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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$endgroup$








  • 1




    $begingroup$
    Is it supposed to be $b_e=frac{g^{(e)}(a)}{e!}$ ? If so, look up the Cauchy Product Formula.
    $endgroup$
    – robjohn♦
    4 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











$endgroup$








  • 1




    $begingroup$
    Is it supposed to be $b_e=frac{g^{(e)}(a)}{e!}$ ? If so, look up the Cauchy Product Formula.
    $endgroup$
    – robjohn♦
    4 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 4 hours ago







Conor

















asked 5 hours ago









ConorConor

556




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  • 1




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
















  • 1




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










1




1




$begingroup$
Is it supposed to be $b_e=frac{g^{(e)}(a)}{e!}$ ? If so, look up the Cauchy Product Formula.
$endgroup$
– robjohn♦
4 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♦
4 hours ago












1 Answer
1






active

oldest

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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









$endgroup$













  • $begingroup$
    In words: the coefficients of the product of two power series is the convolution of the coefficients of the factors.
    $endgroup$
    – J. M. is not a mathematician
    33 mins ago











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









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









$endgroup$













  • $begingroup$
    In words: the coefficients of the product of two power series is the convolution of the coefficients of the factors.
    $endgroup$
    – J. M. is not a mathematician
    33 mins ago
















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









$endgroup$













  • $begingroup$
    In words: the coefficients of the product of two power series is the convolution of the coefficients of the factors.
    $endgroup$
    – J. M. is not a mathematician
    33 mins ago














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 5 hours ago









Michael BiroMichael Biro

11.3k21831




11.3k21831












  • $begingroup$
    In words: the coefficients of the product of two power series is the convolution of the coefficients of the factors.
    $endgroup$
    – J. M. is not a mathematician
    33 mins ago


















  • $begingroup$
    In words: the coefficients of the product of two power series is the convolution of the coefficients of the factors.
    $endgroup$
    – J. M. is not a mathematician
    33 mins ago
















$begingroup$
In words: the coefficients of the product of two power series is the convolution of the coefficients of the factors.
$endgroup$
– J. M. is not a mathematician
33 mins ago




$begingroup$
In words: the coefficients of the product of two power series is the convolution of the coefficients of the factors.
$endgroup$
– J. M. is not a mathematician
33 mins ago


















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