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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 $a_k = frac{f^{(k)}(a)}{k!}$ 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

share|cite|improve this question

edited 4 hours ago

Conor

asked 5 hours ago

ConorConor

556

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Is it supposed to be $b_e=frac{g^{(e)}(a)}{e!}$ ? 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 $a_k = frac{f^{(k)}(a)}{k!}$ 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

share|cite|improve this question

edited 4 hours ago

Conor

asked 5 hours ago

ConorConor

556

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Is it supposed to be $b_e=frac{g^{(e)}(a)}{e!}$ ? 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

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

oldest

votes

4

$begingroup$

Your intuition is good.

Multiplying the series gives an n-th term coefficient of

$$c_n = a_0b_n + a_1b_{n-1} + dots + a_{n-1}b_1 + a_nb_0= sum_{i=0}^n a_i b_{n-i}$$

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

$$c_n = frac{(fg)^{(n)}(a)}{n!} = frac{sum_{i=0}^n binom{n}{i}f^{(i)}(a)g^{(n-i)}(a)}{n!} = sum_{i=0}^n frac{f^{(i)}(a)}{i!} frac{g^{(n-i)}(a)}{(n-i)!} = sum_{i=0}^n a_i b_{n-i}$$

share|cite|improve this answer

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
  47 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

$$c_n = a_0b_n + a_1b_{n-1} + dots + a_{n-1}b_1 + a_nb_0= sum_{i=0}^n a_i b_{n-i}$$

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

$$c_n = frac{(fg)^{(n)}(a)}{n!} = frac{sum_{i=0}^n binom{n}{i}f^{(i)}(a)g^{(n-i)}(a)}{n!} = sum_{i=0}^n frac{f^{(i)}(a)}{i!} frac{g^{(n-i)}(a)}{(n-i)!} = sum_{i=0}^n a_i b_{n-i}$$

share|cite|improve this answer

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
  47 mins ago
  

  
  
  
  

add a comment | 

4

$begingroup$

Your intuition is good.

Multiplying the series gives an n-th term coefficient of

$$c_n = a_0b_n + a_1b_{n-1} + dots + a_{n-1}b_1 + a_nb_0= sum_{i=0}^n a_i b_{n-i}$$

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

$$c_n = frac{(fg)^{(n)}(a)}{n!} = frac{sum_{i=0}^n binom{n}{i}f^{(i)}(a)g^{(n-i)}(a)}{n!} = sum_{i=0}^n frac{f^{(i)}(a)}{i!} frac{g^{(n-i)}(a)}{(n-i)!} = sum_{i=0}^n a_i b_{n-i}$$

share|cite|improve this answer

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
  47 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

$$c_n = frac{(fg)^{(n)}(a)}{n!} = frac{sum_{i=0}^n binom{n}{i}f^{(i)}(a)g^{(n-i)}(a)}{n!} = sum_{i=0}^n frac{f^{(i)}(a)}{i!} frac{g^{(n-i)}(a)}{(n-i)!} = sum_{i=0}^n a_i b_{n-i}$$

share|cite|improve this answer

answered 5 hours ago

Michael BiroMichael Biro

11.3k21831

$endgroup$

share|cite|improve this answer

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