Root of unity filterEvaluation of ratio of two binomial expressionCounting some vanishing polynomials in $mathbbZ_n[X]$Sum of every $k$th binomial coefficient.Sixth Root of UnityRoot of unity paradox$x^n+a_n-1x^n-1+cdots +a_1x+a_0=0$ has real coefficients which satisfy $0<a_0 le a_1 le cdots le a_n-1 le 1$ prove that $z$ is a rootWhat is the sum of ALL of the nth powers of the qth roots of unity?Confusion regarding logic in paper, “A NOTE ON THE INVERSION OF POWER SERIES,” published in the AMS journalProve that if B(0) = 0, then A(B(x)) is a formal power seriesFinding a polynomial with rational coefficients that has the reciprocal of a complex number as its rootfind $a_1+a_3+a_5+cdots+a_37+a_39$
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Root of unity filter
Evaluation of ratio of two binomial expressionCounting some vanishing polynomials in $mathbbZ_n[X]$Sum of every $k$th binomial coefficient.Sixth Root of UnityRoot of unity paradox$x^n+a_n-1x^n-1+cdots +a_1x+a_0=0$ has real coefficients which satisfy $0<a_0 le a_1 le cdots le a_n-1 le 1$ prove that $z$ is a rootWhat is the sum of ALL of the nth powers of the qth roots of unity?Confusion regarding logic in paper, “A NOTE ON THE INVERSION OF POWER SERIES,” published in the AMS journalProve that if B(0) = 0, then A(B(x)) is a formal power seriesFinding a polynomial with rational coefficients that has the reciprocal of a complex number as its rootfind $a_1+a_3+a_5+cdots+a_37+a_39$
$begingroup$
Can some one help me understand the technique called "Root of unity filter" . I just know how to use it. It's as follow:
For series $f(x)=a_0+a_1x+a_2x^2 cdots a_nx^n$ we need to find the sum of coefficient of terms in which the power is a multiple of any number say $k$ for finding the same we have $omega $ as $mathrmk^th$ of unity and write
$$ dfracf(1)+f(omega)+f(omega ^2) cdots (omega^k-1)k=(a_0 + a_k + a_2k cdots)$$
please help me understand why and how this works , I tried googling but didn't get any satisfactory answer
combinatorics complex-numbers binomial-theorem
asked May 4 at 9:16
Advil SellAdvil Sell
1509
$endgroup$
add a comment |
$begingroup$
Can some one help me understand the technique called "Root of unity filter" . I just know how to use it. It's as follow:
For series $f(x)=a_0+a_1x+a_2x^2 cdots a_nx^n$ we need to find the sum of coefficient of terms in which the power is a multiple of any number say $k$ for finding the same we have $omega $ as $mathrmk^th$ of unity and write
$$ dfracf(1)+f(omega)+f(omega ^2) cdots (omega^k-1)k=(a_0 + a_k + a_2k cdots)$$
please help me understand why and how this works , I tried googling but didn't get any satisfactory answer
combinatorics complex-numbers binomial-theorem
asked May 4 at 9:16
Advil SellAdvil Sell
1509
$endgroup$
$begingroup$
Never heard of this, so I'm just curious. But what's the utility of this theorem when we know the polynomial directly? Isn't it better to directly sum the coefficients than to evaluate the entire polynomial at every root of unity up to the $k$th?
$endgroup$
– Allawonder
May 4 at 9:43
$begingroup$
I was doing a question which is as follow : $$(1+x)(1+x^2) cdots (1+x^2070)$$ we have to find the sum of coefficients of $x^9k$ to get the answer here we can easily find the answer using this tool , can you suggest another method @Allawonder
$endgroup$
– Advil Sell
May 4 at 9:50
$begingroup$
I now see how it may be used for facility in some cases. I had been simply thinking of a polynomial strictly in the form $sum a_nx^n.$
$endgroup$
– Allawonder
May 4 at 9:52
add a comment |
$begingroup$
Can some one help me understand the technique called "Root of unity filter" . I just know how to use it. It's as follow:
For series $f(x)=a_0+a_1x+a_2x^2 cdots a_nx^n$ we need to find the sum of coefficient of terms in which the power is a multiple of any number say $k$ for finding the same we have $omega $ as $mathrmk^th$ of unity and write
$$ dfracf(1)+f(omega)+f(omega ^2) cdots (omega^k-1)k=(a_0 + a_k + a_2k cdots)$$
please help me understand why and how this works , I tried googling but didn't get any satisfactory answer
combinatorics complex-numbers binomial-theorem
asked May 4 at 9:16
Advil SellAdvil Sell
1509
$endgroup$
Can some one help me understand the technique called "Root of unity filter" . I just know how to use it. It's as follow:
For series $f(x)=a_0+a_1x+a_2x^2 cdots a_nx^n$ we need to find the sum of coefficient of terms in which the power is a multiple of any number say $k$ for finding the same we have $omega $ as $mathrmk^th$ of unity and write
$$ dfracf(1)+f(omega)+f(omega ^2) cdots (omega^k-1)k=(a_0 + a_k + a_2k cdots)$$
please help me understand why and how this works , I tried googling but didn't get any satisfactory answer
combinatorics complex-numbers binomial-theorem
combinatorics complex-numbers binomial-theorem
asked May 4 at 9:16
Advil SellAdvil Sell
1509
asked May 4 at 9:16
Advil SellAdvil Sell
1509
asked May 4 at 9:16
Advil SellAdvil Sell
1509
asked May 4 at 9:16
Advil SellAdvil Sell
1509
asked May 4 at 9:16
Advil SellAdvil Sell
1509
1509
$begingroup$
Never heard of this, so I'm just curious. But what's the utility of this theorem when we know the polynomial directly? Isn't it better to directly sum the coefficients than to evaluate the entire polynomial at every root of unity up to the $k$th?
$endgroup$
– Allawonder
May 4 at 9:43
$begingroup$
I was doing a question which is as follow : $$(1+x)(1+x^2) cdots (1+x^2070)$$ we have to find the sum of coefficients of $x^9k$ to get the answer here we can easily find the answer using this tool , can you suggest another method @Allawonder
$endgroup$
– Advil Sell
May 4 at 9:50
$begingroup$
I now see how it may be used for facility in some cases. I had been simply thinking of a polynomial strictly in the form $sum a_nx^n.$
$endgroup$
– Allawonder
May 4 at 9:52
add a comment |
$begingroup$
Never heard of this, so I'm just curious. But what's the utility of this theorem when we know the polynomial directly? Isn't it better to directly sum the coefficients than to evaluate the entire polynomial at every root of unity up to the $k$th?
$endgroup$
– Allawonder
May 4 at 9:43
$begingroup$
I was doing a question which is as follow : $$(1+x)(1+x^2) cdots (1+x^2070)$$ we have to find the sum of coefficients of $x^9k$ to get the answer here we can easily find the answer using this tool , can you suggest another method @Allawonder
$endgroup$
– Advil Sell
May 4 at 9:50
$begingroup$
I now see how it may be used for facility in some cases. I had been simply thinking of a polynomial strictly in the form $sum a_nx^n.$
$endgroup$
– Allawonder
May 4 at 9:52
$begingroup$
Never heard of this, so I'm just curious. But what's the utility of this theorem when we know the polynomial directly? Isn't it better to directly sum the coefficients than to evaluate the entire polynomial at every root of unity up to the $k$th?
$endgroup$
– Allawonder
May 4 at 9:43
$begingroup$
Never heard of this, so I'm just curious. But what's the utility of this theorem when we know the polynomial directly? Isn't it better to directly sum the coefficients than to evaluate the entire polynomial at every root of unity up to the $k$th?
$endgroup$
– Allawonder
May 4 at 9:43
$begingroup$
I was doing a question which is as follow : $$(1+x)(1+x^2) cdots (1+x^2070)$$ we have to find the sum of coefficients of $x^9k$ to get the answer here we can easily find the answer using this tool , can you suggest another method @Allawonder
$endgroup$
– Advil Sell
May 4 at 9:50
$begingroup$
I was doing a question which is as follow : $$(1+x)(1+x^2) cdots (1+x^2070)$$ we have to find the sum of coefficients of $x^9k$ to get the answer here we can easily find the answer using this tool , can you suggest another method @Allawonder
$endgroup$
– Advil Sell
May 4 at 9:50
$begingroup$
I now see how it may be used for facility in some cases. I had been simply thinking of a polynomial strictly in the form $sum a_nx^n.$
$endgroup$
– Allawonder
May 4 at 9:52
$begingroup$
I now see how it may be used for facility in some cases. I had been simply thinking of a polynomial strictly in the form $sum a_nx^n.$
$endgroup$
– Allawonder
May 4 at 9:52
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Theorem: (Root of Unity Filter)
Define $omega=e^2pi i/n$ for a positive integer $n$. For any polynomial $F(x)=a_0+a_1x+a_2x^2+dots$(where we take $a_k=0$ if $k>deg(F)$), the sum $a_0+a_n+a_2n+...$ is given by $$a_0+a_n+a_2n+dots=frac1n(F(1)+F(omega)+dots+F(omega^n-1)$$
Proof: Let $s_k=1+omega^k+dots+omega^(n-1)k$
If $n$ divides $k$, then $omega^k=1$ and so $s_k=1+1+1dots+1=n$ otherwise $s_k=frac1-omega^nk1-omega^k=0$. So
$(F(1)+F(omega)+dots+F(omega^n-1)=a_0s_0+a_1s_1+a_2s_2+dots=n(a_0+a_n+a_2n+dots)$
Divide the both sides of the equation by $n$ and the proof is complete.
Source of my knowledge: http://zacharyabel.com/papers/Multi-GF_A06_MathRefl.pdf
There are some examples also which may help you. Please have a look at Problem $2 $ on page $3$.
answered May 4 at 9:35
StammeringMathematicianStammeringMathematician
3,0511326
$endgroup$
add a comment |
$begingroup$
Hint: This technique is called series multisection.
Some instructive examples can be found in H.S. Wilf's book generatingfunctionology (2.4.5) to (2.4.9).
Setting $omega_r=expleft(frac2pi irright), rgeq 1$ an integral value, the formula
beginalign*
sum_k=0^leftlfloorfracn-arrightrfloorbinomna+krx^a+kr=frac1rsum_k=1^rleft(omega_r^kright)^-aleft(1+xomega_r^kright)^n, qquad 0leq aleq n,aleq r-1
endalign*
is stated as (6.20) in Binomial Identities Derived from Trigonometric and Exponential Series by H.W. Gould.
An application proving a binomial identity can be found e.g. in this MSE answer.
answered May 5 at 20:34
Markus ScheuerMarkus Scheuer
65.3k461155
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
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votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Theorem: (Root of Unity Filter)
Define $omega=e^2pi i/n$ for a positive integer $n$. For any polynomial $F(x)=a_0+a_1x+a_2x^2+dots$(where we take $a_k=0$ if $k>deg(F)$), the sum $a_0+a_n+a_2n+...$ is given by $$a_0+a_n+a_2n+dots=frac1n(F(1)+F(omega)+dots+F(omega^n-1)$$
Proof: Let $s_k=1+omega^k+dots+omega^(n-1)k$
If $n$ divides $k$, then $omega^k=1$ and so $s_k=1+1+1dots+1=n$ otherwise $s_k=frac1-omega^nk1-omega^k=0$. So
$(F(1)+F(omega)+dots+F(omega^n-1)=a_0s_0+a_1s_1+a_2s_2+dots=n(a_0+a_n+a_2n+dots)$
Divide the both sides of the equation by $n$ and the proof is complete.
Source of my knowledge: http://zacharyabel.com/papers/Multi-GF_A06_MathRefl.pdf
There are some examples also which may help you. Please have a look at Problem $2 $ on page $3$.
answered May 4 at 9:35
StammeringMathematicianStammeringMathematician
3,0511326
$endgroup$
add a comment |
$begingroup$
Theorem: (Root of Unity Filter)
Define $omega=e^2pi i/n$ for a positive integer $n$. For any polynomial $F(x)=a_0+a_1x+a_2x^2+dots$(where we take $a_k=0$ if $k>deg(F)$), the sum $a_0+a_n+a_2n+...$ is given by $$a_0+a_n+a_2n+dots=frac1n(F(1)+F(omega)+dots+F(omega^n-1)$$
Proof: Let $s_k=1+omega^k+dots+omega^(n-1)k$
If $n$ divides $k$, then $omega^k=1$ and so $s_k=1+1+1dots+1=n$ otherwise $s_k=frac1-omega^nk1-omega^k=0$. So
$(F(1)+F(omega)+dots+F(omega^n-1)=a_0s_0+a_1s_1+a_2s_2+dots=n(a_0+a_n+a_2n+dots)$
Divide the both sides of the equation by $n$ and the proof is complete.
Source of my knowledge: http://zacharyabel.com/papers/Multi-GF_A06_MathRefl.pdf
There are some examples also which may help you. Please have a look at Problem $2 $ on page $3$.
answered May 4 at 9:35
StammeringMathematicianStammeringMathematician
3,0511326
$endgroup$
add a comment |
$begingroup$
Theorem: (Root of Unity Filter)
Define $omega=e^2pi i/n$ for a positive integer $n$. For any polynomial $F(x)=a_0+a_1x+a_2x^2+dots$(where we take $a_k=0$ if $k>deg(F)$), the sum $a_0+a_n+a_2n+...$ is given by $$a_0+a_n+a_2n+dots=frac1n(F(1)+F(omega)+dots+F(omega^n-1)$$
Proof: Let $s_k=1+omega^k+dots+omega^(n-1)k$
If $n$ divides $k$, then $omega^k=1$ and so $s_k=1+1+1dots+1=n$ otherwise $s_k=frac1-omega^nk1-omega^k=0$. So
$(F(1)+F(omega)+dots+F(omega^n-1)=a_0s_0+a_1s_1+a_2s_2+dots=n(a_0+a_n+a_2n+dots)$
Divide the both sides of the equation by $n$ and the proof is complete.
Source of my knowledge: http://zacharyabel.com/papers/Multi-GF_A06_MathRefl.pdf
There are some examples also which may help you. Please have a look at Problem $2 $ on page $3$.
answered May 4 at 9:35
StammeringMathematicianStammeringMathematician
3,0511326
$endgroup$
Theorem: (Root of Unity Filter)
Define $omega=e^2pi i/n$ for a positive integer $n$. For any polynomial $F(x)=a_0+a_1x+a_2x^2+dots$(where we take $a_k=0$ if $k>deg(F)$), the sum $a_0+a_n+a_2n+...$ is given by $$a_0+a_n+a_2n+dots=frac1n(F(1)+F(omega)+dots+F(omega^n-1)$$
Proof: Let $s_k=1+omega^k+dots+omega^(n-1)k$
If $n$ divides $k$, then $omega^k=1$ and so $s_k=1+1+1dots+1=n$ otherwise $s_k=frac1-omega^nk1-omega^k=0$. So
$(F(1)+F(omega)+dots+F(omega^n-1)=a_0s_0+a_1s_1+a_2s_2+dots=n(a_0+a_n+a_2n+dots)$
Divide the both sides of the equation by $n$ and the proof is complete.
Source of my knowledge: http://zacharyabel.com/papers/Multi-GF_A06_MathRefl.pdf
There are some examples also which may help you. Please have a look at Problem $2 $ on page $3$.
answered May 4 at 9:35
StammeringMathematicianStammeringMathematician
3,0511326
edited May 4 at 9:49
answered May 4 at 9:35
StammeringMathematicianStammeringMathematician
3,0511326
answered May 4 at 9:35
StammeringMathematicianStammeringMathematician
3,0511326
answered May 4 at 9:35
StammeringMathematicianStammeringMathematician
3,0511326
3,0511326
add a comment |
add a comment |
$begingroup$
Hint: This technique is called series multisection.
Some instructive examples can be found in H.S. Wilf's book generatingfunctionology (2.4.5) to (2.4.9).
Setting $omega_r=expleft(frac2pi irright), rgeq 1$ an integral value, the formula
beginalign*
sum_k=0^leftlfloorfracn-arrightrfloorbinomna+krx^a+kr=frac1rsum_k=1^rleft(omega_r^kright)^-aleft(1+xomega_r^kright)^n, qquad 0leq aleq n,aleq r-1
endalign*
is stated as (6.20) in Binomial Identities Derived from Trigonometric and Exponential Series by H.W. Gould.
An application proving a binomial identity can be found e.g. in this MSE answer.
answered May 5 at 20:34
Markus ScheuerMarkus Scheuer
65.3k461155
$endgroup$
add a comment |
$begingroup$
Hint: This technique is called series multisection.
Some instructive examples can be found in H.S. Wilf's book generatingfunctionology (2.4.5) to (2.4.9).
Setting $omega_r=expleft(frac2pi irright), rgeq 1$ an integral value, the formula
beginalign*
sum_k=0^leftlfloorfracn-arrightrfloorbinomna+krx^a+kr=frac1rsum_k=1^rleft(omega_r^kright)^-aleft(1+xomega_r^kright)^n, qquad 0leq aleq n,aleq r-1
endalign*
is stated as (6.20) in Binomial Identities Derived from Trigonometric and Exponential Series by H.W. Gould.
An application proving a binomial identity can be found e.g. in this MSE answer.
answered May 5 at 20:34
Markus ScheuerMarkus Scheuer
65.3k461155
$endgroup$
add a comment |
$begingroup$
Hint: This technique is called series multisection.
Some instructive examples can be found in H.S. Wilf's book generatingfunctionology (2.4.5) to (2.4.9).
Setting $omega_r=expleft(frac2pi irright), rgeq 1$ an integral value, the formula
beginalign*
sum_k=0^leftlfloorfracn-arrightrfloorbinomna+krx^a+kr=frac1rsum_k=1^rleft(omega_r^kright)^-aleft(1+xomega_r^kright)^n, qquad 0leq aleq n,aleq r-1
endalign*
is stated as (6.20) in Binomial Identities Derived from Trigonometric and Exponential Series by H.W. Gould.
An application proving a binomial identity can be found e.g. in this MSE answer.
answered May 5 at 20:34
Markus ScheuerMarkus Scheuer
65.3k461155
$endgroup$
Hint: This technique is called series multisection.
Some instructive examples can be found in H.S. Wilf's book generatingfunctionology (2.4.5) to (2.4.9).
Setting $omega_r=expleft(frac2pi irright), rgeq 1$ an integral value, the formula
beginalign*
sum_k=0^leftlfloorfracn-arrightrfloorbinomna+krx^a+kr=frac1rsum_k=1^rleft(omega_r^kright)^-aleft(1+xomega_r^kright)^n, qquad 0leq aleq n,aleq r-1
endalign*
is stated as (6.20) in Binomial Identities Derived from Trigonometric and Exponential Series by H.W. Gould.
An application proving a binomial identity can be found e.g. in this MSE answer.
answered May 5 at 20:34
Markus ScheuerMarkus Scheuer
65.3k461155
answered May 5 at 20:34
Markus ScheuerMarkus Scheuer
65.3k461155
answered May 5 at 20:34
Markus ScheuerMarkus Scheuer
65.3k461155
answered May 5 at 20:34
Markus ScheuerMarkus Scheuer
65.3k461155
65.3k461155
add a comment |
add a comment |
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$begingroup$
Never heard of this, so I'm just curious. But what's the utility of this theorem when we know the polynomial directly? Isn't it better to directly sum the coefficients than to evaluate the entire polynomial at every root of unity up to the $k$th?
$endgroup$
– Allawonder
May 4 at 9:43
$begingroup$
I was doing a question which is as follow : $$(1+x)(1+x^2) cdots (1+x^2070)$$ we have to find the sum of coefficients of $x^9k$ to get the answer here we can easily find the answer using this tool , can you suggest another method @Allawonder
$endgroup$
– Advil Sell
May 4 at 9:50
$begingroup$
I now see how it may be used for facility in some cases. I had been simply thinking of a polynomial strictly in the form $sum a_nx^n.$
$endgroup$
– Allawonder
May 4 at 9:52