Taylor series of product of two functions
$begingroup$
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 23 hours ago
ConorConor
606
$endgroup$
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 23 hours ago
ConorConor
606
$endgroup$
1
$begingroup$
Is it supposed to be $b_e=frac{g^{(e)}(a)}{e!}$ ? If so, look up the Cauchy Product Formula.
$endgroup$
– robjohn♦
22 hours ago
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 23 hours ago
ConorConor
606
$endgroup$
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 23 hours ago
ConorConor
606
asked 23 hours ago
ConorConor
606
edited 22 hours ago
Conor
asked 23 hours ago
ConorConor
606
asked 23 hours ago
ConorConor
606
asked 23 hours ago
ConorConor
606
606
1
$begingroup$
Is it supposed to be $b_e=frac{g^{(e)}(a)}{e!}$ ? If so, look up the Cauchy Product Formula.
$endgroup$
– robjohn♦
22 hours ago
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♦
22 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♦
22 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♦
22 hours ago
add a comment |
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
$$c_n = frac{(fg)^{(n)}(a)}{n!} = frac{sum_{i=0}^n binom{n}{i}f^{(i)}(a)g^{(n-i)}(a)}{n!} = sum_{i=0}^n frac{f^{(i)}(a)}{i!} frac{g^{(n-i)}(a)}{(n-i)!} = sum_{i=0}^n a_i b_{n-i}$$
answered 23 hours ago
Michael BiroMichael Biro
11.3k21831
$endgroup$
$begingroup$
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
18 hours ago
add a comment |
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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
$$c_n = frac{(fg)^{(n)}(a)}{n!} = frac{sum_{i=0}^n binom{n}{i}f^{(i)}(a)g^{(n-i)}(a)}{n!} = sum_{i=0}^n frac{f^{(i)}(a)}{i!} frac{g^{(n-i)}(a)}{(n-i)!} = sum_{i=0}^n a_i b_{n-i}$$
answered 23 hours ago
Michael BiroMichael Biro
11.3k21831
$endgroup$
$begingroup$
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
18 hours ago
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 23 hours ago
Michael BiroMichael Biro
11.3k21831
$endgroup$
$begingroup$
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
18 hours ago
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 23 hours ago
Michael BiroMichael Biro
11.3k21831
$endgroup$
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 23 hours ago
Michael BiroMichael Biro
11.3k21831
answered 23 hours ago
Michael BiroMichael Biro
11.3k21831
answered 23 hours ago
Michael BiroMichael Biro
11.3k21831
answered 23 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
18 hours ago
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 not a mathematician
18 hours 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
18 hours 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
18 hours ago
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♦
22 hours ago