Mixing
Finding a closed form for $I_n=int_0^1 prod_{l=1}^nleft[x^2-frac{l^2}{n^2}right]dx$
$begingroup$
I'm now dealing with this integral and trying to give a closed form to it:
$$I_n=int_0^1 prod_{l=1}^nleft[x^2-dfrac{l^2}{n^2}right]dx$$
Where the first few values are given:
$$I_1=int_0^1 (x^2-1) dx=-dfrac23$$
$$I_2=int_0^1 left(x^2-dfrac14right)(x^2-1)dx=dfrac1{30}$$
$$I_3=int_0^1 left(x^2-frac19right)left(x^2-frac49right)(x^2-1)dx=-dfrac{136}{8505}$$
I've tried to expand the polynomial, but I could merely give a general expression for the first few and the last few coefficients, and they become complicated very fast. I have searched the Internet; did I miss something that I couldn't find them because I don't know the Theorem's name?
calculus integration
edited Jan 18 at 15:59
Blue
48.6k870156
asked Jan 18 at 13:10
kelvin hong 方kelvin hong 方
66018
$endgroup$
add a comment |
$begingroup$
I'm now dealing with this integral and trying to give a closed form to it:
$$I_n=int_0^1 prod_{l=1}^nleft[x^2-dfrac{l^2}{n^2}right]dx$$
Where the first few values are given:
$$I_1=int_0^1 (x^2-1) dx=-dfrac23$$
$$I_2=int_0^1 left(x^2-dfrac14right)(x^2-1)dx=dfrac1{30}$$
$$I_3=int_0^1 left(x^2-frac19right)left(x^2-frac49right)(x^2-1)dx=-dfrac{136}{8505}$$
I've tried to expand the polynomial, but I could merely give a general expression for the first few and the last few coefficients, and they become complicated very fast. I have searched the Internet; did I miss something that I couldn't find them because I don't know the Theorem's name?
calculus integration
edited Jan 18 at 15:59
Blue
48.6k870156
asked Jan 18 at 13:10
kelvin hong 方kelvin hong 方
66018
$endgroup$
add a comment |
$begingroup$
I'm now dealing with this integral and trying to give a closed form to it:
$$I_n=int_0^1 prod_{l=1}^nleft[x^2-dfrac{l^2}{n^2}right]dx$$
Where the first few values are given:
$$I_1=int_0^1 (x^2-1) dx=-dfrac23$$
$$I_2=int_0^1 left(x^2-dfrac14right)(x^2-1)dx=dfrac1{30}$$
$$I_3=int_0^1 left(x^2-frac19right)left(x^2-frac49right)(x^2-1)dx=-dfrac{136}{8505}$$
I've tried to expand the polynomial, but I could merely give a general expression for the first few and the last few coefficients, and they become complicated very fast. I have searched the Internet; did I miss something that I couldn't find them because I don't know the Theorem's name?
calculus integration
edited Jan 18 at 15:59
Blue
48.6k870156
asked Jan 18 at 13:10
kelvin hong 方kelvin hong 方
66018
$endgroup$
I'm now dealing with this integral and trying to give a closed form to it:
$$I_n=int_0^1 prod_{l=1}^nleft[x^2-dfrac{l^2}{n^2}right]dx$$
Where the first few values are given:
$$I_1=int_0^1 (x^2-1) dx=-dfrac23$$
$$I_2=int_0^1 left(x^2-dfrac14right)(x^2-1)dx=dfrac1{30}$$
$$I_3=int_0^1 left(x^2-frac19right)left(x^2-frac49right)(x^2-1)dx=-dfrac{136}{8505}$$
I've tried to expand the polynomial, but I could merely give a general expression for the first few and the last few coefficients, and they become complicated very fast. I have searched the Internet; did I miss something that I couldn't find them because I don't know the Theorem's name?
calculus integration
calculus integration
edited Jan 18 at 15:59
Blue
48.6k870156
asked Jan 18 at 13:10
kelvin hong 方kelvin hong 方
66018
edited Jan 18 at 15:59
Blue
48.6k870156
asked Jan 18 at 13:10
kelvin hong 方kelvin hong 方
66018
edited Jan 18 at 15:59
Blue
48.6k870156
edited Jan 18 at 15:59
Blue
48.6k870156
edited Jan 18 at 15:59
Blue
48.6k870156
48.6k870156
asked Jan 18 at 13:10
kelvin hong 方kelvin hong 方
66018
asked Jan 18 at 13:10
kelvin hong 方kelvin hong 方
66018
asked Jan 18 at 13:10
kelvin hong 方kelvin hong 方
66018
66018
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$I_n=frac{1}{n^{2n}}int_0^1prod_{l=1}^{n}[(xn)^2-l^2]dxqquad x=frac{y}{n}$$
$$I_n=frac{1}{n^{2n+1}}int_0^nprod_{l=1}^{n}[y^2-l^2]dy$$
We can expand the product, without knowing the coefficients for now as following;
$$prod_{l=1}^{n}[y^2-l^2]=sum_{j=0}^n(-1)^{n-j}y^{2j}t(n,j)$$
$$I_n=frac{1}{n^{2n+1}}sum_{j=0}^n(-1)^{n-j}t(n,j)int_0^ny^{2j}dy$$
$$I_n=frac{1}{n^{2n+1}}sum_{j=0}^nfrac{(-1)^{n-j}t(n,j)n^{2j+1}}{2j+1}$$
$$I_n=frac{(-1)^n}{n^{2n}}sum_{j=0}^nfrac{(-1)^{j}t(n,j)n^{2j}}{2j+1}$$
Now back to the coefficients, ive found it on OEIS A008955
they are called the central factorial numbers. I think this is the closest we can get for a closed form.
answered Jan 18 at 17:29
Andrew KovácsAndrew Kovács
1264
$endgroup$
$begingroup$
Since this (I assume) can be computed in a finite number of steps, it counts as a closed form $rightarrow $ (+1).
$endgroup$
– clathratus
Jan 18 at 18:45
$begingroup$
Here's a (actually kinda bad) reference for central factorial. mathworld.wolfram.com/CentralFactorial.html Its the best I could find so far
$endgroup$
– clathratus
Jan 18 at 18:47
$begingroup$
Thank you very much, I think the Central Factorial numbers are what I want.
$endgroup$
– kelvin hong 方
Jan 19 at 2:32
add a comment |
$begingroup$
You can relate the product to the rising factorials $x^overline{n} = x(x+1)dotsb (x+n-1) = sum_{k=0}^n begin{bmatrix}n\kend{bmatrix} x^k$ where the coefficients are Stirling's numbers of the first kind.
begin{align}
I_n
&= int_0^1 prod_{l=1}^n left(x^2 - frac{l^2}{n^2}right);dx
= int_0^1 prod_{l-1 =k=0}^{n-1} frac{xn - k + 1}{n}cdot prod_{k=0}^{n-1} frac{xn + k + 1}{n};dx\
&= frac{(-1)^n }{n^{2n}}int_0^1 (-xn - 1)^{overline{n}}(xn +1)^{overline{n}};dx\
%
&= frac{(-1)^n }{n^{2n}}int_0^1 sum_{i=0}^nsum_{j=0}^n begin{bmatrix}n\iend{bmatrix}
begin{bmatrix}n\jend{bmatrix} (-xn - 1)^i (xn + 1)^j;dx\
%
&= frac{(-1)^n }{n^{2n}} sum_{i=0}^nsum_{j=0}^n (-1)^i begin{bmatrix}n\iend{bmatrix}
begin{bmatrix}n\jend{bmatrix} int_0^1 (xn + 1)^{i+j} ;dx\
%
&= frac{(-1)^n }{n^{2n}} sum_{i=0}^nsum_{j=0}^n (-1)^i begin{bmatrix}n\iend{bmatrix}
begin{bmatrix}n\jend{bmatrix} frac{(n+1)^{i+j+1}-1}{n(i+j+1)}\
%
&= sum_{i=0}^nsum_{j=0}^n (-1)^{i+n}begin{bmatrix}n\iend{bmatrix}
begin{bmatrix}n\jend{bmatrix} frac{(n+1)^{i+j+1}-1}{n^{2n+1}(i+j+1)}\
%
end{align}
answered Jan 19 at 8:43
adfriedmanadfriedman
3,171169
$endgroup$
$begingroup$
I think the sum is too complicated, and the Stirling numbers of the first kind are also hard to compute. But this does give me a new way to deal with this problem. Thank you.
$endgroup$
– kelvin hong 方
Jan 19 at 12:14
$begingroup$
It was likely obvious to you that, as per the other answer, you can expand the product as an $2n$-degree polynomial. I felt that doing a OEIS coefficient search and presenting that, without even proving the connection, fell short of an answer.
$endgroup$
– adfriedman
Jan 19 at 18:58
$begingroup$
Yes, to evaluate this, we still need computer, I think the simplest form is the sum, can be impossible to represented as a term.
$endgroup$
– kelvin hong 方
Jan 20 at 4:05
add a comment |
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2 Answers
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2 Answers
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$begingroup$
$$I_n=frac{1}{n^{2n}}int_0^1prod_{l=1}^{n}[(xn)^2-l^2]dxqquad x=frac{y}{n}$$
$$I_n=frac{1}{n^{2n+1}}int_0^nprod_{l=1}^{n}[y^2-l^2]dy$$
We can expand the product, without knowing the coefficients for now as following;
$$prod_{l=1}^{n}[y^2-l^2]=sum_{j=0}^n(-1)^{n-j}y^{2j}t(n,j)$$
$$I_n=frac{1}{n^{2n+1}}sum_{j=0}^n(-1)^{n-j}t(n,j)int_0^ny^{2j}dy$$
$$I_n=frac{1}{n^{2n+1}}sum_{j=0}^nfrac{(-1)^{n-j}t(n,j)n^{2j+1}}{2j+1}$$
$$I_n=frac{(-1)^n}{n^{2n}}sum_{j=0}^nfrac{(-1)^{j}t(n,j)n^{2j}}{2j+1}$$
Now back to the coefficients, ive found it on OEIS A008955
they are called the central factorial numbers. I think this is the closest we can get for a closed form.
answered Jan 18 at 17:29
Andrew KovácsAndrew Kovács
1264
$endgroup$
$begingroup$
Since this (I assume) can be computed in a finite number of steps, it counts as a closed form $rightarrow $ (+1).
$endgroup$
– clathratus
Jan 18 at 18:45
$begingroup$
Here's a (actually kinda bad) reference for central factorial. mathworld.wolfram.com/CentralFactorial.html Its the best I could find so far
$endgroup$
– clathratus
Jan 18 at 18:47
$begingroup$
Thank you very much, I think the Central Factorial numbers are what I want.
$endgroup$
– kelvin hong 方
Jan 19 at 2:32
add a comment |
$begingroup$
$$I_n=frac{1}{n^{2n}}int_0^1prod_{l=1}^{n}[(xn)^2-l^2]dxqquad x=frac{y}{n}$$
$$I_n=frac{1}{n^{2n+1}}int_0^nprod_{l=1}^{n}[y^2-l^2]dy$$
We can expand the product, without knowing the coefficients for now as following;
$$prod_{l=1}^{n}[y^2-l^2]=sum_{j=0}^n(-1)^{n-j}y^{2j}t(n,j)$$
$$I_n=frac{1}{n^{2n+1}}sum_{j=0}^n(-1)^{n-j}t(n,j)int_0^ny^{2j}dy$$
$$I_n=frac{1}{n^{2n+1}}sum_{j=0}^nfrac{(-1)^{n-j}t(n,j)n^{2j+1}}{2j+1}$$
$$I_n=frac{(-1)^n}{n^{2n}}sum_{j=0}^nfrac{(-1)^{j}t(n,j)n^{2j}}{2j+1}$$
Now back to the coefficients, ive found it on OEIS A008955
they are called the central factorial numbers. I think this is the closest we can get for a closed form.
answered Jan 18 at 17:29
Andrew KovácsAndrew Kovács
1264
$endgroup$
$begingroup$
Since this (I assume) can be computed in a finite number of steps, it counts as a closed form $rightarrow $ (+1).
$endgroup$
– clathratus
Jan 18 at 18:45
$begingroup$
Here's a (actually kinda bad) reference for central factorial. mathworld.wolfram.com/CentralFactorial.html Its the best I could find so far
$endgroup$
– clathratus
Jan 18 at 18:47
$begingroup$
Thank you very much, I think the Central Factorial numbers are what I want.
$endgroup$
– kelvin hong 方
Jan 19 at 2:32
add a comment |
$begingroup$
$$I_n=frac{1}{n^{2n}}int_0^1prod_{l=1}^{n}[(xn)^2-l^2]dxqquad x=frac{y}{n}$$
$$I_n=frac{1}{n^{2n+1}}int_0^nprod_{l=1}^{n}[y^2-l^2]dy$$
We can expand the product, without knowing the coefficients for now as following;
$$prod_{l=1}^{n}[y^2-l^2]=sum_{j=0}^n(-1)^{n-j}y^{2j}t(n,j)$$
$$I_n=frac{1}{n^{2n+1}}sum_{j=0}^n(-1)^{n-j}t(n,j)int_0^ny^{2j}dy$$
$$I_n=frac{1}{n^{2n+1}}sum_{j=0}^nfrac{(-1)^{n-j}t(n,j)n^{2j+1}}{2j+1}$$
$$I_n=frac{(-1)^n}{n^{2n}}sum_{j=0}^nfrac{(-1)^{j}t(n,j)n^{2j}}{2j+1}$$
Now back to the coefficients, ive found it on OEIS A008955
they are called the central factorial numbers. I think this is the closest we can get for a closed form.
answered Jan 18 at 17:29
Andrew KovácsAndrew Kovács
1264
$endgroup$
$$I_n=frac{1}{n^{2n}}int_0^1prod_{l=1}^{n}[(xn)^2-l^2]dxqquad x=frac{y}{n}$$
$$I_n=frac{1}{n^{2n+1}}int_0^nprod_{l=1}^{n}[y^2-l^2]dy$$
We can expand the product, without knowing the coefficients for now as following;
$$prod_{l=1}^{n}[y^2-l^2]=sum_{j=0}^n(-1)^{n-j}y^{2j}t(n,j)$$
$$I_n=frac{1}{n^{2n+1}}sum_{j=0}^n(-1)^{n-j}t(n,j)int_0^ny^{2j}dy$$
$$I_n=frac{1}{n^{2n+1}}sum_{j=0}^nfrac{(-1)^{n-j}t(n,j)n^{2j+1}}{2j+1}$$
$$I_n=frac{(-1)^n}{n^{2n}}sum_{j=0}^nfrac{(-1)^{j}t(n,j)n^{2j}}{2j+1}$$
Now back to the coefficients, ive found it on OEIS A008955
they are called the central factorial numbers. I think this is the closest we can get for a closed form.
answered Jan 18 at 17:29
Andrew KovácsAndrew Kovács
1264
edited Jan 18 at 23:33
answered Jan 18 at 17:29
Andrew KovácsAndrew Kovács
1264
answered Jan 18 at 17:29
Andrew KovácsAndrew Kovács
1264
answered Jan 18 at 17:29
Andrew KovácsAndrew Kovács
1264
1264
$begingroup$
Since this (I assume) can be computed in a finite number of steps, it counts as a closed form $rightarrow $ (+1).
$endgroup$
– clathratus
Jan 18 at 18:45
$begingroup$
Here's a (actually kinda bad) reference for central factorial. mathworld.wolfram.com/CentralFactorial.html Its the best I could find so far
$endgroup$
– clathratus
Jan 18 at 18:47
$begingroup$
Thank you very much, I think the Central Factorial numbers are what I want.
$endgroup$
– kelvin hong 方
Jan 19 at 2:32
add a comment |
$begingroup$
Since this (I assume) can be computed in a finite number of steps, it counts as a closed form $rightarrow $ (+1).
$endgroup$
– clathratus
Jan 18 at 18:45
$begingroup$
Here's a (actually kinda bad) reference for central factorial. mathworld.wolfram.com/CentralFactorial.html Its the best I could find so far
$endgroup$
– clathratus
Jan 18 at 18:47
$begingroup$
Thank you very much, I think the Central Factorial numbers are what I want.
$endgroup$
– kelvin hong 方
Jan 19 at 2:32
$begingroup$
Since this (I assume) can be computed in a finite number of steps, it counts as a closed form $rightarrow $ (+1).
$endgroup$
– clathratus
Jan 18 at 18:45
$begingroup$
Since this (I assume) can be computed in a finite number of steps, it counts as a closed form $rightarrow $ (+1).
$endgroup$
– clathratus
Jan 18 at 18:45
$begingroup$
Here's a (actually kinda bad) reference for central factorial. mathworld.wolfram.com/CentralFactorial.html Its the best I could find so far
$endgroup$
– clathratus
Jan 18 at 18:47
$begingroup$
Here's a (actually kinda bad) reference for central factorial. mathworld.wolfram.com/CentralFactorial.html Its the best I could find so far
$endgroup$
– clathratus
Jan 18 at 18:47
$begingroup$
Thank you very much, I think the Central Factorial numbers are what I want.
$endgroup$
– kelvin hong 方
Jan 19 at 2:32
$begingroup$
Thank you very much, I think the Central Factorial numbers are what I want.
$endgroup$
– kelvin hong 方
Jan 19 at 2:32
add a comment |
$begingroup$
You can relate the product to the rising factorials $x^overline{n} = x(x+1)dotsb (x+n-1) = sum_{k=0}^n begin{bmatrix}n\kend{bmatrix} x^k$ where the coefficients are Stirling's numbers of the first kind.
begin{align}
I_n
&= int_0^1 prod_{l=1}^n left(x^2 - frac{l^2}{n^2}right);dx
= int_0^1 prod_{l-1 =k=0}^{n-1} frac{xn - k + 1}{n}cdot prod_{k=0}^{n-1} frac{xn + k + 1}{n};dx\
&= frac{(-1)^n }{n^{2n}}int_0^1 (-xn - 1)^{overline{n}}(xn +1)^{overline{n}};dx\
%
&= frac{(-1)^n }{n^{2n}}int_0^1 sum_{i=0}^nsum_{j=0}^n begin{bmatrix}n\iend{bmatrix}
begin{bmatrix}n\jend{bmatrix} (-xn - 1)^i (xn + 1)^j;dx\
%
&= frac{(-1)^n }{n^{2n}} sum_{i=0}^nsum_{j=0}^n (-1)^i begin{bmatrix}n\iend{bmatrix}
begin{bmatrix}n\jend{bmatrix} int_0^1 (xn + 1)^{i+j} ;dx\
%
&= frac{(-1)^n }{n^{2n}} sum_{i=0}^nsum_{j=0}^n (-1)^i begin{bmatrix}n\iend{bmatrix}
begin{bmatrix}n\jend{bmatrix} frac{(n+1)^{i+j+1}-1}{n(i+j+1)}\
%
&= sum_{i=0}^nsum_{j=0}^n (-1)^{i+n}begin{bmatrix}n\iend{bmatrix}
begin{bmatrix}n\jend{bmatrix} frac{(n+1)^{i+j+1}-1}{n^{2n+1}(i+j+1)}\
%
end{align}
answered Jan 19 at 8:43
adfriedmanadfriedman
3,171169
$endgroup$
$begingroup$
I think the sum is too complicated, and the Stirling numbers of the first kind are also hard to compute. But this does give me a new way to deal with this problem. Thank you.
$endgroup$
– kelvin hong 方
Jan 19 at 12:14
$begingroup$
It was likely obvious to you that, as per the other answer, you can expand the product as an $2n$-degree polynomial. I felt that doing a OEIS coefficient search and presenting that, without even proving the connection, fell short of an answer.
$endgroup$
– adfriedman
Jan 19 at 18:58
$begingroup$
Yes, to evaluate this, we still need computer, I think the simplest form is the sum, can be impossible to represented as a term.
$endgroup$
– kelvin hong 方
Jan 20 at 4:05
add a comment |
$begingroup$
You can relate the product to the rising factorials $x^overline{n} = x(x+1)dotsb (x+n-1) = sum_{k=0}^n begin{bmatrix}n\kend{bmatrix} x^k$ where the coefficients are Stirling's numbers of the first kind.
begin{align}
I_n
&= int_0^1 prod_{l=1}^n left(x^2 - frac{l^2}{n^2}right);dx
= int_0^1 prod_{l-1 =k=0}^{n-1} frac{xn - k + 1}{n}cdot prod_{k=0}^{n-1} frac{xn + k + 1}{n};dx\
&= frac{(-1)^n }{n^{2n}}int_0^1 (-xn - 1)^{overline{n}}(xn +1)^{overline{n}};dx\
%
&= frac{(-1)^n }{n^{2n}}int_0^1 sum_{i=0}^nsum_{j=0}^n begin{bmatrix}n\iend{bmatrix}
begin{bmatrix}n\jend{bmatrix} (-xn - 1)^i (xn + 1)^j;dx\
%
&= frac{(-1)^n }{n^{2n}} sum_{i=0}^nsum_{j=0}^n (-1)^i begin{bmatrix}n\iend{bmatrix}
begin{bmatrix}n\jend{bmatrix} int_0^1 (xn + 1)^{i+j} ;dx\
%
&= frac{(-1)^n }{n^{2n}} sum_{i=0}^nsum_{j=0}^n (-1)^i begin{bmatrix}n\iend{bmatrix}
begin{bmatrix}n\jend{bmatrix} frac{(n+1)^{i+j+1}-1}{n(i+j+1)}\
%
&= sum_{i=0}^nsum_{j=0}^n (-1)^{i+n}begin{bmatrix}n\iend{bmatrix}
begin{bmatrix}n\jend{bmatrix} frac{(n+1)^{i+j+1}-1}{n^{2n+1}(i+j+1)}\
%
end{align}
answered Jan 19 at 8:43
adfriedmanadfriedman
3,171169
$endgroup$
$begingroup$
I think the sum is too complicated, and the Stirling numbers of the first kind are also hard to compute. But this does give me a new way to deal with this problem. Thank you.
$endgroup$
– kelvin hong 方
Jan 19 at 12:14
$begingroup$
It was likely obvious to you that, as per the other answer, you can expand the product as an $2n$-degree polynomial. I felt that doing a OEIS coefficient search and presenting that, without even proving the connection, fell short of an answer.
$endgroup$
– adfriedman
Jan 19 at 18:58
$begingroup$
Yes, to evaluate this, we still need computer, I think the simplest form is the sum, can be impossible to represented as a term.
$endgroup$
– kelvin hong 方
Jan 20 at 4:05
add a comment |
$begingroup$
You can relate the product to the rising factorials $x^overline{n} = x(x+1)dotsb (x+n-1) = sum_{k=0}^n begin{bmatrix}n\kend{bmatrix} x^k$ where the coefficients are Stirling's numbers of the first kind.
begin{align}
I_n
&= int_0^1 prod_{l=1}^n left(x^2 - frac{l^2}{n^2}right);dx
= int_0^1 prod_{l-1 =k=0}^{n-1} frac{xn - k + 1}{n}cdot prod_{k=0}^{n-1} frac{xn + k + 1}{n};dx\
&= frac{(-1)^n }{n^{2n}}int_0^1 (-xn - 1)^{overline{n}}(xn +1)^{overline{n}};dx\
%
&= frac{(-1)^n }{n^{2n}}int_0^1 sum_{i=0}^nsum_{j=0}^n begin{bmatrix}n\iend{bmatrix}
begin{bmatrix}n\jend{bmatrix} (-xn - 1)^i (xn + 1)^j;dx\
%
&= frac{(-1)^n }{n^{2n}} sum_{i=0}^nsum_{j=0}^n (-1)^i begin{bmatrix}n\iend{bmatrix}
begin{bmatrix}n\jend{bmatrix} int_0^1 (xn + 1)^{i+j} ;dx\
%
&= frac{(-1)^n }{n^{2n}} sum_{i=0}^nsum_{j=0}^n (-1)^i begin{bmatrix}n\iend{bmatrix}
begin{bmatrix}n\jend{bmatrix} frac{(n+1)^{i+j+1}-1}{n(i+j+1)}\
%
&= sum_{i=0}^nsum_{j=0}^n (-1)^{i+n}begin{bmatrix}n\iend{bmatrix}
begin{bmatrix}n\jend{bmatrix} frac{(n+1)^{i+j+1}-1}{n^{2n+1}(i+j+1)}\
%
end{align}
answered Jan 19 at 8:43
adfriedmanadfriedman
3,171169
$endgroup$
You can relate the product to the rising factorials $x^overline{n} = x(x+1)dotsb (x+n-1) = sum_{k=0}^n begin{bmatrix}n\kend{bmatrix} x^k$ where the coefficients are Stirling's numbers of the first kind.
begin{align}
I_n
&= int_0^1 prod_{l=1}^n left(x^2 - frac{l^2}{n^2}right);dx
= int_0^1 prod_{l-1 =k=0}^{n-1} frac{xn - k + 1}{n}cdot prod_{k=0}^{n-1} frac{xn + k + 1}{n};dx\
&= frac{(-1)^n }{n^{2n}}int_0^1 (-xn - 1)^{overline{n}}(xn +1)^{overline{n}};dx\
%
&= frac{(-1)^n }{n^{2n}}int_0^1 sum_{i=0}^nsum_{j=0}^n begin{bmatrix}n\iend{bmatrix}
begin{bmatrix}n\jend{bmatrix} (-xn - 1)^i (xn + 1)^j;dx\
%
&= frac{(-1)^n }{n^{2n}} sum_{i=0}^nsum_{j=0}^n (-1)^i begin{bmatrix}n\iend{bmatrix}
begin{bmatrix}n\jend{bmatrix} int_0^1 (xn + 1)^{i+j} ;dx\
%
&= frac{(-1)^n }{n^{2n}} sum_{i=0}^nsum_{j=0}^n (-1)^i begin{bmatrix}n\iend{bmatrix}
begin{bmatrix}n\jend{bmatrix} frac{(n+1)^{i+j+1}-1}{n(i+j+1)}\
%
&= sum_{i=0}^nsum_{j=0}^n (-1)^{i+n}begin{bmatrix}n\iend{bmatrix}
begin{bmatrix}n\jend{bmatrix} frac{(n+1)^{i+j+1}-1}{n^{2n+1}(i+j+1)}\
%
end{align}
answered Jan 19 at 8:43
adfriedmanadfriedman
3,171169
answered Jan 19 at 8:43
adfriedmanadfriedman
3,171169
answered Jan 19 at 8:43
adfriedmanadfriedman
3,171169
answered Jan 19 at 8:43
adfriedmanadfriedman
3,171169
3,171169
$begingroup$
I think the sum is too complicated, and the Stirling numbers of the first kind are also hard to compute. But this does give me a new way to deal with this problem. Thank you.
$endgroup$
– kelvin hong 方
Jan 19 at 12:14
$begingroup$
It was likely obvious to you that, as per the other answer, you can expand the product as an $2n$-degree polynomial. I felt that doing a OEIS coefficient search and presenting that, without even proving the connection, fell short of an answer.
$endgroup$
– adfriedman
Jan 19 at 18:58
$begingroup$
Yes, to evaluate this, we still need computer, I think the simplest form is the sum, can be impossible to represented as a term.
$endgroup$
– kelvin hong 方
Jan 20 at 4:05
add a comment |
$begingroup$
I think the sum is too complicated, and the Stirling numbers of the first kind are also hard to compute. But this does give me a new way to deal with this problem. Thank you.
$endgroup$
– kelvin hong 方
Jan 19 at 12:14
$begingroup$
It was likely obvious to you that, as per the other answer, you can expand the product as an $2n$-degree polynomial. I felt that doing a OEIS coefficient search and presenting that, without even proving the connection, fell short of an answer.
$endgroup$
– adfriedman
Jan 19 at 18:58
$begingroup$
Yes, to evaluate this, we still need computer, I think the simplest form is the sum, can be impossible to represented as a term.
$endgroup$
– kelvin hong 方
Jan 20 at 4:05
$begingroup$
I think the sum is too complicated, and the Stirling numbers of the first kind are also hard to compute. But this does give me a new way to deal with this problem. Thank you.
$endgroup$
– kelvin hong 方
Jan 19 at 12:14
$begingroup$
I think the sum is too complicated, and the Stirling numbers of the first kind are also hard to compute. But this does give me a new way to deal with this problem. Thank you.
$endgroup$
– kelvin hong 方
Jan 19 at 12:14
$begingroup$
It was likely obvious to you that, as per the other answer, you can expand the product as an $2n$-degree polynomial. I felt that doing a OEIS coefficient search and presenting that, without even proving the connection, fell short of an answer.
$endgroup$
– adfriedman
Jan 19 at 18:58
$begingroup$
It was likely obvious to you that, as per the other answer, you can expand the product as an $2n$-degree polynomial. I felt that doing a OEIS coefficient search and presenting that, without even proving the connection, fell short of an answer.
$endgroup$
– adfriedman
Jan 19 at 18:58
$begingroup$
Yes, to evaluate this, we still need computer, I think the simplest form is the sum, can be impossible to represented as a term.
$endgroup$
– kelvin hong 方
Jan 20 at 4:05
$begingroup$
Yes, to evaluate this, we still need computer, I think the simplest form is the sum, can be impossible to represented as a term.
$endgroup$
– kelvin hong 方
Jan 20 at 4:05
add a comment |
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