Learning how combinatorial expressions relate to integration with complex numbers
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I was playing around with some combinatorial expressions and tried to simplify the expression $binom{k^2+k}{k^2-k}$. I got stuck at the point $prod_{m=1}^{m=2k} frac{k^2-k+m}{m}$ so I plugged the expression $binom{k^2+k}{k^2-k}$ into Wolfram Alpha and got the following equivalent integral representations 
I would like to know what resources I should use to learn about how this simplification occurs. Currently I am knowledgeable about single-variable Calculus, multi-variable Calculus, and differential equations, the latter two to a much lesser extent.
calculus
asked Nov 27 '18 at 19:16
blue applesblue apples
233
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add a comment |
$begingroup$
I was playing around with some combinatorial expressions and tried to simplify the expression $binom{k^2+k}{k^2-k}$. I got stuck at the point $prod_{m=1}^{m=2k} frac{k^2-k+m}{m}$ so I plugged the expression $binom{k^2+k}{k^2-k}$ into Wolfram Alpha and got the following equivalent integral representations
I would like to know what resources I should use to learn about how this simplification occurs. Currently I am knowledgeable about single-variable Calculus, multi-variable Calculus, and differential equations, the latter two to a much lesser extent.
calculus
asked Nov 27 '18 at 19:16
blue applesblue apples
233
$endgroup$
add a comment |
$begingroup$
I was playing around with some combinatorial expressions and tried to simplify the expression $binom{k^2+k}{k^2-k}$. I got stuck at the point $prod_{m=1}^{m=2k} frac{k^2-k+m}{m}$ so I plugged the expression $binom{k^2+k}{k^2-k}$ into Wolfram Alpha and got the following equivalent integral representations
I would like to know what resources I should use to learn about how this simplification occurs. Currently I am knowledgeable about single-variable Calculus, multi-variable Calculus, and differential equations, the latter two to a much lesser extent.
calculus
asked Nov 27 '18 at 19:16
blue applesblue apples
233
$endgroup$
I was playing around with some combinatorial expressions and tried to simplify the expression $binom{k^2+k}{k^2-k}$. I got stuck at the point $prod_{m=1}^{m=2k} frac{k^2-k+m}{m}$ so I plugged the expression $binom{k^2+k}{k^2-k}$ into Wolfram Alpha and got the following equivalent integral representations
I would like to know what resources I should use to learn about how this simplification occurs. Currently I am knowledgeable about single-variable Calculus, multi-variable Calculus, and differential equations, the latter two to a much lesser extent.
calculus
calculus
asked Nov 27 '18 at 19:16
blue applesblue apples
233
asked Nov 27 '18 at 19:16
blue applesblue apples
233
asked Nov 27 '18 at 19:16
blue applesblue apples
233
asked Nov 27 '18 at 19:16
blue applesblue apples
233
asked Nov 27 '18 at 19:16
blue applesblue apples
233
233
add a comment |
add a comment |
1 Answer
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The right-hand side is $sum_{j=0}^{k(k+1)}frac{1}{2pi}int_{-pi}^{pi}binom{k(k+1)}{j}exp i(j+k-k^2)tdt$. But any integer $l$ satisfies $$frac{1}{2pi}int_{-pi}^piexp iltdt=delta_{l0},$$so the only surviving term has $j=k^2-k$.
answered Nov 27 '18 at 19:28
J.G.J.G.
25.2k22539
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@kjetilbhalvorsen It looks like a complete expression for the right-hand side to me. What's missing?
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– J.G.
Nov 27 '18 at 21:00
1
$begingroup$
Writing a binomial as a complex integral is classical, and is a powerful method of investigation/proof. See for example math.stackexchange.com/q/1215985
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– Jean Marie
Nov 27 '18 at 22:22
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I was more looking for resources that would help me understand where that expression comes from because as of right now, it's completely alien to me. As I said in my original post, I am knowledgeable about single-variable Calculus, multi-variable Calculus, and Differential Equations, though the latter two to a much lesser extent. What should I look at to be able to understand this myself?
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– blue apples
Nov 28 '18 at 14:41
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The right-hand side is $sum_{j=0}^{k(k+1)}frac{1}{2pi}int_{-pi}^{pi}binom{k(k+1)}{j}exp i(j+k-k^2)tdt$. But any integer $l$ satisfies $$frac{1}{2pi}int_{-pi}^piexp iltdt=delta_{l0},$$so the only surviving term has $j=k^2-k$.
answered Nov 27 '18 at 19:28
J.G.J.G.
25.2k22539
$endgroup$
$begingroup$
@kjetilbhalvorsen It looks like a complete expression for the right-hand side to me. What's missing?
$endgroup$
– J.G.
Nov 27 '18 at 21:00
1
$begingroup$
Writing a binomial as a complex integral is classical, and is a powerful method of investigation/proof. See for example math.stackexchange.com/q/1215985
$endgroup$
– Jean Marie
Nov 27 '18 at 22:22
$begingroup$
I was more looking for resources that would help me understand where that expression comes from because as of right now, it's completely alien to me. As I said in my original post, I am knowledgeable about single-variable Calculus, multi-variable Calculus, and Differential Equations, though the latter two to a much lesser extent. What should I look at to be able to understand this myself?
$endgroup$
– blue apples
Nov 28 '18 at 14:41
add a comment |
$begingroup$
The right-hand side is $sum_{j=0}^{k(k+1)}frac{1}{2pi}int_{-pi}^{pi}binom{k(k+1)}{j}exp i(j+k-k^2)tdt$. But any integer $l$ satisfies $$frac{1}{2pi}int_{-pi}^piexp iltdt=delta_{l0},$$so the only surviving term has $j=k^2-k$.
answered Nov 27 '18 at 19:28
J.G.J.G.
25.2k22539
$endgroup$
$begingroup$
@kjetilbhalvorsen It looks like a complete expression for the right-hand side to me. What's missing?
$endgroup$
– J.G.
Nov 27 '18 at 21:00
1
$begingroup$
Writing a binomial as a complex integral is classical, and is a powerful method of investigation/proof. See for example math.stackexchange.com/q/1215985
$endgroup$
– Jean Marie
Nov 27 '18 at 22:22
$begingroup$
I was more looking for resources that would help me understand where that expression comes from because as of right now, it's completely alien to me. As I said in my original post, I am knowledgeable about single-variable Calculus, multi-variable Calculus, and Differential Equations, though the latter two to a much lesser extent. What should I look at to be able to understand this myself?
$endgroup$
– blue apples
Nov 28 '18 at 14:41
add a comment |
$begingroup$
The right-hand side is $sum_{j=0}^{k(k+1)}frac{1}{2pi}int_{-pi}^{pi}binom{k(k+1)}{j}exp i(j+k-k^2)tdt$. But any integer $l$ satisfies $$frac{1}{2pi}int_{-pi}^piexp iltdt=delta_{l0},$$so the only surviving term has $j=k^2-k$.
answered Nov 27 '18 at 19:28
J.G.J.G.
25.2k22539
$endgroup$
The right-hand side is $sum_{j=0}^{k(k+1)}frac{1}{2pi}int_{-pi}^{pi}binom{k(k+1)}{j}exp i(j+k-k^2)tdt$. But any integer $l$ satisfies $$frac{1}{2pi}int_{-pi}^piexp iltdt=delta_{l0},$$so the only surviving term has $j=k^2-k$.
answered Nov 27 '18 at 19:28
J.G.J.G.
25.2k22539
answered Nov 27 '18 at 19:28
J.G.J.G.
25.2k22539
answered Nov 27 '18 at 19:28
J.G.J.G.
25.2k22539
answered Nov 27 '18 at 19:28
J.G.J.G.
25.2k22539
25.2k22539
$begingroup$
@kjetilbhalvorsen It looks like a complete expression for the right-hand side to me. What's missing?
$endgroup$
– J.G.
Nov 27 '18 at 21:00
1
$begingroup$
Writing a binomial as a complex integral is classical, and is a powerful method of investigation/proof. See for example math.stackexchange.com/q/1215985
$endgroup$
– Jean Marie
Nov 27 '18 at 22:22
$begingroup$
I was more looking for resources that would help me understand where that expression comes from because as of right now, it's completely alien to me. As I said in my original post, I am knowledgeable about single-variable Calculus, multi-variable Calculus, and Differential Equations, though the latter two to a much lesser extent. What should I look at to be able to understand this myself?
$endgroup$
– blue apples
Nov 28 '18 at 14:41
add a comment |
$begingroup$
@kjetilbhalvorsen It looks like a complete expression for the right-hand side to me. What's missing?
$endgroup$
– J.G.
Nov 27 '18 at 21:00
1
$begingroup$
Writing a binomial as a complex integral is classical, and is a powerful method of investigation/proof. See for example math.stackexchange.com/q/1215985
$endgroup$
– Jean Marie
Nov 27 '18 at 22:22
$begingroup$
I was more looking for resources that would help me understand where that expression comes from because as of right now, it's completely alien to me. As I said in my original post, I am knowledgeable about single-variable Calculus, multi-variable Calculus, and Differential Equations, though the latter two to a much lesser extent. What should I look at to be able to understand this myself?
$endgroup$
– blue apples
Nov 28 '18 at 14:41
$begingroup$
@kjetilbhalvorsen It looks like a complete expression for the right-hand side to me. What's missing?
$endgroup$
– J.G.
Nov 27 '18 at 21:00
$begingroup$
@kjetilbhalvorsen It looks like a complete expression for the right-hand side to me. What's missing?
$endgroup$
– J.G.
Nov 27 '18 at 21:00
1
1
$begingroup$
Writing a binomial as a complex integral is classical, and is a powerful method of investigation/proof. See for example math.stackexchange.com/q/1215985
$endgroup$
– Jean Marie
Nov 27 '18 at 22:22
$begingroup$
Writing a binomial as a complex integral is classical, and is a powerful method of investigation/proof. See for example math.stackexchange.com/q/1215985
$endgroup$
– Jean Marie
Nov 27 '18 at 22:22
$begingroup$
I was more looking for resources that would help me understand where that expression comes from because as of right now, it's completely alien to me. As I said in my original post, I am knowledgeable about single-variable Calculus, multi-variable Calculus, and Differential Equations, though the latter two to a much lesser extent. What should I look at to be able to understand this myself?
$endgroup$
– blue apples
Nov 28 '18 at 14:41
$begingroup$
I was more looking for resources that would help me understand where that expression comes from because as of right now, it's completely alien to me. As I said in my original post, I am knowledgeable about single-variable Calculus, multi-variable Calculus, and Differential Equations, though the latter two to a much lesser extent. What should I look at to be able to understand this myself?
$endgroup$
– blue apples
Nov 28 '18 at 14:41
add a comment |
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