Difference Between Generating Functions And Exponential Generating Functions [duplicate]
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This question already has an answer here:
What is the need of exponential generating functions on combinatorial problems?
2 answers
What is the difference between generating functions and exponential generating functions?
I mean, we can solve a combinatorics sum by using just simple generating functions.
Or simply, for which parts in combinatorics do we use simple generating functions and in which parts we use exponential generating functions?
combinatorics
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marked as duplicate by Hans Lundmark, amWhy, darij grinberg, Math1000, Shailesh Jan 16 at 0:01
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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$begingroup$
This question already has an answer here:
What is the need of exponential generating functions on combinatorial problems?
2 answers
What is the difference between generating functions and exponential generating functions?
I mean, we can solve a combinatorics sum by using just simple generating functions.
Or simply, for which parts in combinatorics do we use simple generating functions and in which parts we use exponential generating functions?
combinatorics
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marked as duplicate by Hans Lundmark, amWhy, darij grinberg, Math1000, Shailesh Jan 16 at 0:01
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
What is the need of exponential generating functions on combinatorial problems?
2 answers
What is the difference between generating functions and exponential generating functions?
I mean, we can solve a combinatorics sum by using just simple generating functions.
Or simply, for which parts in combinatorics do we use simple generating functions and in which parts we use exponential generating functions?
combinatorics
$endgroup$
This question already has an answer here:
What is the need of exponential generating functions on combinatorial problems?
2 answers
What is the difference between generating functions and exponential generating functions?
I mean, we can solve a combinatorics sum by using just simple generating functions.
Or simply, for which parts in combinatorics do we use simple generating functions and in which parts we use exponential generating functions?
This question already has an answer here:
What is the need of exponential generating functions on combinatorial problems?
2 answers
combinatorics
combinatorics
asked Jan 15 at 11:08
SaeeSaee
387
387
marked as duplicate by Hans Lundmark, amWhy, darij grinberg, Math1000, Shailesh Jan 16 at 0:01
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Hans Lundmark, amWhy, darij grinberg, Math1000, Shailesh Jan 16 at 0:01
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
add a comment |
2 Answers
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Aside from the obvious - exponential generating functions divide by a factorial, and ordinary generating functions don't - both sorts of generating functions are transform methods. One key thing transform methods do? They turn convolution into pointwise multiplication. The difference comes in there, in what sort of convolution this applies to.
For ordinary generating functions, that convolution is
$$(a*b)_n = sum_{j=0}^n a_j b_{n-j}$$
corresponding to the usual translation-invariant counting measure on the nonnegative integers. There are a great many situations in which this appears.
For exponential generating functions, that convolution is
$$(a*b)_n = sum_{j=0}^n binom{n}{j} a_j b_{n-j}$$
This form, reminiscent of the binomial theorem, is something that occasionally comes up in combinatorics - and when it does, generating functions aren't going to work well unless we make that adjustment.
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So, can you just give me an example for which you need exponential generating functions and tell me why do you need to use exponential generating functions and not simple generating functions. Thanks
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– Saee
Jan 15 at 15:40
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Hint: Answer moved to the MSE post Deep understanding on exponential generating function.
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add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Aside from the obvious - exponential generating functions divide by a factorial, and ordinary generating functions don't - both sorts of generating functions are transform methods. One key thing transform methods do? They turn convolution into pointwise multiplication. The difference comes in there, in what sort of convolution this applies to.
For ordinary generating functions, that convolution is
$$(a*b)_n = sum_{j=0}^n a_j b_{n-j}$$
corresponding to the usual translation-invariant counting measure on the nonnegative integers. There are a great many situations in which this appears.
For exponential generating functions, that convolution is
$$(a*b)_n = sum_{j=0}^n binom{n}{j} a_j b_{n-j}$$
This form, reminiscent of the binomial theorem, is something that occasionally comes up in combinatorics - and when it does, generating functions aren't going to work well unless we make that adjustment.
$endgroup$
$begingroup$
So, can you just give me an example for which you need exponential generating functions and tell me why do you need to use exponential generating functions and not simple generating functions. Thanks
$endgroup$
– Saee
Jan 15 at 15:40
add a comment |
$begingroup$
Aside from the obvious - exponential generating functions divide by a factorial, and ordinary generating functions don't - both sorts of generating functions are transform methods. One key thing transform methods do? They turn convolution into pointwise multiplication. The difference comes in there, in what sort of convolution this applies to.
For ordinary generating functions, that convolution is
$$(a*b)_n = sum_{j=0}^n a_j b_{n-j}$$
corresponding to the usual translation-invariant counting measure on the nonnegative integers. There are a great many situations in which this appears.
For exponential generating functions, that convolution is
$$(a*b)_n = sum_{j=0}^n binom{n}{j} a_j b_{n-j}$$
This form, reminiscent of the binomial theorem, is something that occasionally comes up in combinatorics - and when it does, generating functions aren't going to work well unless we make that adjustment.
$endgroup$
$begingroup$
So, can you just give me an example for which you need exponential generating functions and tell me why do you need to use exponential generating functions and not simple generating functions. Thanks
$endgroup$
– Saee
Jan 15 at 15:40
add a comment |
$begingroup$
Aside from the obvious - exponential generating functions divide by a factorial, and ordinary generating functions don't - both sorts of generating functions are transform methods. One key thing transform methods do? They turn convolution into pointwise multiplication. The difference comes in there, in what sort of convolution this applies to.
For ordinary generating functions, that convolution is
$$(a*b)_n = sum_{j=0}^n a_j b_{n-j}$$
corresponding to the usual translation-invariant counting measure on the nonnegative integers. There are a great many situations in which this appears.
For exponential generating functions, that convolution is
$$(a*b)_n = sum_{j=0}^n binom{n}{j} a_j b_{n-j}$$
This form, reminiscent of the binomial theorem, is something that occasionally comes up in combinatorics - and when it does, generating functions aren't going to work well unless we make that adjustment.
$endgroup$
Aside from the obvious - exponential generating functions divide by a factorial, and ordinary generating functions don't - both sorts of generating functions are transform methods. One key thing transform methods do? They turn convolution into pointwise multiplication. The difference comes in there, in what sort of convolution this applies to.
For ordinary generating functions, that convolution is
$$(a*b)_n = sum_{j=0}^n a_j b_{n-j}$$
corresponding to the usual translation-invariant counting measure on the nonnegative integers. There are a great many situations in which this appears.
For exponential generating functions, that convolution is
$$(a*b)_n = sum_{j=0}^n binom{n}{j} a_j b_{n-j}$$
This form, reminiscent of the binomial theorem, is something that occasionally comes up in combinatorics - and when it does, generating functions aren't going to work well unless we make that adjustment.
answered Jan 15 at 11:34


jmerryjmerry
9,4231124
9,4231124
$begingroup$
So, can you just give me an example for which you need exponential generating functions and tell me why do you need to use exponential generating functions and not simple generating functions. Thanks
$endgroup$
– Saee
Jan 15 at 15:40
add a comment |
$begingroup$
So, can you just give me an example for which you need exponential generating functions and tell me why do you need to use exponential generating functions and not simple generating functions. Thanks
$endgroup$
– Saee
Jan 15 at 15:40
$begingroup$
So, can you just give me an example for which you need exponential generating functions and tell me why do you need to use exponential generating functions and not simple generating functions. Thanks
$endgroup$
– Saee
Jan 15 at 15:40
$begingroup$
So, can you just give me an example for which you need exponential generating functions and tell me why do you need to use exponential generating functions and not simple generating functions. Thanks
$endgroup$
– Saee
Jan 15 at 15:40
add a comment |
$begingroup$
Hint: Answer moved to the MSE post Deep understanding on exponential generating function.
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add a comment |
$begingroup$
Hint: Answer moved to the MSE post Deep understanding on exponential generating function.
$endgroup$
add a comment |
$begingroup$
Hint: Answer moved to the MSE post Deep understanding on exponential generating function.
$endgroup$
Hint: Answer moved to the MSE post Deep understanding on exponential generating function.
edited Jan 16 at 12:20
answered Jan 15 at 20:50


Markus ScheuerMarkus Scheuer
61.8k457147
61.8k457147
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