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
$begingroup$
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
asked Mar 26 at 1:16
ConorConor
576
$endgroup$
add a comment |
$begingroup$
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
asked Mar 26 at 1:16
ConorConor
576
$endgroup$
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
add a comment |
$begingroup$
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
asked Mar 26 at 1:16
ConorConor
576
$endgroup$
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
asked Mar 26 at 1:16
ConorConor
576
asked Mar 26 at 1:16
ConorConor
576
edited Mar 26 at 2:08
Conor
asked Mar 26 at 1:16
ConorConor
576
asked Mar 26 at 1:16
ConorConor
576
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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
$$c_n = frac{(fg)^{(n)}(a)}{n!} = frac{sum_{i=0}^n binom{n}{i}f^{(i)}(a)g^{(n-i)}(a)}{n!} = sum_{i=0}^n frac{f^{(i)}(a)}{i!} frac{g^{(n-i)}(a)}{(n-i)!} = sum_{i=0}^n a_i b_{n-i}$$
answered Mar 26 at 1:33
Michael BiroMichael Biro
11.7k21931
$endgroup$
$begingroup$
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
add a comment |
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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
$$c_n = frac{(fg)^{(n)}(a)}{n!} = frac{sum_{i=0}^n binom{n}{i}f^{(i)}(a)g^{(n-i)}(a)}{n!} = sum_{i=0}^n frac{f^{(i)}(a)}{i!} frac{g^{(n-i)}(a)}{(n-i)!} = sum_{i=0}^n a_i b_{n-i}$$
answered Mar 26 at 1:33
Michael BiroMichael Biro
11.7k21931
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$$c_n = frac{(fg)^{(n)}(a)}{n!} = frac{sum_{i=0}^n binom{n}{i}f^{(i)}(a)g^{(n-i)}(a)}{n!} = sum_{i=0}^n frac{f^{(i)}(a)}{i!} frac{g^{(n-i)}(a)}{(n-i)!} = sum_{i=0}^n a_i b_{n-i}$$
answered Mar 26 at 1:33
Michael BiroMichael Biro
11.7k21931
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$$c_n = frac{(fg)^{(n)}(a)}{n!} = frac{sum_{i=0}^n binom{n}{i}f^{(i)}(a)g^{(n-i)}(a)}{n!} = sum_{i=0}^n frac{f^{(i)}(a)}{i!} frac{g^{(n-i)}(a)}{(n-i)!} = sum_{i=0}^n a_i b_{n-i}$$
answered Mar 26 at 1:33
Michael BiroMichael Biro
11.7k21931
$endgroup$
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
$$c_n = frac{(fg)^{(n)}(a)}{n!} = frac{sum_{i=0}^n binom{n}{i}f^{(i)}(a)g^{(n-i)}(a)}{n!} = sum_{i=0}^n frac{f^{(i)}(a)}{i!} frac{g^{(n-i)}(a)}{(n-i)!} = sum_{i=0}^n a_i b_{n-i}$$
answered Mar 26 at 1:33
Michael BiroMichael Biro
11.7k21931
answered Mar 26 at 1:33
Michael BiroMichael Biro
11.7k21931
answered Mar 26 at 1:33
Michael BiroMichael Biro
11.7k21931
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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