present asg 2(2)- math phylosphy

8/7/2019 present asg 2(2)- math phylosphy

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INDIAINDIAPresented by: GROUP 5Presented by: GROUP 5


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CONTENTSCONTENTS

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PELL EQUATION inPELL EQUATION in BhaskaraBhaskara IIII

Pells equation is a diophantine equation of the

form:

x,y Z, where n is a given natural number

which is not a square. This equation is solve inpositive integer for x and y where always

assume for given n is positive. Which we can

also rewrite as:

or


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Solving Pell's EquationSolving Pell's EquationIdentity 1

To aid in solving x-ny=1, the following identity

is helpful:

(b - na)(d - nc) = (bd s nac) - n(bc s ad)

From this we see that ifb-na andd-nc are

both.Then,

(bd s nac) - n(bc s ad) = 1

In other words, if(a,b) and(c,d) are solutions to

Pell's equation then so are (bc s ad, bd s nac)


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Brahmagupta'sBrahmagupta's LemmaLemma((The Method of CompositionThe Method of Composition))

If(a, b) and(c, d) are integer solutions of "Pell

type equations" of the form

na + k = b and nc + k' = d

then (bc ad, bd nac) are both integersolutions of the "Pell type equation"

nx + kk' = y

Using the Methodof Composition, if(a,b)

satisfies Pell's equation, then sodoes

(2ab, b+na)Which is obtainedby composing

(a,b) with itself.


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Solving Pell Equations usingSolving Pell Equations usingBrahmagupta'sBrahmagupta's MethodMethod

From Brahmagupta's Lemma, if(a,b) is a

solution to nx + k = y, then composing (a,b)

with itself gives you(2ab, b+na) as a solution to

nx + k = y,Anddividing through by k, so

Then become,

is a solution to the Pell Equation nx + 1 = y


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For most values of k this isn't helpful, becausex and y aren't integers, but when k is 1, 2,

or, with a little more work,

4, this idea helpsa whole lot.

When k=2,

Since (a,b) is a solution to nx + k = y, wehave


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So,

(ab, b-1) is a solution to the related PellEquation nx + 1 = y.


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EXAMPLEEXAMPLE

Solve 23x+1=y.Start by observing that a solution to the relatedPell type equation

23x+2=y is (1,5), so(5,24) is a solution to

23x+1=y.

When k=-2,

Since (a,b) is a solution to nx + k = y, we knowthat,

=


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EXAMPLEEXAMPLE

So ,

(-ab, -b-1) is a solution to the related PellEquation nx + 1 = y, and so(ab, b+1) is also asolution.


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EXAMPLEEXAMPLE

Solve 83x+1=y.

Start by observing that a solution

to the related Pell type equation

83x-2=y is (1,9), so(9,82) is asolution to 83x+1=y.

**DIY


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CHAKRAVALACHAKRAVALA

METHODMETHOD((cyliccylic method)method)


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The chakravala method is a cyclicalgorithm to solve inderteminate quadraticequations,including Pells equation forminimum integers xand y.

This methodwas developed in India and itsroots can be tracedback to the 5th century,summarizedby Aryabhata, which was laterdeveloped further by Brahmagupta,Jayadeva and Bhaskara.


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Jayadeva (9th century) and Bhaskara (12th

century) offered the first complete solutionto this equation using the chakravalamethod

This was not solved in Europe until the timeoflagrange in 1767. Lagrange's methodhowever, requires the calculation of 21successive convergents of the continuedfraction for the square root of 61, whilethe chakravala method is much simpler.


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This method is also contains traces of mathematicalinduction. Bhaskara II has discovered this cyclic

method algorithm to produce a solution of Pells

equation starting off from any close pair (a,b)

For example:

Then,

Assume a, b and k are coprime.Then for any m, (1,m)

satisfied the Pell equation


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Observe that solution to nx2 +(m2-n) = y2

is (1,m)

e.g:

7 x2+2= y2 is (1,3) because 2 = 32- 7

Now suppose we are looking for a solution to

nx2 + 1= y2 and we can find a close pair(a,b)

that solves nx2 + k = y2.In other words, ax2 + k= b2


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Now we compose (1,m) and (a,b) to see that

(am+b, bm+na) is a solution to

nx2 +(m2-n) = y2,

In other words n(am+b)2+(m2-n)k=(bm+na)2.

Dividing by k we see that

((am+b)/k ,(bm+na)/k) solves nx2+(m2-n)/k = y2


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nn = 61= 61

TheThe nn == 6161 case (determining an integer solution satisfyingcase (determining an integer solution satisfyingaa22 61 61bb22 = 1,= 1,We start with a solutionWe start with a solution aa22 61 61bb22 == kk ,,

In this case we can letIn this case we can let bb =1, thus, since=1, thus, since , we, wehave the triplehave the triple

((aa,,bb,,kk) = (8,1,3)) = (8,1,3)Composing it withComposing it with((mm,1,,1,mm22 61) 61) gives the triple,gives the triple, [8[8mm + 61,8 ++ 61,8 + mm,3(,3(mm22 61)] 61)]

which is scaleddown (orwhich is scaleddown (or BhaskaraBhaskara Lemmas is directly used) toLemmas is directly used) toget:get:


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nn = 67= 67Suppose we are to solveSuppose we are to solve xx22 67 67yy22 = 1= 1 forfor xxandand yy..

We start with a solutionWe start with a solution

aa22

67 67bb22

== kkfor anyfor any kk foundby any means;foundby any means;

In this case we can letIn this case we can let bb =1, thus producing=1, thus producing . At. At

each step, we find

aneach step, we find

an mm >> 0 such0 suchthatthat kk dividesdivides aa ++ bmbm, and, and is minimal. We thenis minimal. We then

updateupdate aa,, bb, and, and kk toto respectively.respectively.


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First iteration:First iteration:

We haveWe have aa = 8,= 8, bb = 1,= 1, kk == 3. We want a positive3. We want a positiveintegerinteger mm such thatsuch that kk dividesdivides a+mba+mb, i.e. 3divides 8+m,, i.e. 3divides 8+m,andand is minimal. The first condition impliesis minimal. The first condition impliesthatthat mm is of the form3is of the form3tt+1 (i.e. 1, 4, 7, 11, etc.), and+1 (i.e. 1, 4, 7, 11, etc.), and

among suchamong such mm, the minimalvalue is attained for, the minimalvalue is attained for mm=7.=7.Replacing (Replacing (aa,, bb,, kk) with) with ((am+Nbam+Nb)/|k|)/|k|, (, (aa ++ bmbm)/|)/|kk|, and|, and((mm22 NN)/)/kk, we get the newvalues, we get the newvalues ..

That is, we have the new solution:That is, we have the new solution:

At this point, one roundof the cyclic algorithm isAt this point, one roundof the cyclic algorithm iscomplete.complete.

|m2 67|


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Second iteration:Second iteration:

We now repeat the process. We haveWe now repeat the process. We have aa == 41,41, bb == 5,5, kk == 6.6.We want anWe want an mm >> 0 such that0 such that kk dividesdivides aa ++ mbmb, i.e. 6 divides, i.e. 6 divides4141 ++ 55mm, and |, and |mm22 67| is minimal. The first condition67| is minimal. The first condition

implies thatimplies that mm is of the form 6is of the form 6tt ++ 5 (i.e. 5, 11, 17,), and5 (i.e. 5, 11, 17,), andamong suchamong such mm, |, |mm22 67| is minimal for67| is minimal for mm == 5. This leads5. This leadsto the newto the new solutionsolutionaa == (41(4155 ++ 67675)/6, etc.:5)/6, etc.:

Third iteration:Third iteration:For 7 todivide 90+11For 7 todivide 90+11mm, we must have, we must have mm = 2= 2 ++ 77tt (2, 9,(2, 9,16,) and among such16,) and among such mm, we pick, we pick mm=9.=9.


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Final solution:Final solution:

At this point, we could continue with the cyclic methodAt this point, we could continue with the cyclic method(and it would end, after seven iterations), but since the(and it would end, after seven iterations), but since the

rightright--hand side is amonghand side is among ss1,1, ss2,2, ss4, we can also use4, we can also use

Brahmagupta'sBrahmagupta's observationdirectly. Composing theobservation

directly. Composing thetriple (221, 27, 2) with itself, we gettriple (221, 27, 2) with itself, we get

that is, we have the integer solution:that is, we have the integer solution:

This equation approximatesThis equation approximates (as 48842/5967) to(as 48842/5967) towithin a margin of aboutwithin a margin of about ..


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RATIONAL TRIANGLESRATIONAL TRIANGLES

BRAHMAGUPTA

S THEOREM

DEFINITION

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DEFINITION OF RATIONAL TRIANGLESDEFINITION OF RATIONAL TRIANGLES

Somos As a triangle such that all three sides

measured relative to each other.

Conway and Guy (1996)

A triangle that all of whose sides arerational numbers and all ofwhose angles

are rational numbers ofdegrees. The onlysuch triangle is the equilateral triangle(Conway and Guy 1996).

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RIGHTRIGHT

TRIANGLESTRIANGLES

HERONIANHERONIAN

TRIANGLESTRIANGLES

LATTICELATTICE

TRIANGLESTRIANGLES

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RATIONALRATIONALTRIANGLTRIANGL

ESES


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RIGHT TRIANGLESRIGHT TRIANGLES

Hypotenus

Side

Side

A right triangles is aA right triangles is a

rational triangle ifrational triangle if

and only ifand only if all sixall sixtrigonometricallytrigonometrically

ratiosratios of theof thecomplementarycomplementaryacute anglesacute angles areare

rational.rational.


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Koblitz (1993)

Defined a congruent number as an integer that isequal to the area of a rational right triangle. Thediscovery that a right triangle of unit leg lengthhas an irrational hypotenuse (having a length

equal to a value now known as Pythagoras'sconstant) showed that not all triangles arerational.

It can be proved, using the inscribed circle, that

the tangent of its half angles are rational numbers.


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A Heronian triangle is a triangle having rational

side lengths andrational area.

The simplest example of a Heronian triangle is aright triangle with ratioof sides 3:4:5. In fact, anyrational right triangle is a Heronian triangle.

The triangles are so namedbecause such trianglesare related toHeron's formula .

(1)

Giving a triangle area in terms of its side lengths a,

b,c andsemiperimeter, s= (a+b+c)/2 . Finding aHeronian triangle is therefore equivalent to solvingthe Diophantine equation

(2)

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HERONIAN TRIANGLESHERONIAN TRIANGLES


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The complete set of solutions for integer Heronian

triangles (the three side lengths and area can bemultipliedby their least common multiple tomakethem allintegers) were foundby Euler (Buchholz1992; Dickson 2005, p. 193), andparametric

versions were given by Brahmagupta andCarmichael(1952) as

(3) a= n(m2+k2)

(4) b= m(n2 + k2)

(5) c=(m+n)(mn-k2

)(6) s= mn(m+n)

(7) = kmn(m+n)(mn-k)

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This produces one member of each similarity class of

Heronian triangles for any integers m,n,andk suchthat GCD(m,n,k)=1, mn>k2m2n/(2m+n) , andmn1(Buchholz 1992).

The first few integer Heronian triangles sortedbyincreasing maximal side lengths, are ((3, 4, 5), (5, 5,6), (5, 5, 8), (6, 8, 10), (10, 10, 12), (5, 12, 13), (10,13, 13), (9, 12, 15), (4, 13, 15), (13, 14, 15), (10,10, 16), ...

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A triangle with rational coordinates has

rational area, and if the sides are rational,

then the triangles is Heronian.

Conversely, a Heronian triangles vertices canbe given rational coordinates. This is proven

by analysis of the reduction of the case

integer sides which leads to the triangle

realization as an integer Lattice Triangle.

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LATTICE TRIANGLESLATTICE TRIANGLES


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Alattice triangles is a triangles whosevertices are lattice points

Alattice point means a point in the planewith the integer coordinates.

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LATTICE TRIANGLESLATTICE TRIANGLES

Area of Triangles

(different of cross products of the x and y (different of cross products of the x and ycoordinates of two points)coordinates of two points)

Area of Triangles

(different of cross products of the x and y (different of cross products of the x and ycoordinates of two points)coordinates of two points)


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Parameterization ofRational Triangles

If a cyclic quadrilateral has diagonals that

are perpendicular to each other, then theperpendicular line drawn from the point ofintersection of the diagonals to any side ofthe quadrilateral always bisects the

opposite side

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BRAHMAGUPTAS THEOREMBRAHMAGUPTAS THEOREM


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Let A, B, C andD be four points on a circle such that the linesAC andBD are perpendicular.

Denote the intersection ofAC andBD by M. Drop the

perpendicular fromM to the line

BC, calling the intersectionE.

Let Fbe the intersection of the line EM and the edge AD.

Then, the theorem states that F is in the middle ofAD.

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A

B

C

D

M

E

F

BrahmaguptasBrahmaguptas Theorem states that AF = FDTheorem states that AF = FD


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DerivationofRational Triangles

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a b

c1 c2

h

a,b,c = sides

h = altitude

By using Pythagoras Theorem,By using Pythagoras Theorem,the two right angled triangles are like this:the two right angled triangles are like this:

a2 = c1

2 + h2a2 = c1

2 + h2

a2 = c22 + h2a2 = c22 + h2


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Hence, byHence, by substractionsubstraction::

a2 b2 = c12 + c22 = (c1 - c2)(c1 + c2)= (c1 - c2 ) ca2 b2 = c12 + c22 = (c1 - c2)(c1 + c2)= (c1 - c2 ) c

So, we solve and get the rational form;So, we solve and get the rational form;

(c1

- c2

) = a2-b2

c

(c1

- c2

) = a2-b2

c

we notice that;we notice that;

c1 + c2 = cc1 + c2 = c


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Hence, we have:Hence, we have:

c1 = (a2-b2/c + c)c1 = (a2-b2/c + c)

andand

c2

= (c - a2-b

2)

c

c2

= (c - a2-b

2)

c

From theFrom the DiophantusDiophantus Method for any right Method for any rightangled triangles with hypotenuse 1 has sideangled triangles with hypotenuse 1 has side

of the formof the form

for some rational t.for some rational t.

,,1 - t2

1 + t22t

1 - t21


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Biography ofBiography ofBRAHMAGUPTABRAHMAGUPTA


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BRAHMAGUPTABRAHMAGUPTA

Was born in 598 A.D anddies in 665 A.D

Born in Bhillamala(now Bhinmal), in

Gijurat

His father named Jisnugupta,wrote important workson mathematics and astronomy.

Become court astronomer toKing Vyaghramukha ofthe Chapa Dynasty


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He was the headof the astronomical

observatory at Ujjain, andduring his

tenure there wrote four texts on mathematics andastronomy:

The Cadamekela in 624,

The Brahmasphutasiddhanta (The Opening ofthe Universe), in 628,

The Khandakhadyaka in 665

The Durkeamynarda in 672.


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Brahmasphutasiddhata as the corrected

version of the astronomical text about

arithmetic and algebra.

The work was written in 25 chapters at

Bhillamala (Bhinmal).

The Khandakhadyaka is a hand-book onastronomical calculations where he effectivelyused algebra for the first time in calculation in

this book


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CONTIBUTIONOF

BRAHMAGUPTA

Arithmetic

Algebra

TrigonometryGeometry

DiophantineAnalysis


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Zero

The first person who framed the rules ofoperationfor zero

Here Brahmagupta states that rules for arithmeticon negative numbers and zero are quite close tothe modern understanding.

ARITHMETIC


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ARI E IHe also givesarithmeticalrules in termsoffortunes

and debts :-

e.g:

A debt minus zero I a debt

A fortune minus zero is a fortune

Zero minus zero is a zero

A debt subtracted from zero is fortune

Give arithmetical rules in terms of fortunes (+ve number) and

debts(-ve number)


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Series:

Givethe sum ofthe s uareand ubes ofthe firstn

integer.It is imp

ortantt

ono

tehere rahmaguptafound the result interms ofthe s m ofthefirst n integers, rather than interms ofn as is themodern practice

Square: Cube:

ARITHMETIC


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Brahmaguptagavethe solutionofthe

general linear equation inchapterofBr mas tasi anta

Linear Equation

AL E RA


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General Quadratic

Equation

ALGEBRA


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Pythagorean Triple (chapter12)

The height of a mountain multipliedby a givenmultiplier is the distance to a city; it is noterased. When it is dividedby the multiplierincreasedby two it is the leapofone of the two

who make the same journey

DIOPHANTINE ANALYSIS


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DIOPHANTINE ANALYSIS

For a given length m and an arbitrary multiplier xLet, and

Pells Equation


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In Chapter 2of his Brahmasphutasiddhanta,

entitled Planetary True Longitudes,Brahmagupta presents a sine table.

The Brahmagupta Interpolation formula is aspecial case of the Newton-Stirling.

TRIGONOMETRY


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Interpolation formula to the second-order,

which used in 665 to calculate the valuesof sine at different intervals

TRIGONOMETRY


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Brahmaguptas formula.

a) Most famous in his Cyclic Quadrilaterals.b) Given the lengths of the sides of any cyclic

quadrilateral, Brahmagupta gave anapproximate and an exact formula for the

figure's area

GEOMETRY


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By letting,

The exact area is

GEOMETRY


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GEOMETRYGEOMETRY

Triangles

The base decreased and increasedby thedifference between the squares of the sides

dividedby the base; when dividedby twothey are the True segments. The

perpendicular [altitude] is the square- rootfrom the square of a side diminished by thesquare of its segment


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GEOMETRYGEOMETRY

Pi,( )The diameter and the squareofthe radius[each] multiplied by 3 are [respectively] thepracticalcircumferenceand thearea [ofacircle]. Theaccurate [values] arethe square-

roots from the squares ofthosetwo multipliedbyten. se 3 as practical valueof and as anaccurate valueof

INDIANINDIAN


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BHASKARA1BHASKARA1

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INDIANINDIAN

MATHEMATICIANSMATHEMATICIANS

Bhaskara I (c. 600 680) expanded thework ofAryabhata in his books titledMahabhaskariya, Aryabhatiya-bhashya andLaghu-bhaskariya.Was born in KeralaMost important methematicalcontribution:

the representation of numbers in

a positional system.wrote the first work with full andsystematic use of the decimalnumber system.


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He produced:

Solutions of in

determinate equations.eg:2x = y

Indeterminate equations cannot be directlysolved from the given information

A rational approximation of the sinefunction.

A formula for calculating the sine of anacute angle without the use of a table,

correct to twodecimalplaces.

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BHASKARAIIBHASKARAII

Bhskara II (1114 1185) was a

mathematician-astronomer whowrote anumber of important treatises, namely theSiddhanta Shiromani, Lilavati, Bijaganita,Gola Addhaya, Griha Ganitam andKaran

Kautoohal. Was born near the Bijjada Bida ( now

Bijapur district, Karnataka state, SouthIndia) in the Deshatha Brahmin family.

Lived in Sahyadri, Western Maharashtra.

Headof astronomicalobservatory at Ujjain,mathematical centre of ancient India.

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Also known as Bhaskara IIor asBhaskaracharya, this latter name meaning

"Bhaskara the Teacher". Since he is known in India as Bhaskaracharya

we will refer to him throughout this article bythat name. Bhaskaracharya's father was a

Brahman named Mahesvara that is anastrologer. This happened frequently in Indiansociety with generations of a family beingexcellent mathematicians andoften acting as

teachers toother family members. A number of his contributions were later

transmitted to the Middle East and Europe.

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Given that he was building on theknowledge and understanding of

Brahmagupta it is not surprising thatBhaskaracharya understood about zeroand negative numbers. However hisunderstanding went further even than that

ofBrahmagupta. Examples, he knew that x2 = 9 had two

solutions. He also gave the formula

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Bhaskaracharya stud

ied

Pell's equationusing the chakravala method.

Chakravala method-solving indeterminatequadratic equations of the form ax2+bx+c=y

px2 + 1 = y2 for p = 8, 11, 32, 61 and 67. When p = 61 he found the solutions x=

226153980, y= 1776319049.

When p = 67 he found the solutions x=5967, y= 48842. He studiedmanyDiophantine problems.

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SiddhantasiromaniSiddhantasiromani Twoparts

The first on mathematical astronomy with thesecondpart on the sphere.

Focus on astronomy

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BijaganitaBijaganita Seed counting or Root Extraction

Focus on Algebra

ContributionContribution

LilavatiLilavati The Beautiful Focus on Mathematics


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VasanabhasyaVasanabhasya ofof MitaksaraMitaksara Bhaskaracharyas own commentary on the

Siddhantasiromani

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KaranakutuhalaKaranakutuhala//BrahmatulyaBrahmatulya Calculation ofAstronomical Wonders

A simplifiedversion of theSiddhantasiromani

VivaranaVivarana A

commentary on theShishyadhividdhidatantra of Lalla

Focus on Algebra


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LilavatiLilavati

Focus on mathematics Bhaskaracharya gave twomethods of

multiplication in his Lilavati.

We followIfrah who explains these two

methods due to Bhaskaracharya.

Tomultiply 325 by 243 Bhaskaracharyawrites the numbers thus:

24324324332 5-------------------

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Nowworking with the rightmost of thethree sums he computed 5 times 3 then 5

times 2missing out the 5 times 4 which hedidlast andwrote beneath the others oneplace to the left. Note that this avoidsmaking the "carry" in ones head.

24324324332 5-------------------

101520-------------------

67


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Now add the 1015 and20 sopositioned andwrite the answer under the secondlinebelow the sum next to the left.

24324324332 5-------------------101520-------------------1215

68


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Work out the middle sum as the right-handone, again avoiding the "carry", and add

themwriting the answer below the 1215but displacedone place to the left.

243243243

32 5-------------------4 6 10158 20

-------------------1215486

69

Fi ll k h l f i h


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Finally work out the left most sum in thesame way and again place the resultingaddition one place to the left under the 486.

24324324332 5-------------------

6 9 4 6 101512 8 20-------------------1215

486729-------------------

70

Fi ll k t th l ft t i th


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Finally work out the left most sum in thesame way and again place the resultingaddition one place to the left under the 486.

24324324332 5-------------------

6 9 4 6 101512 8 20-------------------1215

486729-------------------

71

Fi ll dd th th b b l th


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Finally add the three numbers below thesecondline toobtain the answer 78975.

24324324332 5-------------------6 9 4 6 1015

12 8 20-------------------1215486

729-------------------78975

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Despite avoiding the "carry" in the firststages, of course one is still facedwith the"carry" in this final addition. The secondofBhaskaracharya's methods proceeds as

follows:

325243

--------

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Multiply the bottom number by the topnumber starting with the left-most digit andproceeding towards the right. Displace eachrowone place to start one place further

right than the previous line. First step

325243

--------729

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Second step Third step, then add

325 325243 243-------- --------

729 729486 4861215

325 --------

243 78975--------

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Siddh t i iSiddh t i i


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Amathematical astronomy text similar in layouttomany other Indian astronomy texts of this andearlier periods.

First part (twelve chapters) cover the topics of:

mean longitudes of the planets; truelongitudes of the planets; the three problemsofdiurnal rotation; syzygies; lunar eclipses;solar eclipses; latitudes of the planets; risingsand settings; the moon's crescent;

conjunctions of the planets with each other;conjunctions of the planets with the fixedstars; and the patas of the sun andmoon.

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SiddhantasiromaniSiddhantasiromani


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Secondpart (thirteen chapters on thesphere) cover the topics of:

praise of study of the sphere; nature ofthe sphere; cosmography and geography;planetary mean motion; eccentricepicyclic modelof the planets; the

armillary sphere; spherical trigonometry;ellipse calculations; first visibilities of theplanets; calculating the lunar crescent;astronomical instruments; the seasons;

andproblems of astronomicalcalculations.

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present asg 2(2)- math phylosphy

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