Exponentiate generating functions as formal power series
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In my discrete math class, we are studying generating functions. We learned that
$$
e^x = sum_{i = 0}^{infty} frac{x^i}{i!},
$$
which is certainly an identity in calculus. However, in the ring of formal power series, is there any sort of concept of exponentiation? Suppose we have $F, G in R[[X]].$ Is there any sort of a concept or definition for
$$
F^G?
$$
I am mostly looking for a reason why the formal power series $e$ (i.e., the power series whose constant coefficient is $e$ and all remaining coefficients are 0), when raised to the power of the power series $x$, satisfies the above identity. A generalization would be appreciated as well, for the $F^G$ case.
abstract-algebra combinatorics generating-functions formal-power-series
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up vote
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favorite
In my discrete math class, we are studying generating functions. We learned that
$$
e^x = sum_{i = 0}^{infty} frac{x^i}{i!},
$$
which is certainly an identity in calculus. However, in the ring of formal power series, is there any sort of concept of exponentiation? Suppose we have $F, G in R[[X]].$ Is there any sort of a concept or definition for
$$
F^G?
$$
I am mostly looking for a reason why the formal power series $e$ (i.e., the power series whose constant coefficient is $e$ and all remaining coefficients are 0), when raised to the power of the power series $x$, satisfies the above identity. A generalization would be appreciated as well, for the $F^G$ case.
abstract-algebra combinatorics generating-functions formal-power-series
New contributor
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In my discrete math class, we are studying generating functions. We learned that
$$
e^x = sum_{i = 0}^{infty} frac{x^i}{i!},
$$
which is certainly an identity in calculus. However, in the ring of formal power series, is there any sort of concept of exponentiation? Suppose we have $F, G in R[[X]].$ Is there any sort of a concept or definition for
$$
F^G?
$$
I am mostly looking for a reason why the formal power series $e$ (i.e., the power series whose constant coefficient is $e$ and all remaining coefficients are 0), when raised to the power of the power series $x$, satisfies the above identity. A generalization would be appreciated as well, for the $F^G$ case.
abstract-algebra combinatorics generating-functions formal-power-series
New contributor
In my discrete math class, we are studying generating functions. We learned that
$$
e^x = sum_{i = 0}^{infty} frac{x^i}{i!},
$$
which is certainly an identity in calculus. However, in the ring of formal power series, is there any sort of concept of exponentiation? Suppose we have $F, G in R[[X]].$ Is there any sort of a concept or definition for
$$
F^G?
$$
I am mostly looking for a reason why the formal power series $e$ (i.e., the power series whose constant coefficient is $e$ and all remaining coefficients are 0), when raised to the power of the power series $x$, satisfies the above identity. A generalization would be appreciated as well, for the $F^G$ case.
abstract-algebra combinatorics generating-functions formal-power-series
abstract-algebra combinatorics generating-functions formal-power-series
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Alex Thibodeau
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Look for the idea of substitution of formal power series. This is called composition in this Wikipedia article.
To be more specific with respect to $F^G$, this can be done for example in the case $G=sum_{i=1}^infty g_i X^i$ and $F=1+H$ with $H=sum_{i=1}^infty h_i X^i$ by considering the formal power series $exp(G ln(1+H))$ where $ln(1+H):=sum_{i=1}^infty (-1)^{i-1}frac{H^i}{i}$.
Right, but do we just arbitrarily define $mathrm{ln}(1 + H)$ like that because it's convenient and familiar? Or is there a different reason?
– Alex Thibodeau
yesterday
This definition is chosen since then $exp(ln(1+H))=1+H$.
– Jens Schwaiger
yesterday
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Look for the idea of substitution of formal power series. This is called composition in this Wikipedia article.
To be more specific with respect to $F^G$, this can be done for example in the case $G=sum_{i=1}^infty g_i X^i$ and $F=1+H$ with $H=sum_{i=1}^infty h_i X^i$ by considering the formal power series $exp(G ln(1+H))$ where $ln(1+H):=sum_{i=1}^infty (-1)^{i-1}frac{H^i}{i}$.
Right, but do we just arbitrarily define $mathrm{ln}(1 + H)$ like that because it's convenient and familiar? Or is there a different reason?
– Alex Thibodeau
yesterday
This definition is chosen since then $exp(ln(1+H))=1+H$.
– Jens Schwaiger
yesterday
add a comment |
up vote
0
down vote
Look for the idea of substitution of formal power series. This is called composition in this Wikipedia article.
To be more specific with respect to $F^G$, this can be done for example in the case $G=sum_{i=1}^infty g_i X^i$ and $F=1+H$ with $H=sum_{i=1}^infty h_i X^i$ by considering the formal power series $exp(G ln(1+H))$ where $ln(1+H):=sum_{i=1}^infty (-1)^{i-1}frac{H^i}{i}$.
Right, but do we just arbitrarily define $mathrm{ln}(1 + H)$ like that because it's convenient and familiar? Or is there a different reason?
– Alex Thibodeau
yesterday
This definition is chosen since then $exp(ln(1+H))=1+H$.
– Jens Schwaiger
yesterday
add a comment |
up vote
0
down vote
up vote
0
down vote
Look for the idea of substitution of formal power series. This is called composition in this Wikipedia article.
To be more specific with respect to $F^G$, this can be done for example in the case $G=sum_{i=1}^infty g_i X^i$ and $F=1+H$ with $H=sum_{i=1}^infty h_i X^i$ by considering the formal power series $exp(G ln(1+H))$ where $ln(1+H):=sum_{i=1}^infty (-1)^{i-1}frac{H^i}{i}$.
Look for the idea of substitution of formal power series. This is called composition in this Wikipedia article.
To be more specific with respect to $F^G$, this can be done for example in the case $G=sum_{i=1}^infty g_i X^i$ and $F=1+H$ with $H=sum_{i=1}^infty h_i X^i$ by considering the formal power series $exp(G ln(1+H))$ where $ln(1+H):=sum_{i=1}^infty (-1)^{i-1}frac{H^i}{i}$.
answered yesterday
Jens Schwaiger
1,404127
1,404127
Right, but do we just arbitrarily define $mathrm{ln}(1 + H)$ like that because it's convenient and familiar? Or is there a different reason?
– Alex Thibodeau
yesterday
This definition is chosen since then $exp(ln(1+H))=1+H$.
– Jens Schwaiger
yesterday
add a comment |
Right, but do we just arbitrarily define $mathrm{ln}(1 + H)$ like that because it's convenient and familiar? Or is there a different reason?
– Alex Thibodeau
yesterday
This definition is chosen since then $exp(ln(1+H))=1+H$.
– Jens Schwaiger
yesterday
Right, but do we just arbitrarily define $mathrm{ln}(1 + H)$ like that because it's convenient and familiar? Or is there a different reason?
– Alex Thibodeau
yesterday
Right, but do we just arbitrarily define $mathrm{ln}(1 + H)$ like that because it's convenient and familiar? Or is there a different reason?
– Alex Thibodeau
yesterday
This definition is chosen since then $exp(ln(1+H))=1+H$.
– Jens Schwaiger
yesterday
This definition is chosen since then $exp(ln(1+H))=1+H$.
– Jens Schwaiger
yesterday
add a comment |
Alex Thibodeau is a new contributor. Be nice, and check out our Code of Conduct.
Alex Thibodeau is a new contributor. Be nice, and check out our Code of Conduct.
Alex Thibodeau is a new contributor. Be nice, and check out our Code of Conduct.
Alex Thibodeau is a new contributor. Be nice, and check out our Code of Conduct.
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