Crthuky

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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 $f(x)=sum_k=0^infty frac2^-kk+1(x-1)^k$Smoothness of Taylor polynomials coefficients as function of position of expansion?Taylor series with initial value of infinityMisunderstanding about Taylor series

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
1

active

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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
  11 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
  11 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
  11 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