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How do I notate a polynomial with Stirling coefficients and what properties do I need to prove it?
Stirling Binomial PolynomialWhat distribution do the rows of the Stirling numbers of the second kind approach?Why do nth roots (radicals) have closed forms whilst other polynomial roots do not?Properties of polynomials that are polynomial conditions on the coefficientsAn integral involving the Gamma functionStirling numbers of the second kind, general formulaFind the polynomial of the fifth degree with real coefficients such that…Second degree polynomials in one variable (with integer coefficients) and limiting behavior of the number of prime values they takeHow to solve for the roots of a 4th degree polynomial with complex coefficients?Partial sum Stirling number of the first kind with factorial and exponential
$begingroup$
I have a group of polynomials where each term increases in degree and has coefficients that appear as Stirling numbers of the second kind:
$1: 1 \
2: 1+x \
3: 1+3x+x^2 \
4: 1+7x+6x^2+x^3 \
5: 1+15x+25x^2+10x^3+x^4 ...\$
How can I represent consecutive polynomials in closed form and what properties of Stirling numbers and/or polynomials do I need to prove the nth polynomial with induction?
sequences-and-series polynomials stirling-numbers
$endgroup$
add a comment |
$begingroup$
I have a group of polynomials where each term increases in degree and has coefficients that appear as Stirling numbers of the second kind:
$1: 1 \
2: 1+x \
3: 1+3x+x^2 \
4: 1+7x+6x^2+x^3 \
5: 1+15x+25x^2+10x^3+x^4 ...\$
How can I represent consecutive polynomials in closed form and what properties of Stirling numbers and/or polynomials do I need to prove the nth polynomial with induction?
sequences-and-series polynomials stirling-numbers
$endgroup$
add a comment |
$begingroup$
I have a group of polynomials where each term increases in degree and has coefficients that appear as Stirling numbers of the second kind:
$1: 1 \
2: 1+x \
3: 1+3x+x^2 \
4: 1+7x+6x^2+x^3 \
5: 1+15x+25x^2+10x^3+x^4 ...\$
How can I represent consecutive polynomials in closed form and what properties of Stirling numbers and/or polynomials do I need to prove the nth polynomial with induction?
sequences-and-series polynomials stirling-numbers
$endgroup$
I have a group of polynomials where each term increases in degree and has coefficients that appear as Stirling numbers of the second kind:
$1: 1 \
2: 1+x \
3: 1+3x+x^2 \
4: 1+7x+6x^2+x^3 \
5: 1+15x+25x^2+10x^3+x^4 ...\$
How can I represent consecutive polynomials in closed form and what properties of Stirling numbers and/or polynomials do I need to prove the nth polynomial with induction?
sequences-and-series polynomials stirling-numbers
sequences-and-series polynomials stirling-numbers
asked yesterday
Vane VoeVane Voe
246
246
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1 Answer
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$begingroup$
Hint: These polynomials are called Touchard polynomials
beginalign*
T_n(x)=sum_k=0n brace kx^kqquad ngeq 0
endalign*
and you might want to use
beginalign*
T_n+1(x)=xsum_k=0^nbinomnkT_k(x)
endalign*
for induction.
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: These polynomials are called Touchard polynomials
beginalign*
T_n(x)=sum_k=0n brace kx^kqquad ngeq 0
endalign*
and you might want to use
beginalign*
T_n+1(x)=xsum_k=0^nbinomnkT_k(x)
endalign*
for induction.
$endgroup$
add a comment |
$begingroup$
Hint: These polynomials are called Touchard polynomials
beginalign*
T_n(x)=sum_k=0n brace kx^kqquad ngeq 0
endalign*
and you might want to use
beginalign*
T_n+1(x)=xsum_k=0^nbinomnkT_k(x)
endalign*
for induction.
$endgroup$
add a comment |
$begingroup$
Hint: These polynomials are called Touchard polynomials
beginalign*
T_n(x)=sum_k=0n brace kx^kqquad ngeq 0
endalign*
and you might want to use
beginalign*
T_n+1(x)=xsum_k=0^nbinomnkT_k(x)
endalign*
for induction.
$endgroup$
Hint: These polynomials are called Touchard polynomials
beginalign*
T_n(x)=sum_k=0n brace kx^kqquad ngeq 0
endalign*
and you might want to use
beginalign*
T_n+1(x)=xsum_k=0^nbinomnkT_k(x)
endalign*
for induction.
answered yesterday
Markus ScheuerMarkus Scheuer
62.5k459149
62.5k459149
add a comment |
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