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Taylor series of product of two functions


Proving an inequality with Taylor polynomialsIntuition behind Taylor/Maclaurin SeriesUse of taylor series in convergenceRunge Phenomena and Taylor ExpansionWhat is the justification for taylor series for functions with one or no critical points?Marsden's definition of Taylor SeriesFind the Taylor series of $f(x)=sum_k=0^infty frac2^-kk+1(x-1)^k$Smoothness of Taylor polynomials coefficients as function of position of expansion?Taylor series with initial value of infinityMisunderstanding about Taylor series













4












$begingroup$


let $f$ and $g$ be infinitley differentiable functions and $a_k = fracf^(k)(a)k!$ and $b_e = fracg^(e)(a)e!$ 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.










share|cite|improve this question











$endgroup$







  • 1




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
















4












$begingroup$


let $f$ and $g$ be infinitley differentiable functions and $a_k = fracf^(k)(a)k!$ and $b_e = fracg^(e)(a)e!$ 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.










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Is it supposed to be $b_e=fracg^(e)(a)e!$ ? 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 $a_k = fracf^(k)(a)k!$ and $b_e = fracg^(e)(a)e!$ 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.










share|cite|improve this question











$endgroup$




let $f$ and $g$ be infinitley differentiable functions and $a_k = fracf^(k)(a)k!$ and $b_e = fracg^(e)(a)e!$ 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















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




556







  • 1




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













  • 1




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








1




1




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





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











1 Answer
1






active

oldest

votes


















4












$begingroup$

Your intuition is good.



Multiplying the series gives an n-th term coefficient of



$$c_n = a_0b_n + a_1b_n-1 + dots + a_n-1b_1 + a_nb_0= sum_i=0^n a_i b_n-i$$



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



$$c_n = frac(fg)^(n)(a)n! = fracsum_i=0^n binomnif^(i)(a)g^(n-i)(a)n! = sum_i=0^n fracf^(i)(a)i! fracg^(n-i)(a)(n-i)! = sum_i=0^n a_i b_n-i$$






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
    13 mins ago










Your Answer





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1 Answer
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oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

Your intuition is good.



Multiplying the series gives an n-th term coefficient of



$$c_n = a_0b_n + a_1b_n-1 + dots + a_n-1b_1 + a_nb_0= sum_i=0^n a_i b_n-i$$



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



$$c_n = frac(fg)^(n)(a)n! = fracsum_i=0^n binomnif^(i)(a)g^(n-i)(a)n! = sum_i=0^n fracf^(i)(a)i! fracg^(n-i)(a)(n-i)! = sum_i=0^n a_i b_n-i$$






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
    13 mins ago















4












$begingroup$

Your intuition is good.



Multiplying the series gives an n-th term coefficient of



$$c_n = a_0b_n + a_1b_n-1 + dots + a_n-1b_1 + a_nb_0= sum_i=0^n a_i b_n-i$$



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



$$c_n = frac(fg)^(n)(a)n! = fracsum_i=0^n binomnif^(i)(a)g^(n-i)(a)n! = sum_i=0^n fracf^(i)(a)i! fracg^(n-i)(a)(n-i)! = sum_i=0^n a_i b_n-i$$






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
    13 mins ago













4












4








4





$begingroup$

Your intuition is good.



Multiplying the series gives an n-th term coefficient of



$$c_n = a_0b_n + a_1b_n-1 + dots + a_n-1b_1 + a_nb_0= sum_i=0^n a_i b_n-i$$



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



$$c_n = frac(fg)^(n)(a)n! = fracsum_i=0^n binomnif^(i)(a)g^(n-i)(a)n! = sum_i=0^n fracf^(i)(a)i! fracg^(n-i)(a)(n-i)! = sum_i=0^n a_i b_n-i$$






share|cite|improve this answer









$endgroup$



Your intuition is good.



Multiplying the series gives an n-th term coefficient of



$$c_n = a_0b_n + a_1b_n-1 + dots + a_n-1b_1 + a_nb_0= sum_i=0^n a_i b_n-i$$



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



$$c_n = frac(fg)^(n)(a)n! = fracsum_i=0^n binomnif^(i)(a)g^(n-i)(a)n! = sum_i=0^n fracf^(i)(a)i! fracg^(n-i)(a)(n-i)! = sum_i=0^n a_i b_n-i$$







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











  • $begingroup$
    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
    13 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
    13 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
13 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
13 mins ago

















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