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












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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


















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