 Open Access
 Total Downloads : 207
 Authors : Parcha Kalyani, P.S. Rama Chandra Rao
 Paper ID : IJERTV2IS120047
 Volume & Issue : Volume 02, Issue 12 (December 2013)
 Published (First Online): 18122013
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Cubic Spline Interpolation with New Conditions on Mo and Mn
1Parcha Kalyani and P.S. Rama Chandra Rao2
1,2Kakatiya Institute of Technology and Sciences, Warangal506009India
Abstract
In this communication, we defined new conditions on M0 and Mn of cubic spline interpolation. With the defined conditions, an attempt is made to investigate the accuracy of the estimation of the dependent variable at particular values of the independent variable. With these new conditions we observed that the error is reduced considerably compared to the other types of conditions on M0 and Mn.

Introduction
Any function which would effectively correlate the data would be difficult to obtain and highly unwieldy. To this end, the idea of the cubic spline was developed. Using this process, a series of unique cubic polynomials are fitted between each of the data points, with the stipulation that the curve obtained were continuous and appear smooth. These cubic splines can then be used to determine rates of change and cumulative change over an interval.
The first mathematical reference to splines was made in the year 1946 in an interesting paper by Schoenberg (Schoenberg [1]),which is probably the first place that the word spline is used in connection with smooth, piecewise polynomial approximation. Generally I.J. Schoenberg is regarded as the father ofsplines, particularly on account of his pioneering paper [2]. However, the ideas have their roots in the aircraft and shipbuilding industries. Splines are types of curves, originally developed for shipbuilding in the days before computer modelling. Naval architects needed a way to draw a smooth curve through a set of points. The solution was to place metal weights (called knots) at the control points, and bend a thin metal or wooden beam (called a spline) through the weights. Through the advent of computers, splines have gained more importance. They were first used as a replacement for polynomials in interpolation and then as a tool to construct smooth and flexible shapes in computer graphics.
In late 1960s, there were no more than a handful of articles mentioning spline functions by name. Some of the papers which have made great contributions in the development of splines include (Loscalzo and Talbot [3], Maclaren [4], Rubin and Khosla[5],Sastry[6], Schoenberg [2]). Convergence properties of the cubic spline method are discussed
by Ahlberg and Nilson [7]. Univariate splines were studied intensely in the 60s, and by the mid70s they were sufficiently well understood to permit a fairly comprehensive treatment in books from. Some of the books which discuss splines thoroughly include (Ahlberg et al. [8], deBoor [9], Prenter [10], Schumaker [11], Shikin and Plis [12], Spath [13]). Some of the earliest papers using spline functions for the smooth approximate solution of ordinary and partial differential equations (PDEs) include (Albasiny and Hoskins [14], Bickely [15], Crank and Gupta [16], Jain and Aziz [17], Jain and Aziz [18], Rubin and Khosla [19], Usmani [20], Usmani and Sakai [21], Usmani and Warsi [22], Rama Chandra Rao [23], Kalyani and Rama Chandra Rao [24]) . These papers demonstrate the approximate methods of solving second, third, fourth, fifth order linear boundary value problems (BVPs) and solution of elliptic and parabolic equations by spline functions of various degrees.
Today, there are number of research articles published on this subject, and yet it remains an active research area. In these papers various techniques are used such as quadratic, cubic, quartic, quintic, sextic, septic and higher degree splines, and have been discussed for the numerical solution of linear and nonlinear BVPs. A survey of recent spline techniques for solving boundary value problems in ordinary differential equations using cubic, quintic and sextic polynomial andnon polynomial splines are given in Kumar and Srivastava [25].Splines have many applications in the numerical solution of a variety of problems in mathematics and engineering. Some of them are, fitting of curves, function approximation, solution of integrodifferential equations, optimal control problems, computeraided geometric design, and wavelets and so on. Also, these are useful to solve different problems in atomic and molecular physics, and are used extensively at Boeing and throughout much of the industrial world. The main task of cubic spline interpolation techniques is to find the spline function.We will discuss splines which interpolate equally spaced data points, by taking M0 as the slope of the line joining the initial point and next immediate point to it and Mn is taken as the slope of the line joining last point and its immediate preceding point, and compared with natural cubic splines by taking different step
lengths. Computed errors with Endpointslope and Type I conditions. It is observed that the error is reduced with endpointslope conditions. Further by taking h is smaller erroris minimized, probably by taking the slopes at the initial and terminalpoints since the spline is known to be the rate of change of tangent (curvature).

Cubic Splines
Suppose xi, yi fori = 0,1,2, . . , n be the set of points of known or unknown y = f x ,
wherea = x0 < x1 < x2 < < xn = b and
hi = xi xi1, i = 1,2, . , n.(1)
The cubic spline
si x defined in the interval xi1, xi will satisfy the properties
si(x) is almost a cubic in each subinterval
1, , = 1,2, , ,
si xi = yi, i = 0,1,2, . , n,
si x , si x and s"i x are continuous in
x0, xn ,
The spline function can be obtain from the
following equation given by [6]
This spline type includes the stipulation that the second derivative be equal to zeroat the endpoints.That is M0 = Mn = 0(6.1)
This results in the spline extending as a line outside
the endpoints.

Type (Parabolic Runout Spline)
The parabolic spline imposes the condition that the second derivative at the endpoints, M0 and Mn be equal to M1 and Mn1 respectively.
That is M0 = M1 , Mn = Mn1(6.2)

Type (Cubic Runout Spline)

This type of spline has the most extreme endpoint behaviour. It assigns M0 to be 2M1 M2and Mn to be 2Mn1 Mn2i.e.
M0 = 2M1 M2 , Mn = 2Mn1 Mn2 (6.3)
There are many other types of interpolating spline
curves, such as the periodic spline and the clamped Spline. The one compared with this work which we have chosen to examine; is not intrinsically superior to, or more widely used than these other types of splines.
i
i
When M s are known, eq. (2) gives the required
i
i
s x = 1 xix 3 M
hi 6
xxi1 3
+ M +
+ M +
i1 6 i
cubic spline in the subinterval xi1, xi .
3 Numerical results
yi1hi26Mi1xix+yihi26Mixxi1(2)
where si xi = mi, and s"i xi = Mi
In this equation, the spline second derivativesM"i, are still not known. We use the condition of continuity of si x to obtain the recurrence relation
hi Mi1 + 1 hi + hi+1 Mi + hi+1 Mi+1 = yi+1yi
We consider certain problems with known functions which facilitate to study the accuracy of estimation using the endpointslope conditions given by (5) and (6). Further, the resultsobtained through (5) and (6) are compared with the results of Type I. The results are shown in the tabular form. The approximate values by endpointslope conditions, by Type I conditions and exact values are shown graphically.
6 3 6
hi+1
Example1.
yi yi1 i = 1,2, , n 1 (3)
hi
We consider a function y defined on [1, 6] and suppose that the data of x and y is as follows
For equal intervals we have hi = hi+1 = h and eq.
(4) simplifies to
p
p
Mi1 + 4Mi + Mi+1 = 6 yi+1 2yi + yi1, i=1,2,.,n1.(4)
Equations (4) constitute a system of n 1
i
i
equations in n + 1 unknowns M0, M1 , . . Mn .To obtain a solution for M s, we have to impose two new conditions.
2.1 Conditions on
The following conditions are defined on
M0 and Mn
0
0
M = Y1Y0(5)
X1 X0
x0 = 1, x1 = 1.5, x2 = 2, x3 = 2.5, x4 = 3, x5 = 3.5, x6 = 4, x7 = 4.5, x8 = 5, x9 = 5.5, x10 = 6
(7) and y0 = 3, y1 = 9.09375, y2 = 33,
y3 = 98.15625, y4 = 243, y5 = 524.71
y6 = 1023, y7 = 1843.781, y8 = 3123,
y9 = 5030.344, y10 = 7773(8)
From (4) we have a system of equations in
i
i
M s (for i = 1 to 9)
M0 + 4M1 + M2 = 427.5 M1 + 4M2 + M3
= 990
M2 + 4M3 + M4 = 1912.5 M3 + 4M4 + M5 = 3285
M4 + 4M5 + M6 = 5197.5 M5 + 4M6 + M7
= 7740
Mn = yn yn1 (6)
M + 4M + M = 11002.5
Xn Xn1
We call these conditions as Endpointslope
6 7
M7 + 4M
8
8 + M9
= 15075
conditions.
The following types of conditions are specified in
[24] forM0 and Mn .2.2 Type (Natural cubic splines)
M8 + 4M9 + M10 = 20047.5
Cubic spline with endpointslope conditions
From (5) and (6) we have
M0 = 12.1875, M10 = 5485.313(9)
Substituting (9) in the above system of equations, andsolving we getM1 = 65.16567, M2 = 154.6498, M3 = 306.2352, M4 =
532.9094, M5 = 847.1271, M6 =
1276.082, M7 = 1788.544, M8 = 2572.242, M9 =
2997.487 (10)
From (2) and from the equations (7) – (10)
we get the interpolating polynomial in 1 x 1.5
which is
s1 x = 4.0625 1.5 x 3 + 21.72189 x 1 3 + 4.984375 1.5 x + 12.75703 x 1 (11)
Proceeding as the method described above, interpolated functions with endpointslope conditions are obtained in the subintervals xi1, xi , for i = 1,2 ,10 and they are shown in the Table.1.
Cubic spline with Type I
i
i
From (6.1) we have M0 = 0, M10 = 0. (12) Substituting (12) in the system of equations in M s for i = 1 to 9 , and solving the obtained system we get
M1 = 68.46756, M2 = 153.6297, M3 = 307.0135, M4 = 530.8165. M5 =
854.7203, M6 = 1247.802, M7 = 1894.071, M8 =
2178.414, M9 = 4467.271(13)
From (2) and from the equations (7), (8), (12)
and(13)we get the interpolating polynomial in
1 x 1.5 which is
S1 x = 22.82252 x 1 3 + 6 1.5 x +
16.60206, y7 = 20.90321, y8 = 25.69897,
y9 = 30.99036, y10 = 36.77815 (17)
From (4) we have a system of equations in
i
i
M s (for i = 1 to 9)
M0 + 4M1 + M2 = 10.77234 M1 + 4M2 + M3
= 11.32731
M2 + 4M3 + M4 = 11.57451 M3 + 4M4 + M5 = 11.70637
M4 + 4M5 + M6 = 11.78508 M5 + 4M6 + M7
= 11.83585
M6 + 4M7 + M8 = 11.87052 M7 + 4M8 + M9 = 11.89524 M8 + 4M9 + M10 = 11.9135
Cubic spline with endpointslope conditions
From (5) and (6) we have
M0 = 2.852183, M10 = 11.57558. (18)
Substituting (18) in the above system of equations
and solving the obtained system we getM1 = 1.484023, M2 = 1.984066, M3 = 1.907023, M4 =
1.96235, M5 = 1.949946, M6 = 2.022947, M7 =
1.794117, M8 = 2.671104, M9 =
0.5833(19)From (2) and from the equations (16)
– (19) we get the interpolating polynomial in
1 x 1.5 which is
s1 x = 0.950728 1.5 x 3 + 0.494674 x 13+1.7623181.5x+4.728514x1The
interpolated functions with endpointslope
conditions in the intervals x , x , fori =
12.48187 x 1 (14)
The interpolated functions of Type I. in the intervals xi1, xi , for i = 1,2 ,10. are given inthe Table.2.Thedata considered follows the function
y x = x5 x + 3(15)We consider the values of y x in intervals of 0.05 from x = 1 to 1.5 and then interpolate for x using the cubic spline with endpointslope conditions (5), (6), and by natural cubic splines (6.1).The cubic spline values obtained by thesetwo types of conditions in the interval [1, 1.5] are shown in the Table.3 with their corresponding errors and exact values (15). A comparison is givenin Fig.1and comparison of errors for Ex.1 isshown in Fig.2. The cubic spline values obtained by these two types of conditions in
the intervals xi1, xi , for i = 1,2 ,10 are shown in the Table.4 with their corresponding errors and exact values.
Example 2
Suppose the data of
xi, yi for i = 0,1,2,3,4,5,6,7,8,9 (n = 10) is as given below with interval of differencing h = 0.5 x0 = 1, x1 = 1.5, x2 = 2, x3 = 2.5, x4 = 3, x5 = 3.5, x6 = 4, x7 = 4.5, x8 = 5, x9 = 5.5, x10 =
6 (16)
y0 = 1, y1 = 2.426091, y2 = 4.30103, y3 = 6.64794, y4 = 9.477121, y5 = 2.79407, y6 =
i1 i
1,2 ,10. are shown in the Table 5.
Cubic spline with Type I
From (6.1) we have M0 = 0, M10 = 0 (20) Taking (20) in the above system of equations and solving the obtained system we get
M1 = 2.24834, M2 = 1.778982, M3
= 1.963041, M4
= 1.943364. M5
= 1.969874, M6
= 1.962222, M7 = 2.01709,
M8 = 1.83994, M9 = 2.51839(21)
Proceeding as in example1 we get the interpolating
polynomial in 1,1.5 which is
s1 x = 0.749447 x 1 3 + 2 1.5 x + 4.664821 x 1 (22)
The tabulated function for the given data is
y(x) = x2 + log x (23)
The interpolated functions are shown in the Table.6 in the intervals xi1, xi , for i = 1,2 ,10.we consider the values of y x in intervals of 0.05 from x = 1 to 1.5 and then interpolate for x using the cubic spline with endpointslope condition (5) and (6), and by natural cubic splines(6.1). The cubic spline values obtained by these two types of conditions in the interval [1, 1.5] are shown in Table.7 with their corresponding errors and exact values (23). A comparison is given in Fig.3 and error graphs by two types of conditions at n=10 for
Ex.2 are shown in Fig.4 respectively. The cubic spline values obtained by these two types of
conditions in the intervals xi1, xi , for i = 1,2 ,10 are shown in Table.8 with their corresponding errors and exact values. The absolute errors with endpointslope conditions for examples 1 and 2 at h=0.5 are compared with Type I conditions in Table 9.
4 Conclusions
In the present work we applied the cubic spline interpolation method with different types of
conditions on M0 and Mn to approximate the functions at different values of x in examples 1 and 2.The results are given in the tabular for, and shown graphically. From the numericalcomputations it is observed that the error in estimation of y is considerably reduced by the endpointslope conditions. Further, it is observed that as h is decreasing the estimate is very close to
the exactvalue.
Table 1: Spline functions with endpointslope conditions in the corresponding intervals
Interval Cubic spline function
[1,1.5] 4.0625 (1.5x)3+21.72189 (x1)3+4.98437(1.5x)+12.75703(x1) [1.5,2] 21.72189 (2x)3+51.54993 (x1.5)3+12.75703(2x)+53.11252(x1.5) [2,2.5] 51.54993 (2.5x)3+102.0784 (x2)3+53.11252(2.5x)+170.7929(x2) [2.5,3] 102.0784(3x)3+177.6365(x2.5)3+170.7929(3x)+441.5909(x2.5) [3,3.5] 177.6365(3.5x)3+282.3757(x3)3+441.5909(3.5x)+978.8436(x3) [3.5,4] 282.3757(4x)3+425.3607(x3.5)3+978.8436(4x)+1939.66(x3.5)[4,4.5] 425.3607(4.5x)3+596.1813(x4)3+1939.66(4.5x)+3538.517(x4)
[4.5,5] 596.1813(5x)3+857.414(x4.5)3+3538.517(5x)+6031.647(x4.5) [5,5.5] 857.414(5.5x)3+999.1624(x5)3+6031.647(5.5x)+9810.897(x5) [5.5,6] 999.1624(6x)3+1828.438(x5.5)3+9810.897(6x)+15088.89(x5.5)Solution
Solution
Endpointslope Type I
Exact
x values
Fig.1: Comparison of approximate values and exact values for Ex.1
Table 2: Spline functions with Type I conditions in the corresponding intervals
Interval Cubic spline function
[1,1.5] 11.41126 (x1)3+3(1.5x) + 6.240935 (x1) [1.5,2] 11.41126 (2x)3 +25.60496 (x1.5)3+6.240935 (2x)+ 26.59876 (x1.5) [2,2.5] 25.60496 (2.5x)3 +51.16891 (x2)3+26.59876 (2.5x)+ 85.36402 (x2) [2.5,3] 51.16891 (3x)3+88.46942 (x2.5)3+85.36402 (3x)+ 220.8826 (x2.5) [3,3.5] 88.46942 (3.5x)3+142.4534 (x3)3+220.8826 (3.5x)+ 489.1054 (x3) [3.5,4] 142.4534 (4x)3+207.967 (x3.5)3+489.1054 (4x)+ 971.0082 (x3.5)[4,4.5] 207.967 (4.5x)3+315.6785 (x4)3+971.0082 (4.5x)+ 1764.862 (x4)
[4.5,5] 315.6785 (5x)3+363.0691 (x4.5)3+1764.862 (5x)+ 3032.233 (x4.5) [5,5.5] 363.0691 (5.5x)3+744.5452 (x5)3+3032.233 (5.5x)+ 4844.207 (x5) [5.5,6] 744.5452 (6x) 3 +4844.207 (6x) + 7773 (x5.5)Table 3: Approximate values, exact values and errors of Ex 1 in [1, 1.5]
Using End – point slope condition Using Type I condition
x
Values of y 
Values of y 

1.05 
3.226282 
3.253731 
0.02745 
3.226282 
3.326946 
0.10066 
1.1 
3.51051 
3.551175 
0.04066 
3.51051 
3.67101 
0.1605 
1.15 
3.861357 
3.905576 
0.04422 
3.861357 
4.049307 
0.18795 
1.2 
4.28832 
4.330181 
0.04186 
4.28832 
4.478954 
0.19063 
1.25 
4.801758 
4.838232 
0.03647 
4.801758 
4.977069 
0.17531 
1.3 
5.41293 
5.442974 
0.03004 
5.41293 
5.560769 
0.14784 
1.35 
6.134033 
6.157653 
0.02362 
6.134033 
6.24717 
0.11314 
1.4 
6.97824 
6.995512 
0.01727 
6.97824 
7.053389 
0.07515 
1.45 
7.959734 
7.969796 
0.01006 
7.959734 
7.996544 
0.03681 
Values of y 
Values of y 

1.05 
3.226282 
3.253731 
0.02745 
3.226282 
3.326946 
0.10066 
1.1 
3.51051 
3.551175 
0.04066 
3.51051 
3.67101 
0.1605 
1.15 
3.861357 
3.905576 
0.04422 
3.861357 
4.049307 
0.18795 
1.2 
4.28832 
4.330181 
0.04186 
4.28832 
4.478954 
0.19063 
1.25 
4.801758 
4.838232 
0.03647 
4.801758 
4.977069 
0.17531 
1.3 
5.41293 
5.442974 
0.03004 
5.41293 
5.560769 
0.14784 
1.35 
6.134033 
6.157653 
0.02362 
6.134033 
6.24717 
0.11314 
1.4 
6.97824 
6.995512 
0.01727 
6.97824 
7.053389 
0.07515 
1.45 
7.959734 
7.969796 
0.01006 
7.959734 
7.996544 
0.03681 
Exact values of y
Approximate
Error
Exact
Approximate
values of y Error
x values
Error
Error
Endpointslope Type I
Fig.2: Comparison of errors for Ex.1
Table 4: Approximate values, exact values and errors of Ex. 1
Using End – point slope condition Using Type I condition
x Exact Approximate 
Exact Approximate 

values of y Error 
values of y Error 

Values of y 
Values of y 

1.1 
3.51051 
3.551175 
0.04066 
3.51051 
3.67101 
0.1605 
1.6 
11.88576 
11.85581 
0.029946 
11.88576 
11.82435 
0.061409 
2.3 
65.06343 
65.02889 
0.03454 
65.06343 
65.03072 
0.032712 
2.6 
119.2138 
119.1869 
0.026858 
119.2138 
119.1943 
0.019454 
3.1 
286.1915 
286.1718 
0.01969 
286.1915 
286.1362 
0.055321 
3.7 
692.7396 
692.6121 
0.127504 
692.7396 
692.8865 
0.14693 
4.3 
1468.784 
1468.987 
0.20247 
1468.784 
1467.694 
1.090045 
4.8 
2546.24 
2545.117 
1.122665 
2546.24 
2549.941 
3.7012 
5.2 
3799.84 
3802.817 
2.97649 
3799.84 
3788.541 
11.29925 
5.6 
5504.718 
5499.023 
5.695105 
5504.718 
5525.268 
20.55 
Table 5: Spline functions with endpointslope conditions in the corresponding intervals
Interval Cubic Spline function
[1,1.5] 0.950728(1.5x)3+0.494674(x1)3+1.762318(1.5x)+4.728514(x1) [1.5,2] 0.494674 (2x)3+0.661355 (x1.5)3+4.728514 (2x)+ 8.436721 (x1.5) [2,2.5] 0.661355(2.5x)3+0.635674(x2)3+8.436721(2.5x)+ 13.13696 (x2) [2.5,3] 0.635674 (3x)3+0.654117 (x2.5)3+13.13696 (3x)+ 18.79071 (x2.5) [3,3.5] 0.654117 (3.5x)3+0.649982 (x3)3+18.79071 (3.5x)+ 25.42564 (x3) [3.5,4] 0.649982(4x)3+0.674316 (x3.5)3+25.42564 (4x)+ 33.03554 (x3.5) [4,4.5] 0.674316 (4.5x)3+0.598039 (x4)3+33.03554 (4.5x)+ 41.65692 (x4) [4.5,5] 0.598039 (5x)3+0.890368 (x4.5)3+41.65692 (5x)+ 51.17535 (x4.5) [5,5.5] 0.890368 (5.5x)30.19443 (x5)3+51.17535 (5.5x)+ 62.02933 (x5) [5.5,6] 0.19443 (6x)3+3.858526 (x5.5)3+62.02933 (6x)+ 72.59167 (x5.5)Table 6: Spline functions with Type I conditions in the corresponding intervals
Interval Cubic spline function
[1,1.5] 0.749447(x1)3+(1.5x)+4.664821(x1) [1.5,2] 0.749447(2x)3+0.592994(x1.5)3+4.6648212x)+8.453811(x1.5) [2,2.5] 0.592994(2.5x)3+0.654347(x2)3+8.453811(2.5x)+13.13229(x2) [2.5,3] 0.654347(3x)3+0.647788(x2.5)3+13.13229(3x)+18.7923(x2.5) [3,3.5] 0.647788(3.5x)3+0.656625 (x3)3+18.7923(3.5x)+25.42398(x3) [3.5,4] 0.656625(4x)3+0.654074(x3.5)3+25.42398(4x)+33.0406(x3.5) [4,4.5] 0.654074(4.5x)3+0.672363(x4)3+33.0406(4.5x)+41.63833(x4) [4.5,5] 0.672363(5x)3+0.613313(x4.5)3+41.63833(5x)+51.24461(x4.5) [5,5.5] 0.613313(5.5x)30.839463(x5)3+51.24461(5.5x)+61.77086(x5)[5.5,6] 0.839463(6x)3+61.77086 (6x)+73.5563(x5.5)
Table 7: Approximate values, exact values and errors of Ex 2 in [1, 1.5]
Using endpointslope conditions Using Type I. conditions
x Exact Approximate 
Exact Approximate 

values of y Error 
values of y Error 

Values of y 
Values of y 

1.05 
1.123689 
1.116166 
0.007524 
1.123689 
1.133335 
0.00965 
1.1 
1.251393 
1.23912 
0.012273 
1.251393 
1.267232 
0.01584 
1.15 
1.383198 
1.36852 
0.014677 
1.383198 
1.402253 
0.01905 
1.2 
1.519181 
1.504025 
0.015156 
1.519181 
1.53896 
0.01978 
1.25 
1.65941 
1.645292 
0.014118 
1.65941 
1.677915 
0.01851 
1.3 
1.803943 
1.79198 
0.011964 
1.803943 
1.819681 
0.01574 
1.35 
1.952834 
1.943745 
0.009088 
1.952834 
1.96482 
0.01199 
1.4 
2.106128 
2.100247 
0.005881 
2.106128 
2.113893 
0.00776 
1.45 
2.263868 
2.261143 
0.002725 
2.263868 
2.267463 
0.00359 
solution
solution
Endpointslope Type I
exact
x values
Fig.3: Comparison of approximate values and exact values for Ex.2
x values
y values
y values
endpointslope Type I
Fig.4: Error graph by two types of conditions at n=10 for Ex.2
Table 8: Approximate values, exact values and errors for Ex 2 in , , =
, ,
Using End – point slope condition Using Type I condition
x Exact
Values of y 
Values of y 

1.4 
2.106128 
2.100247 
0.005881 
2.106128 
2.113893 
0.00776 
1.9 
3.888754 
3.890361 
0.00161 
3.888754 
3.886708 
0.002046 
2.4 
6.140211 
6.139801 
0.00041 
6.140211 
6.14077 
0.00056 
2.8 
8.287158 
8.287353 
0.00019 
8.287158 
8.286872 
0.000286 
3.1 
10.10136 
10.10136 
1.2E06 
10.10136 
10.10143 
7E05 
3.8 
15.01978 
15.0192 
0.000587 
15.01978 
15.01989 
0.00011 
4.1 
17.42278 
17.42366 
0.00088 
17.42278 
17.42261 
0.000177 
4.9 
24.7002 
24.69341 
0.006784 
24.7002 
24.7016 
0.00141 
5.1 
26.71757 
26.72986 
0.01229 
26.71757 
26.71502 
0.002548 
5.7 
33.24587 
33.15275 
0.093122 
33.24587 
33.26518 
0.01931 
Values of y 
Values of y 

1.4 
2.106128 
2.100247 
0.005881 
2.106128 
2.113893 
0.00776 
1.9 
3.888754 
3.890361 
0.00161 
3.888754 
3.886708 
0.002046 
2.4 
6.140211 
6.139801 
0.00041 
6.140211 
6.14077 
0.00056 
2.8 
8.287158 
8.287353 
0.00019 
8.287158 
8.286872 
0.000286 
3.1 
10.10136 
10.10136 
1.2E06 
10.10136 
10.10143 
7E05 
3.8 
15.01978 
15.0192 
0.000587 
15.01978 
15.01989 
0.00011 
4.1 
17.42278 
17.42366 
0.00088 
17.42278 
17.42261 
0.000177 
4.9 
24.7002 
24.69341 
0.006784 
24.7002 
24.7016 
0.00141 
5.1 
26.71757 
26.72986 
0.01229 
26.71757 
26.71502 
0.002548 
5.7 
33.24587 
33.15275 
0.093122 
33.24587 
33.26518 
0.01931 
values of y
Approximate
Error
Exact
Approximate
values of y Error
Table 9: The absolute errors with two types of conditions
Ex 7.1 Ex 7. 2
condition 
condition 

1.2 
0.04186 
0.19063 
0.015156 
0.01978 
1.25 
0.03647 
0.17531 
0.014118 
0.01851 
1.3 
0.03004 
0.14784 
0.011964 
0.01574 
1.35 
0.02362 
0.11314 
0.009088 
0.01199 
1.4 
0.01727 
0.07515 
0.005881 
0.00776 
1.45 
0.01006 
0.03681 
0.002725 
0.00359 
condition 
condition 

1.2 
0.04186 
0.19063 
0.015156 
0.01978 
1.25 
0.03647 
0.17531 
0.014118 
0.01851 
1.3 
0.03004 
0.14784 
0.011964 
0.01574 
1.35 
0.02362 
0.11314 
0.009088 
0.01199 
1.4 
0.01727 
0.07515 
0.005881 
0.00776 
1.45 
0.01006 
0.03681 
0.002725 
0.00359 
x
Endpointslope
Type I condition Endpointslope
Type I condition
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