Curve Fitting

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UNIT V Curve Fitting and Solution of Equation 5.1 CURVE FITTING In many branches of applied mathematics and engineering sciences we come across experiments and problems, which involve two variables. For example, it is known that the speed v of a ship varies with the horsepower p of an engine according to the formula = + 3 . p a bv Here a and b are the constants to be determined. For this purpose we take several sets of readings of speeds and the corresponding horsepowers. The problem is to find the best values for a and b using the observed values of v and . p Thus, the general problem is to find a suitable relation or law that may exist between the variables x and y from a given set of observed values = ( , ), 1, 2, ..........,. i i x y i n Such a relation connecting x and y is known as empirical law. For above example, x = v and y = p. The process of finding the equation of the curve of best fit, which may be most suitable for predicting the unknown values, is known as curve fitting. Therefore, curve fitting means an exact relationship between two variables by algebraic equations. There are following methods for fitting a curve. I. Graphic method II. Method of group averages III. Method of moments IV. Principle of least square. Out of above four methods, we will only discuss and study here principle of least square. 5.2 PRINCIPLE OF LEAST SQUARES The graphical method has the drawback in that the straight line drawn may not be unique but principle of least squares provides a unique set of values to the constants and hence suggests a curve of best fit to the given data. The method of least square is probably the most systematic procedure to fit a unique curve through the given data points.

Transcript of Curve Fitting

UNIT VCurve Fitting and Solution of Equation5.1 CURVEFITTINGIn many branches of applied mathematics and engineering sciences we come across experiments andproblems, which involve two variables. For example, it is known that the speedvof a ship varies withthe horsepower p of an engine according to the formula+3. p a bvHereaandbare the constants tobedetermined.Forthispurposewetakeseveralsetsofreadingsofspeedsandthecorrespondinghorsepowers. The problem is to find the best values foraand b using the observed values ofvand . pThus, the general problem is to find a suitable relation or law that may exist between the variablesxand y from a given set of observed values( , ), 1, 2,.........., .i ix y i nSuch a relation connectingxand yis known as empirical law. For above example, x = v and y = p.Theprocessoffindingtheequationofthecurveofbestfit,whichmaybemostsuitableforpredictingtheunknownvalues,isknownascurvefitting.Therefore,curvefittingmeansanexactrelationshipbetweentwovariablesbyalgebraicequations.Therearefollowingmethodsforfittingacurve.I. GraphicmethodII. Method of group averagesIII. Method of momentsIV. Principle of least square.Out of above four methods, we will only discuss and study here principle of least square.5.2 PRINCIPLEOFLEASTSQUARESThe graphical method has the drawback in that the straight line drawn may not be unique but principleof least squares provides a unique set of values to the constants and hence suggests a curve of best fit tothegivendata.Themethodofleastsquareisprobablythemostsystematicproceduretofitauniquecurve through the given data points.CURVEFITTINGANDSOLUTIONOFEQUATION 383Let the curve + + +2 1.................my a bx cx kx ...(1)be fitted to the set ofndata points 1 1( , ) x y , 2 2( , ) x y , 3 3( , ) x y ,,( , ).n nx yAt ( )ix xthe observed(or experimental) value of the ordinate is yi = PiMi and the corresponding value on the fitting curve ( ) iis+ + + 2........mi i i i ia bx cx kx L M whichistheexpectedorcalculatedvalue.Thedifferenceoftheobservedandtheexpectedvalueis(say)i i i i iPM L M e thisdifferenceiscallederrorat ( )ix xclearly some of the error 1 2 3, , , ....., , .......,i ne e e e ewill be positive and other negative. To make all errorspositive we square each of the errors i.e. + + + + + +2 2 2 2 21 2 3........ ......i nS e e e e ethe curve of best fit isthat for which e's are as small as possible i.e. S, the sum of the square of the errors is a minimum thisis known as the principle of least square. The theoretical values for x1, x2, ..., xn may be 1 2, ,...... .ny y y5.3 FITTINGOFSTRAIGHTLINELet a straight line + y a bx ...(1)which is fitted to the given data points 1 1( , ) x y , 2 2( , ) x y , 3 3( , ) x y ,,( , )n nx y .Let 1ybe the theoretical value forx1 then 11 1e y y +1 1 1( ) e y a bx 2 21 1 1( ) e y a bxNow we have+ + + +2 2 2 21 2 3............nS e e e e

21niiS e

21( )ni iiS y a bxBy the principle of least squares, the value of S is minimum therefore,

0Sa...(2)384 A TEXTBOOKOFENGINEERINGMATHEMATICSIIIand

0Sb...(3)On solving equations (2) and (3), and dropping the suffix, we have + y na b x ...(4) + 2xy a x b x ...(5)The equation (3) and (4) are known as normal equations.On solving equations (3) and (4), we get the value ofaand. b Putting the value ofaandb inequation (1), we get the equation of the line of best fit.5.4 FITTINGOFPARABOLALet a parabola + +2y a bx cx ...(1)which is fitted to a given data 1 1( , ) x y , 2 2( , ) x y , 3 3( , ) x y ,,( , )n nx y .Let y be the theoretical value for 1xthen 1 1e y y + +21 1 1 1( ) e y a bx cx 2 2 21 1 1 1( ) e y a bx cxNow we have

21niiS e

2 21( )ni i iiS y a bx cxBy the principle of least squares, the value ofSis minimum, therefore

0Sa,

0Sband

0Sc...(2)Solvingequation(2)anddroppingsuffix,wehave + + 2y na b x c x ...(3) + + 2 3xy a x b x c x ...(4) + + 2 2 3 4xy a x b x c x ...(5)The equation (3), (4) and (5) are known as normal equations.On solving equations (3), (4) and (5), we get the values of a,b andc. Putting the values of a, band c in equation (1), we get the equation of the parabola of best fit.5.5 CHANGEOFSCALEWhen the magnitude of the variable in the given data is large number then calculation becomes verymuch tedious then problem is further simplified by taking suitable scale when the value ofxare givenat equally spaced intervals.CURVEFITTINGANDSOLUTIONOFEQUATION 385Let h be the width of the interval at which the values ofxare given and let the origin ofxandy be taken at the point 0 0, x y respectively, then putting

0( ) x xuh and 0v y yIfmis odd then,u = ) ( intervalterm) (middlehx But if m is even then,u = (interval)21term) middle two of (middle xExample 1: Find the best-fit values of a and b so that + y a bxfits the data given in the table.: 0 1 2 3 4: 1 1.8 3.3 4.5 6.3xySol.Let the straight line is + y a bx ...(1) 220 1 0 01 1.8 1.8 12 3.3 6.6 43 4.5 13.5 94 6.3 25.2 1610 16.9 47.1 30x y xy xx y xy xNormalequationsare,+ y na b x ...(2) + 2xy a x b x ...(3)Here 5, n 10, x 16.9, y 47.1, xy 230 xPutting these values in normal equations, we get + 16.9 5 10 a b + 47.1 10 30 a bOn solving these two equations, we get 0.72 a ,1.33. bSorequiredline+ 0.72 1.33 . y xAns.386 A TEXTBOOKOFENGINEERINGMATHEMATICSIIIExample 2: Fit a straight line to the given data regardingxas the independent variable.1 2 3 4 5 61200 900 600 200 110 50xySol. Let the straight line obtained from the given data by + y a bx ...(1)Then the normal equations are+ y na b x ...(2) + 2xy a x b x ...(3)221 1200 1 12002 900 4 18003 600 9 18004 200 16 8005 110 25 5506 50 36 30021 91 6450 3060x y x xyx x xy y Putting all values in the equations (2) and (3), we get + 3060 6 21 a b + 6450 21 91 a bSolving these equations, we get1361.97 aand 243.42 bHence the fitted equation is 1361.97 243.42 . y x Ans.Example 3: Fit a straight line to the following data:71 68 73 69 67 65 66 6769 72 70 70 68 67 68 64xySol.Herewefromthefollowingtable:x y xy x271 69 4899 504168 72 4896 462473 70 5110 532969 70 4830 476167 68 4556 448965 67 4355 422566 68 4488 435667 64 4288 4489Sx = 546 Sy = 548 Sxy = 37422 Sx2 = 37314CURVEFITTINGANDSOLUTIONOFEQUATION 387Let the equation of straight line to be fitted bey = a + bx ...(1)And the normal equations arey an b x+ ...(2)xy 2a x b x + ...(3) 8a + 546b = 548546a + 37314b = 37422Solving these equations, we geta =39.5454,b=0.4242Hencefrom(1)y = 39.5454 + 0.4242x. Ans.Example 4:Find the least square polynomial approximation of degree two to the data. 0 1 2 3 44 1 4 11 20xyAlso compute the least error.Sol. Let the equation of the polynomial be+ +2y a bx cx ...(1)2 2 3 42 2 3 40 4 0 0 0 0 01 1 1 1 1 1 12 4 8 4 16 8 163 11 33 9 99 27 814 20 80 16 320 64 25610 30 120 30 434 100 354x y xy x xy x xx y xy x xy x x

The normal equations are, + + 2y na b x c x ...(2) + + 2 3xy a x b x c x ...(3) + + 2 2 3 4xy a x b x c x ...(4)Heren=5, 10, x 30, y 120, xy 230, x 2434, xy 3100, x

4354. xPutting all these values in (2), (3) and (4), we get + + 30 5 10 30 a b c...(5) + + 120 10 30 100 a b c...(6) + + 434 30 100 354 a b c...(7)388 A TEXTBOOKOFENGINEERINGMATHEMATICSIIIOnsolvingtheseequations,weget 4 a , 2 b ,1. c Thereforerequiredpolynomialis + +24 2 y x x ,errors=0. Ans.Example 5: Fit a second degree curve of regression ofy onx to the following data:1 2 3 46 11 18 27xySol.Weformthefollowingtable:x y x2x3x4xy x2y1 6 1 1 1 6 62 11 4 8 16 22 443 18 9 27 81 54 1624 27 16 64 256 108 432Sx = 10 Sy = 62 Sx2 = 30 Sx3 = 100 Sx4 = 354 Sxy = 190 Sx2y = 644The equation of second degree parabola is given byy = a + bx + cx2...(1)And the normal equations arey 2an b x c x+ + ...(2)xy 2 3a x b x c x + + ...(3)2xy 2 3 4a x b x c x + + ...(4)4 10 30 6210 30 100 190 3, 2, 130 100 354 644a b ca b c a b ca b c+ ++ + ,+ + Hence y = 3 + 2x + x2. Ans.Example 6: By the method of least squares, find the straight line that best fits the following data:1 2 3 4 5 6 714 27 40 55 68 77 85xySol. The equation of line isy = a + bx ...(1)The normal equations arey an b x + ...(2)and xy 2a x b x + ...(3)CURVEFITTINGANDSOLUTIONOFEQUATION 389Nowwefromthefollowingtable:x y xy x21 14 14 12 27 54 43 40 120 94 55 220 165 68 340 256 77 462 367 85 595 49Sx = 28 Sy = 356 Sxy = 1805 Sx2 = 140\ Fromequations(2)and(3),weget7 28 a b += 35628 140 a b += 1805On solving these equations, we geta =3.5714b =13.6071\ y =3.5714+13.6071x. Ans.Example 7: Find the least squares fit of the form 20 1y a a x+to the following data1 0 1 22 5 3 0xySol.We have y20 1a a x+By principle of least squaress 220 1(i iiy a a x +sa 2 20 12 ( ( ) 0i i iiy a a x x + 2xy 2 40 1a x a x + (Dropsuffix) ...(1)390 A TEXTBOOKOFENGINEERINGMATHEMATICSIIIand0sa 20 12 ( ) 0i iy a a x + y 20 1a n a x+ (Dropsuffix) ...(2)Now,weformthefollowingtable:x y x2x2y x41 2 1 2 10 5 0 0 01 3 1 3 12 0 4 0 16Sx = 2 Sy = 10 Sx2 = 6 Sx2y = 5 Sx4 = 18Fromequations(1)and(2),weget0 16 18 a a + 5 ...(3)and0 14 6 a a +10 ...(4)Solving the equations (3) and (4), we geta1=1.111,a0 = 4.166\ The equation is given byy =4.1661.111x2. Ans.Example8:Fitasecond-degreeparabolatothefollowingdatatakingxastheindependentvariable.1 2 3 4 5 6 7 8 92 6 7 8 10 11 11 10 9xySol. The equation of second-degree parabola is given by + +2y a bx cxand the normal equationsare: 22 32 2 3 4

y na b x c xxy a x b x c xxy a x b x c x + + + +, + + ...(1)Here 9 n . The various sums are appearing in the table as follows:CURVEFITTINGANDSOLUTIONOFEQUATION 391

2 2 3 42 2 3 41 2 2 1 2 1 12 6 12 4 24 8 163 7 21 9 63 27 814 8 32 16 128 64 2565 10 50 25 250 125 6256 11 66 36 396 216 12967 11 77 49 539 343 24018 10 80 64 640 512 40969 9 81 81 729 729 656145 74 421 284 2771 2025 15333x y xy x xy x xx y xy x xy x xPuttingthesevaluesof , x , y 2, x, xy 2, x y 3x and 4x inequation(1)andsolving the equations for a, b and c, we get 0.923 a; 3.520 b; 0.267 c.Hence the fitted equation is + 20.923 3.53 0.267 . y x xAns.Example 9: Show that the line of fit to the following data is given by + 0.7 11.28. y x0 5 10 15 20 2512 15 17 22 24 30xySol.Heren = 6 (even)Let 0 012.5, 5, 20 (say) x h yThen,

12.52.5xuand 20, v ywe get 220 12 5 8 40 255 15 3 5 15 910 17 1 3 3 115 22 1 2 2 120 24 3 4 12 925 30 5 10 50 250 0 122 70x y u v uv uu v uv uThe normal equations are,0 6 0 0 a b a+ 122 0 70 a b + 1.743 b392 A TEXTBOOKOFENGINEERINGMATHEMATICSIIIThus line of fit is 1.743 . v uor12.5020 (1.743) 0.69 8.7152.5xy x _ ,or+ 0.7 11.285 y x . Ans.Example 10: Fit a second-degree parabola to the following data by least squares method. 1929 1930 1931 1932 1933 1934 1935 1936 1937352 356 357 358 360 361 361 360 359xySol.Taking01933, x 0357 ythen

0( ) x xuhHere h = 1Taking 0u x x and 0, v y ytherefore, 1933 u x and 357 v y2 2 3 41933 3571929 4 352 5 20 16 80 64 2561930 3 356 1 3 9 9 27 811931 2 357 0 0 4 0 8 161932 1 358 1 1 1 1 1 11933 0 360 3 0 0 0 0 01934 1 361 4 4 1 4 1 11935 2 361 4 8 4 16 8 161936 3 360 3 9 9 27 27 811937 4 359 2 8 16 32 64 2560 11Totalx u x y v y uv u uv u uu v 2 2 3 451 60 9 0 708 uv u uv u u Then the equation + +2y a bx cxis transformed to + +2v A Bu Cu ...(1)Normal equations are: + + 29 v A B u C u + 11 9 60 A C + + 2 3uv A u B u C u 17/ 20 B + + 2 2 3 4uv A u B u C u + 9 60 708 A COn solving these equations, we get 6943231A , 170.8520B and 2470.27924C 23 0.85 0.27 v u u+ 2357 3 0.85( 1933) 0.27( 1933) y x x + 21010135.08 1044.69 0.27 . y x x + Ans.CURVEFITTINGANDSOLUTIONOFEQUATION 393Example 11:Fit second degree parabola to the following:0 1 2 3 41 1.8 1.3 2.5 6.3xySol. Here 5 n (odd) therefore02 x , h = 1, y0 = 0 (say)Now let 2 u x , v yand the curve of fit be + +2. v a bu cu 2 2 3 40 1 2 1 2 4 4 8 161 1.8 1 1.8 1.8 1 1.8 1 12 1.3 0 1.3 0 0 0 0 03 2.5 1 2.5 2.5 1 2.5 1 14 6.3 2 6.3 12.6 4 25.2 8 16Total 0 12.9 11.3 10 33.5 0 34x y u v uv u uv u uHence the normal equations are, + + 25 v a b u c u + + 2 3uv a u b u c u + + 2 2 3 4uv a u b u c uOn putting the values of u ,v etc. from the table in these, we get + 12.9 5 10 , a c11.3 10 , b+ 33.5 10 34 . a cOn solving these equations, we get a = 1.48, b = 1.13 and c = 0.55Therefore the required equation is + +21.48 1.13 0.55 . v u uAgain substituting 2 u xand v = y, we get + + 21.48 1.13( 2) 0.55( 2) y x xor +21.42 1.07 0.55 . y x x Ans.5.6 FITTINGOFANEXPONENTIALCURVESuppose an exponential curve of the form

bxy aeTaking logarithm on both the sides, we get +10 10 10log log log y a bx ei.e., + Y A Bx...(1)where 10 10 10log , log and log . Y y A a B b e394 A TEXTBOOKOFENGINEERINGMATHEMATICSIIIThenormalequationsfor(1)are, + Y nA B x + 2xY A x B xOn solving the above two equations, we get A and Bthena = antilog A,10logBbe5.7 FITTING OF THE CURVE y = ax + bx2Error of estimate for ith point( , )i ix y is 2( )i i i ie y ax bxWehave,

21niiS e

2 21( )ni i iiy ax bxBy the principle of least square, the value of S is minimum

0Sa and

0SbNow

0Sa

212( )( ) 0ni i i iiy ax bx xor

+ 2 31 1 1n n ni i i ii i ix y a x b x...(1)and

0Sb

2 212( )( ) 0ni i i iiy ax bx xor

+ 2 3 41 1 1n n ni i i ii i ixy a x b x...(2)Droppingthesuffixi from(1)and(2),thenthenormalequationsare, + 2 3xy a x b x + 2 3 4xy a x b xCURVEFITTINGANDSOLUTIONOFEQUATION 3955.8 FITTINGOF THECURVE xbax y + Error of estimate for ith point( , )i ix y is ( )i i iibe y axxWehave,

21niiS e

21( )ni ii iby axxBy the principle of least square, the value ofS is minimum

0Sa and

0SbNow

0Sa

12( )( ) 0ni i ii iby ax xxor + 21 1n ni i ii ix y a x nb...(1)and

0Sb

112( )( ) 0ni ii i iby axx xor + 21 11n nii i i iyna bx x...(2)Droppingthesuffixi from(1)and(2),thenthenormalequationsare, + 2xy nb a x + 21 yna bxxwherenis the number of pair of values of x and y.396 A TEXTBOOKOFENGINEERINGMATHEMATICSIII5.9 FITTINGOF THECURVE 01cy = + c xxError of estimate for ith point( , )i ix y is 01( )i i iice y c xxWehave,

21niiS e

2 011( )ni ii icy c xxBy the principle of least square, the value of S is minimum

00Sc and

10ScNow

00Sc

01112( )( ) 0ni ii i icy c xx xor

+ 0 121 1 11 1n n nii i i i i iyc cx x x...(1)and

10Sc

0112( )( ) 0ni i ii icy c x xxor

+ 0 11 1 11n n ni i ii i i iy x c c xx...(2)Droppingthesuffixi from(1)and(2),thenthenormalequationsare, + 0 121 1 yc cxx x0 11. y x c c xx + Example12: Find the curve of best fit of the type

bxy ae to the following data by the methodof least squares:1 5 7 9 1210 15 12 15 21xySol.Thecurvetobefittedisbxy ae or + Y A Bx,where10log , Y y 10log , A a and

10log . B b eCURVEFITTINGANDSOLUTIONOFEQUATION 397Thereforethenormalequationsare: + 5 Y A B x + 2xY A x B x

2102log1 10 1.0000 1 15 15 1.1761 25 5.88057 12 1.0792 49 7.55449 15 1.1761 81 10.584912 21 1.3222 144 15.866434 5.7536 300 40.8862x y Y y x xYx Y x xYSubstitutingthevaluesof , x etc.andcalculatedbymeansofabovetableinthenormalequations, we get + 5.7536 5 34 A Band + 40.8862 34 300 A BOn solving these equations, we obtain, 0.9766 A; 0.02561 BTherefore 10antilog 9.4754 a A ; 100.059logBbeHence the required curve is0.0599.4754xy e . Ans.Example 13: For the data given below, find the equation to the best fitting exponential curve oftheform .bxy ae1 2 3 4 5 61.6 4.5 13.8 40.2 125 300xySol.bxy aeOntakinglogboththesides, + log log log y a bx e whichisoftheform + , Y A Bx where log , Y ylog A a and log . B b e

22log1 1.6 0.2041 1 0.20412 4.5 0.6532 4 1.30643 13.8 1.1399 9 3.41974 40.2 1.6042 16 6.41685 125 2.0969 25 10.48456 300 2.4771 36 14.862621 8.1754 91 36.6941x y Y y x xYx Y x xY398 A TEXTBOOKOFENGINEERINGMATHEMATICSIIINormal equations are:+ 6 Y A B x + 2xY A x B xTherefore from these equations, we have + 8.1754 6 21 A B + 36.6941 21 91 A B0.2534, 0.4617 A BTherefore, antilog antilog( 0.2534) antilog(1.7466) 0.5580. a Aand 0.46171.0631log 0.4343BbeHence required equation is1.06310.5580xy e . Ans.Example14:Giventhefollowingexperimentalvalues:0 1 2 32 4 10 15xyFit by the method of least squares a parabola of the type +2. y a bxSol. Error of estimate for ith point( , )i ix yis 2( )i i ie y a bxBy the principle of least squares, the values ofaandbare such that2 2 21 1( )n ni i ii iS e y a bx isminimum.Therefore normal equations are given by + 20Sy na b xa...(1)and + 2 2 40Sxy a x b xb...(2) 2 2 42 2 40 2 0 0 01 4 1 4 12 10 4 40 163 15 9 135 8131 14 179 98Totalx y x xy xy x xy xHeren = 4.CURVEFITTINGANDSOLUTIONOFEQUATION 399From(1)and(2),+ 31 4 14 a b and+ 179 14 98 a bSolving foraand b, we get 2.71 a and 1.44 bHence the required curve is+22.71 1.44 . y xAns.Example 15: By the method of least square, find the curve +2y ax bxthat best fits the followingdata:1 2 3 4 51.8 5.1 8.9 14.1 19.8xySol.Error of estimate for ith point( , )i ix yis 2( )i i i ie y ax bxBy the principle of least squares, the values ofaand b are such that2 2 21 1( )n ni i i ii iS e y ax bx isminimum.Therefore normal equations are given by2 31 1 10n n ni i i ii i iSx y a x b xa

+ and2 3 41 1 10n n ni i i ii i iSxy a x b xb

+ Dropping the suffix i, normal equations are + 2 3xy a x b x ...(1)and+ 2 3 4xy a x b x ...(2)

2 3 4 22 3 4 21 1.8 1 1 1 1.8 1.82 5.1 4 8 16 10.2 20.43 8.9 9 27 81 26.7 80.14 14.1 16 64 256 56.4 225.65 19.8 25 125 625 99 49555 225 979 194.1 822.9Totalx y x x x xy xyx x x xy xySubstituting these values in equations (1) and (2), we get + 194.1 55 225 a b and + 822.9 225 979 a b 83.851.5255aand 317.40.49664bHence required parabolic curve is +21.52 0.49 y x x . Ans.400 A TEXTBOOKOFENGINEERINGMATHEMATICSIIIExample 16: Fit an exponential curve of the formxy ab to the following data:1 2 3 4 5 6 7 81.0 1.2 1.8 2.5 3.6 4.7 6.6 9.1xySol.xy abtakes the form + Y A Bx , where log Y y ; log A aand log . B bHence the normal equations are given by + Y nA B xand+ 2xY A x x .

22log1 1.0 0.0000 0.000 12 1.2 0.0792 0.1584 43 1.8 0.2553 0.7659 94 2.5 0.3979 1.5916 165 3.6 0.5563 2.7815 256 4.7 0.6721 4.0326 367 6.6 0.8195 5.7365 498 9.1 0.9590 7.6720 6436 30.5 3.7393 22.7385 204x y Y y xY xx y Y xY xPutting the values in the normal equations, we obtain + 3.7393 8 36 A B and + 22.7385 36 204 A B 0.1407 B and0.1656 A antilog 1.38 b Bandantilog 0.68. a AThus the required curve of best fit is(0.68)(1.38) .xy Ans.Example 17: Fit a curvexy ab to the following data:2 3 4 5 6144 172.8 207.4 248.8 298.5xySol. Given equationxy ab reduces to + Y A Bxwhere log , Y ylog A aand log . B bThe normal equations are, + log log log y n a b x + 2log log log x y a x b xThecalculationsof , xlog , y2x and log x y aresubstituteinthefollowingtabularform.CURVEFITTINGANDSOLUTIONOFEQUATION 4012log log2 144 4 2.1584 4.31683 172.8 9 2.2375 6.71254 207.4 16 2.3168 9.26725 248.8 25 2.3959 11.97956 298.5 36 2.4749 14.849420 90 11.5835 47.1254x y x y x yPutting these values in the normal equations, we have + 11.5835 5log 20log a b47.1254 20log 90log . a b+Solvingtheseequationsandtakingantilog,wehavea=100,b=1.2approximate.Thereforeequation of the curve is y = 100(1.2)x. Ans.Example 18: Derive the least square equations for fitting a curve of the type +2( / ) y ax b xtoa set of n points. Hence fit a curve of this type to the data.1 2 3 41.51 0.99 3.88 7.66xySol. Let thenpoints are given by 1 1( , ), x y2 2( , ), x y3 3( , ), x y,( , ).n nx yThe error of estimatefor the ith point( , )i ix y is 2[ ( / )].i i i ie y ax b xBy the principle of least square, the values ofaand b are such so that the sum of the square oferrorS,viz.,

_ , 22 21ni i ii ibS e y axxisminimum.Therefore the normal equations are given by 0, 0S Sa bor

+ 2 41 1 1n n ni i i ii i iy x a x b x and

+ 21 1 11n n niii i i i iya x bx xThese are the required least square equations.2 4 221 11 1.51 1 1 1 1 1.51 1.512 0.99 4 16 0.5 0.25 3.96 0.4953 3.88 9 81 0.3333 0.1111 34.92 1.29334 7.66 16 256 0.25 0.0625 122.56 1.094310 354 1.4236 159.93 1.1933yx y x x yxx x x 402 A TEXTBOOKOFENGINEERINGMATHEMATICSIIIPutting the values in the above least square equations, we get + 159.93 354 10 a b and1.1933 10 1.4236 . a b+Solving these, we get a = 0.509 and b = 2.04.Therefore, the equation of the curve fitted to the above data is 22.040.509 y xx.Ans.Example 19:Fitthecurve pv k tothefollowingdata:2(kg/cm ) 0.5 1 1.5 2 2.5 3(litres) 1620 1000 750 620 520 460pvSol.Given pv k _ ,1/1/ 1/kv k ppTaking log both the sides, we get 1 1log log log v k pwhich is of the form + Y A BXwhere log , Y vlog , X p 1log A kand 1. B

20.5 1620 0.30103 3.20952 0.96616 0.090621 1000 0 3 0 01.5 750 0.17609 2.87506 0.50627 0.031012 620 0.30103 2.79239 0.84059 0.090622.5 520 0.39794 2.716 1.08080 0.158363 460 0.47712 2.66276 1.27046 0.227641.05115Totalp v X Y XY XX Y 217.25573 2.73196 0.59825 XY XHere n = 6Normalequationsare, + 17.25573 6 1.05115 A B + 2.73196 1.05115 0.59825 A BOn solving these, we get 2.99911 A and 0.70298 BCURVEFITTINGANDSOLUTIONOFEQUATION 4031 11.422520.70298 BAgainlog 4.26629 k Aantilog(4.26629) 18462.48 kHence, the required curve is1.4225218462.48. pvAns.Example 20: For the data given below, find the equation to the best fitting exponential curve oftheform .bxy ae1 2 3 4 5 61.6 4.5 13.8 40.2 125 300xySol.Given y = aebx, taking log we get +10log log log y a bx e which is of the + , Y A Bxwherelog Y y , log A aand10log . B ePutthevaluesinthefollowingtabularform,alsotransfertheoriginofx seriesto3,sothat 3 u x.2log 31 1.6 0.204 2 0.408 42 4.5 0.653 1 0.653 13 13.8 1.140 0 0 04 40.2 1.604 1 1.604 15 125.0 2.094 2 4.194 46 300 2.477 3 7.431 9Total 8.175 3 12.168 19x y y Y x u uY u In case. , Y A Bu+then normal equations are given byY nA B u+ 8.175 6 3 A B+...(1)2uY A u B u+ 12.168 3 19 A B+...(2)Solving(1)and(2),weget1.13 A and0.46 B Thus equation is + 1.13 0.46 , Y ui.e. + 1.13 0.46( 3) Y xor 0.46 0.25 Y xWhich giveslog 0.25 ai.e.antilog( 0.25) antilog(1.75) 0.557 10.461.06log 0.4343Bbe

Hence, the required equation of the curve is 1.06(0.557) .xy eAns.404 A TEXTBOOKOFENGINEERINGMATHEMATICSIIIPROBLEM SET 5.11. Fit a straight line to the given data regardingxas the independent variable:1 2 3 4 6 82.4 3.1 3.5 4.2 5.0 6.0xy + [ 2.0253 0.502 ] y x Ans.2. Fit a straight line + y a bxto the following data by the method of least square:0 1 3 6 81 3 2 5 4xy+ [ 1.6 0.38 ] x Ans.3. Find the least square approximation of the form +2y a bx for the following data:0 0.1 0.2 0.3 0.4 0.51 1.01 0.99 0.85 0.81 0.75xy 2[ 1.0032 1.1081 ] y x Ans.4. Fit a second degree parabola to the following data:0.0 1.0 2.01.0 6.0 17.0xy + +2[ 1 2 3 ] y x x Ans.5. Fit a second degree parabola to the following data:1.0 1.5 2.0 2.5 3.0 3.5 4.01.1 1.3 1.6 2.0 2.7 3.4 4.1xy +2[ 1.04 0.193 0.243 ] y x x Ans.6. Fit a second degree parabola to the following data by the least square method:1 2 3 4 51090 1220 1390 1625 1915xy + +2[ 27.5 40.5 1024] y x x Ans.7. Fit a parabola + +2y a bx cxto the following data:2 4 6 8 103.07 12.85 31.47 57.38 91.29xy +2[ 0.34 0.78 0.99 ] y x x Ans.8. Determine the constantsaand b by the method of least squares such thatbxy aefits thefollowingdata:2 4 6 8 104.077 11.084 30.128 81.897 222.62xy

0.50001[ 1.49989 ]xy e Ans.9. Fit a least square geometric curveby ax to the following data:1 2 3 4 50.5 2 4.5 8 12.5xy

1.9977[ 0.5012 ] y x Ans.CURVEFITTINGANDSOLUTIONOFEQUATION 40510. A person runs the same race track for five consecutive days and is timed as follows:( ) 1 2 3 4 5( ) 15.3 15.1 15 14.5 14DayxTimeyMake a least square fit to the above data using a function+ +2b caxx.26.7512 4.4738 13.0065 yxx 1 + + 1 ]Ans.11. Usethemethodofleastsquarestofitthecurve +01cy c xxtothefollowingtableofvalues:0.1 0.2 0.4 0.5 1 221 11 7 6 5 6xy1.97327 3.28182 y xx 1 + 1 ]Ans.12. Using the method of least square to fit a parabola + +2y a bx cxin the following data: ( , ) : ( 1, 2),(0, 0),(0,1),(1,2) x y21 3

2 2y x 1 + 1 ]Ans.13. ThepressureofthegascorrespondingtovariousvolumesV ismeasured,givenbythefollowingdata:32(cm ) 50 60 70 90 100(kgcm ) 64.7 51.3 40.5 25.9 78VpFit the data to the equation pV c . [Ans.0.28997167.78765 pV ]14. Employthemethodofleastsquarestofitaparabolay=a+bx+cx2inthefollowingdata:(x,y):(1,2),(0,0),(0,1),(1,2) [Ans.y=0.5+1.5x2 ]15. Fit a second degree parabola in the following data: [U.T.U.2008]0.0 1.0 2.0 3.0 4.01.0 4.0 10.0 17.0 30.0xy[Ans. y = 1 + 2x + 3x2]16. Fit at least square quadratic curve to the following data:1 2 3 41.7 1.8 2.3 3.2xy,estimatey(2.4)[Ans. y = 2 0.5x + 0.2x2 and y(2.4) = 1.952]406 A TEXTBOOKOFENGINEERINGMATHEMATICSIII17. Fit an exponential curve by least squares1 2 5 10 20 30 40 5098.2 91.7 81.3 64.0 36.4 32.6 17.1 11.3xyEstimate y when x = 25. [Ans.y=100(0.96)x,y(25)=33.9]18. Fit the curve by ax +to the following data1 2 3 43 1.5 6 7.5xyEstimate y whenx = 2.25.1.7 1.3 , (2.25) 4.625 y yx 1 + 1 ]Ans.5.10 POLYNOMIALIf1 20 1 2 1( ) ..............n n nn nf x ax a x a x a x a + + + + +Then the above relation is called the polynominal of nth order in x.5.10.1 Degree of PolynomialThe highest power of x occurring in the given polynomial is called degree of polynomial.The constant 0c cxis called a polynomial of degree zero.The polynomial+ ( ) , 0 f x ax ba is of degree one and is called a linear polynomial.The polynomial+ + 2( ) , 0 f x ax bx c ais of degree two and is called a quadratic polynomial.The polynomial+ + + 3 2( ) , 0 f x ax bx cx da is of degree three and is called a cubic polynomial.Thepolynomial+ + + + 4 3 2( ) , 0 f x ax bx cx dx ea isofdegreefourandiscalledbiquadraticpolynomial.5.11 DESCARTESRULEOFSIGNSThe number of positive roots of the equation f (x) = 0 cannot exceed the number of changes of sign inf (x), and the number of negative roots cannot exceed the number of changes of sign of f (x).Existence of imaginary roots: If an equation of the nth degree has at most p positive roots andat most q negative roots, then it has at least + ( ) n p q imaginary roots.Example1:ApplyDescartesRuleofsignstodiscussthenatureoftherootsoftheequation+ + 4 215 7 11 0. x x xSol. The given equation is + + 4 2( ) 15 7 11 0 f x x x xSigns of( ) f x are + ++ [from + to ]CURVEFITTINGANDSOLUTIONOFEQUATION 407It has one change of sign and hence it must have one +ve root. + + 4 2( ) ( ) 15( ) 7( ) 11 0 f x x x xor + 4 2( ) 15 7 11 f x x x x= 0 Signs of ( ) f xare + + [from + to ]It has only one change in sign and hence it must have one ve root.Thus the equation has two real roots , one +ve and one ve and hence the other two roots must beimaginary.Example 2:Show that + 7 4 33 2 1 0 x x xhas at least four imaginary roots.Sol.The given equation is( ) f x = + 7 4 33 2 1 0 x x x[from+ to or to +] Signs of( ) f x + + ( ) 0 f xhas 3 changes of sign. Therefore, it cannot have more than three positiveroots.Also ( ) f x= + 7 4 3( ) 3( ) 2( ) 1 0 x x xor 7 4 3( ) 3 2 1 0 f x x x xSigns of( ) f xare ( ) 0 f xhas no changes in sign. Therefore the given equation has no negative root.Thus the given equation cannot have more than 3 + 0 = 3 real roots. But the given equation has7 roots. Hence the given equation has 7 3 = 4 imaginary roots.Example 3: Find the least positive number of imaginary roots of the equation + + + 9 5 4 2( ) 1 0 f x x x x xSol.The given equation is + + + 9 5 4 2( ) 1 0 f x x x x xSigns of ( ) f x are + ++ + ( ) 0 f xhas two changes of signs, and hence 2 is the max. number of +ve root. + + + 9 5 4 2( ) ( ) ( ) ( ) ( ) 1 0 f x x x x x or + + + + 9 5 4 2( ) 1 0 f x x x x xSigns of ( ) f x are + + + + ( ) 0 f x has only one changes of sign and hence it has only one ve root or ( ) 0 f xhas onlyoneveroot.Thus the max. number of real roots is 2 + 1=3 and the equation being of9th degree and it willhave at least 9 3 = 6 imaginary roots.408 A TEXTBOOKOFENGINEERINGMATHEMATICSIII5.12 CARDONSMETHODCase 1. When Cubic is of the Form3x +qx +r =0The given cubic is+ + 30 x qx r ...(1)Letx = u + v be a root of(1)Cubing , + + ++ +3 3 3 3 33 ( ) 3 x u v uv u v u v uvxor + 3 3 33 ( ) 0 x uvx u v...(2)Comparing (1) and (2), 13uv qor 3 3 3127u v q and+3 3u v r3 3and u v are the roots of the equation + + 2 3 3 3 3( ) 0 t u v t uvor + 2 31027t rt q ...(3)Solving(3), +

324272qr rtLet + + + 3 32 23 34 427 27;2 2q qr r r ru vNow, the three cube roots of u3 are 2, , uu u and those of 3vare 2, , , vv v where 2and areimaginary cube root of unity.Since+ x u vTo find x, we have to add a cube root of u3 and a cube of v3 in such a manner that their productis real.The three values of x are 2 2, , u v u v u v + + + 3( 1) Example 4: Use Cardons method to solve + 327 54 0. x xSol.Let + x u vCubing, ++ + ++ +3 3 3 3 3 3( ) 3 ( ) 3 x u v u v uv u v u v uvx + 3 3 33 ( ) 0 x uvx u vComparing with the given equation,we get 3 39 729 uv u v(on cubing)and +3 354 u vCURVEFITTINGANDSOLUTIONOFEQUATION 409\ u3 and v3 are the root of + + 2 3 3 3 3( ) 0 t u v t u v+ + + 2 254 729 0 ( 27) 0 t t t 27, 27 tLet3 327 and 27 u vSothat,2 23, 3 , 3 and3, 3 , 3 u v To find x, we have to add a cube root of u3 and a cube root of v3 in such a way that their productis real.\2 2( 3 3), ( 3 3 ), ( 3 3 ) x 3( 1) 32 26, 3( ), 3( ) 6, 3, 3 + + 2( 1 0) + + 3Hence the required roots are 6, 3, 3. Ans.Example 5: Solve by Cardons method 315 126 0. x x Sol. Let + x u vCubing, ++ + ++ +3 3 3 3 3 3( ) 3 ( ) 3 x u v u v uv u v u v uvx + 3 3 33 ( ) 0 x uvx u vComparing with the given equation, we get 3 35 125 uv u v(on cubing)and+ 3 3126 u v3 3and u v are the roots of + + 2 3 3 3 3( ) 0 t u v t uv + 2126 125 0 t t ( 1)( 125) 0 t t 1,125 tLet3 31 and125 u vSothat,2 21, , and 5, 5 , 5 u v To find x, we have to add a cube root of u3 and a cube root of v3 in such a way that their productis real.\2 21 5, 5 , 5 x+ + + ( )31 3 _ _ _ _ + + + +, , , , , , , , 1 3 1 3 1 3 1 36, 5 , 52 2 2 2i i i i + 6, 3 2 3 , 3 2 3 6, 3 2 3 i i iHence, the required roots are 6, 3 2 3 . i Ans.410 A TEXTBOOKOFENGINEERINGMATHEMATICSIIIExample 6: Solve 36 9 0 x x by Cardons method.Sol.Let+ x u vCubing, ++ + +3 3 3 3( ) 3 ( ) x u v u v uv u v + +3 33 u v uvx + 3 3 33 ( ) 0 x uvx u vComparing,weget 3 32 8 uv u v(on cubing)and+ 3 39 u v3 3, u v are the roots of + + 2 3 3 3 3( ) 0 t u v t u v + 29 8 0 t t ( 1)( 8) 0 1,8 t t tLet3 31and8 u vso that

2 21, , and2, 2 , 2 u v To find x, we have to add a cube root of u3 and a cube root of v3 in suchthat their product is real.

2 21 2, 2 , 2 x+ + + 3( 1) 3 _ _ _ _ + + + +, , , , , , , , 1 3 1 3 1 3 1 33, 2 , 22 2 2 2i i i i + 3 3 3 3 3 33, , 3, .2 2 2i i ix Ans.Example 7:Solve the cubic equation 318 35 0 x xSol.Let+ x u vCubing, ++ + +3 3 3 3( ) 3 ( ) x u v u v uv u v + +3 33 u v uvx + 3 3 33 ( ) 0 x uvx u vComparing,weget 3 36 216 uv u v(oncubing)and+ 3 335 u v3 3andu vare the roots of + + 2 3 3 3 3( ) 0 t u v t uv + 235 216 0 ( 8)( 27) 0 8, 27 t t t t tLet3 38& 27 u vso that 2 22, 2 , 2 and 3, 3 , 3 u v CURVEFITTINGANDSOLUTIONOFEQUATION 411To find x, we have to add a cube root of u3 and a cube root of v3 in such that their product is real. 2 22 3, 2 3 , 2 3 x+ + + (31 3)1 3 1 3 1 3 1 35, 2 3 , 2 32 2 2 2i i i i _ _ _ _ + + + +, ,, , , , , , 5 3 5 35, ,2 2i i +

5 35, .2ix Ans.Case 2.When the Cubic Equation is of the Form 3 20 1 2 3a +a +a +a = 0 x x xThen,firstofallweremovethetermcontainingx2.Thisisdonebydiminishingtherootsofthegivenequationby 10anaorSumofroot/No.ofroots. Where n is 3. We proceed with the help of following examples:Example 8:Solve by Cardons method + + + 3 26 9 4 0 x x xSol.Equationing+ + + 3 26 9 4 0 x x x...(1)Equating with+ + + 3 20 1 2 30, ax a x a x a we get ...(2) 0 11, 6 a aThen 10623 3aha

2. hNow remove the x2 terms ,we have 2 1 6 9 42 8 21 4 1 (2)2 41 2 ( 3)21 (0)The transformed equation is + 33 2 0 y y ...(3)where+ 2. y xLet + y u v be a solution of (3), then we get 3 3 33 ( ) 0 y uvy u v +...(4)Equating(3)and(4),weget 1 uv+3 3 3 31 and 2 u v u v412 A TEXTBOOKOFENGINEERINGMATHEMATICSIIILet us consider an equation whose roots are 3 3and u v + 20 t st ps = sum of roots, p = product of roots+ + + 2 22 1 0 ( 1) 0 t t t 1, 1 tLet3 31 and 1 u v so that 2 21, , and 1, , u v \2 2( 1 1), ( ),( ) y u v+ 2,1,1 2( 1 0) + + 3But 2 x y 2 2,1 2,1 2 4, 1, 1 4, 1, 1 x . Ans.Example 9:Solve byCardonsmethod + 3 215 33 847 0. x x xSol. The given equation + 3 215 33 847 0 x x x ...(1)Equating with+ + + 3 20 1 2 30, ax a x a x a we get0 11, 15 a a\ 101553 3ahaNow remove the x2 term, using synthetic division, we have( )( )( ) 5 1 15 33 8475 50 4151 10 83 4325 251 5 10851 0Transformedequationis + 3108 432 0 y y ...(2)where 5. y xLet+ y u v\ + 3 3 33 ( ) 0 y uvy u v ...(3) (oncubing)CURVEFITTINGANDSOLUTIONOFEQUATION 413Comparing(2)and(3), 3 3 636 (6) uv u vand+3 3432 u v3uand 3v are the roots of+ + 2 6432 (6) 0 t t + + + 2 3 6 22. (6) (6) 0 ( 216) 0 216, 216 t t t tLet3 3216 and216 u v so that2 26, 6 , 6 and 6, 6 , 6 u v 2 2( 6 6),( 6 6 ), ( 6 6 ) y u v+ =12, 6, 6 221 01 _+ + + ,3But+ + + + 5 12 5, 6 5, 6 5 x y\7, 11, 11. xAns.Example10:Solve 3 23 12 16 0 x x x + + Sol. Here 10 1031, 3 13 3aa a ha Now remove the termx2, using synthetic division1 1 3 12 161 2 101 2 10 (26)1 11 1 (9)11 (0)The transformed equation is3( 1) 9( 1) 26 x x + + = 0Let y = x 1, then y3 + 9y + 26 = 0 ...(1)Let y = u + v ...(2)3 3 33 ( ) ( ) y uv y u v + = 0Comparing(1)and(2),wegetuv3 33 and26 u v + u3v3= 27414 A TEXTBOOKOFENGINEERINGMATHEMATICSIIINow let us have an equation whose roots are u3 and v32( 26) ( 27) t t +20 26 27 0 t t + ( 27)( 1) t t + = 0 t3 31, 27 . ., 1 and27 i e u v So that u2 31, ,and3, 3 , 3 v y u v+2 2(1 3), ( 3 ), ( 3 ) 1 3 1 3 1 3 1 32, 3 , 32 2 2 2i i i i _ _+ + , , , , , , 2, (1 2 3) , (1 2 3) i i + \ x = y + 1 x ( 2 1), (1 2 3 1), (1 2 3 1) i i + + + +1, (2 2 3), (2 2 3) i i + 1, 2(1 3). i Ans.Example 11:Solve + 3 26 6 5 0 x x x by Cardons method.Sol. The given equation + 3 26 6 5 0 x x x ...(1)Comparing with + + + 3 20 1 2 30, ax a x a x a we get0 11, 6 a a\ 10623 3ahaNow remove the x2 term, using synthetic division 2 1 6 6 52 8 41 4 2 ( 9)2 41 2 ( 6)21 (0)Transformed equation is 36 9 0 y y ...(2)CURVEFITTINGANDSOLUTIONOFEQUATION 415where 2 y xLet y = u +v be the solution of (2), then + + +3 3 33 ( ) y u v uvu v + +3 3 33 ( ) y u v uv y + 3 3 33 ( ) ( ) 0 y uv y u v ...(3)Comparing(2)and(3),wehave 3 33 6 2 8 uv uv uvand+ 3 39 u vNow let us have an equation whose roots are 3 3and u v + 20 t st p ; s = sum of roots, p = product of the roots. + 29 8 0 t t1,8 t i.e.,3 31and 8 u vso that 2 21, , and2,2 ,2 u v \+ y u v2 2(1 2),( 2 ),( 2 )+ + + 1 3 1 3 1 3 1 33, 2 , 22 2 2 2i i i i _ _ _ _+ + + +, , , , , , , , +

3 3 3 33, , .2 2i iBut+ 2 x y +

1 3 1 35, ,2 2i i 15, (1 3).2i Ans.Example 12: Solve the equation + 3 23 3 1 0 x x xby Cardons method.Sol. Here, 0 11, 3 a a\ 10313 3aha416 A TEXTBOOKOFENGINEERINGMATHEMATICSIIINow remove the 2xterm, using synthetic division, we have 1 1 3 3 11 2 11 2 1 (0)1 11 1 (0)11 (0)Transformed equation is 30 y ...(1)where1. y xFrom(1),0, 0, 0 y +1 1,1,1 x yHence the required roots are 1, 1, 1. Ans.Example 13: Solve the cubic equation+ + 3 216 20 0. x x xSol. Given equation is+ + 3 216 20 0 x x xHere,0 11, 1 a a\ 1013 3ahaNow remove the x2 term, using synthetic division, we have( ) _ , _ ,1/ 3 1 1 16 201461/ 3 2/ 927146 6861 2/ 39 271/ 3 1/ 91471 1/ 391/ 31 0Transformed equation is + 3147 68609 27y y ...(1)where +1.3y xLet+ y u vCURVEFITTINGANDSOLUTIONOFEQUATION 417Cubing, + +3 3 33 y u v uvy + 3 3 33 ( ) 0 y uvy u v ...(2)Comparingwith(1),weget _ + ,63 3 3 349 7 686and 9 3 27uv uv u v3 3and u v are the roots of _+ + ,62686 7027 3t t _ _+ + , ,3 627 72 03 3t t _ _ _+ , , , , , 23 3 37 7 70 ,3 3 3t tLet _ _ , ,3 33 37 7 and 3 3u vSo that2 27 7 7 7 7 7, ,and, ,3 3 3 3 3 3u v 2 27 7 7 7 7 7 14 7 7, , , ,3 3 3 3 3 3 3 3 3y u v _ _ _ + , , ,Now, 1 14 1 7 1 7 1, , 5,2, 23 3 3 3 3 3 3x yHence required roots are 5, 2, 2. Ans.Example 14: Solve the cubic equation + + 3 26 12 32 0. x x xSol.Given equation is+ + 3 26 12 32 0 x x xHere,0 11, 6 a a\ 10623 3ahaNow remove thex2 term, using synthetic division method, we have 2 1 6 12 322 8 401 4 20 (72)2 41 2 ( 24)21 (0)418 A TEXTBOOKOFENGINEERINGMATHEMATICSIIITransformedequationis + 324 72 0 y y ...(1)where+ 2 y xLet+ y u vCubing,+ +3 3 33 y u v uvy + 3 3 33 ( ) 0 y uvy u v ...(2)Comparing(1)and(2),weget +3 3 3 38 512 and72 uv u v u v3 3 andu v are the roots of+ + 272 512 0 t t 8, 64 tLet3 38 and64 u vSo that2 22, 2 , 2 & 4, 4 , 4 u v 2 2( 2 4),( 2 4 ),( 2 4 ) y u v+ =6,3+i 3,3 3 iNow, + 2 8,1 3,1 3 x y i iHence the required roots are 8,1 3. iAns.Example 15: Solve x3 + 3ax2 +3(a2 bc)x + a3 + b3 + c3 3abc = 0 by Cardons method.Sol.Given equation is x3 + 3ax2 + 3(a2 bc)x + a3 + b3 +c3 3abc = 0Here a0 = 1, a1 = 3a\h = 103. 3.1a aan aNow remove the x2 term, using synthetic division, we have2 3 3 32 32 3 321 3 3( ) 32 31 2 3 ( )1 31 (0)a a a bc a b c abca a a abca a bc b ca aa bca + + + +