NPTEL C OURSE ON
M ATHEMATICS IN I NDIA :
F ROM V EDIC PERIOD TO M ODERN TIMES
Lecture 20
Lı̄lāvatı̄ of Bhāskarācārya I
M. S. Sriram
University of Madras, Chennai.
Outline
◮
Introduction, Importance of Lı̄lāvatı̄
◮
Arithmetical operations: Inversion method, rule of
supposition
◮
Solution of quadratic equations
◮
Mixtures
◮
Combinations, progressions.
◮
Plane figures: Application of Right Triangles.
Introduction
Bhāsakarācārya: Perhaps the most well known name among the
ancient Indian astronomer-mathematicians. Designated as
Bhāskara-II to differentiate him from his earlier namesake, who
lived in the seventh century CE (Bhāskara-I).
According to his own statement in his Golādhyāya, was born in
saka 1036 or 1114 CE. He also adds that he came from
Vijjalavı̄da near the Sahyādri mountain. → Bijapur? S.B. Dixit →
Bhāskara’s original home was Pāt.an in (Khāndeś) → Inscription
lists his grandfather, father, son and grandson: Manoratha,
Maheśvara, Laks.mı̄dhara, Cangadeva respectively.
Bhāskara’s Lı̄lāvatı̄ :A standard work of Indian Mathematics: Patı̄
or Patı̄gan.ita: elementary mathematics covering arithmetic,
algebra, geometry and mensuration. Still used as a textbook in
Sanskrit institutions in India. Composed around 1150 CE.
Commentaries and other works
Many commentaries: Parameśvara (about 1430), Gan.es.a
(Buddhivilāsinı̄) (1545), Sūryadāsa, Munı̄śvara (about 1635), and
Rāmakr.s.n.a (1687). According to R. C. Gupta, a well known
historian of Indian mathematics, the best traditional commentary
is Kriyākramakarı̄ (C. 1534) of Śaṅkara Vāriyar and
Mahis.amaṅgala Nārāyan.a (who compiled this after the demise of
Śaṅkara). Number of commentaries and versions in regional
languages, Kannada, Telugu, etc., Three Persian translations
are known: Abul-Fayd Faydi (1587 CE). My lectures based on
Colebrook’s translation (1817) reprinted along with the original
Sanskrit text and commentaries with notes by H. C. Banerji.
Others works of Bhāskara: Bı̄jagan.ita (Algebra), one of the most
important treatises on Indian algebra. Siddhāntaśiroman. i:
comprising Grahagan. ita and Golādhyāya also very popular.
Other work Karan.akutūhala (epoch 1183 CE).
Operations of arithmatic
Lı̄lāvatı̄ : Rules and Examples in about 270 Verses.
Chapter 1 on weights and measures, money denominations.
Numeration upto 1014 .
Eight operations of arithmetic.
i) Addition and subtraction
ii) Multiplication
a) a1 . . . an × B. Multiply an by A, an−1 by A etc. and add (taking
into account place values).
b) A(b + c) = Ab + Ac.
B
c) A × B = Ax × , where x is a factor of B.
x
d) ab = a(b ± c ∓ c) = a(b ± c) ∓ ac.
iii Division: Remove common factors.
Squaring , Square roots etc.
Verses 18-19. Squaring
i)
(a1 + · · · an )2
=
a2n + 2an (a1 + · · · + an−1 )
+a2n−1 + 2an−1 (a1 + · · · + an−2 )
+···
+a22 + 2a2 a1
+a21 .
ii) (a + b)2 = a2 + 2ab + b 2 .
iii) a2 = (a + b)(a − b) + b 2 .
Verses 21-22: Square root and example.
Verses 23-26: Cube and example.
Verse 27-28: Cube root: Standard Indian method.
Multipulation with fractions
a c
ad cb
, =
,
for addition, subtraction etc.,
b d
bd db
Verse 32-33:
Fractions:
c
a
× × ··· =
b d
a
c
a e a
c
a
+
+ × +
+ ×
b d
b
f b d
b
=
Division of fractions.
a
b
c
d
=
ad
bc
ac...
bd ....
g ha
h b
+
c
a e
× + (
d
b
f
i
) ...
a(d + c)(e + f )(g + h) . . .
bdfh . . .
Operations with zero
Verses 44-45 :
ya;ea;gea KMa [ea;pa;sa;mMa va;ga;Ra;d;Ea KMa Ka;Ba;a;Æa:$a;ta;ea .=+a;a;ZaH Á
Ka;h:=.H .~ya;a;t,a Ka;gua;NaH KMa Ka;gua;Na;a;(
ãÉa Zea;Sa;
a;va;Da;Ea Á Á 44
ãÉa;ntya;(
ÁÁ
ZUa;nyea gua;Na;k
e
.$a;a;tea KMa h;a:=+(
ãÉea;t,a :pua;na;~ta;d;a .=+a;a;ZaH Á
x +:ta O;;va :℄ea;ya;~ta;TEa;va Kea;na;ea;a;na;ta;(
A;
a;va;k
ãÉa yua;taH Á Á 45
ÁÁ
“In addition, zero makes the sum equal to the additive.
In involution and (evolution) the result is zero. A definite
quantity , divided by zero, is the submultiple of nought.
The product of zero is nought: but it must be retained
as a multiple of zero, if any operations impend. Zero
having become a multiplier, should nought afterwards
become a divisor, the definite quantity must be
understood to be unchanged. So likewise any quantity,
to which zero is added, or from which it is subtracted,
(is unaltered).
Operations with zero
a±0 = a,
a
= Kha-hara:
0
a fraction with zero for its denominator.
( In Bı̄jagan.ita. Gan.eśa: Infinity).
a×0=0
But it must retained as a multiple of zero, if further operations,
impend. “Zero having become a multiplier, should nought
afterwards become a divisor, the definite quantity must be
understood to be unchanged”.
a×
.
0
= a?
0
(But
0
is indeterminate )
0
Inversion method
Inversion method to find the solution of a problem in Verse 47-48 :
x +:a;ta;m,a Á
Ce+.dM gua;NMa gua;NMa Ce+.dM va;ga mUa;lM :pa;dM k
+.NMa .~vMa .~va;mxa;NMa k
u +:ya;Ra;d, dx;Zyea .=+a;a;Za;pra;Æa;sa:;d
Äâ ;yea Á Á 47
ÁÁ
A;Ta .~va;Ma;Za;a;a;Da;k+:ea;nea tua l+.va;a;Q.•a;ea;nea h:=+ea h:=.H Á
x +:ta;~ta:ˆa ;
a;va;l+.ea;mea Zea;Sa;mua;€+:va;t,a Á Á 48 Á Á
AM;Za;~tva;
a;va;k
“To investigate a quantity, one being given, make the divisor a
multiplicator; and the multiplier, a divisor; the square, a root; and
the root, a square; turn the negative into positive and the positive
into negative. If a quantity is to be increased or diminished by its
own proportional part, let the (lower) denominator, being the
increased or diminished by its numerator, becomes the (corrected)
denominator, and the numerator remain uncharged*; and then
proceed with the other operations of inversion, as before directed.”
(* If we have 1 +
a
a+b
=
, in the inversion, it will be
b
b
b
.)
a+b
Inversion method: Example
Example in Verse 49 :
é åé
È
/ / a;Ba:=+a;nva;taH
/ /
.~va;.a:=+NEa;BRa;€+:~ta;taH .sa;‹a;Æa;BaH
ya;a;~:a.
/ / îÅé:a;a;~:a;Æ
ú
.~va:ˆyMa;Zea;na ;
a;va;va;Æa:$Ra;taH .~va;gua;a;Na;ta;ea h;a;na;ea ;
a;d
õ ;pa:*.
ãÁa;a;Za;ta;a Á
å
ta;n}.Ua;le;Y;;yua;tea &+.tea . a d;Za;Æa;Ba:$a;Ra;tMa d
õ ;yMa b.rUa;
a;h tMa
ú
.=+a;a;ZMa :vea;a;tsa
/ / / ;
a;h . a:*.
ãÁa;l+.a;Æa;[a ;
a;va;ma;l+.Ma
va;a;mea ;
a;va;l+.ea;ma;
a;k
Ò +:ya;a;m,a Á Á 49
ÁÁ
“Pretty girl, with tremulous eyes, if thou know the
correct method of inversion, tell me the number, which
multiplied by three and added to three quarters of the
product, and divided by seven, and reduced by
subtraction of half a third part of the quotient and then
multiplied into itself, and having fiftytwo subtracted from
the product and the square root of the remainder
extracted, and eight added and the sum divided by ten
yields two (2).”
Inversion method: Example


1 

10 

Answer:
v
2

u
u
u


u
3
1 1


×3
×
x
8+u
)
(1
−
)
(1
−
−52
+


u

| {z 3 } 7 | {z 4 }
t
2
3
x=
q
(2 × 10 − 8)2 + 52 ×
7
4



=2


3 4 1
× × = 28.
2 7 3
Rule of false position
Verse 50:
o+.‰e+Za;k+:a;l+.a;pa;va;
a;d;:=+a;a;ZaH
[ua;NNa;ea &+.ta;eMa.Y;ZEa .=+
a;h;ta;ea yua;ta;ea va;a
I+.;a;h;tMa dx;;ma;nea;na Ba;€M .=+a;a;Za;BRa;vea;t,a :pra;ea;€+:Æa;ma;ta;a;;k+:mRa
Á
Á Á 50 Á Á
“Any number assumed at pleasure is treated as specified in
the particular question, being multiplied and divided, raised
or diminished by fractions; then the given quantity, being
multiplied by the assumed and divided by that (which has
been found) yields th number sought. This is called the
process of supposition.”
Let some operations be done on the number x , represent it as A. Let it
yield A(x ) = y . Take any x ′ and do the same operations. Let it yield y ′ ,
A(x ′ ) = y ′ . Then
x′
x=y× ′
(y )
Rule of false position
Example in Verse 51.
ú È
îå Á
*.a.*+H
.~va;
a:ˆa;Ba;a;ga;ea;na;ea d;Za;Ba;€;H .sa;ma;a;nva;taH
/ /
Á
:pa:ãÁ
.=+a;a;Za:ˆyMa;Za;a;DRa;pa;a;dEH .~ya;a;t,a k+:ea .=+a;a;Za;d
õùÅ;aRU;na;sa;‹a;a;taH Á Á 51
ÁÁ
” What is that number, which multiplied by 5 and having the third
part of the product subtracted and the remainder divided by ten,
and one-third, a half and a quarter of the original quantity added,
gives two less than seventy?”
Solution: Multiplier 5. Subtractive
1
3
of itself. Divisor 10. Additive
the quantity.
1 1 1
, , , of
3 2 4
1
1
)15 = 10;
× 10 = 1;
3
10
1
1
1
17
17
1 + 3( + + ) =
; This is y ′ . So, y ′ =
.
3
2
4
4
4
Given 68 = y. Take x ′ = 3; 5 × 3 = 15, (1 −
∴, x = y ×
x′
3
= 68 × 17 = 48.
y′
4
One gets x = 48, whatever value one chooses for x ′ . y ′ will be
correspondingly different.
Saṅkraman.a
Rule of concurrence or ‘Saṅkraman.a’.
Verse 55:
ya;ea;ga;ea.Y;nta:=e+Na;ea;na;yua;ta;ea.Y;a;DRa;ta;~ta;Ea
ñÍ
.=+a;Za;a .smxa;ta;Ea .sa;*:
öÅ " +.ma;Na;a;K.ya;mea;ta;t,a
Á Á 55 Á Á
“The sum with the difference added or subtracted,
being halved, gives the two quantities. This is termed
Concurrence.”.
If x + y = k and x − y = l where k and l are given quantities.
x=
1
1
(k + l) and y = (k − l)
2
2
Verse 57. If x − y and x 2 − y 2 are given, find x + y =
Then x and y are found by Saṅkraman.a.
x2 − y2
.
x −y
Rational Squares
Verse 59 and 60: Examples of generation of rational squares:
Consider the pairs
a)
x1 =
1
1 1
(8n2 − 1), x2 = { (8n2 − 1)}2 + 1,
2n
2 2n
b)
x1 =
1
+ n, x2 = 1,
2n
c)
x1 = 8n4 + 1, x2 = 8n3 .
For all these pairs, x12 + x22 − 1 are squares.
Solution of Quadratic equations
Next, Bhāskara gives the solution of a quadratic equation .
Verse 62-63: Suppose we have
√
x ±a x =b
q
√
a
Then x = ∓ + b + (a/2)2 and x =
2
a
∓ +
2
2
q
2
b + (a/2)
.
(At least here, only one root is being talked about)
√
c
Similarly, given: x + x ± a x = b, we obtain
d
√
a x
b
x±
c =
c.
1±
1±
d
d
√
Then proceed as before to find x and then x.
Quadratic equations: Example
An example in Verse 67.
:pa;a;TRaH k+:NRa;va;Da;a;ya ma;a;gRa;Na;ga;NMa k
Òu +.;d
Äâ ;ea .=+Nea .sa;nd;Dea
ta;~ya;a;DeRa;na ;a;na;va;a;yRa
ta;.C+=+ga;NMa mUa;lE+.(
ãÉa;tua;Æa;BRa;hR;ya;a;n,a Á Á
ñ
Za;yMa :Sa;
a:Â
*.+.=+Te
a;Sua;Æa;Ba;a;~:a;Æ
/// a;Ba:=+
a;pa
å ÅÅ
C+ˆMa ;Dva:$Ma k+:a;mRua;k
M
;a;.a;.Ce+.d;a;~ya ;a;Za:=.H Za:=e+Na k+:a;ta .tea ya;a;na:$Rua;naH .sa;nd;Dea Á Á 76
ÁÁ
“The Son of Pr.tha, irritated in fight, shot a quiver of
arrows to slay Karn.a. With half his arrows he parried
those of his antagonist; with four times the square root
of the quiver-full, he killed his horses; with six arrows,
he slew Śalya; with three he demolished the umbrella,
standard and bow; and with one, he cut off the head of
the foe. How many were the arrows which Arjuna let
fly?”
Solution of Quadratic equations
Let x denote the number of arrows.
√
1
+4 x +6+3+1=x
2
√
1
x − 4 x − 10 = 0
2
√
∴ x − 8 x − 20 = 0
√
x = 10; x = 100.
√
√
(Here x = −2 is also a solution. But in the problem x has to
be +ve.)
Rule of 3, 5, 7, 9, ...
Investigation of mixtures
Rule in Verse 90.
é åé
ÈÅ
A;Ta :pra;a;NEa;gRua;a;Na;ta;aH .~va;k+:a;l+.aH :pra;ta;a;ta;k+:a;l; îé:a;P+.l+.e
a:;d
Äâx ;ta;a;~tea Á
È
îå Á
.~va;ya;ea;ga;Ba;€+:a;(
ãÉa ;
a;va;Æa;ma;(ra;a;na.*+;aH
:pra;yua;€+:Ka;Nq+.a;a;na :pxa;Ta;g,a Ba;va;a;nta
//// Á Á
“The arguments taken into their respective times one
divided by the fruit taken into the elapsed times; the
several quotients, divided by their sum, and multiplied
by the mixed quantity, and the parts as severally lent.”
Mixtures
Let x, y, z be the portions lent at r1 , r2 , r3 percent per month and
let I = common interest in t1 , t2 , t3 months respectively.
Let x + y + z = a, a given quantity. Then,
y × r2 × t2
z × r3 × t3
x × r1 × t1
=
=
= I.
100
100
100
∴ x : y : z ::
x=
100 × 1 100 × 1 100 × 1
:
:
.
r1 × t1
r2 × t2
r3 × t3
a
100 × 1
×
100
×
1
100
× 1 100 × 1
r1 × t1
+
+
r1 × t1
r2 × t2
r3 × t3
Here: Argument 100, Time: 1, fruit: r, elapsed time : t1 .
Similar expressions for y and z.
Mixtures: Example
Example in Verse 91:
ú
ya;t,a :pa:*.
ãÁa;k+:
a:ˆa;k+:.a;tua;Sk+:P+.le+.na d:†Ma
Ka;NqE+.a;~:a;Æ
/ / a;Ba;gRa;Na;k ;a;na;Sk+:Za;tMa :Sa;qU+.na;m,a Á
ú
ma;a;sea;Sua .sa;‹a;d;Za;pa:*.
ãÁa;sua tua;ya;ma;a;‹Ma
Ka;Nq+ˆa;yea.Y;
a;pa ;
a;h :P+.lM va;d Ka;Nq+.sMa;K.ya;a;m,a Á Á 91
ÁÁ
“The sum of six less than a hundred nis.kas being lent in
three proportion at interest of 5, 3, and 4 percent, an
equal interest was obtained on three portions in 7, 10
and 5 months respectively. Tell mathematician, the
amount of each portion.”
Solutions
Here a = 94, r1 = 5, r2 = 3, r3 = 4; t1 = 7, t2 = 10, t3 = 5..
Now r1 t1 = 35, r2 t2 = 30, r3 t3 = 20..
∴
100 100 100
(6 + 7 + 10.5)
235
+
+
=
× 100 =
.
35
30
20
210
21
∴ x (portion 1) =
y (portion 2) =
z(portion 3) =
100
35
× 94
235
21
100
30
100
21
×
× 94 = 24,
35
235
× 94
=
21
100
×
× 94 = 28,
30
235
× 94
=
100
21
×
× 94 = 42.
20
235
235
21
100
20
=
235
21
24 × 5 × 7
840
2
=
=8 .
100
100
5
He considers the filling of a cistern from n fountains : Same as in
Gan.itasārasaṅgraha
Common interest = I =
Another kind of problem involving mixtures
Let there be a mixture of n items. Let the relative proportion of
type i in the Mixture = βi .
∴ Relative proportion = β1 : β2 : · · · βn .
β P
(Fraction of type i in the mixture = βi′ = P i ; βi′ = 1).
βi
Let the price of type i be xi per measure (volume or weight).
Let the total amount of the mixture = A (volume or weight).
∴ Total price =
X
(Aβi′ ) xi
=
A
P
xβ
P i i = Mixed Sum = X ,
βi
Formal solution
As the amount of item i in the mixture is Aβi′ , and xi is the price per
measure.
P
X
βi
∴ A= P
.
x i βi
Portion × Mixed sum
βi X
=
∴ Amount (Measure) of type i = βi′ A = P
.
x i βi
Sum of quotients
X
(as βi = βi′
βi )
.
(Price of i) × Portion × Mixed Sum
x i βi × X
.
=
Price of item of type i = P
Sum of quotients
x i βi
.
Mixed quantities: Example
Example in Verse 98.
k+:pRUa:=+~ya va:=+~ya ;a;na;Sk+:yua;ga;le+.nEa;k
M :pa;lM :pra;a;pya;tea
;vEa;Zya;a;na;nd;na;.a;nd;na;~ya . a :pa;lM dÒ+mma;a;;Ba;a;gea;na . ea;t,a Á
A;;Ma;Zea;na ta;Ta;a.Y;ga:=+eaH :pa;l+.d;lM ;a;na;Sk
e +:Na mea :de;
a;h ta;a;n,a
Ba;a;gEa;=e+k+:k+:Sa;ea;q+.Za;a;;k+:Æa;ma;tEa;DRUa;pMa ;a;.a;k
+:a;Sa;Ra;}ya;h;m,a Á Á 98
ÁÁ
“If a pala of best camphor may be had for two nisk ′s
(=32 drammas) and a pala of sandalwood for the eighth
part of a dramma and half a pala of alae wood also for
eighth of a dramma, good merchant, give me the value
for one nis.ka (=16 drammas) in the proportions of 1, 16,
and 8, for I wish to prepare a perfume.”
Solution
Here proportion:
Prices:
X
(1)
Camphor:
32
(x1 )
(2)
Sandalwood:
1 81
(x2 )
(3)
Alaewood
= 1 : 16 : 8
1
4
(x3 )
1
1
× 16 + × 8 = 36.
8
4
Total price = Mixed amount = 16 drammas = X =16.
xi βi = 32 × 1 +
Quotient, q =
16
Mixed sum
X
4
=
= P
= .
xi βi
Sum of quotients
36
9
βi X
βi X
4
= βi , Price = xi P
= xi × Amount.
Amounts of item i = P
9
xi βi
xi βi
4
4
64
4
32
∴ Amount of Camphor: 1× , Sandalwood = 16× =
, Alaewood = 8× =
9
9
9
9
9
128
1 64
8
1 32
8
4
, Sandal: ×
= , Alaewood: = ×
= .
Prices : Camphor: 32× =
9
9
8 9
9
4 9
9
Merchants, trading. Alligation: Mixture of gold of different quantities.
Section six on combinations
Verse 130-132.
ñÍ
v.ya;~ta;a Ba;a:$ya;aH :pxa;Ta;a;#~Ta;tE
////////////////// aH Á
O;;k+:a;dù;aÅe ;k+:ea:†a:=+a;*:+.a
öÅ
:pa:=.H :pUa;veRa;Na .sMa;gua;Nya ta;tpa:=+~tea;na .tea;na ..a Á Á 130 Á Á
O;;k+:
a;d
õ :ˆya;a;
a;d;Bea;d;aH
.~yuaH I+.dM .sa;a;Da;a:=+NMa .smxa;ta;m,a Á
C+.nd;a;(
ãÉa:†yua:†a:=e C+.nd;~yua;pa;ya;ea;ga;ea.Y;~ya
mxa;Sa;a;va;h;na;Bea;d;a;d;Ea
ta;
a;d
õ ;d;a;m,a Á Á 131 Á Á
Ka;Nq+.mea:=+Ea ..a ;a;Za;pa;k
e Á
;vEa;dù;a;k
Å e
.=+sa;Bea;d;a;yea ta;Ša;ea;€M ;
a;va;~txa;tea;BRa;ya;a;t,a Á Á 132
ÁÁ
“Let the figures of one upwards, differing by one, put in the
inverse order, be divided by the same (arithmetical
sequence) in direct order; and let the subsequent be
multiplied by the preceding, and the next following by the
foregoing (result). The several results are the changes, ones,
twos, threes, etc. This is termed a general rule.”
Combinations
It serves in prosody, for those versed therein, to find the
variations of metre; in the arts (as in architecture) to
compute the changes upon apertures (of a building);
and (in music) the scheme of musical permutations; in
medicine, the combinations of different savours. For
fear of prolixity, this is not (fully) set forth.”
n(n − 1)(n − 2)(n − r ) . . . (n − r + 1)
n!
n n · (n − 1)
,
,...,
=
1
1·2
1 · 2 · 3···r
r !(n − r )!
Out of n: 1 at a time: 2 at a time: 3 at a time, ... , r at a time.
Combinations: Example
Example in Verse 134.
O;;k+:
a;d
õ :ˆya;a;
a;d;mUa;Sa;a;va;h;na;Æa;ma;a;ta;ma;h;ea
b.rUa;
a;h mea BUa;Æa;ma;Ba;tRuaH h;}yeRa
Å
É
ì;+[Na;Za;a;l+.a;
.=+}yea.Y;;mUa;Sea ..a;tua:=+
a;va:=+a;.a;tea (
a;va;Za;a;le Á
O;;k+:
a;d
õ :ˆya;a;
a;d;yua;€”+.a;a
å ÅÉ
ma;Dua:=+k+:f
u ;k+:Sa;a;ya;a;}ì++.k+:[a;a:=+
a;ta;€
E ;H O;;k+:a;sma;n,
///////// a
ú
:Sa;q"+.sEaH .~yua;gRa;Na;k k+:a;ta va;d v.ya:ê
*+;ne
ÁÁ a v.ya;a;€+:Bea;d;aH Á Á
“In a pleasant, spacious and elegant edifice with eight doors,
constructed by a skillful architect, as a palace for the lord of
the land, tell me combinations of doors taken one, two, three,
etc. Say, mathematician, how many are the combinations in
one composition, with ingredients of six different tastes
sweet, pungent, astringent, sour, salt and bitter, taking them
by ones, twos, threes etc.”
87654321
12345678
Example contd.
8×7
= 28, 3 at a time
No. of ways: 1 at a time = 81 = 8, 2 at a time = 1×2
8×7×6×5
8×7×6
= 1×2×3 = 56, 4 at a time = 1×2×3×4 = 70. 5 at a time
8×7×6×5×4
8×7×6×5×4×3
= 1×2×3×4×5
= 56, 6 at a time = 1×2×3×4×5××6
= 28, 7 at a time
8×7×6×5×4×3×2×1
8×7×6×5×4×3×2
= 8.
= 1×2×3×4×5×6×7 = 8, 8 at a time = 1×2×3×4×5×6×7×8
Total no. of variations = 28 − 1 = 255.
Combinations of 6 different tastes:
654321
123456
No. of different types of compositions: 6, 15, 20, 15, 6, 1. Total
= 26 − 1 = 63.
Chapter 5. Progressions.
P P 2 P r (r +1) P 3
Standard results on
r, r ,
r , etc., Arithmetic and
2 ,
Geometric Progressions.
Number of verses with a fixed number of syllables
Verse 130-131:
:pa;a;d;a;[a:=+Æa;ma;ta;ga;.Ce gua;Na;va;gRa;P+.lM ..a;yea ;
a;d
õ ;gua;Nea Á
ñÍ
É +;a;a
ta;d
.sa;ma;vxa:†a;a;na;Ma .sa;*
õ ;ga;eRa va;gRa;va;gRa;(
ãÉa Á
Ë
ù
Á
,
.~va;~va;pa;d;ea;na;Ea .~ya;a;ta;Ma A;DRa;sa;ma;a;na;Ma ..a ;
a;va;Sa;ma;a;Na;a;m,a Á Á 130 131
ÁÁ
“The number of syllables in a verse being taken for the
period, and the increase twofold, the produce of
multiplication and squaring (as above directed) will be
the number of (variations) of like verses. Its square and
square’s square, less their respective roots, will be (the
variations of) alternately similar and of dissimilar
verses.”
This rule refers specifically to the example in Verse 132.
Example
Example 132.
.sa;ma;a;na;a;ma;DRa;tua;ya;a;na;Ma ;
a;va;Sa;ma;a;Na;Ma :pxa;Ta;k
,
:pxa;Ta;k
, Á
ñÍ
É +;a;M
vxa:†a;a;na;Ma va;d mea .sa;*
u ;pC+.nd;Æa;sa dÒu+ta;m,a Á Á 140
Ë
ù a A;nua;
Á
ÁÁ
“Tell me directly the number (of varieties) of like,
alternating-like, and dissimilar Verses respectively in
the metre named anus.tup.”

x
x
4 caran.as 
.x
x
◮
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
8 Syllables
8
8
8
Total 32 Syllables.
Variations of like Verses. (All the caran.as are alike.)
= 28 = 256.
Example
◮
◮
Alternatively, alike caran.as. To find the number of these, we
find the number of varieties of the syllables in two caran.as
= 216 = 65536. If we place each one of these under itself,
we get all the cases included in “alternatively like”. But of
these, the number of cases of “all like” are also included.
Therefore number of cases of “only alternatively like”
= 216 − 28 = 65536 − 256 = 65280.
Total no. of variations = 232 . Subtracting the case of “all
like” and “alternatively like”, Number of dissimilar Verses
= 232 − 216 . Note that the “dissimilar” only means it
excludes “all like” and “alternatively like” and does not mean
that all the caran.as are dissimilar. For instance, it includes
cases in which “first two caran.as are alike, as also last two,
etc.”
Plane Figures
Chapter 6 is on this topic.
As expected, begins by considering Right Triangles.
Upright
Diagonal
side
A right triangle
Concentrates on Rational triangles. One type is described in
Verse 139.
È
îå Á
x +:tyEa;k+:
a;va;yua;€+:ya;a;‹a;m,a Á
I+.e;na Bua:$a;ea.Y;sma;a;d, ;
a;d
a I+.;~ya k
õ ;gua;Nea;na;a;na.*+;a;t,
k+:ea;a;fH :pxa;Tua;k
, .sea;;gua;Na;a;Bua:$a;ea;na;a k+:Na;eRa Ba;vea;t,a ˆya;(ra;Æa;ma;dM tua .$a;a;tya;m,a
ÁÁ
A Rational right triangle
“A side is put. From this multiplied by twice some assumed
number, and divided by one less than the square of the assumed
number, the upright is obtained. This, being set apart, is
multiplied by the arbitrary number, and the side as put is
subtracted; the remainder will be the hypotenuse. Such a
triangle is termed right-angled.”
2an
n 2 −1
a(n 2 +1)
(n 2 −1)
a
2an
yields a right triangle
n2 − 1
a(n2 + 1)
2an
×n−a=
.
with hypotenuse = 2
n −1
n2 − 1
It is stated that side = a, upright =
Explanation by Sūryadāsa in his commentary
Consider two triangles. The first triangle has side = n2 − 1, upright = 2n and
diagonal n2 + 1. Correct as (n2 + 1)2 = (n2 − 1)2 + (2n)2 . Let the second
triangle be similar to the above, with side = a.
2an
n 2 −1
?
n2 + 1
2n
n2 − 1
a
(a)
(b)
Similar right triangles
The upright of the second triangle =
a
× 2n.
n2 − 1
Now, hypotenuse of the first triangle
= upright × n − side = 2n × n − (n2 − 1) = n2 + 1.
Then, hypotenuse of the second triangle
2an
a(n2 + 1)
= Upright × n − side = 2
×n−a=
.
n −1
n2 − 1
References
1.H.T.Colebrooke, Algebra with Arithmetic and Mensuration from
the Sanskrit of Brahmagupta and Bhaskara, London 1817; Rep.
Sharada Publishing House, New Delhi, 2006.
2. Bhāskarācārya’s Lı̄lāvatı̄ with Colebrooke’s translation and
notes by H.C.Banerji, Second Edition,The Book Company,
Calcutta, 1927; Rep. Asian Educational Series, New Delhi, 1993.
3.Lı̄lāvatı̄ of Bhāsakarācārya II: Ed. with Bhāskara’s Vāsanā and
Buddhivilāsanı̄ of Gan.eśa Daivajña by V.G.Apte, 2 Vols., Pune,
1937.
4.Lı̄lāvatı̄ of Bhāsakarācārya , A Treatise of Mathematics of Vedic
Tradition, Tr. by K.S.Patwardhan, S.M.Naimpally and S.L.Singh,
Motilal Banarsidass, Delhi, 2001.
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