combinatorics - generating functions Announcing the arrival of Valued Associate #679: Cesar...
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combinatorics - generating functions
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Generating functions word problem(balloons)Combinations of a Multi-set with generating functionsgenerating function combinatorics solutionLearning about generating functions and sequences.Stuck on Generating FunctionsGenerating Functions for MultinomialsFind the number of solutions of the equation using generating functionsGenerating functions $a_{n} =n^{2}$Generating Functions To Deal WithProve Fibonacci Identity using generating functions
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I need help making an OGF for $1 + x^i + x^{2i}+...+x^{ki}$. I already know how to verify that $1 +x +x^2+...+x^k$ can be written by $({1-x^{k+1}})/({1-x})$. I'm wondering if there is any correlation between the two..?
Any help would be greatly appreciated. Thanks!
combinatorics generating-functions
asked Mar 25 at 23:21
Jeffery RiceJeffery Rice
134
$endgroup$
add a comment |
$begingroup$
I need help making an OGF for $1 + x^i + x^{2i}+...+x^{ki}$. I already know how to verify that $1 +x +x^2+...+x^k$ can be written by $({1-x^{k+1}})/({1-x})$. I'm wondering if there is any correlation between the two..?
Any help would be greatly appreciated. Thanks!
combinatorics generating-functions
asked Mar 25 at 23:21
Jeffery RiceJeffery Rice
134
$endgroup$
$begingroup$
Hint: If you wee to substitute $z=x^i$, what would your GF look like then?
$endgroup$
– awkward
Mar 25 at 23:26
add a comment |
$begingroup$
I need help making an OGF for $1 + x^i + x^{2i}+...+x^{ki}$. I already know how to verify that $1 +x +x^2+...+x^k$ can be written by $({1-x^{k+1}})/({1-x})$. I'm wondering if there is any correlation between the two..?
Any help would be greatly appreciated. Thanks!
combinatorics generating-functions
asked Mar 25 at 23:21
Jeffery RiceJeffery Rice
134
$endgroup$
I need help making an OGF for $1 + x^i + x^{2i}+...+x^{ki}$. I already know how to verify that $1 +x +x^2+...+x^k$ can be written by $({1-x^{k+1}})/({1-x})$. I'm wondering if there is any correlation between the two..?
Any help would be greatly appreciated. Thanks!
combinatorics generating-functions
combinatorics generating-functions
asked Mar 25 at 23:21
Jeffery RiceJeffery Rice
134
asked Mar 25 at 23:21
Jeffery RiceJeffery Rice
134
asked Mar 25 at 23:21
Jeffery RiceJeffery Rice
134
asked Mar 25 at 23:21
Jeffery RiceJeffery Rice
134
asked Mar 25 at 23:21
Jeffery RiceJeffery Rice
134
134
$begingroup$
Hint: If you wee to substitute $z=x^i$, what would your GF look like then?
$endgroup$
– awkward
Mar 25 at 23:26
add a comment |
$begingroup$
Hint: If you wee to substitute $z=x^i$, what would your GF look like then?
$endgroup$
– awkward
Mar 25 at 23:26
$begingroup$
Hint: If you wee to substitute $z=x^i$, what would your GF look like then?
$endgroup$
– awkward
Mar 25 at 23:26
$begingroup$
Hint: If you wee to substitute $z=x^i$, what would your GF look like then?
$endgroup$
– awkward
Mar 25 at 23:26
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: make the substitution $u=x^i$.
answered Mar 25 at 23:27
Peter ForemanPeter Foreman
8,3351321
$endgroup$
add a comment |
$begingroup$
Notice that
$$1 + x^i + x^{2i} + x^{3i} + ... + x^{ki} = 1 + (x^i)^1 + (x^i)^2 + (x^i)^3 + ... + (x^i)^k$$
This is a finite geometric series with ratio $x^i$, and thus
$$1 + x^i + x^{2i} + x^{3i} + ... + x^{ki} = 1 + (x^i)^1 + (x^i)^2 + (x^i)^3 + ... + (x^i)^k = frac{1 - (x^i)^{k+1}}{1 - x^i}$$
answered Mar 25 at 23:28
Eevee TrainerEevee Trainer
10.6k31842
$endgroup$
$begingroup$
Thank you so much. That makes sense, it's just a simple substitution!
$endgroup$
– Jeffery Rice
Mar 25 at 23:38
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: make the substitution $u=x^i$.
answered Mar 25 at 23:27
Peter ForemanPeter Foreman
8,3351321
$endgroup$
add a comment |
$begingroup$
Hint: make the substitution $u=x^i$.
answered Mar 25 at 23:27
Peter ForemanPeter Foreman
8,3351321
$endgroup$
add a comment |
$begingroup$
Hint: make the substitution $u=x^i$.
answered Mar 25 at 23:27
Peter ForemanPeter Foreman
8,3351321
$endgroup$
Hint: make the substitution $u=x^i$.
answered Mar 25 at 23:27
Peter ForemanPeter Foreman
8,3351321
answered Mar 25 at 23:27
Peter ForemanPeter Foreman
8,3351321
answered Mar 25 at 23:27
Peter ForemanPeter Foreman
8,3351321
answered Mar 25 at 23:27
Peter ForemanPeter Foreman
8,3351321
8,3351321
add a comment |
add a comment |
$begingroup$
Notice that
$$1 + x^i + x^{2i} + x^{3i} + ... + x^{ki} = 1 + (x^i)^1 + (x^i)^2 + (x^i)^3 + ... + (x^i)^k$$
This is a finite geometric series with ratio $x^i$, and thus
$$1 + x^i + x^{2i} + x^{3i} + ... + x^{ki} = 1 + (x^i)^1 + (x^i)^2 + (x^i)^3 + ... + (x^i)^k = frac{1 - (x^i)^{k+1}}{1 - x^i}$$
answered Mar 25 at 23:28
Eevee TrainerEevee Trainer
10.6k31842
$endgroup$
$begingroup$
Thank you so much. That makes sense, it's just a simple substitution!
$endgroup$
– Jeffery Rice
Mar 25 at 23:38
add a comment |
$begingroup$
Notice that
$$1 + x^i + x^{2i} + x^{3i} + ... + x^{ki} = 1 + (x^i)^1 + (x^i)^2 + (x^i)^3 + ... + (x^i)^k$$
This is a finite geometric series with ratio $x^i$, and thus
$$1 + x^i + x^{2i} + x^{3i} + ... + x^{ki} = 1 + (x^i)^1 + (x^i)^2 + (x^i)^3 + ... + (x^i)^k = frac{1 - (x^i)^{k+1}}{1 - x^i}$$
answered Mar 25 at 23:28
Eevee TrainerEevee Trainer
10.6k31842
$endgroup$
$begingroup$
Thank you so much. That makes sense, it's just a simple substitution!
$endgroup$
– Jeffery Rice
Mar 25 at 23:38
add a comment |
$begingroup$
Notice that
$$1 + x^i + x^{2i} + x^{3i} + ... + x^{ki} = 1 + (x^i)^1 + (x^i)^2 + (x^i)^3 + ... + (x^i)^k$$
This is a finite geometric series with ratio $x^i$, and thus
$$1 + x^i + x^{2i} + x^{3i} + ... + x^{ki} = 1 + (x^i)^1 + (x^i)^2 + (x^i)^3 + ... + (x^i)^k = frac{1 - (x^i)^{k+1}}{1 - x^i}$$
answered Mar 25 at 23:28
Eevee TrainerEevee Trainer
10.6k31842
$endgroup$
Notice that
$$1 + x^i + x^{2i} + x^{3i} + ... + x^{ki} = 1 + (x^i)^1 + (x^i)^2 + (x^i)^3 + ... + (x^i)^k$$
This is a finite geometric series with ratio $x^i$, and thus
$$1 + x^i + x^{2i} + x^{3i} + ... + x^{ki} = 1 + (x^i)^1 + (x^i)^2 + (x^i)^3 + ... + (x^i)^k = frac{1 - (x^i)^{k+1}}{1 - x^i}$$
answered Mar 25 at 23:28
Eevee TrainerEevee Trainer
10.6k31842
answered Mar 25 at 23:28
Eevee TrainerEevee Trainer
10.6k31842
answered Mar 25 at 23:28
Eevee TrainerEevee Trainer
10.6k31842
answered Mar 25 at 23:28
Eevee TrainerEevee Trainer
10.6k31842
10.6k31842
$begingroup$
Thank you so much. That makes sense, it's just a simple substitution!
$endgroup$
– Jeffery Rice
Mar 25 at 23:38
add a comment |
$begingroup$
Thank you so much. That makes sense, it's just a simple substitution!
$endgroup$
– Jeffery Rice
Mar 25 at 23:38
$begingroup$
Thank you so much. That makes sense, it's just a simple substitution!
$endgroup$
– Jeffery Rice
Mar 25 at 23:38
$begingroup$
Thank you so much. That makes sense, it's just a simple substitution!
$endgroup$
– Jeffery Rice
Mar 25 at 23:38
add a comment |
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$begingroup$
Hint: If you wee to substitute $z=x^i$, what would your GF look like then?
$endgroup$
– awkward
Mar 25 at 23:26