Contribution of Mahaviracharya in the

Author(s): R S Lal
Publisher: Z_Deshbhushanji_Maharaj_Abhinandan_Granth_012045.pdf
Catalog link: https://jainqq.org/explore/250066/1

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Contribution of Mahaviracharya in the development of theory of Series In the present paper, an attempt has been made to summarize some of the salient features of the work of the great ancient Indian Mathematician Mahaviracharya (850 A.D.) on the development of theory of series as evinced from his renowned mathematical text Ganita Sarasangraha. No doubt his predecessors Aryabhata I (476 A.D.) and Brahmagupta (599 A.D.) had their contributions to the subject, yet Mahaviracharya can be named as the first amongst them who put the subject elaborately using lucid methods and charming language. The text GSS consists of nine chapters but it is only chapters II, III and VI which contain the sutras regarding series. In chapters II the A.P. and G P. are given in detail. For example, the following sutra gives the sum of the A.P. whose first term, common difference and number of terms are known.1 rUpeNono gaccho dalIkRtaH pracayatADito mizraH / prabhaveNa padAbhyastaH saGkalitaM bhavati sarveSAm // Algebraically if a first term, d=common difference and n-number of terms and s=the sum of the series then terms then The above formula has been given in three ways234. In the following sutra the method is given to find out the number of terms of the series if the first term, common difference and the sum of the series be known." aSTosa guNaguNAyutara vizeSakRtimahitAt / H mUlaM cayayutamatimA caya gaccha: I! Symbolically, if a first term, d=common difference, S=sum of the series and n=number of 1. GSS 2. GSS 3. GSS 4. GSS 3. GSS jana prAcya vidyAeM Note:-For references: See Ganita Sarasangraha by Sh. L. C. Jain Sloka 61 Ch. 2 2 2 n = [(n-1) d+2a] S= n= 2 2 V(2a-d)2+8d2s-2a+d 2d2 p. 20 20 21 21 22 Dr. R. S. LAL 62 63 64 69 63

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Apart from the above formula, methods are given 1. 2. 3. to find the common difference and the first term if the remaining term are known. Quite a good number of examples are also given whose solution by the above formula can easily be done. Three rules giving stanzas for splitting up into the component elements) such as sum of the series in A. P.) as is combined with the first term (af 1997 or with the common difference (uttara mizradhana) or with the number of terms (gacchamizradhana) or with all these (sarvamizradhana) are given below: uttaradhanena rahitaM gacchenaikena saMyutena hRtam / mizraghanaM prabhavaHsthAditi gaNaka ziromaNe viddhi / / "O crest jewel of calculators, understand that misradhana diminished by the Uttardhana and (then) divided by the number of terms increased by one, gives rise to the first term." Symbolically, if S=sum, a=first term, d=C. D. and n=number of terms then 2 S'_ n (1-1) a = -- -- n+1 Now in the second stanza where S=S+a. AdidhanonaM mitraM rUponapadArdhaguNitagacchena / sakena hRtaM pracayo gacchavidhAnAspadaM mukhe sake / / "The misradhana diminished by the adidhana, and then divided by the quantity obtained by the) addition of one to the (product of the number of terms multiplied by the half of the number of terms lessened by one (gives rise to the common difference). (In splitting of the number of terms from the misradhana) the (required) number of terms (is obtained) in accordance with the rule for obtaining the number of terms, provided that the first term is taken to be increased by one (so as to cause a corresponding increase in all the terms)".! lesenthana the crequired number of terms (is obtained in acoma Algebraically if S=S+d=Uttardhana and na=adidhana then de s'- na n(n-1) +1 1. GSS GSS GSS - int no no 8. GSS mAcAryaratta zrI dezabhUSaNa jI mahArAja abhinandana grantha

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And in the third stanza mizrAdapanIteSTau mukhagacchau pracayamizravidhilabdhaH / yo rAziH sa cayaH syAtkaraNamidaM sarvasaMyoge / "The misradhan is diminished by the first term and the number of terms, both (of these) being optionally chosen : (then) that quantity, which is obtained (from this difference) by applying the rule for (splitting up) the Uttarmisradhana happens to be the common difference (required here). This is the method of work in (splitting up) the all combined (misradhana)".1 herey applying the rule for Symbolically, if S=S+n S=a+ (a+d) + (a+2d) + ......to n terms. then S= (a +1) + (a +1+d) + (a +1+2d) +......to n terms = [2 (2+1) +(n-1)] which is a quadratic equation and hence n can be found. Now according to the above rule, a and n can be chosen in any way. This method is the same as the previous one. Example : dvitrikapaMcadazAmA catvAriMzanmukhAdimizradhanam / tatra prabhavaM pracayaM gaccha sarva ca me brUhi // Forty, exceeded by 2, 3, 5 and 10, represents (in order) the adimisradhana and the other (misradhanas). Tell me what (respectively) in these cases happens to be the first term, the common difference, the number of terms and all (these three)." This means (i) find a when s'=42, d=3, n-5. (ii) find d when s'=43, a=2, n=5. (iii) find n when S=45, a=2, d=3, and (iv) find a, d, n when S=50. From the formulae given above the results can be obtained easily. In the following sutra the rule is given for finding, in relation to two (series), the number of terms wherein are optionally chosen their mutually interchanged first terms and common difference as also their sums which may be equal or (one of which may be) twice, thrice, half or one-third or any such (multiple or fraction of the other) :3 vyekAtmahato gaccha: sveSTaghno dviguNitAnyapadahInaH / mukhamAtmonAnyakRtivikeSTapadadhAtavarjitA pracayaH / / 1. GSS 2. GSS 3. GSS 2 2 jaina prAcya vidyAe~

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"The number of terms (in one series), multiplied by itself as lessened by one and then multiplied by the chosen ratio between the sums of the two series, and then diminished by twice the number of terms in the other series gives (rise to the interchangeable) first term of one (of the series). The square of the (number of terms in the) other (series) diminished (again) by the product of two (times the) chosen (ratio) and the number of terms (in the first series) gives (rise to the interchangeable common difference (of that series)." Symbolically if S, S, be the sums, a, a, the first terms and d, d, the common differences of the two given series then a1 = Sa and did. Now if - r and n, n, be the respective number of terms in the S S two series, then according to the above formula a=n (n-1) x r-2n, and Example1: and In relation to two men (whose wealth is measured) respectively by the sums of two series in A.P. having 5 and 8 for the number of terms, the first term and the common difference and both these series be interchangeable (in relation to each other), the sums (of the series) being equal or the sum (of one of them) being twice, thrice, or any such (multiple of that of the other), 'O arithmetician, give out the (value of these) sums and the interchangeable first term and common difference after calculating (them all) well." Solution : If S=S, then r=1 so in the above case where n=5 and n,-8 we have Then and which proves that 1. 66 GSS d-(n,)-n,-2rn 2 paMcASTagacchapuMsorvyastaprabhavottare samAnayanam / dviguNAdidhanaM vA hi tvaM gaNaka vigaNayya // a=n (n-1) x 1-20, =5 (5-1) x 12 x 8 2016 4 d (D1) D1 - 2 rn - (8)282 X 1 X 5 64 8 10 r= - s=2(2 x 4 (2 x 4+ (5-1) x 46) 16) = 5 (4+92) = 480 8 S-(2 x 46+ (8-1)x4)-4 (9+28) - 480 S =1. S 27 9 46 87 AcArya ratna zrI dezabhUSaNa jI mahArAja abhinandana grantha

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Partial Sums The sum of any part of a series is known as the partial sum of the series. In the following verse, the method is given for finding the partial sum of a given series :1 "(Take) the chosen off number of terms as combined with the total number of terms (in the series) and (take) also your own chosen off number of terms (simply) diminish (each of) these (resulting products) and these (resulting quantities) when multiplied by the remaining number of terms (respectively), give rise to the sum of the remainder series and to the sum of the chosen off part of the series (in order)." Symbolically. Vyutkalita-S and the sum of the chosen part=S, sapadeSTaM svaSTamapi vyakaM dalitaM yAtaM samukham / zeSeSTa gaccha guNitaM vyutkalitaM sveSTavittaM ca // [Pd+a] p. where p is the number of terms of the chosen part of the series. Another form of the same formula is given in a different verse.2 Example: In the following sutra is given the rule for finding the sum of a series in arithmetic progression in which the common difference is either positive or negative : 1. GSS 2. GSS 3. GSS 4. GSS jana prAcya vidyAeM [maeul 2=1 "The first term is either decreased or increased by the product of the negative or the positive common difference and the quantity obtained by halving the number of terms in the series as diminished by one. (Then), this is (further) multiplied by the number of terms of the series and (thus), the sum of series of terms in arithmetical progression with positive or negative common difference is obtained." 1 Symbolically, S("="+a 11d+a) n where a, d, n and S have their usual meanings. 2 2 6 6 -d+a] (n-p) vye kApa donAdhikacayaghAtonAnvitaH punaH prabhavaH / gacchAbhyasto hInAdhikacayasamudAya saMkalitam // 32 33 165 165 caturuttaradaza cAdirhInaca yastrINi paJca gacchaH kim / dvAvAdivRddhicayaH SaT padamaSTI ghanaM bhavedatra || 106 107 290 291 67

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The first term is 14, the negative common difference is 3 and the number of terms is 5 : the first term is 2 ; the positive common difference is 6 and the number of terms is 8. What is the sum of the series in (each of) these cases ? Solution: (i) a=14, d= -3, n=5. .: S=$(${1x (3) + 14 ] =51-6 +14) = 5 x 8 = 40 (ii) a = 2, d = 6, n=8 .: S = 8 [8;1x (6) + 27-8 (21 + 2) = 8 x 23 = 184 In the following sutra the rule is given for finding the time of arrival of two persons at a common terminus when one, who is moving (with successive velocities representable) in arithmetical progression and another moving with steady unchanging velocity, may meet together again (after starting at the same instant of time):1 dhruvagatirAdivihInazcayadalabhaktaH sarUpakaH kAlaH / dviguNo mArgastadgatiyogahRto yogakAlaH syAt // "The unchanging velocity is diminished by the first term (of the velocities in series in A. P.) and is (then) divided by the half of the common difference. On adding one to the resulting quantity), the required time (of meeting) is arrived at. (Where two persons travel in opposite directions, each with a definite velocity) twice (the average distance to be covered by either of them) is the (whole) way (to be travelled). This when divided by the sum of their velocities gives rise to the time of (their) meeting." Symbolically, if V = the unchanging velocity a = first term of changing vel. d = common difference t = time taken. then t = (V-a Example? : kazcinnaraH prayAti tribhirAdA uttaraistathASTAbhiH / niyatagatirekaviMzatiranayoH kaH prAptakAlaH syAt // "A certain person goes with velocity 3 in the beginning increased (regularly) by 8 as the successive) C. D. The steady unchanging velocity (of another person) is 21. What may be the time of their meeting (again if they start from the same place, at the same time, and move in the same direction)?" 1. GSS 2. GSS 6 6 173 174 319 320 AcAryaratna zrI vezabhUSaNa jI mahArAja abhinandana panya

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Solution : V = 21, 4 = 3, d= 8 then 1 = (V-a) = 2 +1 = (21-3) = q + 1 = 18 - 4+1=1 In the following stanza the rule is given for arriving at the time and distance of meeting together (when two persons start from the same place, at the same time and travel) with (varying) velocities in A. P.1 ubhayorAdyoH zeSazcayazeSahRtodvisaGguNaH saikaH / yugapatprayANayoH syAnmArge tu samAgamaH kAlaH // "The difference between the first two terms divided by the difference between the two common differences when multiplied by two and increased by one, gives rise to the time of coming together on the way by the two persons travelling, simultaneously (with two series of velocities varying in A. P.)" Symbolically, if a, a, be the velocities in beginning and d, d, be their respective common differences then the time of meeting is given by d d2 x 2 + 1 di-dg The same formula has been given in another stanza' too. Examples: paMcAdyaSTottarataH prathamo nAtha dvitIyanaraH / AdiH paJcaghnanava pracayo hIno'STa yogakAlaH kaH / / "The first man travels with velocity beginning with 5, and increased (successively) by 8 as the common difference. In the case of the second person, the starting velocity is 45, and the common difference is minus 8. What is the time of meeting ?" Solution :twe 5-45 e X 2+1= 5+1 = 6. tures me to the In the following sutra the rule is given for arriving at the number of bricks to be found in structures made up of layers (of bricks one over the other). taravargoM rUponastribhivibhaktastareNa sNgunnitH| tarasaMkalite sveSTapratADite mizrataH sAram / / 3221 1. GSS 2. GSS 3. GSS 4. GSS 6 6 6 6 174 175 175 176 3247 3257 3301 jana prAcya vidyAeM

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"The square of the number of layers is diminished by one, divided by three, and (then) multiplied by the number of layers. On adding (to quantity so obtained) the product, obtained by multiplying the arbitrarily chosen number (representing) the bricks in (the topmost layer) by the sum of the (natural numbers beginning with one and going upto the given) number of layers, the required answer is obtained". Symbolically, if n be the number of layers, and a number arbitrarily chosen representing the bricks in the topmost layer, then Example: Total number of bricks = 2=1 n2-1 3 "There is constructed an equilateral quadrilateral structure consisting of 5 layers. The topmost layer is made up of one brick. O' you, who know the calculation tell me how many bricks there are (in all)". Solution : n = 5, a= 52-1 3 So total number of bricks 25-1 3 40+15 = 55 bricks. 1. GSS 2. GSS :*o paJcataraMkenAgraM vyvghttit| gaNitavinmizre / samacaturazrazra DhI katISTakAH syurmamAcakSva // = = Xn+ax 6 2 = 1. X 5 + 1 x Now we shall consider the work of Mahavira on Geometrical progressions. In the following sutra is given the rule for finding gunadhana (T) and the sum of a G. P. if the first term, common ratio and the number of terms of the series are known :2 x 5+ The product of the first term with the common ratio multiplied to itself as many times as the number of terms gives the gunadhana. It be known that the gunadhana lessened by the first term and divided by one less than the number of terms gives the gunasankalita. Symbolically, if n = the number of terms a = first term common ratio then and gunadhana ar" = (n+1)th term. and gunasankalita (sum of the series) =S= 177 28 5 x 6 2 n (n+1) 2 padamitaguNahRtamitaprabhavaH svAdguNadhanaM tadAcUnam / ekonaguNavibhaktaM guNasaGkalita vijAnIyAt // 5 (5+1) 2 3311 93 ar"-a 7-1 AcArya ratna zrI vezabhUSaNa jI mahArAja abhinandana grantha

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In the following sutra another rule is given to find out the sum of a series in G. P.1 samadalaviSamakharUpo guNaguNito vargatADito gacchaH / rUponaH prabhavaghno vyekottarabhAjitaH sAram // "The number of terms in the series is caused to be marked (in a separate column) by zero and by one (respectively) corresponding to the even (value) which is halved and to the uneven (value from which one is substracted till by continuing these processes zero is ultimately reached), then this (representative series made up of zero and one is used in order from the last one thcrc in, so that this one multiplicd by the common ratio is again) multiplied by the common ratio (wherever one happens to be the denoting item) and multiplied so as to obtain the square (wherever zero happen in the invins iv...). This the result of this (operation) is diminished by one and (is then) multiplicd by the first term, and (is then) divided by the common ratio lessened by one it becomes the sum of the series)." Example : svarNa dvayaM gRhItvA triguNadhanaM pratipura samArjayati / yaH puruSo'STana garyAM tasya kiyadvittamAcakSva / / *Having obtained 2 gold coins (in some city), a man goes on from city to city, earning (everywhere) three times (of what he earned immediately before). Say how much he will make in the eighth city." Solution : Here n = 7, r = 3, a = 2 7 an odd number, hence one is subtracted from it and also it is denoted by one. 2-1 = 6 = an even number, hence it is divided by 2 and 0 denotes it hy w = 3 = an odd number, it is diminished by one and I denotes it h br brdm -- 1 = 2 = an even number, it is divided by 2 and 0 denotes it = 1 - an odd number, it is diminished by one and 1 denotes it 1-1=0 =, where the operation ends. Now the whole is put in the side column. Since in the column, 1 is in the last hence it is 11 multiplied by the common ratio 3, then comes zero so 3 is squared and we get 39, then comes 1 above lo it so it is multiplied by >> i.e. we get 33, then comes zero above it so it is squared and we get 39. then in the end there is one above it so it is multiplied by 3 and get 3?. So the guradhana arn 2 37 = 2 X 2187 = 4374 coins, will be the amount obtained by the man in eighth city. The rules for finding out the last term and the sum of series in G. P. have also been given in nza3 There are other sutras in which rules have been given to find out the first term, common ratio and the number of terms of the series in G. P. 4, 5, 6, 7. 94 90 1. GSS 2. GSS 3. GSS 4. GSS GSS 6. GSS 7. GSS 2 2 2 2 98 101 103 jana prAcya vidyAeM

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In the following sutras the rules have been given to find out the sum of a series in geometrical progression, wherein the terms are either increased or decreased in a specified manner by a given known quantity). gaNacitiranyAdihatA vipadAdhikahInasaMguNA bhaktA / vyekaguNenAnyA phalarahitA hone'dhike tu phlyuktaa|| Algebraically, if S = sum of the series, a = first term n=number of terms, r = common ratio and m = the quantity to be added or subtracted from each term of the series in G. P., and S' = the sum of the series in G. P., then S - sum of the resulting series - + S' r- 1 Proof : Theorem : Let S-a + (ar + m) + [(ar $ m)r m + ... to n terms and S'= a + ar + ar2 + ar3 + ... to n terms Now S =la + ar + arl + ar3 + ... to n terms] + m[(r+p2 + ... to n-1 terms)] + m[(r + pa + 3 + ... to n-2 terms)] + ... + m - 2 = 1 +m">=1+may + .. + m i -1-1) + (p-2-1) + (pm-3-1) + ... + (r--1)] + + mir(r+re+r+-+pm-1) - 1 (n-1)] -s+ , [ *} - (n-1)] + [ 1-n+1] + + 1. GSS 172 3 14 AcAryaratna zrI dezabhUSaNa jI mahArAja abhinandana anya

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which can be generalised as S = + -1 +S Now we shall discuss the contribution of Mahavira in the development of the series which can be put in another category called miscellaneous series. This work no doubt, is quite voluminous and it can be said without any hesitation that no other Hindu mathematician contributed so much. In the following stanza a rule is given for finding the sum of the squares of natural numbers. He has not given any formula for the sum of natural numbers like others. sekeSTakRtivighnA saikeSToneSTadalaguNitA / kRtighanacitisaMghAtastrikabhakto vargasaMkalitam / / Algebraically, if n = number of terms and n(n+1) - sum of first n natural nos. En = sum of the squares of n natural nos. then } [ 2 (n+1)* -- (n + 1) ] = = [ ** + ndeg + " (n!! ]} = = nx(n+1) X(2n+1) In the following sutra a rule is given for finding the sum of the squares of numbers which are in A. P. This is most general form of the rule which can be applied broadly.2 dviguNa konapadottarakRtihatiSaSThAMzamukhacayahatayutiH / vyekapadaghnA mukhakRtisahitA padatADiteSTakRticitikA // Algebraically, if a = first term, d = common diff. n= number of terms and s = sum of the squares of the terms which are in A. P. then S = E[a + (n--1) a 1 = n [{{20 ) d + ad} (n-1 + qe ] which can easily be substantiated by taking LHS. i.e. {a + (n-1) d]' = [(a-d) + 2 nd ( a d) + na da] = n(a-d)2 + 2 d (a-d) En + d2 En2 we know that sn= n(n+1) and 5 ne = n(n+1) (2n+1) 2 Hence by substituting these values we get the result 1. GSS 2. GSS 6 6 167 167 296 208 a seu faang

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s = " [{ 21-12 d_ + ad } (n-1) + a ] which is in its most general form. Another method of the same formula is given inl of the text. Now comes the rule for finding the sum of the cubes of first n natural numbers which has been given to be equal to square of the sum of first n natural numbers.2 gacchArdhavargarAzIrUpAdhikagacchavargasaGguNitaH / dhanasaGkalitaM proktaM gaNite'smin gaNitatattvajJaH / / "The square of half of the number of terms is multiplied by the square of (the number of term increased by one) which gives rise to the sum of cubes of first n natural numbers as stated by mathematicians." Algebraically. En = { n {n}" } In the following stanza he has given a rule for finding out the sum of the cubes of the terms which are in A. P. This formula is in its most general form. cityAdihatirmukhacayazeSaghnA prcynighncitivrge| Adau pracayAdUne viyutA yuktAdhike tu dhanacitikA // Algebraically, if s = sum of terms in A. P. a = first term, d = common difference n = number of terms Sn = sum of the given series then Sn= Z[a + (n-1) d] = Sd + Sa (a-d) where s = , [ 2a + (n-1) d] or specifically (i) when a>d, Sn = + Sa (a-d) + Sad (ii) when aPage #13
Symbolically we can write, if s. = 46,71_, Sapa = (a+a) (a+d+1), etc., Se Sa + Sad+Sated + ... + Sa + (n-11d = [{ (2r-19 dP + + + ad } xn--1) + a (a+10 ] x 2 In the following stanzal a rule has been given for finding of the sum of the series which can be written symbolically in the form 1+(1+2)+(1+2+3)+...+(1+2+3+...+n),ne,n and En. i.e. S = 2n +22 (+1) + ne + ni . sakapadArdhapadAhatiravainihatA padonitA vyaaptaa| saikapadaghnA citiciticitikRtighanasaMyutirbhavati / / Algebraically, n (n+1)X7_.. -x (n+1) S = _ 2 3 which can be proved easily by substituting values En== n (n+1) , An+n) = } En + } En = n (n+1) (2n+1) + n (n+1).. 12 Lastly, in the following stanza a rule has been given for finding out a single formula for the sum of the four above mentioned series. gacchastrirUpasahito gacchacaturbhAgatADita: saikaH / sapadapadakRtivinighno bhavati hi saMghAtasaMkalitam // Symbolically, the above formula takes the form Unt +27deg + ES. + En = [ (n+3) x 2 +1] (+n) 1. Gss 6 2. Gss 6 jana prAcya vidyAeM 170 171 3071 3098 75

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where Son - S, + S, + Sg + Sa + ... + Sn and s = Ar) = xn i.e. Es. = ZEn = [{ 2n-10 + 2 + 1 }(n-)+1 (1+10] " Since a = d = 1 in the formula , [{ (23) + 2 + aa } (n--1)+ au+1)] Examplel : saptakRteH SaTSaSTyAstrayodazAnAM caturdazAnAM ca / paMcAyaviMzatInAM kiM syAta saMghAtasaMkalitam // "What would be the (required) collective sum in relation to the (various) series represented by (each of) 49, 66, 13, 14 and 25 ?" Solution : The above given values are the number of terms in the five series. Hence for the first series in which n = 49. Required sum = ((n+32 " + 1 ) nt + n) = ( (99 +3) x 49 +1 ) (499 + 499 = ( 52 x 49 +1) x 49 (49 + 1) = [13 x 49 + 1] x 49 x 50 = [ 637 +1] x 2450 -- 638 X 2450 - 1563100. 1. Gss 6 _171 310 .. AcAryaratna zrI dezabhUSaNa jI mahArAja abhinandana grantha