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Transform $prod_k=0^n (1+x^2^k)$ to $sum_k=0^n c_kx^k$

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6

0

$begingroup$

I want to transform the following
$$prod_k=0^n (1+x^2^k)$$
to the canonical form $sum_k=0^n c_kx^k$

This is what I got so far
beginalign*
prod_k=0^n (1+x^2^k)= dfracx^2^n-1x-1 (x^2^n+1) \
endalign*
but I don't know how to continue, can someone help me with this?

calculus

share|cite|improve this question

asked Mar 29 at 9:44

D. QaD. Qa

1656

$endgroup$

add a comment | 

6

0

$begingroup$

I want to transform the following
$$prod_k=0^n (1+x^2^k)$$
to the canonical form $sum_k=0^n c_kx^k$

This is what I got so far
beginalign*
prod_k=0^n (1+x^2^k)= dfracx^2^n-1x-1 (x^2^n+1) \
endalign*
but I don't know how to continue, can someone help me with this?

calculus

share|cite|improve this question

asked Mar 29 at 9:44

D. QaD. Qa

1656

$endgroup$

add a comment | 

$begingroup$

I want to transform the following
$$prod_k=0^n (1+x^2^k)$$
to the canonical form $sum_k=0^n c_kx^k$

This is what I got so far
beginalign*
prod_k=0^n (1+x^2^k)= dfracx^2^n-1x-1 (x^2^n+1) \
endalign*
but I don't know how to continue, can someone help me with this?

calculus

share|cite|improve this question

asked Mar 29 at 9:44

D. QaD. Qa

1656

$endgroup$

share|cite|improve this question

asked Mar 29 at 9:44

D. QaD. Qa

1656

share|cite|improve this question

asked Mar 29 at 9:44

D. QaD. Qa

1656

asked Mar 29 at 9:44

D. QaD. Qa

1656

asked Mar 29 at 9:44

D. QaD. Qa

1656

3 Answers
3

active

oldest

votes

3

$begingroup$

Note that $ sum_k=0^n-1 2^k=2^n-1$. Therefore, the highest power of $x$ in $prod_k=0^n-1(1+x^2^k)$ is $2^n-1$. In other words, none of the terms will repeat as we keep multiplying by $1+x^2^n$. Thus, the relation can be obtained using induction.

It is easy to see that for $n=1$
beginequation
prod_k=0^1(1+x^2^k) = (1+x)(1+x^2)= 1+x+x^2+x^3.
endequation
Now, let us assume that $prod_k=0^n-1(1+x^2^k) = sum_k=0^2^n-1 x^k$. Then, we have
beginequation
prod_k=0^n(1+x^2^k) = left(sum_k=0^2^n-1 x^kright)(1+x^2^n) = sum_k=0^2^n-1 x^k+ sum_k=0^2^n-1 x^k+2^n = sum_k=0^2^n+1-1 x^k.
endequation

share|cite|improve this answer

answered Mar 29 at 10:13

Geethu JosephGeethu Joseph

1858

$endgroup$

add a comment | 

3

$begingroup$

You may also approach this from a combinatorical point fo view:

• Expanding would give a sum of $2^n+1$ summands.


• Each summand is a product of powers of $x$ where the exponent corresponds to a uniquely determined sequence $(b_0,b_1,ldots , b_n)$ of binary digits $b_i in 0,1$ for $i=0,ldots ,n$ where $0$ means choose $1 = x^0$ and $1$ means choose $x^2^i$ from the factor $(1+x^2^i)$
  

Hence,

$$prod_k=0^n (1+x^2^k) = sum_(b_0,b_1,ldots , b_n) in 0,1^n+1x^sum_i=0^nb_icdot 2^i= sum_k=0^2^n+1-1x^k$$

share|cite|improve this answer

answered Mar 29 at 10:22

trancelocationtrancelocation

13.6k1828

$endgroup$

add a comment | 

2

$begingroup$

Looks good as

beginalign
prod_k=0^n-1left(1+x^2^kright)&=prod_k=0^n-1frac1-x^2^k+11-x^2^k\ &=frac11-xprod_k=0^n-1left(1-x^2^k+1right)prod_k=0^n-2left(1-x^2^k+1right)^-1\ &=frac1-x^2^n1-x.
endalign

share|cite|improve this answer

answered Mar 29 at 9:48

KevinKevin

5,746823

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But how would I transform it to the canonical form? In a solution I encountered the final solution was $dfrac1-x^2^n 1-x (1+x^2^n )= (1+x+ dots +x^2^n-1 ) (1+x^2^n )$, but I did not understand how does $dfrac1-x^2^n 1-x =(1+x+ dots +x^2^n-1 ) $
  $endgroup$
  – D. Qa
  Mar 29 at 9:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @D.Qa, sum of a geometric progression.
  $endgroup$
  – Martín-Blas Pérez Pinilla
  Mar 29 at 10:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Martín-BlasPérezPinilla Does the same also apply for $(1+x^2^n)=x^2^n+x^2^n+1+ dots + x^2^n+1-1$ ?
  $endgroup$
  – D. Qa
  Mar 29 at 13:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @D.Qa, I think that you have forgotten the denominator and a sign... But your RHS is the sum of a geometric progression.
  $endgroup$
  – Martín-Blas Pérez Pinilla
  Mar 29 at 14:52
  

  
  
  
  

add a comment | 

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3

$begingroup$

Note that $ sum_k=0^n-1 2^k=2^n-1$. Therefore, the highest power of $x$ in $prod_k=0^n-1(1+x^2^k)$ is $2^n-1$. In other words, none of the terms will repeat as we keep multiplying by $1+x^2^n$. Thus, the relation can be obtained using induction.

It is easy to see that for $n=1$
beginequation
prod_k=0^1(1+x^2^k) = (1+x)(1+x^2)= 1+x+x^2+x^3.
endequation
Now, let us assume that $prod_k=0^n-1(1+x^2^k) = sum_k=0^2^n-1 x^k$. Then, we have
beginequation
prod_k=0^n(1+x^2^k) = left(sum_k=0^2^n-1 x^kright)(1+x^2^n) = sum_k=0^2^n-1 x^k+ sum_k=0^2^n-1 x^k+2^n = sum_k=0^2^n+1-1 x^k.
endequation

share|cite|improve this answer

answered Mar 29 at 10:13

Geethu JosephGeethu Joseph

1858

$endgroup$

add a comment | 

3

$begingroup$

Note that $ sum_k=0^n-1 2^k=2^n-1$. Therefore, the highest power of $x$ in $prod_k=0^n-1(1+x^2^k)$ is $2^n-1$. In other words, none of the terms will repeat as we keep multiplying by $1+x^2^n$. Thus, the relation can be obtained using induction.

It is easy to see that for $n=1$
beginequation
prod_k=0^1(1+x^2^k) = (1+x)(1+x^2)= 1+x+x^2+x^3.
endequation
Now, let us assume that $prod_k=0^n-1(1+x^2^k) = sum_k=0^2^n-1 x^k$. Then, we have
beginequation
prod_k=0^n(1+x^2^k) = left(sum_k=0^2^n-1 x^kright)(1+x^2^n) = sum_k=0^2^n-1 x^k+ sum_k=0^2^n-1 x^k+2^n = sum_k=0^2^n+1-1 x^k.
endequation

share|cite|improve this answer

answered Mar 29 at 10:13

Geethu JosephGeethu Joseph

1858

$endgroup$

add a comment | 

$begingroup$

Note that $ sum_k=0^n-1 2^k=2^n-1$. Therefore, the highest power of $x$ in $prod_k=0^n-1(1+x^2^k)$ is $2^n-1$. In other words, none of the terms will repeat as we keep multiplying by $1+x^2^n$. Thus, the relation can be obtained using induction.

It is easy to see that for $n=1$
beginequation
prod_k=0^1(1+x^2^k) = (1+x)(1+x^2)= 1+x+x^2+x^3.
endequation
Now, let us assume that $prod_k=0^n-1(1+x^2^k) = sum_k=0^2^n-1 x^k$. Then, we have
beginequation
prod_k=0^n(1+x^2^k) = left(sum_k=0^2^n-1 x^kright)(1+x^2^n) = sum_k=0^2^n-1 x^k+ sum_k=0^2^n-1 x^k+2^n = sum_k=0^2^n+1-1 x^k.
endequation

share|cite|improve this answer

answered Mar 29 at 10:13

Geethu JosephGeethu Joseph

1858

$endgroup$

share|cite|improve this answer

answered Mar 29 at 10:13

Geethu JosephGeethu Joseph

1858

answered Mar 29 at 10:13

Geethu JosephGeethu Joseph

1858

answered Mar 29 at 10:13

Geethu JosephGeethu Joseph

1858

3

$begingroup$

You may also approach this from a combinatorical point fo view:

• Expanding would give a sum of $2^n+1$ summands.


• Each summand is a product of powers of $x$ where the exponent corresponds to a uniquely determined sequence $(b_0,b_1,ldots , b_n)$ of binary digits $b_i in 0,1$ for $i=0,ldots ,n$ where $0$ means choose $1 = x^0$ and $1$ means choose $x^2^i$ from the factor $(1+x^2^i)$
  

Hence,

$$prod_k=0^n (1+x^2^k) = sum_(b_0,b_1,ldots , b_n) in 0,1^n+1x^sum_i=0^nb_icdot 2^i= sum_k=0^2^n+1-1x^k$$

share|cite|improve this answer

answered Mar 29 at 10:22

trancelocationtrancelocation

13.6k1828

$endgroup$

add a comment | 

3

$begingroup$

You may also approach this from a combinatorical point fo view:

• Expanding would give a sum of $2^n+1$ summands.


• Each summand is a product of powers of $x$ where the exponent corresponds to a uniquely determined sequence $(b_0,b_1,ldots , b_n)$ of binary digits $b_i in 0,1$ for $i=0,ldots ,n$ where $0$ means choose $1 = x^0$ and $1$ means choose $x^2^i$ from the factor $(1+x^2^i)$
  

Hence,

$$prod_k=0^n (1+x^2^k) = sum_(b_0,b_1,ldots , b_n) in 0,1^n+1x^sum_i=0^nb_icdot 2^i= sum_k=0^2^n+1-1x^k$$

share|cite|improve this answer

answered Mar 29 at 10:22

trancelocationtrancelocation

13.6k1828

$endgroup$

add a comment | 

$begingroup$

You may also approach this from a combinatorical point fo view:

• Expanding would give a sum of $2^n+1$ summands.


• Each summand is a product of powers of $x$ where the exponent corresponds to a uniquely determined sequence $(b_0,b_1,ldots , b_n)$ of binary digits $b_i in 0,1$ for $i=0,ldots ,n$ where $0$ means choose $1 = x^0$ and $1$ means choose $x^2^i$ from the factor $(1+x^2^i)$
  

Hence,

$$prod_k=0^n (1+x^2^k) = sum_(b_0,b_1,ldots , b_n) in 0,1^n+1x^sum_i=0^nb_icdot 2^i= sum_k=0^2^n+1-1x^k$$

share|cite|improve this answer

answered Mar 29 at 10:22

trancelocationtrancelocation

13.6k1828

$endgroup$

share|cite|improve this answer

answered Mar 29 at 10:22

trancelocationtrancelocation

13.6k1828

answered Mar 29 at 10:22

trancelocationtrancelocation

13.6k1828

answered Mar 29 at 10:22

trancelocationtrancelocation

13.6k1828

2

$begingroup$

Looks good as

beginalign
prod_k=0^n-1left(1+x^2^kright)&=prod_k=0^n-1frac1-x^2^k+11-x^2^k\ &=frac11-xprod_k=0^n-1left(1-x^2^k+1right)prod_k=0^n-2left(1-x^2^k+1right)^-1\ &=frac1-x^2^n1-x.
endalign

share|cite|improve this answer

answered Mar 29 at 9:48

KevinKevin

5,746823

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But how would I transform it to the canonical form? In a solution I encountered the final solution was $dfrac1-x^2^n 1-x (1+x^2^n )= (1+x+ dots +x^2^n-1 ) (1+x^2^n )$, but I did not understand how does $dfrac1-x^2^n 1-x =(1+x+ dots +x^2^n-1 ) $
  $endgroup$
  – D. Qa
  Mar 29 at 9:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @D.Qa, sum of a geometric progression.
  $endgroup$
  – Martín-Blas Pérez Pinilla
  Mar 29 at 10:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Martín-BlasPérezPinilla Does the same also apply for $(1+x^2^n)=x^2^n+x^2^n+1+ dots + x^2^n+1-1$ ?
  $endgroup$
  – D. Qa
  Mar 29 at 13:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @D.Qa, I think that you have forgotten the denominator and a sign... But your RHS is the sum of a geometric progression.
  $endgroup$
  – Martín-Blas Pérez Pinilla
  Mar 29 at 14:52
  

  
  
  
  

add a comment | 

2

$begingroup$

Looks good as

beginalign
prod_k=0^n-1left(1+x^2^kright)&=prod_k=0^n-1frac1-x^2^k+11-x^2^k\ &=frac11-xprod_k=0^n-1left(1-x^2^k+1right)prod_k=0^n-2left(1-x^2^k+1right)^-1\ &=frac1-x^2^n1-x.
endalign

share|cite|improve this answer

answered Mar 29 at 9:48

KevinKevin

5,746823

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But how would I transform it to the canonical form? In a solution I encountered the final solution was $dfrac1-x^2^n 1-x (1+x^2^n )= (1+x+ dots +x^2^n-1 ) (1+x^2^n )$, but I did not understand how does $dfrac1-x^2^n 1-x =(1+x+ dots +x^2^n-1 ) $
  $endgroup$
  – D. Qa
  Mar 29 at 9:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @D.Qa, sum of a geometric progression.
  $endgroup$
  – Martín-Blas Pérez Pinilla
  Mar 29 at 10:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Martín-BlasPérezPinilla Does the same also apply for $(1+x^2^n)=x^2^n+x^2^n+1+ dots + x^2^n+1-1$ ?
  $endgroup$
  – D. Qa
  Mar 29 at 13:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @D.Qa, I think that you have forgotten the denominator and a sign... But your RHS is the sum of a geometric progression.
  $endgroup$
  – Martín-Blas Pérez Pinilla
  Mar 29 at 14:52
  

  
  
  
  

add a comment | 

$begingroup$

Looks good as

beginalign
prod_k=0^n-1left(1+x^2^kright)&=prod_k=0^n-1frac1-x^2^k+11-x^2^k\ &=frac11-xprod_k=0^n-1left(1-x^2^k+1right)prod_k=0^n-2left(1-x^2^k+1right)^-1\ &=frac1-x^2^n1-x.
endalign

share|cite|improve this answer

answered Mar 29 at 9:48

KevinKevin

5,746823

$endgroup$

share|cite|improve this answer

answered Mar 29 at 9:48

KevinKevin

5,746823

answered Mar 29 at 9:48

KevinKevin

5,746823

answered Mar 29 at 9:48

KevinKevin

5,746823