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.










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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.










    share|cite|improve this question







    New contributor




    Alex Thibodeau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






















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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.










      share|cite|improve this question







      New contributor




      Alex Thibodeau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      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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      asked 2 days ago









      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 $ln(1+H):=sum_{i=1}^infty (-1)^{i-1}frac{H^i}{i}$.






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          • 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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          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 $ln(1+H):=sum_{i=1}^infty (-1)^{i-1}frac{H^i}{i}$.






          share|cite|improve this answer





















          • 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















          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 $ln(1+H):=sum_{i=1}^infty (-1)^{i-1}frac{H^i}{i}$.






          share|cite|improve this answer





















          • 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













          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 $ln(1+H):=sum_{i=1}^infty (-1)^{i-1}frac{H^i}{i}$.






          share|cite|improve this answer












          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 $ln(1+H):=sum_{i=1}^infty (-1)^{i-1}frac{H^i}{i}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • 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










          Alex Thibodeau is a new contributor. Be nice, and check out our Code of Conduct.










           

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