Curve Fitting and Solution of Equation

UNIT V Curve Fttg ad Soluto of Equato 5. CURVE FITTING I ma braches of appled mathematcs ad egeerg sceces we come across epermets ad problems, whch volve two varables. For eample, t s kow that the speed v of a shp vares wth the horsepower p of a ege accordg to the formula p= a+ bv. Here a ad b are the costats to be determed. For ths purpose we take several sets of readgs of speeds ad the correspodg horsepowers. The problem s to fd the best values for a ad b usg the observed values of v ad p. Thus, the geeral problem s to fd a sutable relato or law that ma est betwee the varables ad from a gve set of observed values (, ), =,,...,. Such a relato coectg ad s kow as emprcal law. For above eample, = v ad = p. The process of fdg the equato of the curve of best ft, whch ma be most sutable for predctg the ukow values, s kow as curve fttg. Therefore, curve fttg meas a eact relatoshp betwee two varables b algebrac equatos. There are followg methods for fttg a curve. I. Graphc method II. Method of group averages III. Method of momets IV. Prcple of least square. Out of above four methods, we wll ol dscuss ad stud here prcple of least square. 5. PRINCIPLE OF LEAST SQUARES The graphcal method has the drawback that the straght le draw ma ot be uque but prcple of least squares provdes a uque set of values to the costats ad hece suggests a curve of best ft to the gve data. The method of least square s probabl the most sstematc procedure to ft a uque curve through the gve data pots.

CURVE FITTING AND SOLUTION OF EQUATION 8 Let the curve m = a + b + c +... k...() be ftted to the set of data pots (, ), (, ), (, ),, (, ). At ( = ) the observed (or epermetal) value of the ordate s = P M ad the correspodg value o the fttg curve () s m a+ b + c +... k = LM whch s the epected or calculated value. The dfferece of the observed ad the epected value s PM LM = e (sa) ths dfferece s called error at ( = ) clearl some of the error e, e, e,..., e,..., e wll be postve ad other egatve. To make all errors postve we square each of the errors.e. S= e + e + e +... + e +... + e the curve of best ft s that for whch e's are as small as possble.e. S, the sum of the square of the errors s a mmum ths s kow as the prcple of least square. The theoretcal values for,,..., ma be λ, λ,... λ. 5. FITTING OF STRAIGHT LINE Let a straght le = a+ b...() whch s ftted to the gve data pots (, ), (, ), (, ),, (, ). Let λ be the theoretcal value for the e = λ e = ( a+ b ) e = ( a b ) Now we have S = e + e + e +... + e = = S = e S = ( a b ) B the prcple of least squares, the value of S s mmum therefore, S = a...()

84 A TEXTBOOK OF ENGINEERING MATHEMATICS III ad S =...() b O solvg equatos () ad (), ad droppg the suff, we have = a+ b...(4) = + a b...(5) The equato () ad (4) are kow as ormal equatos. O solvg equatos () ad (4), we get the value of a ad b. Puttg the value of a ad b equato (), we get the equato of the le of best ft. 5.4 FITTING OF PARABOLA Let a parabola = a+ b+ c...() whch s ftted to a gve data (, ), (, ), (, ),, (, ). Let λ be the theoretcal value for the e = e = ( a+ b + c ) e = ( a b c ) Now we have = ( ) = S = e S = a b c B the prcple of least squares, the value of S s mmum, therefore S =, S = ad S = a b c Solvg equato () ad droppg suff, we have λ...() = a+ b+ c...() = a+ b + c...(4) = a + b + c 4...(5) The equato (), (4) ad (5) are kow as ormal equatos. O solvg equatos (), (4) ad (5), we get the values of a,b ad c. Puttg the values of a, b ad c equato (), we get the equato of the parabola of best ft. 5.5 CHANGE OF SCALE Whe the magtude of the varable the gve data s large umber the calculato becomes ver much tedous the problem s further smplfed b takg sutable scale whe the value of are gve at equall spaced tervals.

CURVE FITTING AND SOLUTION OF EQUATION 85 Let h be the wdth of the terval at whch the values of are gve ad let the org of ad be take at the pot, respectvel, the puttg ( ) u = ad v= h If m s odd the, u = But f m s eve the, u = (mddleterm) terval( h) (mddleof two mddleterm) (terval) Eample : Fd the best-ft values of a ad b so that = a+ b fts the data gve the table. : 4 :.8. 4.5 6. Sol. Let the straght le s = a+ b...().8.8. 6.6 4 4.5.5 9 4 6. 5. 6 = = 6.9 = 47. = Normal equatos are, = a+ b...() = a+ b...() Here = 5, =, = 6.9, = 47., = Puttg these values ormal equatos, we get 6.9 = 5a+ b 47. = a+ b O solvg these two equatos, we get So requred le =.7 +.. As. a =.7, b =..

86 A TEXTBOOK OF ENGINEERING MATHEMATICS III Eample : Ft a straght le to the gve data regardg as the depedet varable. 4 5 6 9 6 5 Sol. Let the straght le obtaed from the gve data b = a+ b...() The the ormal equatos are = a+ b...() = a + b...() 9 4 8 6 9 8 4 6 8 5 5 55 6 5 6 9 645 = = 6 = = Puttg all values the equatos () ad (), we get 6 = 6a+ b 645 = a+ 9b Solvg these equatos, we get a = 6.97 ad b = 4.4 Hece the ftted equato s = 6.97 4.4. As. Eample : Ft a straght le to the followg data: Sol. Here we from the followg table: 7 68 7 69 67 65 66 67 69 7 7 7 68 67 68 64 7 69 4899 54 68 7 4896 464 7 7 5 59 69 7 48 476 67 68 4556 4489 65 67 455 45 66 68 4488 456 67 64 488 4489 S = 546 S = 548 S = 74 S = 74

CURVE FITTING AND SOLUTION OF EQUATION 87 Let the equato of straght le to be ftted be = a + b...() Ad the ormal equatos are Σ = a + bσ...() Σ = aσ + bσ...() 8a + 546b = 548 546a + 74b = 74 Solvg these equatos, we get a = 9.5454, b =.44 Hece from () = 9.5454 +.44. As. Eample 4: Fd the least square polomal appromato of degree two to the data. 4 4 4 Also compute the least error. Sol. Let the equato of the polomal be = a+ b+ c...() 4 4 4 8 4 6 8 6 9 99 7 8 4 8 6 64 56 = = = = = 44 = 4 = 54 The ormal equatos are, = a+ b+ c...() = a+ b + c...() = a + b + c 4...(4) Here = 5, =, =, =, =, = 44, =, 4 = 54. Puttg all these values (), () ad (4), we get = 5a+ b+ c...(5) = a+ b+ c...(6) 44 = a+ b+ 54c...(7)

88 A TEXTBOOK OF ENGINEERING MATHEMATICS III O solvg these equatos, we get a = 4, b =, c =. Therefore requred polomal s = 4+ +, errors =. As. Eample 5: Ft a secod degree curve of regresso of o to the followg data: 4 6 8 7 Sol. We form the followg table: 4 6 6 6 4 8 6 44 8 9 7 8 54 6 4 7 6 64 56 8 4 S = S = 6 S = S = S 4 = 54 S = 9 S = 644 The equato of secod degree parabola s gve b = a + b + c...() Ad the ormal equatos are Σ = a + bσ + cσ...() Σ Σ = aσ + bσ + cσ...() 4 = aσ + bσ + cσ...(4) Þ 4a+ b+ c= 6 a+ b+ c= 9 a=, b=, c= a+ b+ 54c= 644 Hece = + +. As. Eample 6: B the method of least squares, fd the straght le that best fts the followg data: 4 5 6 7 4 7 4 55 68 77 85 Sol. The equato of le s = a + b...() The ormal equatos are Σ = a+ bσ...() ad Σ = aσ + bσ...()

CURVE FITTING AND SOLUTION OF EQUATION 89 Now we from the followg table: 4 4 7 54 4 4 9 4 55 6 5 68 4 5 6 77 46 6 7 85 595 49 S = 8 S = 56 S = 85 S = 4 \ From equatos () ad (), we get 7a+ 8b = 56 8a+ 4b = 85 O solvg these equatos, we get a =.574 b =.67 \ =.574 +.67. As. Eample 7: Fd the least squares ft of the form = a + a to the followg data 5 Sol. We have = a + a B prcple of least squares s a s { } = ( a + a { } = ( a + a ( ) = Σ 4 = a Σ + a Σ (Drop suff)...()

9 A TEXTBOOK OF ENGINEERING MATHEMATICS III ad s a { a a } = ( + ) = Σ = a + aσ (Drop suff)...() Now, we form the followg table: 4 5 4 6 S = S = S = 6 S = 5 S 4 = 8 From equatos () ad (), we get 6a + 8a = 5...() ad 4a + 6a =...(4) Solvg the equatos () ad (4), we get a =., a = 4.66 \ The equato s gve b = 4.66.. As. Eample 8: Ft a secod-degree parabola to the followg data takg as the depedet varable. 4 5 6 7 8 9 6 7 8 9 are: Sol. The equato of secod-degree parabola s gve b = a+ b+ c ad the ormal equatos = a+ b + c = a + b + c 4 = a + b + c...() Here = 9. The varous sums are appearg the table as follows:

CURVE FITTING AND SOLUTION OF EQUATION 9 4 6 4 4 8 6 7 9 6 7 8 4 8 6 8 64 56 5 5 5 5 5 65 6 66 6 96 6 96 7 77 49 59 4 4 8 8 64 64 5 496 9 9 8 8 79 79 656 4 = 45 = 74 = 4 = 84 = 77 = 5 = 5 Puttg these values of,,,,, ad 4 equato () ad solvg the equatos for a, b ad c, we get a =.9 ; b =.5 ; c =.67. Hece the ftted equato s =.9 +.5.67. As. Eample 9: Show that the le of ft to the followg data s gve b =.7 +.8. 5 5 5 5 7 4 Sol. Here = 6 (eve) Let =.5, h = 5, = (sa) The,.5 u = ad v=, we get.5 u v uv u 5 8 4 5 5 5 5 5 9 7 5 4 4 9 5 5 5 5 u= v= uv= u = 7 The ormal equatos are, = 6a+ b a = = a+ 7b b =.74

9 A TEXTBOOK OF ENGINEERING MATHEMATICS III or Thus le of ft s v=.74 u..5 = (.74) =.69 8.75.5 or =.7 +.85. As. Eample : Ft a secod-degree parabola to the followg data b least squares method. 99 9 9 9 9 94 95 96 97 5 56 57 58 6 6 6 6 59 ( Sol. Takg = 9, = 57 ) the u = h Here h = Takg u = ad v=, therefore, u = 9 ad v= 57 4 u = 9 v= 57 uv u u v u u 99 4 5 5 6 8 64 56 9 56 9 9 7 8 9 57 4 8 6 9 58 9 6 94 6 4 4 4 95 6 4 8 4 6 8 6 96 6 9 9 7 7 8 97 4 59 8 6 64 56 Total u= v= uv = 5 u = 6 u v = 9 u = 4 u = 78 The the equato = a+ b+ c s trasformed to v= A+ Bu+ Cu...() Normal equatos are: v = 9A+ Bu+ Cu = 9A+ 6C uv = A u + B u + C u B = 7 / = + + 4 694 O solvg these equatos, we get A = =, v= +.85u.7u u v A u B u C u 9= 6A+ 78C 7 B = =.85 ad 57 = +.85( 9).7( 9) = 5.8 + 44.69.7. As. 47 C = =.7 94

CURVE FITTING AND SOLUTION OF EQUATION 9 Eample : Ft secod degree parabola to the followg: 4.8..5 6. Sol. Here = 5 (odd) therefore =, h =, = (sa) Now let u =, v= ad the curve of ft be v= a+ bu+ cu. 4 u v uv u u v u u 4 4 8 6.8.8.8.8...5.5.5.5 4 6. 6..6 4 5. 8 6 Total.9..5 4 Hece the ormal equatos are, = 5 + + uv= au+ bu + cu uv = au + bu + cu 4 O puttg the values of u,v etc. from the table these, we get.9 = 5a+ c,. = b,.5 = a+ 4 c. O solvg these equatos, we get a =.48, b =. ad c =.55 Therefore the requred equato s v=.48 +.u+.55 u. Aga substtutg u = ad v =, we get =.48 +.( ) +.55( ) or =.4.7+.55. As. 5.6 FITTING OF AN EXPONENTIAL CURVE Suppose a epoetal curve of the form = ae Takg logarthm o both the sdes, we get log = log a+ blog e b.e., Y = A+ B...() where Y = log, A= log a ad B= blog e.

94 A TEXTBOOK OF ENGINEERING MATHEMATICS III The ormal equatos for () are, Y = A+ B Y = A + B O solvg the above two equatos, we get A ad B B the a = atlog A, b = log e 5.7 FITTING OF THE CURVE = a + b Error of estmate for th pot (, ) s e = ( a b ) We have, S= e = = ( a b ) = B the prcple of least square, the value of S s mmum S = ad S = a b S Now = a = ( a b )( ) = or ad or = + = = = a b...() S = b a b = = + 4 = = = ( )( ) = a b...() Droppg the suff from () ad (), the the ormal equatos are, = a + b = a + b 4

CURVE FITTING AND SOLUTION OF EQUATION 95 5.8 FITTING OF THE CURVE b = a + Error of estmate for th pot (, ) s b e = ( a ) We have, S = e = b = ( a ) = B the prcple of least square, the value of S s mmum S = ad S = a b S Now = a = b ( a )( ) = or ad = a + b...() = = S = b = b ( a )( ) = or = a + b = = Droppg the suff from () ad (), the the ormal equatos are, = b + a = a + b where s the umber of par of values of ad....()

96 A TEXTBOOK OF ENGINEERING MATHEMATICS III 5.9 FITTING OF THE CURVE c = +c Error of estmate for th pot (, ) s We have, S= e c e = ( c ) = = c = ( c ) B the prcple of least square, the value of S s mmum S = ad S = c c Now or ad S = c ( c )( ) = c = = c + c = = = S = c c ( )( ) = = c = c + c or = = = Droppg the suff from () ad (), the the ormal equatos are, = c + c c c. = +...()...() Eample : Fd the curve of best ft of the tpe = ae to the followg data b the method of least squares: b 5 7 9 5 5 Sol. The curve to be ftted s b = ae or Y = A+ B, where Y = log, A= log a, ad B= blog e.

CURVE FITTING AND SOLUTION OF EQUATION 97 Therefore the ormal equatos are: Y = 5A+ B Y = A + B Y = log Y. 5 5.76 5 5.885 7.79 49 7.5544 9 5.76 8.5849. 44 5.8664 = 4 Y = 5.756 = Y = 4.886 Substtutg the values of, etc. ad calculated b meas of above table the ormal equatos, we get 5.756 = 5A+ 4B ad 4.886 = 4A+ B O solvg these equatos, we obta, A =.9766 ; B =.56 B Therefore a= atlog A = 9.4754 ; b = =.59 log e.59 Hece the requred curve s = 9.4754e. As. Eample : For the data gve below, fd the equato to the best fttg epoetal curve of the form b = ae. Sol. = ae b 4 5 6.6 4.5.8 4. 5 O takg log both the sdes, log = log a+ bloge whch s of the form Y = A+ B, where Y = log, A= log aad B= blog e. Y = log Y.6.4.4 4.5.65 4.64.8.99 9.497 4 4..64 6 6.468 5 5.969 5.4845 6.477 6 4.866 = Y = 8.754 = 9 Y = 6.694

98 A TEXTBOOK OF ENGINEERING MATHEMATICS III Normal equatos are: Y = 6A+ B Y = A + B Therefore from these equatos, we have 8.754 = 6A+ B 6.694 = A+ 9B A=.54, B=.467 Therefore, a= atloga= atlog(.54) = atlog(.7466) =.558. B.467 ad b = = =.6 loge.44.6 Hece requred equato s =.558e. As. Eample 4: Gve the followg epermetal values: 4 5 ad Ft b the method of least squares a parabola of the tpe = a+ b. Sol. Error of estmate for th pot (, ) s e = ( a b ) B the prcple of least squares, the values of a ad b are such that ( ) = = s mmum. S = e = a b Therefore ormal equatos are gve b S = = a + b a S = = a + b b 4...()...() Total Here = 4. 4 4 4 4 4 6 5 9 5 8 4 = = 4 = 79 = 98

CURVE FITTING AND SOLUTION OF EQUATION 99 From () ad (), = 4a+ 4b ad 79 = 4a+ 98b Solvg for a ad b, we get a =.7 ad b =.44 Hece the requred curve s =.7 +.44. As. Eample 5: B the method of least square, fd the curve = a+ b that best fts the followg data: 4 5.8 5. 8.9 4. 9.8 Sol. Error of estmate for th pot (, ) s e = ( a b ) B the prcple of least squares, the values of a ad b are such that ( ) = = s mmum. S = e = a b Therefore ormal equatos are gve b S a b a = = = = = + ad Droppg the suff, ormal equatos are S 4 a b b = = = = = + = a + b...() 4 ad = a + b...() 4.8.8.8 5. 4 8 6..4 8.9 9 7 8 6.7 8. 4 4. 6 64 56 56.4 5.6 5 9.8 5 5 65 99 495 Total = 55 = 5 4 = 979 = 94. = 8.9 Substtutg these values equatos () ad (), we get 8.85 a =.5 55 7.4 ad b =.49 664 94. = 55a+ 5b ad 8.9 = 5a+ 979b Hece requred parabolc curve s =.5+.49. As.

4 A TEXTBOOK OF ENGINEERING MATHEMATICS III Eample 6: Ft a epoetal curve of the form = ab to the followg data: 4 5 6 7 8...8.5.6 4.7 6.6 9. Sol. = ab takes the form Y = A+ B, where Y = log ; A= log a ad B= log b. Hece the ormal equatos are gve b Y = A+ B ad Y = A +. Y = log Y.....79.584 4.8.55.7659 9 4.5.979.596 6 5.6.556.785 5 6 4.7.67 4.6 6 7 6.6.895 5.765 49 8 9..959 7.67 64 = 6 =.5 Y =.79 Y =.785 = 4 Puttg the values the ormal equatos, we obta.79 = 8A+ 6B ad.785 = 6A+ 4B B =.47 ad A =.656 b= atlogb =.8 ad a= atlog A=.68. Thus the requred curve of best ft s = (.68)(.8). As. Eample 7: Ft a curve = ab to the followg data: 4 5 6 44 7.8 7.4 48.8 98.5 form. Sol. Gve equato = ab reduces to Y = A+ B where Y = log, A= log a ad B= log b. The ormal equatos are, log = log a+ log b log = log a + log b The calculatos of, log, ad log are substtute the followg tabular

CURVE FITTING AND SOLUTION OF EQUATION 4 log log 44 4.584 4.68 7.8 9.75 6.75 4 7.4 6.68 9.67 5 48.8 5.959.9795 6 98.5 6.4749 4.8494 9.585 47.54 Puttg these values the ormal equatos, we have.585 = 5loga+ logb 47.54 = loga+ 9 log b. Solvg these equatos ad takg atlog, we have a =, b =. appromate. Therefore equato of the curve s = (.). As. Eample 8: Derve the least square equatos for fttg a curve of the tpe = a + ( b/ ) to a set of pots. Hece ft a curve of ths tpe to the data. 4.5.99.88 7.66 Sol. Let the pots are gve b (, ), (, ), (, ),, (, ). The error of estmate for the th pot (, ) s e = [ a ( b/ )]. B the prcple of least square, the values of a ad b are such so that the sum of the square of error S, vz., or b S = e = a s mmum. = Therefore the ormal equatos are gve b S S =, = a b = 4 a + b ad = a + b = = = These are the requred least square equatos. = = = 4.5.5.5.99 4 6.5.5.96.495.88 9 8.. 4.9.9 4 7.66 6 56.5.65.56.94 54.46 59.9.9

4 A TEXTBOOK OF ENGINEERING MATHEMATICS III Puttg the values the above least square equatos, we get 59.9 = 54a+ b ad.9 = a+.46 b. Solvg these, we get a =.59 ad b =.4..4 Therefore, the equato of the curve ftted to the above data s =.59. As. Eample 9: Ft the curve pv γ = k to the followg data: Sol. Gve pv γ = Takg log both the sdes, we get whch s of the form Y = A+ BX p(kg/cm ).5.5.5 v (ltres) 6 75 6 5 46 k / γ k v= = k p p / γ / γ logv= logk log p γ γ where Y = log v, X = log p, A = logk ad B = γ. γ p v X Y XY X.5 6..95.9666.96.5 75.769.8756.567. 6..799.8459.96.5 5.9794.76.88.586 46.477.6676.746.764 Total X =.55 Y = 7.557 XY =.796 X =.5985 Here = 6 Normal equatos are, 7.557 = 6A+.55B.796 =.55A+.5985B O solvg these, we get A =.999 ad B =.798

CURVE FITTING AND SOLUTION OF EQUATION 4 γ= = =.45 B.798 Aga logk =γ A= 4.669 k = atlog(4.669) = 846.48 Hece, the requred curve s.45 pv = 846.48. As. Eample : For the data gve below, fd the equato to the best fttg epoetal curve of the form b = ae. 4 5 6.6 4.5.8 4. 5 Sol. Gve = ae b, takg log we get log = log a+ blog e whch s of the Y = A+ B, where Y = log, A= loga ad B= log e. Put the values the followg tabular form, also trasfer the org of seres to, so that u=. log = Y = u uy u.6.4.48 4 4.5.65.65.8.4 4 4..64.64 5 5..94 4.94 4 6.477 7.4 9 Total 8.75.68 9 I case Y = A+ Bu., the ormal equatos are gve b Y = A+ B u 8.75 = 6A+ B...() uy A u B u = +.68 A 9 Solvg () ad (), we get A =. ad B =.46 Thus equato s Y =. +.46 u,.e. Y =. +.46( ) or Y =.46.5 Whch gves log a =.5.e. atlog(.5) = atlog(.75) =.557 B.46 b = = =.6 log e.44 = + B...() Hece, the requred equato of the curve s.6 = (.557) e. As.

44 A TEXTBOOK OF ENGINEERING MATHEMATICS III PROBLEM SET 5.. Ft a straght le to the gve data regardg as the depedet varable: 4 6 8.4..5 4. 5. 6. [ As. =.5+.5 ]. Ft a straght le = a+ b to the followg data b the method of least square: 6 8 5 4 [ As..6 +.8 ]. Fd the least square appromato of the form = a+ b for the followg data:....4.5..99.85.8.75 [ As. =..8 ] 4. Ft a secod degree parabola to the followg data:.... 6. 7. [ As. = + + ] 5. Ft a secod degree parabola to the followg data:..5..5..5 4....6..7.4 4. [ As. =.4.9+.4 ] 6. Ft a secod degree parabola to the followg data b the least square method: 4 5 9 9 65 95 [ As. = 7.5 + 4.5+ 4] 7. Ft a parabola = a+ b+ c to the followg data: 4 6 8.7.85.47 57.8 9.9 [ As. =.4.78+.99 ] 8. Determe the costats a ad b b the method of least squares such that b = ae fts the followg data: 4 6 8.5 4.77.84.8 8.897.6 [ As. =.49989 e ] 9. Ft a least square geometrc curve b = a to the followg data: 4 5.9977.5 4.5 8.5 [ As. =.5 ]

CURVE FITTING AND SOLUTION OF EQUATION 45. A perso rus the same race track for fve cosecutve das ad s tmed as follows: Da( ) 4 5 Tme( ) 5. 5. 5 4.5 4 b c Make a least square ft to the above data usg a fucto a + +. As. 6.75 4.478 =.65 + + c. Use the method of least squares to ft the curve = + c to the followg table of values:...4.5.977.88 7 6 5 6 As. = +. Usg the method of least square to ft a parabola = a+ b+ c the followg data: (, ) : (,),(,),(,),(,) As. = +. The pressure of the gas correspodg to varous volumes V s measured, gve b the followg data: V(cm ) 5 6 7 9 p(kgcm ) 64.7 5. 4.5 5.9 78 Ft the data to the equato pv γ = c. [As..8997 pv = 67.78765 ] 4. Emplo the method of least squares to ft a parabola = a + b + c the followg data: (, ): (, ), (, ), (, ), (, ) [As. =.5 +.5 ] 5. Ft a secod degree parabola the followg data: [U.T.U. 8].... 4.. 4.. 7.. [As. = + + ] 6. Ft at least square quadratc curve to the followg data: 4.7.8.., estmate (.4) [As. =.5 +. ad (.4) =.95]

46 A TEXTBOOK OF ENGINEERING MATHEMATICS III 7. Ft a epoetal curve b least squares 5 4 5 98. 9.7 8. 64. 6.4.6 7.. Estmate whe = 5. [As. = (.96), (5) =.9] b 8. Ft the curve = a+ to the followg data 4.5 6 7.5 Estmate whe =.5. As..7 =. +, (.5) = 4.65 5. POLYNOMIAL If ( )... f = a + a + a + + a + a The the above relato s called the polomal of th order. 5.. Degree of Polomal The hghest power of occurrg the gve polomal s called degree of polomal. The costat c= c s called a polomal of degree zero. The polomal f( ) = a+ b, a s of degree oe ad s called a lear polomal. The polomal f ( ) = a + b + c, a s of degree two ad s called a quadratc polomal. The polomal f ( ) = a + b + c + d, a s of degree three ad s called a cubc polomal. 4 The polomal f ( ) = a + b + c + d + e, a s of degree four ad s called bquadratc polomal. 5. DESCARTE S RULE OF SIGNS The umber of postve roots of the equato f () = caot eceed the umber of chages of sg f (), ad the umber of egatve roots caot eceed the umber of chages of sg of f ( ). Estece of magar roots: If a equato of the th degree has at most p postve roots ad at most q egatve roots, the t has at least ( p+ q) magar roots. Eample : Appl Descarte s Rule of sgs to dscuss the ature of the roots of the equato 4 + 5 + 7 =. Sol. The gve equato s f( ) = 4 + 5 + 7 = Sgs of f( ) are + + + [from + to ]

CURVE FITTING AND SOLUTION OF EQUATION 47 It has oe chage of sg ad hece t must have oe +ve root. 4 f( ) = ( ) + 5( ) + 7( ) = or 4 f( ) = + 5 7 = Sgs of f( ) are + + [from + to ] It has ol oe chage sg ad hece t must have oe ve root. Thus the equato has two real roots, oe +ve ad oe ve ad hece the other two roots must be magar. 7 4 Eample : Show that + = has at least four magar roots. Sol. The gve equato s f( ) = 7 4 + = [from + to or to +] Sgs of f( ) + + f( ) = has chages of sg. Therefore, t caot have more tha three postve roots. Also 7 4 f( ) = ( ) ( ) + ( ) = or f( ) = 7 4 = Sgs of f( ) are f( ) = has o chages sg. Therefore the gve equato has o egatve root. Thus the gve equato caot have more tha + = real roots. But the gve equato has 7 roots. Hece the gve equato has 7 = 4 magar roots. Eample : Fd the least postve umber of magar roots of the equato 9 5 4 f( ) = + + + = Sol. The gve equato s f( ) = 9 5 + 4 + + = Sgs of f( ) are + + + + f( ) = has two chages of sgs, ad hece s the ma. umber of +ve root. 9 5 4 f( ) = ( ) ( ) + ( ) + ( ) + = or f( ) = 9 + 5 + 4 + + = Sgs of f( ) are + + + + f( ) = has ol oe chages of sg ad hece t has ol oe ve root or f( ) = has ol oe ve root. Thus the ma. umber of real roots s + = ad the equato beg of 9th degree ad t wll have at least 9 = 6 magar roots.

48 A TEXTBOOK OF ENGINEERING MATHEMATICS III 5. CARDON S METHOD Case. Whe Cubc s of the Form +q+r= The gve cubc s + q+ r =...() Let = u + v be a root of () Cubg, = u + v + uv( u+ v) = u + v + uv or uv ( u + v ) =...() Comparg () ad (), uv = q or uv = q 7 ad u + v = r u ad v are the roots of the equato t ( u + v ) t+ u v = or t + rt q =...() 7 Solvg (), Let q r± r + t = 7 u 4 4q 4q r+ r + r r + = 7 ; v = 7 Now, the three cube roots of u are uu, ω, uω ad those of v are vv, ω, vω, where ωad ω are magar cube root of ut. Sce = u+ v To fd, we have to add a cube root of u ad a cube of v such a maer that ther product s real. The three values of are u+ v, uω+ vω, uω + vω ( ω = ) Eample 4: Use Cardo s method to solve Sol. Let = u+ v 7+ 54 =. Cubg, = ( u+ v) = u + v + uv( u+ v) = u + v + uv uv ( u + v ) = Comparg wth the gve equato,we get uv = 9 u v = 79 (o cubg) ad u + v = 54

CURVE FITTING AND SOLUTION OF EQUATION 49 \ u ad v are the root of t ( u + v ) t+ u v = t + 54t+ 79 = ( t+ 7) = t = 7, 7 Let u = 7 ad v = 7 So that, u=, ω ω, ad v=, ω ω, To fd, we have to add a cube root of u ad a cube root of v such a wa that ther product s real. \ = ( ),( ω ω ),( ω ω ) ( Qω = ) = 6, ( ω+ω ), ( ω +ω ) = 6,, ( Q+ω+ω = ) Hece the requred roots are 6,,. As. Eample 5: Solve b Cardo s method 5 6 =. Sol. Let = u+ v Cubg, = ( u+ v) = u + v + uv( u+ v) = u + v + uv uv ( u + v ) = Comparg wth the gve equato, we get uv = 5 u v = 5 (o cubg) ad u + v = 6 u ad v are the roots of t ( u + v ) t + u v = t 6t + 5 = ( t )( t 5) = t =,5 Let u = ad v = 5 So that, u=, ω, ω ad v= 5, 5 ω, 5ω To fd, we have to add a cube root of u ad a cube root of v such a wa that ther product s real. \ 5, 5, 5 = + ω+ ω ω + ω ( Q ω =) + + = 6, + 5, + 5 = 6,, + = 6, ± Hece, the requred roots are 6, ±. As.

4 A TEXTBOOK OF ENGINEERING MATHEMATICS III Eample 6: Solve 6 9= b Cardo s method. Sol. Let = u+ v Cubg, = ( u+ v) = u + v + uv( u+ v) uv ( u + v ) = Comparg, we get = u + v + uv uv = u v = 8 (o cubg) ad u + v = 9 u, v are the roots of t ( u + v ) t+ u v = t 9t+ 8= ( t )( t 8) = t =,8 Let u = ad v = 8 so that u=, ω, ω ad v=, ω,ω To fd, we have to add a cube root of u ad a cube root of v such that ther product s real. = +, ω+ ω, ω + ω ( Qω = ) + + =, +, + + ± =,, =,. As. Eample 7: Solve the cubc equato 8 5 = Sol. Let = u+ v Cubg, = ( u+ v) = u + v + uv( u+ v) uv ( u + v ) = = u + v + uv Comparg, we get uv = 6 u v = 6 (o cubg) ad u + v = 5 u ad v are the roots of t ( u + v ) t + u v = t 5t+ 6 = ( t 8)( t 7) = t = 8, 7 Let u = 8& v = 7 so that u =, ω,ω ad v=, ω,ω

CURVE FITTING AND SOLUTION OF EQUATION 4 To fd, we have to add a cube root of u ad a cube root of v such that ther product s real. = +, ω+ ω, ω + ω ( Q ω = ) + + = 5, +, + 5 5+ = 5,, 5± = 5,. As. Case. Whe the Cubc Equato s of the Form a +a +a +a = The, frst of all we remove the term cotag. Ths s doe b dmshg the roots of the gve equato b a a or Sum of root/no. of roots. Where s. We proceed wth the help of followg eamples: Eample 8: Solve b Cardo s method + 6 + 9+ 4= Sol. Equatog + 6 + 9 + 4=...() Equatg wth a + a + a+ a = we get...(), a =, a = 6 a 6 The h = = = a h =. Now remove the terms,we have 6 9 4 8 4 () 4 ( ) () The trasformed equato s + =...() where = +. Let = u+ vbe a soluto of (), the we get uv ( u + v ) =...(4) Equatg () ad (4), we get uv = uv = ad u + v =

4 A TEXTBOOK OF ENGINEERING MATHEMATICS III Let us cosder a equato whose roots are u ad v t st+ p = s = sum of roots, p = product of roots t + t+ = ( t+ ) = t =, Let u = ad v = so that u =, ω, ω ad v=, ω, ω \ But = = u+ v= ( ),( ω ω ),( ω ω) =,, ( Q+ω+ω = ) =,, = 4,, = 4,,. As. Eample 9: Solve b Cardo s method 5 + 847 =. Sol. The gve equato 5 + 847 =...(), Equatg wth a + a + a+ a = we get a =, a = 5 a 5 \ h = = = 5 a Now remove the term, usg sthetc dvso, we have 5 5 847 5 5 45 8 4 5 5 5 ( 8) 5 ( ) ( ) Trasformed equato s 8 + 4 =...() where = 5. Let = u+ v \ uv ( u + v ) =...() (o cubg)

CURVE FITTING AND SOLUTION OF EQUATION 4 Comparg () ad (), u ad 6 uv = 6 u v = (6) ad u + v = 4 6 v are the roots of t + 4 t+ (6) = 6 t +. (6) t+ (6) = ( t+ 6) = t= 6, 6 Let u = 6 ad v = 6 so that u= 6, ω ω 6, 6 ad v = 6, ω ω 6, 6 = + = ( 6 6),( 6ω 6 ω ),( 6ω 6 ω) u v =, 6, 6 Qω+ω + = ω+ω = But = + 5= + 5,6+ 5,6+ 5 \ = 7,,. As. Eample : Solve Sol. Here + + 6= a a =, a = h= = = a Now remove the term, usg sthetc dvso 6 (6) (9) () The trasformed equato s ( ) + 9( ) + 6 = Let =, the + 9 + 6 =...() Let = u + v...() uv( ) ( u + v ) = Comparg () ad (), we get u v = 7 uv = ad u + v = 6

44 A TEXTBOOK OF ENGINEERING MATHEMATICS III Now let us have a equato whose roots are u ad v t ( 6) t+ ( 7) ( t+ 7)( t ) = = t + 6t 7= So that t u =, 7 e.., u = ad v = 7 =, ω, ω ad v =, ω, ω = u+ v \ = + = ( ), ( ω ω ), ( ω ω) + + =,, { } { } =, ( + ), ( ) = ( + ), ( + + ), ( + ) =, ( + ), ( ) =, ( ± ). As. Eample : Solve 6 + 6 5= b Cardo s method. Sol. The gve equato 6 + 6 5=...(), Comparg wth a + a + a+ a = we get a =, a = 6 a 6 \ h = = = a Now remove the term, usg sthetc dvso 6 6 5 8 4 4 ( 9) 4 ( 6) () Trasformed equato s 6 9=...()

CURVE FITTING AND SOLUTION OF EQUATION 45 where = Let = u + v be the soluto of (), the = u + v + uv( u+ v) = u + v + uv( ) uv( ) ( u + v ) =...() Comparg () ad (), we have uv = 6 uv = u v = 8 ad u + v = 9 Now let us have a equato whose roots are u ad v t st+ p = ; s = sum of roots, p = product of the roots. t 9t+ 8= \ = u+ v But = + t =,8.e., u = ad v = 8 so that u =, ω, ω ad v=, ω,ω = ( + ),( ω+ ω ),( ω + ω) + + =, +, + + =,,. + = 5,, = 5, ( ± ). As. Eample : Solve the equato + = b Cardo s method. Sol. Here, a =, a = a \ h = = = a

46 A TEXTBOOK OF ENGINEERING MATHEMATICS III Now remove the term, usg sthetc dvso, we have () () () Trasformed equato s =...() where =. From (), =,, = + =,, Hece the requred roots are,,. As. Eample : Solve the cubc equato + 6 + =. Sol. Gve equato s + 6+ = Here, a =, a = a \ h = = a Now remove the term, usg sthetc dvso, we have / 6 / /9 46 7 / 46 686 9 7 / /9 / 47 9 / ( ) 47 686 Trasformed equato s + =...() 9 7 where = +. Let = u+ v

CURVE FITTING AND SOLUTION OF EQUATION 47 Cubg, = u + v + uv uv ( u + v ) =...() Comparg wth (), we get u ad v are the roots of 49 7 686 uv = u v = ad u + v = 9 7 686 7 t + t+ = 7 6 7 7 t + t+ = 7 7 7 t+ = t =, 6 Let u 7 7 = ad v = So that 7 7 7 7 7 7 u=, ω, ω ad v=, ω, ω 7 7 7 7 7 7 4 7 7 = u+ v =,,,, ω ω ω ω = Now, 4 7 7 = =,, = 5,, Hece requred roots are 5,,. As. Eample 4: Solve the cubc equato + 6 + =. Sol. Gve equato s + 6 + = Here, a =, a = 6 \ a 6 h = = = a Now remove the term, usg sthetc dvso method, we have 6 8 4 4 (7) 4 ( 4) () 6

48 A TEXTBOOK OF ENGINEERING MATHEMATICS III Trasformed equato s 4 + 7 =...() where = + Let = u+ v Cubg, = u + v + uv uv ( u + v ) =...() Comparg () ad (), we get uv = 8 u v = 5 ad u + v = 7 u ad v are the roots of t + 7t+ 5 = t = 8, 64 Let u = 8ad v = 64 So that u =, ω ω, & v = 4, ω ω 4, 4 = + = ( 4),( ω 4 ω ),( ω 4 ω) u v = 6, +, Now, = = 8,+, Hece the requred roots are 8, ±. As. Eample 5: Solve + a +(a bc) + a + b + c abc = b Cardo s method. Sol. Gve equato s + a + (a bc) + a + b + c abc = Here a =, a = a a a \ h = = = a a.. Now remove the term, usg sthetc dvso, we have a a ( a bc) a + b + c abc a a a + abc a a bc ( b + c ) a a a bc a ()