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 5 hours ago
ConorConor
556
$endgroup$
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 5 hours ago
ConorConor
556
$endgroup$
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
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 5 hours ago
ConorConor
556
$endgroup$
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 5 hours ago
ConorConor
556
asked 5 hours ago
ConorConor
556
edited 4 hours ago
Conor
asked 5 hours ago
ConorConor
556
asked 5 hours ago
ConorConor
556
asked 5 hours ago
ConorConor
556
556
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
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♦
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
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 4 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
6 mins 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 4 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
6 mins 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 4 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
6 mins 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 4 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 4 hours ago
Michael BiroMichael Biro
11.3k21831
answered 4 hours ago
Michael BiroMichael Biro
11.3k21831
answered 4 hours ago
Michael BiroMichael Biro
11.3k21831
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
6 mins 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
6 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
6 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
6 mins 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♦
4 hours ago