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Taylor series of product of two functions

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Taylor series of product of two functions

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4

1

$begingroup$

let $f$ and $g$ be infinitley differentiable functions and $a_k = fracf^(k)(a)k!$ and $b_e = fracg^(e)(a)e!$ 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

edited 4 hours ago

Conor

asked 5 hours ago

ConorConor

556

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Is it supposed to be $b_e=fracg^(e)(a)e!$ ? If so, look up the Cauchy Product Formula.
  $endgroup$
  – robjohn♦
  4 hours ago
  
  

  
  
  
  

add a comment | 

4

1

$begingroup$

let $f$ and $g$ be infinitley differentiable functions and $a_k = fracf^(k)(a)k!$ and $b_e = fracg^(e)(a)e!$ 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

edited 4 hours ago

Conor

asked 5 hours ago

ConorConor

556

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Is it supposed to be $b_e=fracg^(e)(a)e!$ ? 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 $a_k = fracf^(k)(a)k!$ and $b_e = fracg^(e)(a)e!$ 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

edited 4 hours ago

Conor

asked 5 hours ago

ConorConor

556

$endgroup$

share|cite|improve this question

edited 4 hours ago

Conor

asked 5 hours ago

ConorConor

556

share|cite|improve this question

edited 4 hours ago

Conor

asked 5 hours ago

ConorConor

556

asked 5 hours ago

ConorConor

556

asked 5 hours ago

ConorConor

556

1 Answer
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4

$begingroup$

Your intuition is good.

Multiplying the series gives an n-th term coefficient of

$$c_n = a_0b_n + a_1b_n-1 + dots + a_n-1b_1 + a_nb_0= sum_i=0^n a_i b_n-i$$

which is the same as doing the Taylor series of $fg$ the long way, since

$$c_n = frac(fg)^(n)(a)n! = fracsum_i=0^n binomnif^(i)(a)g^(n-i)(a)n! = sum_i=0^n fracf^(i)(a)i! fracg^(n-i)(a)(n-i)! = sum_i=0^n a_i b_n-i$$

share|cite|improve this answer

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
  8 mins ago
  

  
  
  
  

add a comment | 

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4

$begingroup$

Your intuition is good.

Multiplying the series gives an n-th term coefficient of

$$c_n = a_0b_n + a_1b_n-1 + dots + a_n-1b_1 + a_nb_0= sum_i=0^n a_i b_n-i$$

which is the same as doing the Taylor series of $fg$ the long way, since

$$c_n = frac(fg)^(n)(a)n! = fracsum_i=0^n binomnif^(i)(a)g^(n-i)(a)n! = sum_i=0^n fracf^(i)(a)i! fracg^(n-i)(a)(n-i)! = sum_i=0^n a_i b_n-i$$

share|cite|improve this answer

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
  8 mins ago
  

  
  
  
  

add a comment | 

4

$begingroup$

Your intuition is good.

Multiplying the series gives an n-th term coefficient of

$$c_n = a_0b_n + a_1b_n-1 + dots + a_n-1b_1 + a_nb_0= sum_i=0^n a_i b_n-i$$

which is the same as doing the Taylor series of $fg$ the long way, since

$$c_n = frac(fg)^(n)(a)n! = fracsum_i=0^n binomnif^(i)(a)g^(n-i)(a)n! = sum_i=0^n fracf^(i)(a)i! fracg^(n-i)(a)(n-i)! = sum_i=0^n a_i b_n-i$$

share|cite|improve this answer

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
  8 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

Your intuition is good.

Multiplying the series gives an n-th term coefficient of

$$c_n = a_0b_n + a_1b_n-1 + dots + a_n-1b_1 + a_nb_0= sum_i=0^n a_i b_n-i$$

which is the same as doing the Taylor series of $fg$ the long way, since

$$c_n = frac(fg)^(n)(a)n! = fracsum_i=0^n binomnif^(i)(a)g^(n-i)(a)n! = sum_i=0^n fracf^(i)(a)i! fracg^(n-i)(a)(n-i)! = sum_i=0^n a_i b_n-i$$

share|cite|improve this answer

answered 4 hours ago

Michael BiroMichael Biro

11.3k21831

$endgroup$

share|cite|improve this answer

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