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History of hypergeometric equation
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It is known that Gauss studied hypergeometric equation
$$x(1-x) dfrac d^2ydx^2+(c-(a+b+1)x)dfrac dydx-aby=0$$
I would like to know something about history of this equation:
1) If $a=b=c=0$ then what are the solutions of corresponding equation and did they been studied before Gauss?
2) Why it is called "hypergeometric"?
3) What are contributions of Gauss regarding this equation?
mathematics differential-equations
asked Sep 1 at 10:28
Right
1233
add a comment |Â
up vote
4
down vote
favorite
It is known that Gauss studied hypergeometric equation
$$x(1-x) dfrac d^2ydx^2+(c-(a+b+1)x)dfrac dydx-aby=0$$
I would like to know something about history of this equation:
1) If $a=b=c=0$ then what are the solutions of corresponding equation and did they been studied before Gauss?
2) Why it is called "hypergeometric"?
3) What are contributions of Gauss regarding this equation?
mathematics differential-equations
asked Sep 1 at 10:28
Right
1233
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
It is known that Gauss studied hypergeometric equation
$$x(1-x) dfrac d^2ydx^2+(c-(a+b+1)x)dfrac dydx-aby=0$$
I would like to know something about history of this equation:
1) If $a=b=c=0$ then what are the solutions of corresponding equation and did they been studied before Gauss?
2) Why it is called "hypergeometric"?
3) What are contributions of Gauss regarding this equation?
mathematics differential-equations
asked Sep 1 at 10:28
Right
1233
It is known that Gauss studied hypergeometric equation
$$x(1-x) dfrac d^2ydx^2+(c-(a+b+1)x)dfrac dydx-aby=0$$
I would like to know something about history of this equation:
1) If $a=b=c=0$ then what are the solutions of corresponding equation and did they been studied before Gauss?
2) Why it is called "hypergeometric"?
3) What are contributions of Gauss regarding this equation?
mathematics differential-equations
asked Sep 1 at 10:28
Right
1233
asked Sep 1 at 10:28
Right
1233
asked Sep 1 at 10:28
Right
1233
asked Sep 1 at 10:28
Right
1233
1233
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
8
down vote
accepted
If $a=b=c=0$ we get a first order separable equation for $y'$, viz. $ x[(1-x)y''-y']=0$ whose solution $y'= c(1-x)^-1$ would have been obvious since at least Bernoulli.
Dutka (1984, p. 16):
In contrast to the geometric progression $a, ar, ar^2,dots$ in which each term,
after the first, is obtained by multiplying the preceding term by a constant ratio, Wallis (Arithmetica infinitorum, 1656, Scholium to Prop. 190; A Treatise of Algebra, 1685, pp. 315–316) introduced another type of progression, to which he later gave the name “hypergeometric,†in which the successive multipliers are unequal. (E.g., the factorial sequence, 1, 2, 6, 24,... is a hypergeometrical progression.)Andrews, Askey and Roy (1999, p. 61):
A hypergeometric series is a series $sum c_n$ such that $c_n+1/c_n$ is a rational function of $n$. On factorizing the polynomials in $n$, we obtain
$$
fracc_n+1c_n=frac(n+a_1)cdots(n+a_p)x(n+b_1)cdots(n+b_q)(n+1).
tag2.1.1
$$
The $x$ occurs because the polynomial may not be monic. The factor $(n+1)$ may result from the factorization, or it may not. If not, add it along with the compensating factor $(n+1)$ in the numerator (...) From (2.1.1) we have
$$
sum_n=0^infty c_n
=c_0sum_n=0^inftyfrac(a_1)_ncdots(a_p)_n(b_1)_ncdots(b_q)_nfracx^nn!
=:c_0,_pF_qleft(beginmatrixa_1,dots,a_p\b_1,dots,b_qendmatrix;xright).
tag2.1.2
$$Birkhoff (1973, p. 61):
Kummer (1836) gave the name “hypergeometric function†to $F$.
See Dutka (ibid., pp. 29–33). Necessary context is the prior work of Wallis (pp. 15–17), Newton (17), Stirling (17–20), Euler (20–26), Legendre (26), Pfaff (26–29).
answered Sep 1 at 13:25
Francois Ziegler
5,1121431
add a comment |Â
up vote
6
down vote
1) For $a=b=c=0$ we have equation
$$
x(1-x) dfrac d^2ydx^2-xdfrac dydx=0
$$
with solution $C_1 + C_2log(1-x)$. This would probably not be considered hypergeometric, except in a degenerate sense.
2) Answer (needs reference): In a geometric series $sum a_n$, the ratio $fraca_n+1a_n$ is constant, does not depend on $n$. We generailze that to hypegeomeric series where $fraca_n+1a_n$ is a rational function of $n$. These series are the $_p F_q$ functions, not only the $_2 F_1$ of Gauss.
3) Gauss did a lot on this. But it is beyond the scope of this answer... Perhaps read the bit in Wikipedia for some idea
answered Sep 1 at 12:42
Gerald Edgar
3,103515
Oops sorry... I only saw your answer after hitting “postâ€Â.
– Francois Ziegler
Sep 1 at 13:26
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
If $a=b=c=0$ we get a first order separable equation for $y'$, viz. $ x[(1-x)y''-y']=0$ whose solution $y'= c(1-x)^-1$ would have been obvious since at least Bernoulli.
Dutka (1984, p. 16):
In contrast to the geometric progression $a, ar, ar^2,dots$ in which each term,
after the first, is obtained by multiplying the preceding term by a constant ratio, Wallis (Arithmetica infinitorum, 1656, Scholium to Prop. 190; A Treatise of Algebra, 1685, pp. 315–316) introduced another type of progression, to which he later gave the name “hypergeometric,†in which the successive multipliers are unequal. (E.g., the factorial sequence, 1, 2, 6, 24,... is a hypergeometrical progression.)Andrews, Askey and Roy (1999, p. 61):
A hypergeometric series is a series $sum c_n$ such that $c_n+1/c_n$ is a rational function of $n$. On factorizing the polynomials in $n$, we obtain
$$
fracc_n+1c_n=frac(n+a_1)cdots(n+a_p)x(n+b_1)cdots(n+b_q)(n+1).
tag2.1.1
$$
The $x$ occurs because the polynomial may not be monic. The factor $(n+1)$ may result from the factorization, or it may not. If not, add it along with the compensating factor $(n+1)$ in the numerator (...) From (2.1.1) we have
$$
sum_n=0^infty c_n
=c_0sum_n=0^inftyfrac(a_1)_ncdots(a_p)_n(b_1)_ncdots(b_q)_nfracx^nn!
=:c_0,_pF_qleft(beginmatrixa_1,dots,a_p\b_1,dots,b_qendmatrix;xright).
tag2.1.2
$$Birkhoff (1973, p. 61):
Kummer (1836) gave the name “hypergeometric function†to $F$.
See Dutka (ibid., pp. 29–33). Necessary context is the prior work of Wallis (pp. 15–17), Newton (17), Stirling (17–20), Euler (20–26), Legendre (26), Pfaff (26–29).
answered Sep 1 at 13:25
Francois Ziegler
5,1121431
add a comment |Â
up vote
8
down vote
accepted
If $a=b=c=0$ we get a first order separable equation for $y'$, viz. $ x[(1-x)y''-y']=0$ whose solution $y'= c(1-x)^-1$ would have been obvious since at least Bernoulli.
Dutka (1984, p. 16):
In contrast to the geometric progression $a, ar, ar^2,dots$ in which each term,
after the first, is obtained by multiplying the preceding term by a constant ratio, Wallis (Arithmetica infinitorum, 1656, Scholium to Prop. 190; A Treatise of Algebra, 1685, pp. 315–316) introduced another type of progression, to which he later gave the name “hypergeometric,†in which the successive multipliers are unequal. (E.g., the factorial sequence, 1, 2, 6, 24,... is a hypergeometrical progression.)Andrews, Askey and Roy (1999, p. 61):
A hypergeometric series is a series $sum c_n$ such that $c_n+1/c_n$ is a rational function of $n$. On factorizing the polynomials in $n$, we obtain
$$
fracc_n+1c_n=frac(n+a_1)cdots(n+a_p)x(n+b_1)cdots(n+b_q)(n+1).
tag2.1.1
$$
The $x$ occurs because the polynomial may not be monic. The factor $(n+1)$ may result from the factorization, or it may not. If not, add it along with the compensating factor $(n+1)$ in the numerator (...) From (2.1.1) we have
$$
sum_n=0^infty c_n
=c_0sum_n=0^inftyfrac(a_1)_ncdots(a_p)_n(b_1)_ncdots(b_q)_nfracx^nn!
=:c_0,_pF_qleft(beginmatrixa_1,dots,a_p\b_1,dots,b_qendmatrix;xright).
tag2.1.2
$$Birkhoff (1973, p. 61):
Kummer (1836) gave the name “hypergeometric function†to $F$.
See Dutka (ibid., pp. 29–33). Necessary context is the prior work of Wallis (pp. 15–17), Newton (17), Stirling (17–20), Euler (20–26), Legendre (26), Pfaff (26–29).
answered Sep 1 at 13:25
Francois Ziegler
5,1121431
add a comment |Â
up vote
8
down vote
accepted
up vote
8
down vote
accepted
If $a=b=c=0$ we get a first order separable equation for $y'$, viz. $ x[(1-x)y''-y']=0$ whose solution $y'= c(1-x)^-1$ would have been obvious since at least Bernoulli.
Dutka (1984, p. 16):
In contrast to the geometric progression $a, ar, ar^2,dots$ in which each term,
after the first, is obtained by multiplying the preceding term by a constant ratio, Wallis (Arithmetica infinitorum, 1656, Scholium to Prop. 190; A Treatise of Algebra, 1685, pp. 315–316) introduced another type of progression, to which he later gave the name “hypergeometric,†in which the successive multipliers are unequal. (E.g., the factorial sequence, 1, 2, 6, 24,... is a hypergeometrical progression.)Andrews, Askey and Roy (1999, p. 61):
A hypergeometric series is a series $sum c_n$ such that $c_n+1/c_n$ is a rational function of $n$. On factorizing the polynomials in $n$, we obtain
$$
fracc_n+1c_n=frac(n+a_1)cdots(n+a_p)x(n+b_1)cdots(n+b_q)(n+1).
tag2.1.1
$$
The $x$ occurs because the polynomial may not be monic. The factor $(n+1)$ may result from the factorization, or it may not. If not, add it along with the compensating factor $(n+1)$ in the numerator (...) From (2.1.1) we have
$$
sum_n=0^infty c_n
=c_0sum_n=0^inftyfrac(a_1)_ncdots(a_p)_n(b_1)_ncdots(b_q)_nfracx^nn!
=:c_0,_pF_qleft(beginmatrixa_1,dots,a_p\b_1,dots,b_qendmatrix;xright).
tag2.1.2
$$Birkhoff (1973, p. 61):
Kummer (1836) gave the name “hypergeometric function†to $F$.
See Dutka (ibid., pp. 29–33). Necessary context is the prior work of Wallis (pp. 15–17), Newton (17), Stirling (17–20), Euler (20–26), Legendre (26), Pfaff (26–29).
answered Sep 1 at 13:25
Francois Ziegler
5,1121431
If $a=b=c=0$ we get a first order separable equation for $y'$, viz. $ x[(1-x)y''-y']=0$ whose solution $y'= c(1-x)^-1$ would have been obvious since at least Bernoulli.
Dutka (1984, p. 16):
In contrast to the geometric progression $a, ar, ar^2,dots$ in which each term,
after the first, is obtained by multiplying the preceding term by a constant ratio, Wallis (Arithmetica infinitorum, 1656, Scholium to Prop. 190; A Treatise of Algebra, 1685, pp. 315–316) introduced another type of progression, to which he later gave the name “hypergeometric,†in which the successive multipliers are unequal. (E.g., the factorial sequence, 1, 2, 6, 24,... is a hypergeometrical progression.)Andrews, Askey and Roy (1999, p. 61):
A hypergeometric series is a series $sum c_n$ such that $c_n+1/c_n$ is a rational function of $n$. On factorizing the polynomials in $n$, we obtain
$$
fracc_n+1c_n=frac(n+a_1)cdots(n+a_p)x(n+b_1)cdots(n+b_q)(n+1).
tag2.1.1
$$
The $x$ occurs because the polynomial may not be monic. The factor $(n+1)$ may result from the factorization, or it may not. If not, add it along with the compensating factor $(n+1)$ in the numerator (...) From (2.1.1) we have
$$
sum_n=0^infty c_n
=c_0sum_n=0^inftyfrac(a_1)_ncdots(a_p)_n(b_1)_ncdots(b_q)_nfracx^nn!
=:c_0,_pF_qleft(beginmatrixa_1,dots,a_p\b_1,dots,b_qendmatrix;xright).
tag2.1.2
$$Birkhoff (1973, p. 61):
Kummer (1836) gave the name “hypergeometric function†to $F$.
See Dutka (ibid., pp. 29–33). Necessary context is the prior work of Wallis (pp. 15–17), Newton (17), Stirling (17–20), Euler (20–26), Legendre (26), Pfaff (26–29).
answered Sep 1 at 13:25
Francois Ziegler
5,1121431
edited Sep 1 at 13:45
answered Sep 1 at 13:25
Francois Ziegler
5,1121431
answered Sep 1 at 13:25
Francois Ziegler
5,1121431
answered Sep 1 at 13:25
Francois Ziegler
5,1121431
5,1121431
add a comment |Â
add a comment |Â
up vote
6
down vote
1) For $a=b=c=0$ we have equation
$$
x(1-x) dfrac d^2ydx^2-xdfrac dydx=0
$$
with solution $C_1 + C_2log(1-x)$. This would probably not be considered hypergeometric, except in a degenerate sense.
2) Answer (needs reference): In a geometric series $sum a_n$, the ratio $fraca_n+1a_n$ is constant, does not depend on $n$. We generailze that to hypegeomeric series where $fraca_n+1a_n$ is a rational function of $n$. These series are the $_p F_q$ functions, not only the $_2 F_1$ of Gauss.
3) Gauss did a lot on this. But it is beyond the scope of this answer... Perhaps read the bit in Wikipedia for some idea
answered Sep 1 at 12:42
Gerald Edgar
3,103515
Oops sorry... I only saw your answer after hitting “postâ€Â.
– Francois Ziegler
Sep 1 at 13:26
add a comment |Â
up vote
6
down vote
1) For $a=b=c=0$ we have equation
$$
x(1-x) dfrac d^2ydx^2-xdfrac dydx=0
$$
with solution $C_1 + C_2log(1-x)$. This would probably not be considered hypergeometric, except in a degenerate sense.
2) Answer (needs reference): In a geometric series $sum a_n$, the ratio $fraca_n+1a_n$ is constant, does not depend on $n$. We generailze that to hypegeomeric series where $fraca_n+1a_n$ is a rational function of $n$. These series are the $_p F_q$ functions, not only the $_2 F_1$ of Gauss.
3) Gauss did a lot on this. But it is beyond the scope of this answer... Perhaps read the bit in Wikipedia for some idea
answered Sep 1 at 12:42
Gerald Edgar
3,103515
Oops sorry... I only saw your answer after hitting “postâ€Â.
– Francois Ziegler
Sep 1 at 13:26
add a comment |Â
up vote
6
down vote
up vote
6
down vote
1) For $a=b=c=0$ we have equation
$$
x(1-x) dfrac d^2ydx^2-xdfrac dydx=0
$$
with solution $C_1 + C_2log(1-x)$. This would probably not be considered hypergeometric, except in a degenerate sense.
2) Answer (needs reference): In a geometric series $sum a_n$, the ratio $fraca_n+1a_n$ is constant, does not depend on $n$. We generailze that to hypegeomeric series where $fraca_n+1a_n$ is a rational function of $n$. These series are the $_p F_q$ functions, not only the $_2 F_1$ of Gauss.
3) Gauss did a lot on this. But it is beyond the scope of this answer... Perhaps read the bit in Wikipedia for some idea
answered Sep 1 at 12:42
Gerald Edgar
3,103515
1) For $a=b=c=0$ we have equation
$$
x(1-x) dfrac d^2ydx^2-xdfrac dydx=0
$$
with solution $C_1 + C_2log(1-x)$. This would probably not be considered hypergeometric, except in a degenerate sense.
2) Answer (needs reference): In a geometric series $sum a_n$, the ratio $fraca_n+1a_n$ is constant, does not depend on $n$. We generailze that to hypegeomeric series where $fraca_n+1a_n$ is a rational function of $n$. These series are the $_p F_q$ functions, not only the $_2 F_1$ of Gauss.
3) Gauss did a lot on this. But it is beyond the scope of this answer... Perhaps read the bit in Wikipedia for some idea
answered Sep 1 at 12:42
Gerald Edgar
3,103515
answered Sep 1 at 12:42
Gerald Edgar
3,103515
answered Sep 1 at 12:42
Gerald Edgar
3,103515
answered Sep 1 at 12:42
Gerald Edgar
3,103515
3,103515
Oops sorry... I only saw your answer after hitting “postâ€Â.
– Francois Ziegler
Sep 1 at 13:26
add a comment |Â
Oops sorry... I only saw your answer after hitting “postâ€Â.
– Francois Ziegler
Sep 1 at 13:26
Oops sorry... I only saw your answer after hitting “postâ€Â.
– Francois Ziegler
Sep 1 at 13:26
Oops sorry... I only saw your answer after hitting “postâ€Â.
– Francois Ziegler
Sep 1 at 13:26
add a comment |Â
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