Taylor series of product of two functions












4












$begingroup$


let $f$ and $g$ be infinitley differentiable functions and $a_k = frac{f^{(k)}(a)}{k!}$ and $b_e = frac{g^{(e)}(a)}{e!}$ be cofficients of Taylor Polynomial at $a$ then what would be the coefficients of $fg$.



rather than asking my specific question I asked this general question so other can benefit too



So I think we would need to multiply the two polynomials but that's just my intuition and I don't know how to justify it and I don't think it would be as simple.






analysis taylor-expansion







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$endgroup$












  • $begingroup$
    Is it supposed to be $b_e=frac{g^{(e)}(a)}{e!}$ ? If so, look up the Cauchy Product Formula.
    $endgroup$
    – robjohn♦
    3 hours ago


















4












$begingroup$


let $f$ and $g$ be infinitley differentiable functions and $a_k = frac{f^{(k)}(a)}{k!}$ and $b_e = frac{g^{(e)}(a)}{e!}$ be cofficients of Taylor Polynomial at $a$ then what would be the coefficients of $fg$.



rather than asking my specific question I asked this general question so other can benefit too



So I think we would need to multiply the two polynomials but that's just my intuition and I don't know how to justify it and I don't think it would be as simple.






analysis taylor-expansion







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is it supposed to be $b_e=frac{g^{(e)}(a)}{e!}$ ? If so, look up the Cauchy Product Formula.
    $endgroup$
    – robjohn♦
    3 hours ago
















4












4








4


1



$begingroup$


let $f$ and $g$ be infinitley differentiable functions and $a_k = frac{f^{(k)}(a)}{k!}$ and $b_e = frac{g^{(e)}(a)}{e!}$ be cofficients of Taylor Polynomial at $a$ then what would be the coefficients of $fg$.



rather than asking my specific question I asked this general question so other can benefit too



So I think we would need to multiply the two polynomials but that's just my intuition and I don't know how to justify it and I don't think it would be as simple.






analysis taylor-expansion







share|cite|improve this question











$endgroup$




let $f$ and $g$ be infinitley differentiable functions and $a_k = frac{f^{(k)}(a)}{k!}$ and $b_e = frac{g^{(e)}(a)}{e!}$ be cofficients of Taylor Polynomial at $a$ then what would be the coefficients of $fg$.



rather than asking my specific question I asked this general question so other can benefit too



So I think we would need to multiply the two polynomials but that's just my intuition and I don't know how to justify it and I don't think it would be as simple.






analysis taylor-expansion




analysis taylor-expansion






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 3 hours ago







Conor

















asked 4 hours ago









ConorConor

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556












  • $begingroup$
    Is it supposed to be $b_e=frac{g^{(e)}(a)}{e!}$ ? If so, look up the Cauchy Product Formula.
    $endgroup$
    – robjohn♦
    3 hours ago




















  • $begingroup$
    Is it supposed to be $b_e=frac{g^{(e)}(a)}{e!}$ ? If so, look up the Cauchy Product Formula.
    $endgroup$
    – robjohn♦
    3 hours ago


















$begingroup$
Is it supposed to be $b_e=frac{g^{(e)}(a)}{e!}$ ? If so, look up the Cauchy Product Formula.
$endgroup$
– robjohn♦
3 hours ago






$begingroup$
Is it supposed to be $b_e=frac{g^{(e)}(a)}{e!}$ ? If so, look up the Cauchy Product Formula.
$endgroup$
– robjohn♦
3 hours ago












1 Answer
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$begingroup$

Your intuition is good.



Multiplying the series gives an n-th term coefficient of



$$c_n = a_0b_n + a_1b_{n-1} + dots + a_{n-1}b_1 + a_nb_0= sum_{i=0}^n a_i b_{n-i}$$



which is the same as doing the Taylor series of $fg$ the long way, since



$$c_n = frac{(fg)^{(n)}(a)}{n!} = frac{sum_{i=0}^n binom{n}{i}f^{(i)}(a)g^{(n-i)}(a)}{n!} = sum_{i=0}^n frac{f^{(i)}(a)}{i!} frac{g^{(n-i)}(a)}{(n-i)!} = sum_{i=0}^n a_i b_{n-i}$$






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    1 Answer
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    4












    $begingroup$

    Your intuition is good.



    Multiplying the series gives an n-th term coefficient of



    $$c_n = a_0b_n + a_1b_{n-1} + dots + a_{n-1}b_1 + a_nb_0= sum_{i=0}^n a_i b_{n-i}$$



    which is the same as doing the Taylor series of $fg$ the long way, since



    $$c_n = frac{(fg)^{(n)}(a)}{n!} = frac{sum_{i=0}^n binom{n}{i}f^{(i)}(a)g^{(n-i)}(a)}{n!} = sum_{i=0}^n frac{f^{(i)}(a)}{i!} frac{g^{(n-i)}(a)}{(n-i)!} = sum_{i=0}^n a_i b_{n-i}$$






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      Your intuition is good.



      Multiplying the series gives an n-th term coefficient of



      $$c_n = a_0b_n + a_1b_{n-1} + dots + a_{n-1}b_1 + a_nb_0= sum_{i=0}^n a_i b_{n-i}$$



      which is the same as doing the Taylor series of $fg$ the long way, since



      $$c_n = frac{(fg)^{(n)}(a)}{n!} = frac{sum_{i=0}^n binom{n}{i}f^{(i)}(a)g^{(n-i)}(a)}{n!} = sum_{i=0}^n frac{f^{(i)}(a)}{i!} frac{g^{(n-i)}(a)}{(n-i)!} = sum_{i=0}^n a_i b_{n-i}$$






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        Your intuition is good.



        Multiplying the series gives an n-th term coefficient of



        $$c_n = a_0b_n + a_1b_{n-1} + dots + a_{n-1}b_1 + a_nb_0= sum_{i=0}^n a_i b_{n-i}$$



        which is the same as doing the Taylor series of $fg$ the long way, since



        $$c_n = frac{(fg)^{(n)}(a)}{n!} = frac{sum_{i=0}^n binom{n}{i}f^{(i)}(a)g^{(n-i)}(a)}{n!} = sum_{i=0}^n frac{f^{(i)}(a)}{i!} frac{g^{(n-i)}(a)}{(n-i)!} = sum_{i=0}^n a_i b_{n-i}$$






        share|cite|improve this answer









        $endgroup$



        Your intuition is good.



        Multiplying the series gives an n-th term coefficient of



        $$c_n = a_0b_n + a_1b_{n-1} + dots + a_{n-1}b_1 + a_nb_0= sum_{i=0}^n a_i b_{n-i}$$



        which is the same as doing the Taylor series of $fg$ the long way, since



        $$c_n = frac{(fg)^{(n)}(a)}{n!} = frac{sum_{i=0}^n binom{n}{i}f^{(i)}(a)g^{(n-i)}(a)}{n!} = sum_{i=0}^n frac{f^{(i)}(a)}{i!} frac{g^{(n-i)}(a)}{(n-i)!} = sum_{i=0}^n a_i b_{n-i}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 4 hours ago









        Michael BiroMichael Biro

        11.3k21831




        11.3k21831






























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