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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 $f(x)=sum_{k=0}^infty frac{2^{-k}}{k+1}(x-1)^k$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 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
1
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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 5 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
28 mins ago
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 5 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
28 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 5 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
28 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 5 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 5 hours ago
Michael BiroMichael Biro
11.3k21831
answered 5 hours ago
Michael BiroMichael Biro
11.3k21831
answered 5 hours ago
Michael BiroMichael Biro
11.3k21831
answered 5 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
28 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
28 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
28 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
28 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