51 Slices
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ALLC3-3

Manish Goyal Laxmi Publications PDF

103

ALGEBRAIC AND TRANSCENDENTAL EQUATIONS

= .56 −

FG .56716 − .56 IJ (− .01121)

H .00002619 + .01121K

= .567143

Since, x2 and x3 agree upto four decimal places, the required root correct to three decimal places is 0.567.

(ii) Let

f(x) = xex – 2

Since,

f(.852) = – .00263

and

f(.853) = .001715

Hence, root lies between .852 and .853.

Let

x0 = .852 and x1 = .853

Using method of false position,

RS x − x UV f (x )

T f (x ) − f (x ) W

R .853 − .852 UV (− .00263)

= .852 − S

T.001715 − (− .00263) W

1

x2 = x 0 −

0

1

0

0

= .852605293

Now

f(x2) = – .00000090833

Hence, root lies between .852605293 and .853

Using method of false position,

RS x − x UV f (x )

| Replacing x by x

T f (x ) − f (x ) W

R . 853 − . 852605293 UV (− .00000090833)

= (.852605293) – S

T.001715 − (−.00000090833) W

x3 = x 2 −

1

2

1

2

2

0

2

= 0.852605501

Since, x2 and x3 agree upto 6 decimal places, hence the required root correct to 4 decimal places is 0.8526.

Example 8. (i) Solve x3 – 5x + 3 = 0 by using Regula-Falsi method.

(ii) Use the method of false position to solve x3 – x – 4 = 0.

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ALLC10-3

Manish Goyal Laxmi Publications PDF

593

CURVE-FITTING, CUBIC SPLINE AND APPROXIMATION

Sol. The normal equations to the curve are

Σy = aΣx 2 + bΣx + 5c

Σxy = aΣx 3 + bΣx 2 + cΣx

Σx 2 y = aΣx 4 + bΣx 3 + cΣx 2

and

U|

V|

W

…(1)

The values of Σx, Σx2,...... etc., are calculated by means of the following table : x

y

x2

x3

x4

xy

x2 y

10

12

15

23

20

14

17

23

25

21

100

144

225

529

400

1000

1728

3375

12167

8000

10000

20736

50625

279841

160000

140

204

345

575

420

1400

2448

5175

13225

8400

Σx = 80

Σy = 100

Σx2 = 1398

Σx3 = 26270

Σx4 = 521202

Σxy = 1684 Σx2y = 30648

Substituting the obtained values from the table in normal equation (1), we have

100 = 1398a + 80b + 5c

1684 = 26270a + 1398b + 80c

30648 = 521202a + 26270b + 1398c

On solving,

a = – 0.07, b = 3.03, c = – 8.89

∴ The required equation is y = – 0.07x2 + 3.03x – 8.89.

Example 3. Fit a parabolic curve of regression of y on x to the following data : x:

1.0

1.5

2.0

2.5

3.0

3.5

4.0

y:

1.1

1.3

1.6

2.0

2.7

3.4

4.1

Sol. Here

m = 7 (odd)

x − 2.5

= 2x – 5

0.5

Results in tabular form are

Let

u=

and v = y

x

y

u

v

u2

uv

u2 v

u3

u4

1.0

1.1

–3

1.1

9

– 3.3

9.9

– 27

81

1.5

1.3

–2

1.3

4

– 2.6

5.2

–8

16

2.0

1.6

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ALLC5-1

Manish Goyal Laxmi Publications PDF

5

Numerical Integration and Differentiation

5.1. INTRODUCTION

Consider a function of a single variable y = f(x). If f(x) is defined as an expression, its derivative or integral may often be determined using the techniques of calculus.

However, when f(x) is a complicated function or when it is given in a tabular form, we use numerical methods.

In this chapter, we will discuss numerical methods for approximating the derivative(s) f (r)(x), r ≥ 1 of a given function f(x) and for the evaluation of the integral

z

b

a

f ( x) dx where a, b

may be finite or infinite.

The accuracy attainable by these methods would depend on the given function and the order of the polynomial used. If the polynomial fitted is exact then the error would be theoretically zero. In practice, however, rounding errors will introduce errors in the calculated values.

The error introduced in obtaining derivatives is in general much worse than that introduced in determining integrals.

It may be observed that any errors in approximating a function are amplified while taking the derivative whereas they are smoothed out in integration.

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ALLC10-2

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578

NUMERICAL METHODS AND STATISTICAL TECHNIQUES USING ‘C’

10.17. FITTING OF THE CURVE 2x = ax2 + bx + c

Normal equations are

Σ 2xx2 = aΣx4+ bΣx3 + cΣx2

Σ 2x x = aΣx3 + bΣx2 + cΣx and

Σ 2x = aΣx2 + bΣx + mc where m is number of points (xi, yi)

10.18. FITTING OF THE CURVE y = ae�3x + be�2x

and

Normal equations are

Σye–3x = a Σe–6x + b Σe–5x

Σye–2x = a Σe–5x + b Σe–4x

EXAMPLES

Example 1. Find the curve of best fit of the type y = aebx to the following data by the method of Least squares : x:

1

5

7

9

12 y:

10

15

12

15

21.

Sol. The curve to be fitted is y = aebx or

Y = A + Bx, where Y = log10 y, A = log10 a, and B = b log10 e

∴ The normal equations are ΣY = 5A + BΣx and

ΣxY = AΣx + BΣx2 x

y

Y = log10 y

x2

xy

1

5

7

9

12

10

15

12

15

21

1.0000

1.1761

1.0792

1.1761

1.3222

1

25

49

81

144

1

5.8805

7.5544

10.5849

15.8664

Σx = 34

and

ΣY = 5.7536

Σx2 = 300

ΣxY = 40.8862

Substituting the above values in the normal equations, we get

5.7536 = 5A + 34B

40.8862 = 34A + 300B

On solving A = 0.9766 ; B = 0.02561

∴ a = antilog10 A = 9.4754 ; b =

Hence the required curve is

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ALLC10-1

Manish Goyal Laxmi Publications PDF

10

Curve-Fitting, Cubic

Spline and Approximation

10.1. CURVE FITTING

Let there be two variables x and y which give us a set of n pairs of numerical values

(x1, y1), (x2, y2).......(xn, yn). In order to have an approximate idea about the relationship of these two variables, we plot these n paired points on a graph thus, we get a diagram showing the simultaneous variation in values of both the variables called scatter or dot diagram. From scatter diagram, we get only an approximate non-mathematical relation between two variables.

Curve fitting means an exact relationship between two variables by algebraic equations, infact this relationship is the equation of the curve. Therefore, curve fitting means to form an equation of the curve from the given data. Curve fitting is considered of immense importance both from the point of view of theoretical and practical statistics.

Theoretically, it is useful in the study of correlation and regression. Practically, it enables us to represent the relationship between two variables by simple algebraic expressions e.g., polynomials, exponential or logarithmic functions.

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