Generating function - $1,1,1,1,1,1$












0












$begingroup$


What is the generating function for $1,1,1,1,1,1$?



I know this to be $1 + x +x^2+ x^3+x^4+x^5$



But then I saw this:



$$frac{x^6-1}{x-1} = 1 + x +x^2+ x^3+x^4+x^5$$



How was this equality obtained?



Was it just a random (manual)? or is there any method involved to obtain that fractional part?






generating-functions







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Are you asking why $frac{x^6-1}{x-1}=x^5+x^4+x^3+x^2+x+1$? Multiply both sides by $x-1$, expand the RHS, and things should cancel out to $x^6-1$. In general, $x^a-1=(x-a)(x^{a-1}+x^{a-2}+dots+x^2+x+1)$
    $endgroup$
    – user574848
    Dec 25 '18 at 10:59












  • $begingroup$
    I take it you've checked oeis.org
    $endgroup$
    – onepound
    Dec 25 '18 at 11:00










  • $begingroup$
    thanks user574848, but why there was a need to write that equation ?
    $endgroup$
    – swapnil
    Dec 25 '18 at 11:07








  • 4




    $begingroup$
    Geometric progression?
    $endgroup$
    – Lord Shark the Unknown
    Dec 25 '18 at 11:18










  • $begingroup$
    okay got it, that was a silly doubt didn't notice that there is geometric progression involved. I though it must be related with theorem related to the generating function.
    $endgroup$
    – swapnil
    Dec 25 '18 at 11:21
















0












$begingroup$


What is the generating function for $1,1,1,1,1,1$?



I know this to be $1 + x +x^2+ x^3+x^4+x^5$



But then I saw this:



$$frac{x^6-1}{x-1} = 1 + x +x^2+ x^3+x^4+x^5$$



How was this equality obtained?



Was it just a random (manual)? or is there any method involved to obtain that fractional part?






generating-functions







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Are you asking why $frac{x^6-1}{x-1}=x^5+x^4+x^3+x^2+x+1$? Multiply both sides by $x-1$, expand the RHS, and things should cancel out to $x^6-1$. In general, $x^a-1=(x-a)(x^{a-1}+x^{a-2}+dots+x^2+x+1)$
    $endgroup$
    – user574848
    Dec 25 '18 at 10:59












  • $begingroup$
    I take it you've checked oeis.org
    $endgroup$
    – onepound
    Dec 25 '18 at 11:00










  • $begingroup$
    thanks user574848, but why there was a need to write that equation ?
    $endgroup$
    – swapnil
    Dec 25 '18 at 11:07








  • 4




    $begingroup$
    Geometric progression?
    $endgroup$
    – Lord Shark the Unknown
    Dec 25 '18 at 11:18










  • $begingroup$
    okay got it, that was a silly doubt didn't notice that there is geometric progression involved. I though it must be related with theorem related to the generating function.
    $endgroup$
    – swapnil
    Dec 25 '18 at 11:21














0












0








0





$begingroup$


What is the generating function for $1,1,1,1,1,1$?



I know this to be $1 + x +x^2+ x^3+x^4+x^5$



But then I saw this:



$$frac{x^6-1}{x-1} = 1 + x +x^2+ x^3+x^4+x^5$$



How was this equality obtained?



Was it just a random (manual)? or is there any method involved to obtain that fractional part?






generating-functions







share|cite|improve this question











$endgroup$




What is the generating function for $1,1,1,1,1,1$?



I know this to be $1 + x +x^2+ x^3+x^4+x^5$



But then I saw this:



$$frac{x^6-1}{x-1} = 1 + x +x^2+ x^3+x^4+x^5$$



How was this equality obtained?



Was it just a random (manual)? or is there any method involved to obtain that fractional part?






generating-functions




generating-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 5:57









Eevee Trainer

10.1k31742




10.1k31742










asked Dec 25 '18 at 10:55









swapnilswapnil

335




335








  • 3




    $begingroup$
    Are you asking why $frac{x^6-1}{x-1}=x^5+x^4+x^3+x^2+x+1$? Multiply both sides by $x-1$, expand the RHS, and things should cancel out to $x^6-1$. In general, $x^a-1=(x-a)(x^{a-1}+x^{a-2}+dots+x^2+x+1)$
    $endgroup$
    – user574848
    Dec 25 '18 at 10:59












  • $begingroup$
    I take it you've checked oeis.org
    $endgroup$
    – onepound
    Dec 25 '18 at 11:00










  • $begingroup$
    thanks user574848, but why there was a need to write that equation ?
    $endgroup$
    – swapnil
    Dec 25 '18 at 11:07








  • 4




    $begingroup$
    Geometric progression?
    $endgroup$
    – Lord Shark the Unknown
    Dec 25 '18 at 11:18










  • $begingroup$
    okay got it, that was a silly doubt didn't notice that there is geometric progression involved. I though it must be related with theorem related to the generating function.
    $endgroup$
    – swapnil
    Dec 25 '18 at 11:21














  • 3




    $begingroup$
    Are you asking why $frac{x^6-1}{x-1}=x^5+x^4+x^3+x^2+x+1$? Multiply both sides by $x-1$, expand the RHS, and things should cancel out to $x^6-1$. In general, $x^a-1=(x-a)(x^{a-1}+x^{a-2}+dots+x^2+x+1)$
    $endgroup$
    – user574848
    Dec 25 '18 at 10:59












  • $begingroup$
    I take it you've checked oeis.org
    $endgroup$
    – onepound
    Dec 25 '18 at 11:00










  • $begingroup$
    thanks user574848, but why there was a need to write that equation ?
    $endgroup$
    – swapnil
    Dec 25 '18 at 11:07








  • 4




    $begingroup$
    Geometric progression?
    $endgroup$
    – Lord Shark the Unknown
    Dec 25 '18 at 11:18










  • $begingroup$
    okay got it, that was a silly doubt didn't notice that there is geometric progression involved. I though it must be related with theorem related to the generating function.
    $endgroup$
    – swapnil
    Dec 25 '18 at 11:21








3




3




$begingroup$
Are you asking why $frac{x^6-1}{x-1}=x^5+x^4+x^3+x^2+x+1$? Multiply both sides by $x-1$, expand the RHS, and things should cancel out to $x^6-1$. In general, $x^a-1=(x-a)(x^{a-1}+x^{a-2}+dots+x^2+x+1)$
$endgroup$
– user574848
Dec 25 '18 at 10:59






$begingroup$
Are you asking why $frac{x^6-1}{x-1}=x^5+x^4+x^3+x^2+x+1$? Multiply both sides by $x-1$, expand the RHS, and things should cancel out to $x^6-1$. In general, $x^a-1=(x-a)(x^{a-1}+x^{a-2}+dots+x^2+x+1)$
$endgroup$
– user574848
Dec 25 '18 at 10:59














$begingroup$
I take it you've checked oeis.org
$endgroup$
– onepound
Dec 25 '18 at 11:00




$begingroup$
I take it you've checked oeis.org
$endgroup$
– onepound
Dec 25 '18 at 11:00












$begingroup$
thanks user574848, but why there was a need to write that equation ?
$endgroup$
– swapnil
Dec 25 '18 at 11:07






$begingroup$
thanks user574848, but why there was a need to write that equation ?
$endgroup$
– swapnil
Dec 25 '18 at 11:07






4




4




$begingroup$
Geometric progression?
$endgroup$
– Lord Shark the Unknown
Dec 25 '18 at 11:18




$begingroup$
Geometric progression?
$endgroup$
– Lord Shark the Unknown
Dec 25 '18 at 11:18












$begingroup$
okay got it, that was a silly doubt didn't notice that there is geometric progression involved. I though it must be related with theorem related to the generating function.
$endgroup$
– swapnil
Dec 25 '18 at 11:21




$begingroup$
okay got it, that was a silly doubt didn't notice that there is geometric progression involved. I though it must be related with theorem related to the generating function.
$endgroup$
– swapnil
Dec 25 '18 at 11:21










1 Answer
1






active

oldest

votes


















0












$begingroup$

As noted in the comments to your question, the equation



$$frac{x^6 - 1}{x-1} = 1 + x + x^2 + x^3 + x^4 + x^5$$



comes about as the sum of a finite geometric series. Suppose we have a finite geometric series of ratio $x$. Then it can be shown



$$1 + x + x^2 + ... + x^n = frac{x^{n+1} - 1}{x-1}$$



Take $n=5$ and the equality results.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It should be stated when using the sum of the geometric power that it doesn't hold when $x = 1$.
    $endgroup$
    – user150203
    Jan 8 at 5:36












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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3052007%2fgenerating-function-1-1-1-1-1-1%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

As noted in the comments to your question, the equation



$$frac{x^6 - 1}{x-1} = 1 + x + x^2 + x^3 + x^4 + x^5$$



comes about as the sum of a finite geometric series. Suppose we have a finite geometric series of ratio $x$. Then it can be shown



$$1 + x + x^2 + ... + x^n = frac{x^{n+1} - 1}{x-1}$$



Take $n=5$ and the equality results.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It should be stated when using the sum of the geometric power that it doesn't hold when $x = 1$.
    $endgroup$
    – user150203
    Jan 8 at 5:36
















0












$begingroup$

As noted in the comments to your question, the equation



$$frac{x^6 - 1}{x-1} = 1 + x + x^2 + x^3 + x^4 + x^5$$



comes about as the sum of a finite geometric series. Suppose we have a finite geometric series of ratio $x$. Then it can be shown



$$1 + x + x^2 + ... + x^n = frac{x^{n+1} - 1}{x-1}$$



Take $n=5$ and the equality results.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It should be stated when using the sum of the geometric power that it doesn't hold when $x = 1$.
    $endgroup$
    – user150203
    Jan 8 at 5:36














0












0








0





$begingroup$

As noted in the comments to your question, the equation



$$frac{x^6 - 1}{x-1} = 1 + x + x^2 + x^3 + x^4 + x^5$$



comes about as the sum of a finite geometric series. Suppose we have a finite geometric series of ratio $x$. Then it can be shown



$$1 + x + x^2 + ... + x^n = frac{x^{n+1} - 1}{x-1}$$



Take $n=5$ and the equality results.






share|cite|improve this answer









$endgroup$



As noted in the comments to your question, the equation



$$frac{x^6 - 1}{x-1} = 1 + x + x^2 + x^3 + x^4 + x^5$$



comes about as the sum of a finite geometric series. Suppose we have a finite geometric series of ratio $x$. Then it can be shown



$$1 + x + x^2 + ... + x^n = frac{x^{n+1} - 1}{x-1}$$



Take $n=5$ and the equality results.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 27 '18 at 5:55









Eevee TrainerEevee Trainer

10.1k31742




10.1k31742












  • $begingroup$
    It should be stated when using the sum of the geometric power that it doesn't hold when $x = 1$.
    $endgroup$
    – user150203
    Jan 8 at 5:36


















  • $begingroup$
    It should be stated when using the sum of the geometric power that it doesn't hold when $x = 1$.
    $endgroup$
    – user150203
    Jan 8 at 5:36
















$begingroup$
It should be stated when using the sum of the geometric power that it doesn't hold when $x = 1$.
$endgroup$
– user150203
Jan 8 at 5:36




$begingroup$
It should be stated when using the sum of the geometric power that it doesn't hold when $x = 1$.
$endgroup$
– user150203
Jan 8 at 5:36


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3052007%2fgenerating-function-1-1-1-1-1-1%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Bundesstraße 106

Image
Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
Read more

Verónica Boquete

Image
Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
Read more

Ida-Boy-Ed-Garten

Image
Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
Read more