinleft((2k+1)fracpi6right)(2k+1)^2tag5\
&=-sum_k=0^inftyleft(frac(-1)^k2frac1(2k+1)^2+frac(-1)^k9frac32frac1(2k+1)^2right)tag6\
&=-frac23sum_k=0^inftyfrac(-1)^k(2k+1)^2tag7\[3pt]
&=-frac23mathrmGtag8
endalign
$$
Explanation:
$(5)$: integrate $(3)$ term by term
$(6)$: apply $(4)$; $k$ in the right term is $frack-13$ from $(4)$
$(7)$: collect terms
$(8)$: definition of $mathrmG$
answered Mar 28 at 15:10
robjohn♦robjohn
270k27312640
$endgroup$
$begingroup$
Interesting...+1 is there a name for this technique, or is there any sort of generalization to it?
$endgroup$
– clathratus
Mar 28 at 16:45
$begingroup$
$(1)$-$(3)$ are Fourier Series, $(4)$ is picking out patterns in periodic functions.
$endgroup$
– robjohn♦
Mar 28 at 19:06
add a comment |
$begingroup$
As shown in this answer,
$$
log(2cos(x))=cos(2x)-fraccos(4x)2+fraccos(6x)3-dotstag1
$$
Substituting $xmapstofracpi2-x$ gives
$$
log(2sin(x))=-cos(2x)-fraccos(4x)2-fraccos(6x)3-dotstag2
$$
Subtracting $(1)$ from $(2)$ yields
$$
log(tan(x))=-2left(cos(2x)+fraccos(6x)3+fraccos(10x)5+dotsright)tag3
$$
Notice that $sinleft((2k+1)fracpi6right)$ has period $6$:
$$
beginarrayc
kpmod6&0&1&2&3&4&5\
color#C00sinleft((2k+1)fracpi6right)&color#C00frac12&color#C001&color#C00frac12&color#C00-frac12&color#C00-1&color#C00-frac12\
color#0902k+1&color#0901&color#0903&color#0905&color#0907&color#0909&color#09011\
color#C00frac(-1)^k2&color#C00frac12&color#C00-frac12&color#C00frac12&color#C00-frac12&color#C00frac12&color#C00-frac12\
color#C00frac32(-1)^frack-13&&color#C00frac32&&&color#C00-frac32&&kequiv1pmod3\
color#0903left(2frack-13+1right)&&color#0903&&&color#0909&&kequiv1pmod3 \
endarraytag4
$$
Note that the upper red row is the sum of the two lower red rows
Thus,
$$
beginalign
int_0^pi/12log(tan(x)),mathrmdx
&=-sum_k=0^inftyfracsinleft((2k+1)fracpi6right)(2k+1)^2tag5\
&=-sum_k=0^inftyleft(frac(-1)^k2frac1(2k+1)^2+frac(-1)^k9frac32frac1(2k+1)^2right)tag6\
&=-frac23sum_k=0^inftyfrac(-1)^k(2k+1)^2tag7\[3pt]
&=-frac23mathrmGtag8
endalign
$$
Explanation:
$(5)$: integrate $(3)$ term by term
$(6)$: apply $(4)$; $k$ in the right term is $frack-13$ from $(4)$
$(7)$: collect terms
$(8)$: definition of $mathrmG$
answered Mar 28 at 15:10
robjohn♦robjohn
270k27312640
$endgroup$
As shown in this answer,
$$
log(2cos(x))=cos(2x)-fraccos(4x)2+fraccos(6x)3-dotstag1
$$
Substituting $xmapstofracpi2-x$ gives
$$
log(2sin(x))=-cos(2x)-fraccos(4x)2-fraccos(6x)3-dotstag2
$$
Subtracting $(1)$ from $(2)$ yields
$$
log(tan(x))=-2left(cos(2x)+fraccos(6x)3+fraccos(10x)5+dotsright)tag3
$$
Notice that $sinleft((2k+1)fracpi6right)$ has period $6$:
$$
beginarrayc
kpmod6&0&1&2&3&4&5\
color#C00sinleft((2k+1)fracpi6right)&color#C00frac12&color#C001&color#C00frac12&color#C00-frac12&color#C00-1&color#C00-frac12\
color#0902k+1&color#0901&color#0903&color#0905&color#0907&color#0909&color#09011\
color#C00frac(-1)^k2&color#C00frac12&color#C00-frac12&color#C00frac12&color#C00-frac12&color#C00frac12&color#C00-frac12\
color#C00frac32(-1)^frack-13&&color#C00frac32&&&color#C00-frac32&&kequiv1pmod3\
color#0903left(2frack-13+1right)&&color#0903&&&color#0909&&kequiv1pmod3 \
endarraytag4
$$
Note that the upper red row is the sum of the two lower red rows
Thus,
$$
beginalign
int_0^pi/12log(tan(x)),mathrmdx
&=-sum_k=0^inftyfracsinleft((2k+1)fracpi6right)(2k+1)^2tag5\
&=-sum_k=0^inftyleft(frac(-1)^k2frac1(2k+1)^2+frac(-1)^k9frac32frac1(2k+1)^2right)tag6\
&=-frac23sum_k=0^inftyfrac(-1)^k(2k+1)^2tag7\[3pt]
&=-frac23mathrmGtag8
endalign
$$
Explanation:
$(5)$: integrate $(3)$ term by term
$(6)$: apply $(4)$; $k$ in the right term is $frack-13$ from $(4)$
$(7)$: collect terms
$(8)$: definition of $mathrmG$
answered Mar 28 at 15:10
robjohn♦robjohn
270k27312640
answered Mar 28 at 15:10
robjohn♦robjohn
270k27312640
answered Mar 28 at 15:10
robjohn♦robjohn
270k27312640
answered Mar 28 at 15:10
robjohn♦robjohn
270k27312640
270k27312640
$begingroup$
Interesting...+1 is there a name for this technique, or is there any sort of generalization to it?
$endgroup$
– clathratus
Mar 28 at 16:45
$begingroup$
$(1)$-$(3)$ are Fourier Series, $(4)$ is picking out patterns in periodic functions.
$endgroup$
– robjohn♦
Mar 28 at 19:06
add a comment |
$begingroup$
Interesting...+1 is there a name for this technique, or is there any sort of generalization to it?
$endgroup$
– clathratus
Mar 28 at 16:45
$begingroup$
$(1)$-$(3)$ are Fourier Series, $(4)$ is picking out patterns in periodic functions.
$endgroup$
– robjohn♦
Mar 28 at 19:06
$begingroup$
Interesting...+1 is there a name for this technique, or is there any sort of generalization to it?
$endgroup$
– clathratus
Mar 28 at 16:45
$begingroup$
Interesting...+1 is there a name for this technique, or is there any sort of generalization to it?
$endgroup$
– clathratus
Mar 28 at 16:45
$begingroup$
$(1)$-$(3)$ are Fourier Series, $(4)$ is picking out patterns in periodic functions.
$endgroup$
– robjohn♦
Mar 28 at 19:06
$begingroup$
$(1)$-$(3)$ are Fourier Series, $(4)$ is picking out patterns in periodic functions.
$endgroup$
– robjohn♦
Mar 28 at 19:06
add a comment |
$begingroup$
I understand how to obtain $(1)$, but I do not know how to go from $(1)$ to $(2)$ and then from $(2)$ to $(3)$.
HINT:
In order to arrive at $(2)$, note that $sinleft((2n+1)fracpi6right)$ takes on the values
$$sinleft((2n+1)fracpi6right)=begincasesfrac12 &,n=0,6,12,dots\\
1 &,n=1,7,13,dots\\
1/2&,n=2,8,14,dots\\
-frac12&,n=3,9,15,cdots\\
-1&,n=4,10,16,cdots\\
-frac12&,n=5,11,17,cdots
endcases$$
answered Mar 28 at 3:08
Mark ViolaMark Viola
134k1278176
$endgroup$
$begingroup$
If you hover your cursor above the accept-answer button, it says "Click to accept this answer because it solved your problem or it was most helpful in finding your solution..." The accepted answer was accepted because it was not a mere hint, but a full answer. Don't get me wrong-I appreciate your hint (as of now I am the only up-voter of it), but the accepted answer happens to be more helpful. Also: the first answer isn't always the best.
$endgroup$
– clathratus
Mar 28 at 22:22
$begingroup$
Thank you for the note. Was the "HINT" useful? I really wanted to give you the best answer I could, and thought the hint would have sufficed to show the way forward.
$endgroup$
– Mark Viola
Mar 28 at 22:30
$begingroup$
The hint was somewhat useful. I must admit I can't exactly see any patterns in the sets of values of $n$ which I could use... Perhaps your answer would be more helpful if you showed a little more how to proceed with this information. I really appreciate your dedication to the quality of your answers across the site.
$endgroup$
– clathratus
Mar 29 at 1:27
add a comment |
$begingroup$
I understand how to obtain $(1)$, but I do not know how to go from $(1)$ to $(2)$ and then from $(2)$ to $(3)$.
HINT:
In order to arrive at $(2)$, note that $sinleft((2n+1)fracpi6right)$ takes on the values
$$sinleft((2n+1)fracpi6right)=begincasesfrac12 &,n=0,6,12,dots\\
1 &,n=1,7,13,dots\\
1/2&,n=2,8,14,dots\\
-frac12&,n=3,9,15,cdots\\
-1&,n=4,10,16,cdots\\
-frac12&,n=5,11,17,cdots
endcases$$
answered Mar 28 at 3:08
Mark ViolaMark Viola
134k1278176
$endgroup$
$begingroup$
If you hover your cursor above the accept-answer button, it says "Click to accept this answer because it solved your problem or it was most helpful in finding your solution..." The accepted answer was accepted because it was not a mere hint, but a full answer. Don't get me wrong-I appreciate your hint (as of now I am the only up-voter of it), but the accepted answer happens to be more helpful. Also: the first answer isn't always the best.
$endgroup$
– clathratus
Mar 28 at 22:22
$begingroup$
Thank you for the note. Was the "HINT" useful? I really wanted to give you the best answer I could, and thought the hint would have sufficed to show the way forward.
$endgroup$
– Mark Viola
Mar 28 at 22:30
$begingroup$
The hint was somewhat useful. I must admit I can't exactly see any patterns in the sets of values of $n$ which I could use... Perhaps your answer would be more helpful if you showed a little more how to proceed with this information. I really appreciate your dedication to the quality of your answers across the site.
$endgroup$
– clathratus
Mar 29 at 1:27
add a comment |
$begingroup$
I understand how to obtain $(1)$, but I do not know how to go from $(1)$ to $(2)$ and then from $(2)$ to $(3)$.
HINT:
In order to arrive at $(2)$, note that $sinleft((2n+1)fracpi6right)$ takes on the values
$$sinleft((2n+1)fracpi6right)=begincasesfrac12 &,n=0,6,12,dots\\
1 &,n=1,7,13,dots\\
1/2&,n=2,8,14,dots\\
-frac12&,n=3,9,15,cdots\\
-1&,n=4,10,16,cdots\\
-frac12&,n=5,11,17,cdots
endcases$$
answered Mar 28 at 3:08
Mark ViolaMark Viola
134k1278176
$endgroup$
I understand how to obtain $(1)$, but I do not know how to go from $(1)$ to $(2)$ and then from $(2)$ to $(3)$.
HINT:
In order to arrive at $(2)$, note that $sinleft((2n+1)fracpi6right)$ takes on the values
$$sinleft((2n+1)fracpi6right)=begincasesfrac12 &,n=0,6,12,dots\\
1 &,n=1,7,13,dots\\
1/2&,n=2,8,14,dots\\
-frac12&,n=3,9,15,cdots\\
-1&,n=4,10,16,cdots\\
-frac12&,n=5,11,17,cdots
endcases$$
answered Mar 28 at 3:08
Mark ViolaMark Viola
134k1278176
answered Mar 28 at 3:08
Mark ViolaMark Viola
134k1278176
answered Mar 28 at 3:08
Mark ViolaMark Viola
134k1278176
answered Mar 28 at 3:08
Mark ViolaMark Viola
134k1278176
134k1278176
$begingroup$
If you hover your cursor above the accept-answer button, it says "Click to accept this answer because it solved your problem or it was most helpful in finding your solution..." The accepted answer was accepted because it was not a mere hint, but a full answer. Don't get me wrong-I appreciate your hint (as of now I am the only up-voter of it), but the accepted answer happens to be more helpful. Also: the first answer isn't always the best.
$endgroup$
– clathratus
Mar 28 at 22:22
$begingroup$
Thank you for the note. Was the "HINT" useful? I really wanted to give you the best answer I could, and thought the hint would have sufficed to show the way forward.
$endgroup$
– Mark Viola
Mar 28 at 22:30
$begingroup$
The hint was somewhat useful. I must admit I can't exactly see any patterns in the sets of values of $n$ which I could use... Perhaps your answer would be more helpful if you showed a little more how to proceed with this information. I really appreciate your dedication to the quality of your answers across the site.
$endgroup$
– clathratus
Mar 29 at 1:27
add a comment |
$begingroup$
If you hover your cursor above the accept-answer button, it says "Click to accept this answer because it solved your problem or it was most helpful in finding your solution..." The accepted answer was accepted because it was not a mere hint, but a full answer. Don't get me wrong-I appreciate your hint (as of now I am the only up-voter of it), but the accepted answer happens to be more helpful. Also: the first answer isn't always the best.
$endgroup$
– clathratus
Mar 28 at 22:22
$begingroup$
Thank you for the note. Was the "HINT" useful? I really wanted to give you the best answer I could, and thought the hint would have sufficed to show the way forward.
$endgroup$
– Mark Viola
Mar 28 at 22:30
$begingroup$
The hint was somewhat useful. I must admit I can't exactly see any patterns in the sets of values of $n$ which I could use... Perhaps your answer would be more helpful if you showed a little more how to proceed with this information. I really appreciate your dedication to the quality of your answers across the site.
$endgroup$
– clathratus
Mar 29 at 1:27
$begingroup$
If you hover your cursor above the accept-answer button, it says "Click to accept this answer because it solved your problem or it was most helpful in finding your solution..." The accepted answer was accepted because it was not a mere hint, but a full answer. Don't get me wrong-I appreciate your hint (as of now I am the only up-voter of it), but the accepted answer happens to be more helpful. Also: the first answer isn't always the best.
$endgroup$
– clathratus
Mar 28 at 22:22
$begingroup$
If you hover your cursor above the accept-answer button, it says "Click to accept this answer because it solved your problem or it was most helpful in finding your solution..." The accepted answer was accepted because it was not a mere hint, but a full answer. Don't get me wrong-I appreciate your hint (as of now I am the only up-voter of it), but the accepted answer happens to be more helpful. Also: the first answer isn't always the best.
$endgroup$
– clathratus
Mar 28 at 22:22
$begingroup$
Thank you for the note. Was the "HINT" useful? I really wanted to give you the best answer I could, and thought the hint would have sufficed to show the way forward.
$endgroup$
– Mark Viola
Mar 28 at 22:30
$begingroup$
Thank you for the note. Was the "HINT" useful? I really wanted to give you the best answer I could, and thought the hint would have sufficed to show the way forward.
$endgroup$
– Mark Viola
Mar 28 at 22:30
$begingroup$
The hint was somewhat useful. I must admit I can't exactly see any patterns in the sets of values of $n$ which I could use... Perhaps your answer would be more helpful if you showed a little more how to proceed with this information. I really appreciate your dedication to the quality of your answers across the site.
$endgroup$
– clathratus
Mar 29 at 1:27
$begingroup$
The hint was somewhat useful. I must admit I can't exactly see any patterns in the sets of values of $n$ which I could use... Perhaps your answer would be more helpful if you showed a little more how to proceed with this information. I really appreciate your dedication to the quality of your answers across the site.
$endgroup$
– clathratus
Mar 29 at 1:27
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%2f3165412%2fproof-clarification-int-0-pi-12-log-tan-tdt-frac23-mathrm-g%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
