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NPTEL COURSE ON
MATHEMATICS IN INDIA:FROM VEDIC PERIOD TO MODERN TIMES
LECTURE 37
Proofs in Indian Mathematics - Part 2
K. Ramasubramanian
IIT Bombay
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OutlineProofs in Indian Mathematics - Part 2
Volume of a sphere The principle involved Area of a circle Volume of a slice of a given thickness Total volume
A couple of theorems
Jya-sam.varga-nyaya(Theorem on product of chords) Jya-vargantara-nyaya(diff. of the squares of chords) Jya-sam.varga-nyaya Jyotpatti
The cyclic quadrilateral
Expressing diagonals in terms of sides Area in terms of sides Circumradius in terms of sides
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Volume of a sphereThe principle involved
To find the volume of a sphere, it is divided intolarge number(sayn) of
slices ofequal thickness. In the figure the sphereNASBNis divided into slices by planes parallel
to the equatorial planeAGB.
Then the volume of each slice is
obtained. Volume= Area thickess. Area is obtained by finding theaverage
radius of circlesat the top and bottomof the slice.
Ifdis diameter, then the thickess of the
slice= dn
.
Thus, first we need to obtain an
expression for finding the thearea of a
circle.
Sum up the elementary volumesof the
slices, that will add up to the sphere.
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Volume of a sphereObtaining the area of a circular slice
Figure:Circular slice cut into two pieces.
In the figure above we have indicateda circular slice being turned into a
rectangleby appropriately sectioning it, and inserting one half of thecircular slice into the other.
The length of this rectangular strip correspondsto half the
circumferenceC. If the radius is r, then the area of this slice is
Area=
1
2 C
r
. (1)
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Volume of a sphereObtaining the elementary volume of the sphere (= volume of the circular slice
Let r be the radius of the sphere andCthe circumference of a great
circle on this sphere.. The radius of thej-th sliceinto which the sphere has been divided
intocan be conceived of as the half-chordBj (bhuja).
Now, the circumference of this slice is given by
Cjth slice=C
r
Bj
Hence the area of this circular slice is
Ajth slice= 1
2
C
r
Bj Bj
. Therefore, the elementary volume is given by
V = 1
2
C
r
B2j , (2)
where is the thickness of the slices.
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Volume of a sphereSumming up the elementary volumes of the circular slices)
The volume of the sphere is obtained by finding thesum of theelementary volumesconstituted by the slices:
V =
V =n
j=1
1
2
C
r
B2j (3)
In the above expression, the
thickness can be expressed in
terms of the radius as = 2r
n .
IfBjcan also be expressed in termsofr, then we can get V =V(r).
For this we invoke the
Jya-sara-sam.varga-nyayagiven by
Aryabhat.a.
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Jya-sara-sam.varga-nyayaof Aryabhat.aTheorem on the square of chords
In his Aryabhat. ya, Aryabhat.ahas presents the theorem on theproduct of chords as follows (in half arya):
(Aryabhat.ya, Gan. ita17)
The wordsvargaand qsamvargarefer tosquareandproduct
respectively.
Similarly,dhanusand sararefer toarcandarrowrespectively.
Using modern notations the abovenyayamay be expressed as:
product of saras = Rsine2
DE EB = AE2
C
D E B
A
O
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Volume of a sphereSumming up the elementary volumes of the circular slices)
It was shown that the volume of the sphere may be expressed as:
V =
nj=1
12
Cr
B2j
In the figureAP=PB=Bjis the jthhalf-chord, starting fromN.
Applyingjya-sara-sam.varga-nyaya,we have
B2j = AP PB=NP SP
= 1
2
[(NP+ SP)2 (NP2 +SP2)]
= 1
2[(2r)2 (NP2 +SP2)]. (4)
It may be noted in the figure that thej-th RversineNP=jand itscomplementSP= (nj). Hence, while summing the squares of the
RsinesB2j, bothNP2 andSP2 add to the same result.
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Volume of a sphereSumming up the elementary volumes of the circular slices)
Thus the expression for the volume of the sphere given by
V n
j=1
12
Cr
B2j ,
reduces to
V C
2r1
2 [n.(2r)22r
n2
2.[12
+22
+...n2
] 2r
n
.
It was known to Kerala mathematicians, thatfor largen
12 +22 +...n2 = n3
3 .
Using this, the expression for the volume of the sphere becomes
V =
C
2r
4r3
8
3r3
= C6d2. (5)
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Jyasam.varga-nyayaandJya-vargantara-nyayaTheorem on product of chords and the difference of the square of chords
In the figureDMis the perpendicular
from the vertexDonto the diagonal
ACof the cyclic quadrilateral.
Let us consider the two triangles
AMD and CMD, that is formed by DM
in the triangleADC. It can be easily
seen that
AD2 DC2 =AM2 MC2. (6)
In other words, the difference in the
squares of thejyasAD, DCis equal
to the difference in the square of the
base segments (abadhas)AM, MC.
This result may be noted downfor later useas
jyavargantara= abadhavargantara
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Jyasam.varga-nyayaandJya-vargantara-nyayaTheorem on product of chords and the difference of the square of chords
In order to derive a certain property, we
rewrite (6) as
AD2 DC2 = (AM+MC)(AM MC) (7)
It may be noted that
AM+MC jyaof sum of the arcsAM MC jyaof diff. of the arcs
Denoting the chordsADandDCasJ1 andJ2, and the arcs associated
with them asc1 andc2, we may rewrite (7) as
J21 J22 =Jya(c1+ c2) Jya(c1 c2)
In otherwords,
=
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Jyasam.varga-nyayaandJya-vargantara-nyayaTheorem on product of chords and the difference of the square of chords
Recalling the equation,
J21 J22 =Jya(c1+ c2) Jya(c1 c2) (8)
It can also be shown that
J1J2 = Jya
c1+ c2
2 2
Jya
c1 c2
2 2
(9)
In otherwords,
=
-
The two equations (8) and (9) are equivalent to the trigonometric
relations:
sin2(1) sin2(2) = sin(1+ 2) sin(1 2),
sin(1)sin(2) = sin2
(1+ 2)
2
sin2
(1 2)
2
.
with our convention that c1>
c2
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The cyclic quadrilateral & the third diagonal
Consider the equation
sin 1 sin2 =sin2
(1+ 2)
2
sin2
(1 2)
2
.
If we put 1 = (n+1), and 2 = (n 1)in the above equation,then we immediately obtain the following equation
sin(n+1)= sin2 n sin2
sin(n 1)
This is precisely the equation that is presented in the following
verse given by Sankarain hisKriyakramakar:
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The cyclic quadrilateral & the third diagonal
Notation:
the arcs (AB, BC, CD, DA) c1, c2, c3, c4
the chords (AB, BC, CD, DA)
J1,
J2,
J3,
J4 the diagonals (BD,AC,DG) K1, K2, K3
Having drawn a quadrilateral,obviously we can have only two
diagonals. Gis a point chosen such that
BC=AG.
By swapping any two sides could be adjacent, or oppositewe would be affectingonly oneof the two diagonal of thequadrilateral, while the otherdiagonal remains fixed.
Thisnewdiagonal that getsgenerated due to this swappingof sides is referred to as thethird diagonal or bhavikarn. a
A I
B
CD
G
H
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The elegant results we would like to prove
The three diagonals of the cyclic quadrilateral would be referred to as
1. (is. t.akarn. a) chosen/first diagonal
2. (itarakarn. a) other/second diagonal
3.
(bhavikarn. a) future/third diagonal
The results that we would prove:
1. =
2.
=
3. =
The above results essentially express the product of the diagonals in
terms of the sum of the product of the sidesjyas.
Making use of them we express the diagonals in terms of sides.
Then by making use of yet another result
product of two sides of a triangle
circum-diameter =the altitude, (10)
we show that thearea can be expressed in terms of the diagonals and
in turn, in terms of the sides.
Th k !
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Thanks!
THANK YOU
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