Why is this equality with sums and monomials true? The Next CEO of Stack OverflowMultinomial theorem: Number of elements where all coefficients have even powers..Artin-Chevalley Theorem.Relation between roots an coefficients in a generic equation: $a_0+a_1cdot x+cdots+a_ncdot x^n$Polynomial function in several variablesWhen is $X_1^a_1 cdots X_n^a_n-1$ irreducible?Why should this ideal be maximal?Solving system of equations with sums of odd powerAlgebraic identity involving partitions of indexesTrouble with understanding the definition of cyclic sums?Definition of a valuation of a rational fraction
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Why is this equality with sums and monomials true?
The Next CEO of Stack OverflowMultinomial theorem: Number of elements where all coefficients have even powers..Artin-Chevalley Theorem.Relation between roots an coefficients in a generic equation: $a_0+a_1cdot x+cdots+a_ncdot x^n$Polynomial function in several variablesWhen is $X_1^a_1 cdots X_n^a_n-1$ irreducible?Why should this ideal be maximal?Solving system of equations with sums of odd powerAlgebraic identity involving partitions of indexesTrouble with understanding the definition of cyclic sums?Definition of a valuation of a rational fraction
$begingroup$
Consider $$(x_1 + ... + x_n)^k = sum_alphac_alphax^alpha$$ where $x^alpha = x_1^a_1cdots x_n^a_n$ and $|alpha| = a_1 + ... + a_n$.
Why is this true? Is it something to do with the binomial sum?
polynomials summation binomial-theorem
asked Mar 27 at 18:54
the manthe man
831716
$endgroup$
add a comment |
$begingroup$
Consider $$(x_1 + ... + x_n)^k = sum_alphac_alphax^alpha$$ where $x^alpha = x_1^a_1cdots x_n^a_n$ and $|alpha| = a_1 + ... + a_n$.
Why is this true? Is it something to do with the binomial sum?
polynomials summation binomial-theorem
asked Mar 27 at 18:54
the manthe man
831716
$endgroup$
$begingroup$
Actually, it's multinomial rather than binomial.
$endgroup$
– Robert Israel
Mar 27 at 18:56
1
$begingroup$
Don't you want to say what the $c_alpha$ are?
$endgroup$
– Lord Shark the Unknown
Mar 27 at 19:02
add a comment |
$begingroup$
Consider $$(x_1 + ... + x_n)^k = sum_alphac_alphax^alpha$$ where $x^alpha = x_1^a_1cdots x_n^a_n$ and $|alpha| = a_1 + ... + a_n$.
Why is this true? Is it something to do with the binomial sum?
polynomials summation binomial-theorem
asked Mar 27 at 18:54
the manthe man
831716
$endgroup$
Consider $$(x_1 + ... + x_n)^k = sum_alphac_alphax^alpha$$ where $x^alpha = x_1^a_1cdots x_n^a_n$ and $|alpha| = a_1 + ... + a_n$.
Why is this true? Is it something to do with the binomial sum?
polynomials summation binomial-theorem
polynomials summation binomial-theorem
asked Mar 27 at 18:54
the manthe man
831716
asked Mar 27 at 18:54
the manthe man
831716
asked Mar 27 at 18:54
the manthe man
831716
asked Mar 27 at 18:54
the manthe man
831716
asked Mar 27 at 18:54
the manthe man
831716
831716
$begingroup$
Actually, it's multinomial rather than binomial.
$endgroup$
– Robert Israel
Mar 27 at 18:56
1
$begingroup$
Don't you want to say what the $c_alpha$ are?
$endgroup$
– Lord Shark the Unknown
Mar 27 at 19:02
add a comment |
$begingroup$
Actually, it's multinomial rather than binomial.
$endgroup$
– Robert Israel
Mar 27 at 18:56
1
$begingroup$
Don't you want to say what the $c_alpha$ are?
$endgroup$
– Lord Shark the Unknown
Mar 27 at 19:02
$begingroup$
Actually, it's multinomial rather than binomial.
$endgroup$
– Robert Israel
Mar 27 at 18:56
$begingroup$
Actually, it's multinomial rather than binomial.
$endgroup$
– Robert Israel
Mar 27 at 18:56
1
1
$begingroup$
Don't you want to say what the $c_alpha$ are?
$endgroup$
– Lord Shark the Unknown
Mar 27 at 19:02
$begingroup$
Don't you want to say what the $c_alpha$ are?
$endgroup$
– Lord Shark the Unknown
Mar 27 at 19:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can consider it related to the binomial sum, as any multinomial, can be rewritten as nested binomials. However it's technical name is the multinomial sum. Consider the trinomial power $$(a+b+c)^3$$. It can be rewritten as $$(a+(b+c))^3$$ because of the associative property of addition, which is then a binomial power with $$b+c=d$$ Expanding once, we get:$$a^3+3da^2+3ad^2+d^3$$, which follows the binomial sum of exponents rule. Expanding the powers of d, will also not break the binomial exponent sum rule (d is a binomial after all). Therefore expanding it fully, the whole thing will not break the exponent sum rule. We then get:$$a^3+3ba^2+3ca^2+3ab^2+6abc+3ac^2+b^3+3cb^2+3bc^2+c^3$$
answered yesterday
Roddy MacPheeRoddy MacPhee
573118
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can consider it related to the binomial sum, as any multinomial, can be rewritten as nested binomials. However it's technical name is the multinomial sum. Consider the trinomial power $$(a+b+c)^3$$. It can be rewritten as $$(a+(b+c))^3$$ because of the associative property of addition, which is then a binomial power with $$b+c=d$$ Expanding once, we get:$$a^3+3da^2+3ad^2+d^3$$, which follows the binomial sum of exponents rule. Expanding the powers of d, will also not break the binomial exponent sum rule (d is a binomial after all). Therefore expanding it fully, the whole thing will not break the exponent sum rule. We then get:$$a^3+3ba^2+3ca^2+3ab^2+6abc+3ac^2+b^3+3cb^2+3bc^2+c^3$$
answered yesterday
Roddy MacPheeRoddy MacPhee
573118
$endgroup$
add a comment |
$begingroup$
You can consider it related to the binomial sum, as any multinomial, can be rewritten as nested binomials. However it's technical name is the multinomial sum. Consider the trinomial power $$(a+b+c)^3$$. It can be rewritten as $$(a+(b+c))^3$$ because of the associative property of addition, which is then a binomial power with $$b+c=d$$ Expanding once, we get:$$a^3+3da^2+3ad^2+d^3$$, which follows the binomial sum of exponents rule. Expanding the powers of d, will also not break the binomial exponent sum rule (d is a binomial after all). Therefore expanding it fully, the whole thing will not break the exponent sum rule. We then get:$$a^3+3ba^2+3ca^2+3ab^2+6abc+3ac^2+b^3+3cb^2+3bc^2+c^3$$
answered yesterday
Roddy MacPheeRoddy MacPhee
573118
$endgroup$
add a comment |
$begingroup$
You can consider it related to the binomial sum, as any multinomial, can be rewritten as nested binomials. However it's technical name is the multinomial sum. Consider the trinomial power $$(a+b+c)^3$$. It can be rewritten as $$(a+(b+c))^3$$ because of the associative property of addition, which is then a binomial power with $$b+c=d$$ Expanding once, we get:$$a^3+3da^2+3ad^2+d^3$$, which follows the binomial sum of exponents rule. Expanding the powers of d, will also not break the binomial exponent sum rule (d is a binomial after all). Therefore expanding it fully, the whole thing will not break the exponent sum rule. We then get:$$a^3+3ba^2+3ca^2+3ab^2+6abc+3ac^2+b^3+3cb^2+3bc^2+c^3$$
answered yesterday
Roddy MacPheeRoddy MacPhee
573118
$endgroup$
You can consider it related to the binomial sum, as any multinomial, can be rewritten as nested binomials. However it's technical name is the multinomial sum. Consider the trinomial power $$(a+b+c)^3$$. It can be rewritten as $$(a+(b+c))^3$$ because of the associative property of addition, which is then a binomial power with $$b+c=d$$ Expanding once, we get:$$a^3+3da^2+3ad^2+d^3$$, which follows the binomial sum of exponents rule. Expanding the powers of d, will also not break the binomial exponent sum rule (d is a binomial after all). Therefore expanding it fully, the whole thing will not break the exponent sum rule. We then get:$$a^3+3ba^2+3ca^2+3ab^2+6abc+3ac^2+b^3+3cb^2+3bc^2+c^3$$
answered yesterday
Roddy MacPheeRoddy MacPhee
573118
answered yesterday
Roddy MacPheeRoddy MacPhee
573118
answered yesterday
Roddy MacPheeRoddy MacPhee
573118
answered yesterday
Roddy MacPheeRoddy MacPhee
573118
573118
add a comment |
add a comment |
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$begingroup$
Actually, it's multinomial rather than binomial.
$endgroup$
– Robert Israel
Mar 27 at 18:56
1
$begingroup$
Don't you want to say what the $c_alpha$ are?
$endgroup$
– Lord Shark the Unknown
Mar 27 at 19:02