Taylor series of product of two functions Announcing the arrival of Valued Associate #679:...

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Taylor series of product of two functions



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)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 $f(x)=sum_{k=0}^infty frac{2^{-k}}{k+1}(x-1)^k$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 $a_k = frac{f^{(k)}(a)}{k!}$ 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♦
    Mar 26 at 2:08


















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♦
    Mar 26 at 2:08
















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 Mar 26 at 2:08







Conor

















asked Mar 26 at 1:16









ConorConor

576




576








  • 1




    $begingroup$
    Is it supposed to be $b_e=frac{g^{(e)}(a)}{e!}$ ? If so, look up the Cauchy Product Formula.
    $endgroup$
    – robjohn♦
    Mar 26 at 2:08
















  • 1




    $begingroup$
    Is it supposed to be $b_e=frac{g^{(e)}(a)}{e!}$ ? If so, look up the Cauchy Product Formula.
    $endgroup$
    – robjohn♦
    Mar 26 at 2:08










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♦
Mar 26 at 2:08






$begingroup$
Is it supposed to be $b_e=frac{g^{(e)}(a)}{e!}$ ? If so, look up the Cauchy Product Formula.
$endgroup$
– robjohn♦
Mar 26 at 2:08












1 Answer
1






active

oldest

votes


















4












$begingroup$

Your intuition is good.



Multiplying the series gives an n-th term coefficient of



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



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 a poor mathematician
    Mar 26 at 6:11












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1 Answer
1






active

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



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



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 a poor mathematician
    Mar 26 at 6:11
















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 a poor mathematician
    Mar 26 at 6:11














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 Mar 26 at 1:33









Michael BiroMichael Biro

11.7k21931




11.7k21931












  • $begingroup$
    In words: the coefficients of the product of two power series is the convolution of the coefficients of the factors.
    $endgroup$
    – J. M. is a poor mathematician
    Mar 26 at 6:11


















  • $begingroup$
    In words: the coefficients of the product of two power series is the convolution of the coefficients of the factors.
    $endgroup$
    – J. M. is a poor mathematician
    Mar 26 at 6:11
















$begingroup$
In words: the coefficients of the product of two power series is the convolution of the coefficients of the factors.
$endgroup$
– J. M. is a poor mathematician
Mar 26 at 6:11




$begingroup$
In words: the coefficients of the product of two power series is the convolution of the coefficients of the factors.
$endgroup$
– J. M. is a poor mathematician
Mar 26 at 6:11


















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