Mixing
If $a_0 = -1$ and $sum_{k=0}^nfrac{a_{n-k}}{k+1}=0$, then show $a_n =...
$begingroup$
Bonjour. The following is from IMO 2006:
The sequence $(a_n)$ is defined recursively by $a_0=-1$ and $$sum_{k=0}^{n}{frac{a_{n-k}}{k+1}}=0, ngeq 1. $$
In the shortlist, the commentator said that this relation holds:
$$a_n=int_{1}^{infty}{frac{dt}{t^n({pi}^2+log^2(t-1))}} qquadforall ngeq 1 tag{1}$$
Question: How to show the relation $(1)$?
calculus integration combinatorics complex-analysis analysis
edited Feb 2 at 12:14
Blue
49.6k870158
asked Feb 2 at 12:03
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,061212
$endgroup$
add a comment |
$begingroup$
Bonjour. The following is from IMO 2006:
The sequence $(a_n)$ is defined recursively by $a_0=-1$ and $$sum_{k=0}^{n}{frac{a_{n-k}}{k+1}}=0, ngeq 1. $$
In the shortlist, the commentator said that this relation holds:
$$a_n=int_{1}^{infty}{frac{dt}{t^n({pi}^2+log^2(t-1))}} qquadforall ngeq 1 tag{1}$$
Question: How to show the relation $(1)$?
calculus integration combinatorics complex-analysis analysis
edited Feb 2 at 12:14
Blue
49.6k870158
asked Feb 2 at 12:03
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,061212
$endgroup$
$begingroup$
This is known as the integral representation for Gregory coefficients: en.wikipedia.org/wiki/Gregory_coefficients
$endgroup$
– Zacky
Feb 2 at 23:41
add a comment |
$begingroup$
Bonjour. The following is from IMO 2006:
The sequence $(a_n)$ is defined recursively by $a_0=-1$ and $$sum_{k=0}^{n}{frac{a_{n-k}}{k+1}}=0, ngeq 1. $$
In the shortlist, the commentator said that this relation holds:
$$a_n=int_{1}^{infty}{frac{dt}{t^n({pi}^2+log^2(t-1))}} qquadforall ngeq 1 tag{1}$$
Question: How to show the relation $(1)$?
calculus integration combinatorics complex-analysis analysis
edited Feb 2 at 12:14
Blue
49.6k870158
asked Feb 2 at 12:03
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,061212
$endgroup$
Bonjour. The following is from IMO 2006:
The sequence $(a_n)$ is defined recursively by $a_0=-1$ and $$sum_{k=0}^{n}{frac{a_{n-k}}{k+1}}=0, ngeq 1. $$
In the shortlist, the commentator said that this relation holds:
$$a_n=int_{1}^{infty}{frac{dt}{t^n({pi}^2+log^2(t-1))}} qquadforall ngeq 1 tag{1}$$
Question: How to show the relation $(1)$?
calculus integration combinatorics complex-analysis analysis
calculus integration combinatorics complex-analysis analysis
edited Feb 2 at 12:14
Blue
49.6k870158
asked Feb 2 at 12:03
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,061212
edited Feb 2 at 12:14
Blue
49.6k870158
asked Feb 2 at 12:03
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,061212
edited Feb 2 at 12:14
Blue
49.6k870158
edited Feb 2 at 12:14
Blue
49.6k870158
edited Feb 2 at 12:14
Blue
49.6k870158
49.6k870158
asked Feb 2 at 12:03
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,061212
asked Feb 2 at 12:03
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,061212
asked Feb 2 at 12:03
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,061212
2,061212
$begingroup$
This is known as the integral representation for Gregory coefficients: en.wikipedia.org/wiki/Gregory_coefficients
$endgroup$
– Zacky
Feb 2 at 23:41
add a comment |
$begingroup$
This is known as the integral representation for Gregory coefficients: en.wikipedia.org/wiki/Gregory_coefficients
$endgroup$
– Zacky
Feb 2 at 23:41
$begingroup$
This is known as the integral representation for Gregory coefficients: en.wikipedia.org/wiki/Gregory_coefficients
$endgroup$
– Zacky
Feb 2 at 23:41
$begingroup$
This is known as the integral representation for Gregory coefficients: en.wikipedia.org/wiki/Gregory_coefficients
$endgroup$
– Zacky
Feb 2 at 23:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The claim is equivalent to
$$
left(a*frac1{i+1}right)_n = -1_{n=0}
$$ holds for every $nge 0$, where $$
left( a*frac1{i+1}right)_n =sum_{k=0}^n a_{n-i}frac{1}{i+1}
$$ is the convolution of $(a_i)_{ige 0}$ and $(frac1{i+1})_{ige 0}$ and
$$
a_n =int_{1}^{infty}frac1{t^n}frac{mathrm d t}{{pi}^2+log^2(t-1)},quad nge 1,
$$$$
a_0=-1.
$$
In view of the generating function of $(frac1{i+1})_{ige 0}$, which is $-frac{ln(1-x)}x$, it suffices to prove that for $0<x<1$,
$$
sum_{nge 0} a_n x^n=-1+sum_{nge 1}a_n x^n = frac{x}{ln(1-x)}.
$$ We find that
$$begin{eqnarray}
sum_{n=1}^infty a_n x^n &=& int_{1}^{infty}sum_{n=1}^infty left(frac x tright)^nfrac{mathrm d t}{{pi}^2+log^2(t-1)}\ &=&int_{-infty}^{infty}sum_{n=1}^infty left(frac x {1+e^u}right)^nfrac{e^umathrm d u}{{pi}^2+u^2} quad(t=1+e^u)\
&=& int_{-infty}^{infty}frac{x}{(1-x)+e^u}frac{e^umathrm d u}{{pi}^2+u^2}\
&=&int_{Im z =0}frac{x}{(1-x)+e^z}frac{e^zmathrm d z}{{pi}^2+z^2}.
end{eqnarray}$$ We also observe that
$$
lim_{cto infty}int_{Im z =c}frac{x}{(1-x)+e^z}frac{e^zmathrm d z}{{pi}^2+z^2}=0,
$$ which implies by residue theorem:
$$begin{eqnarray}
int_{Im z =0}frac{x}{(1-x)+e^z}frac{e^zmathrm d z}{{pi}^2+z^2}&=&2pi i sum_{z:Im(z)>0} text{res}_{z}frac{x}{(1-x)+e^z}frac{e^z}{{pi}^2+z^2}\
&=&1+2pi i xsum_{jge 0} frac{1}{left((2j+1)pi i + ln(1-x)right)^2 +pi^2}\
&=&1+xsum_{jge 0} left(frac{1}{(2j+1)pi i + ln(1-x)-pi i} \ -frac{1}{(2j+1)pi i + ln(1-x)+pi i}right)\
&=&1+ frac{x}{ln(1-x)}.
end{eqnarray}$$ (Here, residues are calculated at $z=pi i$ and $(2j+1)pi i +ln(1-x)$ where $jge 0$.)
This shows $$
sum_{nge 0}a_n x^n = frac{x}{ln(1-x)}
$$ and the claim follows.
answered Feb 2 at 17:46
SongSong
18.6k21651
$endgroup$
$begingroup$
Hell yeah! The master key was the use of convolution. Did not see that. Thanks
$endgroup$
– HAMIDINE SOUMARE
Feb 2 at 20:15
add a comment |
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$begingroup$
The claim is equivalent to
$$
left(a*frac1{i+1}right)_n = -1_{n=0}
$$ holds for every $nge 0$, where $$
left( a*frac1{i+1}right)_n =sum_{k=0}^n a_{n-i}frac{1}{i+1}
$$ is the convolution of $(a_i)_{ige 0}$ and $(frac1{i+1})_{ige 0}$ and
$$
a_n =int_{1}^{infty}frac1{t^n}frac{mathrm d t}{{pi}^2+log^2(t-1)},quad nge 1,
$$$$
a_0=-1.
$$
In view of the generating function of $(frac1{i+1})_{ige 0}$, which is $-frac{ln(1-x)}x$, it suffices to prove that for $0<x<1$,
$$
sum_{nge 0} a_n x^n=-1+sum_{nge 1}a_n x^n = frac{x}{ln(1-x)}.
$$ We find that
$$begin{eqnarray}
sum_{n=1}^infty a_n x^n &=& int_{1}^{infty}sum_{n=1}^infty left(frac x tright)^nfrac{mathrm d t}{{pi}^2+log^2(t-1)}\ &=&int_{-infty}^{infty}sum_{n=1}^infty left(frac x {1+e^u}right)^nfrac{e^umathrm d u}{{pi}^2+u^2} quad(t=1+e^u)\
&=& int_{-infty}^{infty}frac{x}{(1-x)+e^u}frac{e^umathrm d u}{{pi}^2+u^2}\
&=&int_{Im z =0}frac{x}{(1-x)+e^z}frac{e^zmathrm d z}{{pi}^2+z^2}.
end{eqnarray}$$ We also observe that
$$
lim_{cto infty}int_{Im z =c}frac{x}{(1-x)+e^z}frac{e^zmathrm d z}{{pi}^2+z^2}=0,
$$ which implies by residue theorem:
$$begin{eqnarray}
int_{Im z =0}frac{x}{(1-x)+e^z}frac{e^zmathrm d z}{{pi}^2+z^2}&=&2pi i sum_{z:Im(z)>0} text{res}_{z}frac{x}{(1-x)+e^z}frac{e^z}{{pi}^2+z^2}\
&=&1+2pi i xsum_{jge 0} frac{1}{left((2j+1)pi i + ln(1-x)right)^2 +pi^2}\
&=&1+xsum_{jge 0} left(frac{1}{(2j+1)pi i + ln(1-x)-pi i} \ -frac{1}{(2j+1)pi i + ln(1-x)+pi i}right)\
&=&1+ frac{x}{ln(1-x)}.
end{eqnarray}$$ (Here, residues are calculated at $z=pi i$ and $(2j+1)pi i +ln(1-x)$ where $jge 0$.)
This shows $$
sum_{nge 0}a_n x^n = frac{x}{ln(1-x)}
$$ and the claim follows.
answered Feb 2 at 17:46
SongSong
18.6k21651
$endgroup$
$begingroup$
Hell yeah! The master key was the use of convolution. Did not see that. Thanks
$endgroup$
– HAMIDINE SOUMARE
Feb 2 at 20:15
add a comment |
$begingroup$
The claim is equivalent to
$$
left(a*frac1{i+1}right)_n = -1_{n=0}
$$ holds for every $nge 0$, where $$
left( a*frac1{i+1}right)_n =sum_{k=0}^n a_{n-i}frac{1}{i+1}
$$ is the convolution of $(a_i)_{ige 0}$ and $(frac1{i+1})_{ige 0}$ and
$$
a_n =int_{1}^{infty}frac1{t^n}frac{mathrm d t}{{pi}^2+log^2(t-1)},quad nge 1,
$$$$
a_0=-1.
$$
In view of the generating function of $(frac1{i+1})_{ige 0}$, which is $-frac{ln(1-x)}x$, it suffices to prove that for $0<x<1$,
$$
sum_{nge 0} a_n x^n=-1+sum_{nge 1}a_n x^n = frac{x}{ln(1-x)}.
$$ We find that
$$begin{eqnarray}
sum_{n=1}^infty a_n x^n &=& int_{1}^{infty}sum_{n=1}^infty left(frac x tright)^nfrac{mathrm d t}{{pi}^2+log^2(t-1)}\ &=&int_{-infty}^{infty}sum_{n=1}^infty left(frac x {1+e^u}right)^nfrac{e^umathrm d u}{{pi}^2+u^2} quad(t=1+e^u)\
&=& int_{-infty}^{infty}frac{x}{(1-x)+e^u}frac{e^umathrm d u}{{pi}^2+u^2}\
&=&int_{Im z =0}frac{x}{(1-x)+e^z}frac{e^zmathrm d z}{{pi}^2+z^2}.
end{eqnarray}$$ We also observe that
$$
lim_{cto infty}int_{Im z =c}frac{x}{(1-x)+e^z}frac{e^zmathrm d z}{{pi}^2+z^2}=0,
$$ which implies by residue theorem:
$$begin{eqnarray}
int_{Im z =0}frac{x}{(1-x)+e^z}frac{e^zmathrm d z}{{pi}^2+z^2}&=&2pi i sum_{z:Im(z)>0} text{res}_{z}frac{x}{(1-x)+e^z}frac{e^z}{{pi}^2+z^2}\
&=&1+2pi i xsum_{jge 0} frac{1}{left((2j+1)pi i + ln(1-x)right)^2 +pi^2}\
&=&1+xsum_{jge 0} left(frac{1}{(2j+1)pi i + ln(1-x)-pi i} \ -frac{1}{(2j+1)pi i + ln(1-x)+pi i}right)\
&=&1+ frac{x}{ln(1-x)}.
end{eqnarray}$$ (Here, residues are calculated at $z=pi i$ and $(2j+1)pi i +ln(1-x)$ where $jge 0$.)
This shows $$
sum_{nge 0}a_n x^n = frac{x}{ln(1-x)}
$$ and the claim follows.
answered Feb 2 at 17:46
SongSong
18.6k21651
$endgroup$
$begingroup$
Hell yeah! The master key was the use of convolution. Did not see that. Thanks
$endgroup$
– HAMIDINE SOUMARE
Feb 2 at 20:15
add a comment |
$begingroup$
The claim is equivalent to
$$
left(a*frac1{i+1}right)_n = -1_{n=0}
$$ holds for every $nge 0$, where $$
left( a*frac1{i+1}right)_n =sum_{k=0}^n a_{n-i}frac{1}{i+1}
$$ is the convolution of $(a_i)_{ige 0}$ and $(frac1{i+1})_{ige 0}$ and
$$
a_n =int_{1}^{infty}frac1{t^n}frac{mathrm d t}{{pi}^2+log^2(t-1)},quad nge 1,
$$$$
a_0=-1.
$$
In view of the generating function of $(frac1{i+1})_{ige 0}$, which is $-frac{ln(1-x)}x$, it suffices to prove that for $0<x<1$,
$$
sum_{nge 0} a_n x^n=-1+sum_{nge 1}a_n x^n = frac{x}{ln(1-x)}.
$$ We find that
$$begin{eqnarray}
sum_{n=1}^infty a_n x^n &=& int_{1}^{infty}sum_{n=1}^infty left(frac x tright)^nfrac{mathrm d t}{{pi}^2+log^2(t-1)}\ &=&int_{-infty}^{infty}sum_{n=1}^infty left(frac x {1+e^u}right)^nfrac{e^umathrm d u}{{pi}^2+u^2} quad(t=1+e^u)\
&=& int_{-infty}^{infty}frac{x}{(1-x)+e^u}frac{e^umathrm d u}{{pi}^2+u^2}\
&=&int_{Im z =0}frac{x}{(1-x)+e^z}frac{e^zmathrm d z}{{pi}^2+z^2}.
end{eqnarray}$$ We also observe that
$$
lim_{cto infty}int_{Im z =c}frac{x}{(1-x)+e^z}frac{e^zmathrm d z}{{pi}^2+z^2}=0,
$$ which implies by residue theorem:
$$begin{eqnarray}
int_{Im z =0}frac{x}{(1-x)+e^z}frac{e^zmathrm d z}{{pi}^2+z^2}&=&2pi i sum_{z:Im(z)>0} text{res}_{z}frac{x}{(1-x)+e^z}frac{e^z}{{pi}^2+z^2}\
&=&1+2pi i xsum_{jge 0} frac{1}{left((2j+1)pi i + ln(1-x)right)^2 +pi^2}\
&=&1+xsum_{jge 0} left(frac{1}{(2j+1)pi i + ln(1-x)-pi i} \ -frac{1}{(2j+1)pi i + ln(1-x)+pi i}right)\
&=&1+ frac{x}{ln(1-x)}.
end{eqnarray}$$ (Here, residues are calculated at $z=pi i$ and $(2j+1)pi i +ln(1-x)$ where $jge 0$.)
This shows $$
sum_{nge 0}a_n x^n = frac{x}{ln(1-x)}
$$ and the claim follows.
answered Feb 2 at 17:46
SongSong
18.6k21651
$endgroup$
The claim is equivalent to
$$
left(a*frac1{i+1}right)_n = -1_{n=0}
$$ holds for every $nge 0$, where $$
left( a*frac1{i+1}right)_n =sum_{k=0}^n a_{n-i}frac{1}{i+1}
$$ is the convolution of $(a_i)_{ige 0}$ and $(frac1{i+1})_{ige 0}$ and
$$
a_n =int_{1}^{infty}frac1{t^n}frac{mathrm d t}{{pi}^2+log^2(t-1)},quad nge 1,
$$$$
a_0=-1.
$$
In view of the generating function of $(frac1{i+1})_{ige 0}$, which is $-frac{ln(1-x)}x$, it suffices to prove that for $0<x<1$,
$$
sum_{nge 0} a_n x^n=-1+sum_{nge 1}a_n x^n = frac{x}{ln(1-x)}.
$$ We find that
$$begin{eqnarray}
sum_{n=1}^infty a_n x^n &=& int_{1}^{infty}sum_{n=1}^infty left(frac x tright)^nfrac{mathrm d t}{{pi}^2+log^2(t-1)}\ &=&int_{-infty}^{infty}sum_{n=1}^infty left(frac x {1+e^u}right)^nfrac{e^umathrm d u}{{pi}^2+u^2} quad(t=1+e^u)\
&=& int_{-infty}^{infty}frac{x}{(1-x)+e^u}frac{e^umathrm d u}{{pi}^2+u^2}\
&=&int_{Im z =0}frac{x}{(1-x)+e^z}frac{e^zmathrm d z}{{pi}^2+z^2}.
end{eqnarray}$$ We also observe that
$$
lim_{cto infty}int_{Im z =c}frac{x}{(1-x)+e^z}frac{e^zmathrm d z}{{pi}^2+z^2}=0,
$$ which implies by residue theorem:
$$begin{eqnarray}
int_{Im z =0}frac{x}{(1-x)+e^z}frac{e^zmathrm d z}{{pi}^2+z^2}&=&2pi i sum_{z:Im(z)>0} text{res}_{z}frac{x}{(1-x)+e^z}frac{e^z}{{pi}^2+z^2}\
&=&1+2pi i xsum_{jge 0} frac{1}{left((2j+1)pi i + ln(1-x)right)^2 +pi^2}\
&=&1+xsum_{jge 0} left(frac{1}{(2j+1)pi i + ln(1-x)-pi i} \ -frac{1}{(2j+1)pi i + ln(1-x)+pi i}right)\
&=&1+ frac{x}{ln(1-x)}.
end{eqnarray}$$ (Here, residues are calculated at $z=pi i$ and $(2j+1)pi i +ln(1-x)$ where $jge 0$.)
This shows $$
sum_{nge 0}a_n x^n = frac{x}{ln(1-x)}
$$ and the claim follows.
answered Feb 2 at 17:46
SongSong
18.6k21651
edited Feb 2 at 21:16
answered Feb 2 at 17:46
SongSong
18.6k21651
answered Feb 2 at 17:46
SongSong
18.6k21651
answered Feb 2 at 17:46
SongSong
18.6k21651
18.6k21651
$begingroup$
Hell yeah! The master key was the use of convolution. Did not see that. Thanks
$endgroup$
– HAMIDINE SOUMARE
Feb 2 at 20:15
add a comment |
$begingroup$
Hell yeah! The master key was the use of convolution. Did not see that. Thanks
$endgroup$
– HAMIDINE SOUMARE
Feb 2 at 20:15
$begingroup$
Hell yeah! The master key was the use of convolution. Did not see that. Thanks
$endgroup$
– HAMIDINE SOUMARE
Feb 2 at 20:15
$begingroup$
Hell yeah! The master key was the use of convolution. Did not see that. Thanks
$endgroup$
– HAMIDINE SOUMARE
Feb 2 at 20:15
add a comment |
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$begingroup$
This is known as the integral representation for Gregory coefficients: en.wikipedia.org/wiki/Gregory_coefficients
$endgroup$
– Zacky
Feb 2 at 23:41