Mixing
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Geometric series and polynomials
Clash Royale CLAN TAG#URR8PPP
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i) Find an expression for the generating function of the sequence with terms
$$d_n=sum_k=0^n k^3$$
using operations on the geometric series $sum_ngeq 0 x^n$
ii) Derive a polynomial (in $n$) expression for $d_n$.
for i) I got $x(1+4x+x^2)/(1-x)^5$
but I'm confused what to do for ii), how does one derive that?
combinatorics
edited 1 hour ago
Robert Z
86.2k1056125
asked 2 hours ago
user600384
92
New contributor
user600384 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
2
down vote
favorite
i) Find an expression for the generating function of the sequence with terms
$$d_n=sum_k=0^n k^3$$
using operations on the geometric series $sum_ngeq 0 x^n$
ii) Derive a polynomial (in $n$) expression for $d_n$.
for i) I got $x(1+4x+x^2)/(1-x)^5$
but I'm confused what to do for ii), how does one derive that?
combinatorics
edited 1 hour ago
Robert Z
86.2k1056125
asked 2 hours ago
user600384
92
New contributor
user600384 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
any tips? 15 views but no responses lol
– user600384
2 hours ago
1
Your post will probably receive more positive feedback if you format it in MathJax throughout (I see you've done the middle sums).
– Clayton
1 hour ago
Plus, I have no clue what $d_n$, $b_n$ even are. This was confusing, so I saw it and left
– Rushabh Mehta
1 hour ago
@user600384 I edited your question for clearness sake. Now you have TWO answers...
– Robert Z
1 hour ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
i) Find an expression for the generating function of the sequence with terms
$$d_n=sum_k=0^n k^3$$
using operations on the geometric series $sum_ngeq 0 x^n$
ii) Derive a polynomial (in $n$) expression for $d_n$.
for i) I got $x(1+4x+x^2)/(1-x)^5$
but I'm confused what to do for ii), how does one derive that?
combinatorics
edited 1 hour ago
Robert Z
86.2k1056125
asked 2 hours ago
user600384
92
New contributor
user600384 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
i) Find an expression for the generating function of the sequence with terms
$$d_n=sum_k=0^n k^3$$
using operations on the geometric series $sum_ngeq 0 x^n$
ii) Derive a polynomial (in $n$) expression for $d_n$.
for i) I got $x(1+4x+x^2)/(1-x)^5$
but I'm confused what to do for ii), how does one derive that?
combinatorics
combinatorics
edited 1 hour ago
Robert Z
86.2k1056125
asked 2 hours ago
user600384
92
New contributor
user600384 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
Robert Z
86.2k1056125
asked 2 hours ago
user600384
92
New contributor
user600384 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
Robert Z
86.2k1056125
edited 1 hour ago
Robert Z
86.2k1056125
edited 1 hour ago
Robert Z
86.2k1056125
86.2k1056125
asked 2 hours ago
user600384
92
New contributor
user600384 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
user600384
92
asked 2 hours ago
user600384
92
92
New contributor
user600384 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user600384 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user600384 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
any tips? 15 views but no responses lol
– user600384
2 hours ago
1
Your post will probably receive more positive feedback if you format it in MathJax throughout (I see you've done the middle sums).
– Clayton
1 hour ago
Plus, I have no clue what $d_n$, $b_n$ even are. This was confusing, so I saw it and left
– Rushabh Mehta
1 hour ago
@user600384 I edited your question for clearness sake. Now you have TWO answers...
– Robert Z
1 hour ago
add a comment |Â
any tips? 15 views but no responses lol
– user600384
2 hours ago
1
Your post will probably receive more positive feedback if you format it in MathJax throughout (I see you've done the middle sums).
– Clayton
1 hour ago
Plus, I have no clue what $d_n$, $b_n$ even are. This was confusing, so I saw it and left
– Rushabh Mehta
1 hour ago
@user600384 I edited your question for clearness sake. Now you have TWO answers...
– Robert Z
1 hour ago
any tips? 15 views but no responses lol
– user600384
2 hours ago
any tips? 15 views but no responses lol
– user600384
2 hours ago
1
1
Your post will probably receive more positive feedback if you format it in MathJax throughout (I see you've done the middle sums).
– Clayton
1 hour ago
Your post will probably receive more positive feedback if you format it in MathJax throughout (I see you've done the middle sums).
– Clayton
1 hour ago
Plus, I have no clue what $d_n$, $b_n$ even are. This was confusing, so I saw it and left
– Rushabh Mehta
1 hour ago
Plus, I have no clue what $d_n$, $b_n$ even are. This was confusing, so I saw it and left
– Rushabh Mehta
1 hour ago
@user600384 I edited your question for clearness sake. Now you have TWO answers...
– Robert Z
1 hour ago
@user600384 I edited your question for clearness sake. Now you have TWO answers...
– Robert Z
1 hour ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
Yes, the generating function is correct: just apply the operator $xfracddx$ to $sum_ngeq 0 x^n=1/(1-x)$ three times ($3$ is the exponent of $k$) and then divide by $(1-x)$ (for $sum_k=0^n$):
$$f(x)=fracx(1+4x+x^2)(1-x)^5$$
Now, in order to find a polynomial formula for $sum_k=0^n k^3$ we have to extract the coefficient of $x^n$,
$$beginalignsum_k=0^n k^3&=[x^n]f(x)=[x^n](x+4x^2+x^3)cdot(1-x)^-5\&=
[x^n-1](1-x)^-5+4[x^n-2](1-x)^-5+
[x^n-3](1-x)^-5\
&=(-1)^n-1binom-5n-1+4(-1)^n-2binom-5n-2+(-1)^n-3binom-5n-3\
&=binomn+34+4binomn+24+binomn+14
=fracn^2(n+1)^24
endalign$$
where we used
$$binom-rk =(-1)^kbinomr+k-1r-1.$$
answered 1 hour ago
Robert Z
86.2k1056125
add a comment |Â
up vote
1
down vote
Consider $$S_n=sum_k=0^n k^3 x^k$$ and write
$$k^3=k(k-1)(k-2)+3k(k-1)+k$$ So
$$S_n=x^3sum_k=0^n k(k-1)(k-2)x^k-3+3x^2sum_k=0^n k(k-1)x^k-2+xsum_k=0^n kx^k-1$$ that is to say
$$S_n=x^3left(sum_k=0^n x^k right)'''+3x^2left(sum_k=0^n x^k right)''+xleft(sum_k=0^n x^k right)'$$
When finished, consider the limit when $xto 1$ to get the beautiful result.
answered 1 hour ago
Claude Leibovici
114k1155129
add a comment |Â
up vote
0
down vote
Using
$$sum_k=0^n x^k = frac1 - x^n+11-x$$
then by applying the operator $delta = x D$, $D = fracddx$, leads to
$$sum_k=0^n k^3 , x^k = fracx(1-x)^4 , ( 1 + 4x + x^2 - (n+1)^3 , x^n + (3n^3 + 6n^2 - 4) , x^n+1 - (3n^3 + 3n^2 - 3n +1) , x^n+2 + n^3 , x^n+3).$$
By taking the limit as $x to 1$ of both sides, which leads to several derivatives f the right hand side by making use of L'Hospital's rule, leads to the desired result
$$sum_k=0^n k^3 = fracn^2 (n+1)^24.$$
A possible polynomial for the second part may be obtained by seeking the generating function for:
$$sum_n=0^infty fracn^2 (n+1)^24 , x^n = fracx(1+4 x + x^2)(1-x)^5$$
which matches a previously stated polynomial by the proposer.
answered 1 hour ago
Leucippus
19.3k102869
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Yes, the generating function is correct: just apply the operator $xfracddx$ to $sum_ngeq 0 x^n=1/(1-x)$ three times ($3$ is the exponent of $k$) and then divide by $(1-x)$ (for $sum_k=0^n$):
$$f(x)=fracx(1+4x+x^2)(1-x)^5$$
Now, in order to find a polynomial formula for $sum_k=0^n k^3$ we have to extract the coefficient of $x^n$,
$$beginalignsum_k=0^n k^3&=[x^n]f(x)=[x^n](x+4x^2+x^3)cdot(1-x)^-5\&=
[x^n-1](1-x)^-5+4[x^n-2](1-x)^-5+
[x^n-3](1-x)^-5\
&=(-1)^n-1binom-5n-1+4(-1)^n-2binom-5n-2+(-1)^n-3binom-5n-3\
&=binomn+34+4binomn+24+binomn+14
=fracn^2(n+1)^24
endalign$$
where we used
$$binom-rk =(-1)^kbinomr+k-1r-1.$$
answered 1 hour ago
Robert Z
86.2k1056125
add a comment |Â
up vote
2
down vote
Yes, the generating function is correct: just apply the operator $xfracddx$ to $sum_ngeq 0 x^n=1/(1-x)$ three times ($3$ is the exponent of $k$) and then divide by $(1-x)$ (for $sum_k=0^n$):
$$f(x)=fracx(1+4x+x^2)(1-x)^5$$
Now, in order to find a polynomial formula for $sum_k=0^n k^3$ we have to extract the coefficient of $x^n$,
$$beginalignsum_k=0^n k^3&=[x^n]f(x)=[x^n](x+4x^2+x^3)cdot(1-x)^-5\&=
[x^n-1](1-x)^-5+4[x^n-2](1-x)^-5+
[x^n-3](1-x)^-5\
&=(-1)^n-1binom-5n-1+4(-1)^n-2binom-5n-2+(-1)^n-3binom-5n-3\
&=binomn+34+4binomn+24+binomn+14
=fracn^2(n+1)^24
endalign$$
where we used
$$binom-rk =(-1)^kbinomr+k-1r-1.$$
answered 1 hour ago
Robert Z
86.2k1056125
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Yes, the generating function is correct: just apply the operator $xfracddx$ to $sum_ngeq 0 x^n=1/(1-x)$ three times ($3$ is the exponent of $k$) and then divide by $(1-x)$ (for $sum_k=0^n$):
$$f(x)=fracx(1+4x+x^2)(1-x)^5$$
Now, in order to find a polynomial formula for $sum_k=0^n k^3$ we have to extract the coefficient of $x^n$,
$$beginalignsum_k=0^n k^3&=[x^n]f(x)=[x^n](x+4x^2+x^3)cdot(1-x)^-5\&=
[x^n-1](1-x)^-5+4[x^n-2](1-x)^-5+
[x^n-3](1-x)^-5\
&=(-1)^n-1binom-5n-1+4(-1)^n-2binom-5n-2+(-1)^n-3binom-5n-3\
&=binomn+34+4binomn+24+binomn+14
=fracn^2(n+1)^24
endalign$$
where we used
$$binom-rk =(-1)^kbinomr+k-1r-1.$$
answered 1 hour ago
Robert Z
86.2k1056125
Yes, the generating function is correct: just apply the operator $xfracddx$ to $sum_ngeq 0 x^n=1/(1-x)$ three times ($3$ is the exponent of $k$) and then divide by $(1-x)$ (for $sum_k=0^n$):
$$f(x)=fracx(1+4x+x^2)(1-x)^5$$
Now, in order to find a polynomial formula for $sum_k=0^n k^3$ we have to extract the coefficient of $x^n$,
$$beginalignsum_k=0^n k^3&=[x^n]f(x)=[x^n](x+4x^2+x^3)cdot(1-x)^-5\&=
[x^n-1](1-x)^-5+4[x^n-2](1-x)^-5+
[x^n-3](1-x)^-5\
&=(-1)^n-1binom-5n-1+4(-1)^n-2binom-5n-2+(-1)^n-3binom-5n-3\
&=binomn+34+4binomn+24+binomn+14
=fracn^2(n+1)^24
endalign$$
where we used
$$binom-rk =(-1)^kbinomr+k-1r-1.$$
answered 1 hour ago
Robert Z
86.2k1056125
edited 47 mins ago
answered 1 hour ago
Robert Z
86.2k1056125
answered 1 hour ago
Robert Z
86.2k1056125
answered 1 hour ago
Robert Z
86.2k1056125
86.2k1056125
add a comment |Â
add a comment |Â
up vote
1
down vote
Consider $$S_n=sum_k=0^n k^3 x^k$$ and write
$$k^3=k(k-1)(k-2)+3k(k-1)+k$$ So
$$S_n=x^3sum_k=0^n k(k-1)(k-2)x^k-3+3x^2sum_k=0^n k(k-1)x^k-2+xsum_k=0^n kx^k-1$$ that is to say
$$S_n=x^3left(sum_k=0^n x^k right)'''+3x^2left(sum_k=0^n x^k right)''+xleft(sum_k=0^n x^k right)'$$
When finished, consider the limit when $xto 1$ to get the beautiful result.
answered 1 hour ago
Claude Leibovici
114k1155129
add a comment |Â
up vote
1
down vote
Consider $$S_n=sum_k=0^n k^3 x^k$$ and write
$$k^3=k(k-1)(k-2)+3k(k-1)+k$$ So
$$S_n=x^3sum_k=0^n k(k-1)(k-2)x^k-3+3x^2sum_k=0^n k(k-1)x^k-2+xsum_k=0^n kx^k-1$$ that is to say
$$S_n=x^3left(sum_k=0^n x^k right)'''+3x^2left(sum_k=0^n x^k right)''+xleft(sum_k=0^n x^k right)'$$
When finished, consider the limit when $xto 1$ to get the beautiful result.
answered 1 hour ago
Claude Leibovici
114k1155129
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Consider $$S_n=sum_k=0^n k^3 x^k$$ and write
$$k^3=k(k-1)(k-2)+3k(k-1)+k$$ So
$$S_n=x^3sum_k=0^n k(k-1)(k-2)x^k-3+3x^2sum_k=0^n k(k-1)x^k-2+xsum_k=0^n kx^k-1$$ that is to say
$$S_n=x^3left(sum_k=0^n x^k right)'''+3x^2left(sum_k=0^n x^k right)''+xleft(sum_k=0^n x^k right)'$$
When finished, consider the limit when $xto 1$ to get the beautiful result.
answered 1 hour ago
Claude Leibovici
114k1155129
Consider $$S_n=sum_k=0^n k^3 x^k$$ and write
$$k^3=k(k-1)(k-2)+3k(k-1)+k$$ So
$$S_n=x^3sum_k=0^n k(k-1)(k-2)x^k-3+3x^2sum_k=0^n k(k-1)x^k-2+xsum_k=0^n kx^k-1$$ that is to say
$$S_n=x^3left(sum_k=0^n x^k right)'''+3x^2left(sum_k=0^n x^k right)''+xleft(sum_k=0^n x^k right)'$$
When finished, consider the limit when $xto 1$ to get the beautiful result.
answered 1 hour ago
Claude Leibovici
114k1155129
edited 1 hour ago
answered 1 hour ago
Claude Leibovici
114k1155129
answered 1 hour ago
Claude Leibovici
114k1155129
answered 1 hour ago
Claude Leibovici
114k1155129
114k1155129
add a comment |Â
add a comment |Â
up vote
0
down vote
Using
$$sum_k=0^n x^k = frac1 - x^n+11-x$$
then by applying the operator $delta = x D$, $D = fracddx$, leads to
$$sum_k=0^n k^3 , x^k = fracx(1-x)^4 , ( 1 + 4x + x^2 - (n+1)^3 , x^n + (3n^3 + 6n^2 - 4) , x^n+1 - (3n^3 + 3n^2 - 3n +1) , x^n+2 + n^3 , x^n+3).$$
By taking the limit as $x to 1$ of both sides, which leads to several derivatives f the right hand side by making use of L'Hospital's rule, leads to the desired result
$$sum_k=0^n k^3 = fracn^2 (n+1)^24.$$
A possible polynomial for the second part may be obtained by seeking the generating function for:
$$sum_n=0^infty fracn^2 (n+1)^24 , x^n = fracx(1+4 x + x^2)(1-x)^5$$
which matches a previously stated polynomial by the proposer.
answered 1 hour ago
Leucippus
19.3k102869
add a comment |Â
up vote
0
down vote
Using
$$sum_k=0^n x^k = frac1 - x^n+11-x$$
then by applying the operator $delta = x D$, $D = fracddx$, leads to
$$sum_k=0^n k^3 , x^k = fracx(1-x)^4 , ( 1 + 4x + x^2 - (n+1)^3 , x^n + (3n^3 + 6n^2 - 4) , x^n+1 - (3n^3 + 3n^2 - 3n +1) , x^n+2 + n^3 , x^n+3).$$
By taking the limit as $x to 1$ of both sides, which leads to several derivatives f the right hand side by making use of L'Hospital's rule, leads to the desired result
$$sum_k=0^n k^3 = fracn^2 (n+1)^24.$$
A possible polynomial for the second part may be obtained by seeking the generating function for:
$$sum_n=0^infty fracn^2 (n+1)^24 , x^n = fracx(1+4 x + x^2)(1-x)^5$$
which matches a previously stated polynomial by the proposer.
answered 1 hour ago
Leucippus
19.3k102869
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Using
$$sum_k=0^n x^k = frac1 - x^n+11-x$$
then by applying the operator $delta = x D$, $D = fracddx$, leads to
$$sum_k=0^n k^3 , x^k = fracx(1-x)^4 , ( 1 + 4x + x^2 - (n+1)^3 , x^n + (3n^3 + 6n^2 - 4) , x^n+1 - (3n^3 + 3n^2 - 3n +1) , x^n+2 + n^3 , x^n+3).$$
By taking the limit as $x to 1$ of both sides, which leads to several derivatives f the right hand side by making use of L'Hospital's rule, leads to the desired result
$$sum_k=0^n k^3 = fracn^2 (n+1)^24.$$
A possible polynomial for the second part may be obtained by seeking the generating function for:
$$sum_n=0^infty fracn^2 (n+1)^24 , x^n = fracx(1+4 x + x^2)(1-x)^5$$
which matches a previously stated polynomial by the proposer.
answered 1 hour ago
Leucippus
19.3k102869
Using
$$sum_k=0^n x^k = frac1 - x^n+11-x$$
then by applying the operator $delta = x D$, $D = fracddx$, leads to
$$sum_k=0^n k^3 , x^k = fracx(1-x)^4 , ( 1 + 4x + x^2 - (n+1)^3 , x^n + (3n^3 + 6n^2 - 4) , x^n+1 - (3n^3 + 3n^2 - 3n +1) , x^n+2 + n^3 , x^n+3).$$
By taking the limit as $x to 1$ of both sides, which leads to several derivatives f the right hand side by making use of L'Hospital's rule, leads to the desired result
$$sum_k=0^n k^3 = fracn^2 (n+1)^24.$$
A possible polynomial for the second part may be obtained by seeking the generating function for:
$$sum_n=0^infty fracn^2 (n+1)^24 , x^n = fracx(1+4 x + x^2)(1-x)^5$$
which matches a previously stated polynomial by the proposer.
answered 1 hour ago
Leucippus
19.3k102869
answered 1 hour ago
Leucippus
19.3k102869
answered 1 hour ago
Leucippus
19.3k102869
answered 1 hour ago
Leucippus
19.3k102869
19.3k102869
add a comment |Â
add a comment |Â
user600384 is a new contributor. Be nice, and check out our Code of Conduct.
user600384 is a new contributor. Be nice, and check out our Code of Conduct.
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any tips? 15 views but no responses lol
– user600384
2 hours ago
1
Your post will probably receive more positive feedback if you format it in MathJax throughout (I see you've done the middle sums).
– Clayton
1 hour ago
Plus, I have no clue what $d_n$, $b_n$ even are. This was confusing, so I saw it and left
– Rushabh Mehta
1 hour ago
@user600384 I edited your question for clearness sake. Now you have TWO answers...
– Robert Z
1 hour ago