Dr. Nirav Vyas numerical method 4.pdf

8/20/2019 Dr. Nirav Vyas numerical method 4.pdf

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Numerical Methods - Interpolation

Unequal Intervals

Dr. N. B. Vyas

Department of Mathematics,Atmiya Institute of Tech. and Science, Rajkot (Guj.)

[email protected]

Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals


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Interpolation

To find the value of y for an x between different x - valuesx0, x1, . . . , xn is called problem of interpolation.



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Interpolation

To find the value of y for an x between different x - values

x0, x1, . . . , xn is called problem of interpolation.

To find the value of y for an x which falls outside the range of x(x < x0 or x > xn) is called the problem of extrapolation.



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Interpolation




Theorem by Weierstrass in 1885, “Every continuous

function in an interval (a,b) can be represented in thatinterval to any desired accuracy by a polynomial. ”



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Interpolation






Let us assign polynomial P n of degree n (or less) that assumesthe given data values



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Interpolation







P n(x0) = y0, P n(x1) = y1, . . ., P n(xn) = yn



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Interpolation







P n(x0) = y0, P n(x1) = y1, . . ., P n(xn) = ynThis polynomial P n is called interpolation polynomial.



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Interpolation







P n(x0) = y0, P n(x1) = y1, . . ., P n(xn) = ynThis polynomial P n is called interpolation polynomial.

x0, x1, . . . , xn is called the nodes ( tabular points, pivotalpoints or arguments).



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Interpolation with unequal intervals

Lagrange’s interpolation formula with unequal intervals:

Let y = f (x) be continuous and differentiable in the interval(a, b).



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Given the set of n + 1 values (x0, y0), (x1, y1), . . . , (xn, yn) of xand y, where the values of x need not necessarily be equallyspaced.



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It is required to find P n(x), a polynomial of degree n such that y

and P

n(x

) agree at the tabulated points.



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It is required to find P n(x), a polynomial of degree n such that yand P

n(x) agree at the tabulated points.



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Lagrange’s Interpolation

This polynomial is given by the following formula:



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y = f (x) ≈ P n(x) = (x − x1)(x − x2) . . . (x − xn)(x0 − x1)(x0 − x2) . . . (x0 − xn) y0



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+

(x

−x0)(x

−x2) . . . (x

−xn)

(x1 − x0)(x1 − x2) . . . (x1 − xn) y1 + . . .



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+

(x

−x0)(x

−x2) . . . (x

−xn)

(x1 − x0)(x1 − x2) . . . (x1 − xn) y1 + . . .+

(x − x0)(x − x1) . . . (x − xn−1)(xn − x0)(xn − x1) . . . (xn − xn−1) yn


’


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+

(x

−x0)(x

−x2) . . . (x

−xn)

(x1 − x0)(x1 − x2) . . . (x1 − xn) y1 + . . .+

(x − x0)(x − x1) . . . (x − xn−1)(xn − x0)(xn − x1) . . . (xn − xn−1) yn

NOTE:

The above formula can be used irrespective of whether the valuesx0, x1, . . . , xn are equally spaced or not.


L ’ I I


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Lagrange’s Inverse Interpolation

In the Lagrange’s interpolation formula y is treated as dependentvariable and expressed as function of independent variable x.


L ’ I I


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Instead if x is treated as dependent variable and expressed as thefunction of independent variable y, then Lagrange’s interpolationformula becomes


L ’ I I


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x = g(y) ≈ P n(y) = (y − y1)(y − y2) . . . (y − yn)(y0 − y1)(y0 − y2) . . . (y0 − yn) x0




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+ (y − y0)(y − y2) . . . (y − yn)(y1 − y0)(y1 − y2) . . . (y1 − yn) x1 + . . .




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+ (y

−y

0)(y

−y

1) . . . (y

−yn−1

)

(yn − y0)(yn − y1) . . . (yn − yn−1) xn




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+ (y

−y

0)(y

−y

1) . . . (y

−yn−1

)

(yn − y0)(yn − y1) . . . (yn − yn−1) xnThis relation is referred as Lagrange’s inverse interpolationformula.


Example


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Example

Ex. Given the table of values:

x 150 152 154 156y = √ x 12.247 12.329 12.410 12.490

Evaluate√

155 using Lagrange’s interpolation formula.


Example


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Example

Sol.: Here x0 = 150, x1 = 152, x2 = 154 and x3 = 156


Example


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Example

Sol.: Here x0 = 150, x1 = 152, x2 = 154 and x3 = 156

y0 = 12

.247,

y1 = 12

.329,

y2 = 12

.410 and

y3 = 12

.490


Example


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Example

Sol.: Here x0 = 150, x1 = 152, x2 = 154 and x3 = 156

y0 = 12

.247,

y1 = 12

.329,

y2 = 12

.410 and

y3 = 12

.490By Lagrange’s interpolation formula,


Example


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Example

Sol.: Here x0 = 150, x1 = 152, x2 = 154 and x3 = 156

y0 = 12

.247,

y1 = 12

.329,

y2 = 12

.410 and

y3 = 12

.490By Lagrange’s interpolation formula,

f (x) ≈ P n(x) = (x − x1)(x − x2)(x − x3)(x0

−x1)(x0

−x2)(x0

−x3)

y0


Example


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Example

Sol.: Here x0 = 150, x1 = 152, x2 = 154 and x3 = 156

y0

= 12.247, y1

= 12.329, y2

= 12.410 and y3

= 12.490

By Lagrange’s interpolation formula,

f (x) ≈ P n(x) = (x − x1)(x − x2)(x − x3)(x0

−x1)(x0

−x2)(x0

−x3)

y0

+ (x − x0)(x − x2)(x − x3)

(x1 − x0)(x1 − x2)(x1 − x3) y1


Example


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Example

Sol.: Here x0 = 150, x1 = 152, x2 = 154 and x3 = 156

y0

= 12.247, y1

= 12.329, y2

= 12.410 and y3

= 12.490


f (x) ≈ P n(x) = (x − x1)(x − x2)(x − x3)(x0

−x1)(x0

−x2)(x0

−x3)

y0

+ (x − x0)(x − x2)(x − x3)(x1 − x0)(x1 − x2)(x1 − x3) y1

+ (x − x0)(x − x1)(x − x3)(x2 − x0)(x2 − x1)(x2 − x3) y2


Example


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Example

Sol.: Here x0 = 150, x1 = 152, x2 = 154 and x3 = 156

y0

= 12.247, y1

= 12.329, y2

= 12.410 and y3

= 12.490


f (x) ≈ P n(x) = (x − x1)(x − x2)(x − x3)(x0

−x1)(x0

−x2)(x0

−x3)

y0

+ (x − x0)(x − x2)(x − x3)(x1 − x0)(x1 − x2)(x1 − x3) y1

+ (x − x0)(x − x1)(x − x3)(x2 − x0)(x2 − x1)(x2 − x3) y2

+ (x − x0)(x − x1)(x − x2)(x3 − x0)(x3 − x1)(x3 − x2) y3


Example


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Sol.: Here x0 = 150, x1 = 152, x2 = 154 and x3 = 156

y0

= 12.247, y1

= 12.329, y2

= 12.410 and y3

= 12.490


f (x) ≈ P n(x) = (x − x1)(x − x2)(x − x3)(x0

−x1)(x0

−x2)(x0

−x3)

y0

+ (x − x0)(x − x2)(x − x3)(x1 − x0)(x1 − x2)(x1 − x3) y1

+ (x − x0)(x − x1)(x − x3)(x2 − x0)(x2 − x1)(x2 − x3) y2

+ (x − x0)(x − x1)(x − x2)(x3 − x0)(x3 − x1)(x3 − x2) y3

for x = 155

∴ f (155) =


Example


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Ex. Compute f (0.4) for the table below by the Lagrange’sinterpolation:

x 0.3 0.5 0.6

f (x) 0.61 0.69 0.72


Example


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Ex. Using Lagrange’s formula, find the form of f (x) for the followingdata:

x 0 1 2 5

f (x) 2 3 12 147


Example


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Ex. Using Lagrange’s formula, find x for y = 7 for the following data:

x 1 3 4y 4 12 19


Example


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Ex. Using Lagrange’s formula, express the function

3x2

+ x + 1(x − 1)(x − 2)(x − 3)


Example


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Sol.: Let us evaluate y = 3x2

+ x + 1 for x = 1, x = 2 and x = 3


Example


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+ x + 1 for x = 1, x = 2 and x = 3These values are x0 = 1, x1 = 2 and x2 = 3 and


Example


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y0 = 5,


Example


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y0 = 5, y1 = 15 and y2 = 31



Example


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y0 = 5, y1 = 15 and y2 = 31


y = (x − x1)(x − x2)(x0 − x1)(x0 − x2) y0 +

(x − x0)(x − x2)(x1 − x0)(x1 − x2) y1


Example


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y0 = 5, y1 = 15 and y2 = 31


y = (x − x1)(x − x2)(x0 − x1)(x0 − x2) y0 +

(x − x0)(x − x2)(x1 − x0)(x1 − x2) y1

+ (x − x0)(x − x1)(x2

−x0)(x2

−x1)

y2


Example


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y0 = 5, y1 = 15 and y2 = 31


y = (x − x1)(x − x2)(x0 − x1)(x0 − x2) y0 +

(x − x0)(x − x2)(x1 − x0)(x1 − x2) y1

+ (x − x0)(x − x1)(x2

−x0)(x2

−x1)

y2

substituting above values, we get

y = 2.5(x − 2)(x − 3) − 15(x − 1)(x − 3) + 15.5(x − 1)(x − 2)


Example


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y0 = 5, y1 = 15 and y2 = 31


y = (x − x1)(x − x2)(x0 − x1)(x0 − x2) y0 +

(x − x0)(x − x2)(x1 − x0)(x1 − x2) y1

+ (x − x0)(x − x1)(x2

−x0)(x2

−x1)

y2

substituting above values, we get

y = 2.5(x − 2)(x − 3) − 15(x − 1)(x − 3) + 15.5(x − 1)(x − 2)


Example


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Thus3x2 + x + 1

(x − 1)(x − 2)(x − 3)


Example


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Thus3x2 + x + 1

(x − 1)(x − 2)(x − 3)

= 2.5(x − 2)(x − 3) − 15(x − 1)(x − 3) + 15.5(x − 1)(x − 2)(x − 1)(x − 2)(x − 3)

= 2.5

(x

−1)

- 15

(x

−2)

+ 15.5

(x

−3)


Error in Interpolation


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Error in Interpolation:

We assume that f (x) has continuous derivatives of order upton + 1 for all x ∈ (a, b). Since, f (x) is approximated by P n(x), theresults contains errors. We define the error of interpolation ortruncation error as




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E (f, x) = f (x) − P n(x) = (x

−x0)(x

−x1) . . . (x

−xn)

(n + 1)! f (n+1)

(ξ )

where min(x0, x1, . . . , xn, x) < ξ < min(x0, x1, . . . , xn, x)




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E (f, x) = f (x) − P n(x) = (x

−x0)(x

−x1) . . . (x

−xn)

(n + 1)! f (n+1)

(ξ )


since, ξ is an unknown, it is difficult to find the value of error.However, we can find a bound of the error. The bound of the

error is obtained as




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E (f, x) = f (x) − P n(x) = (x

−x0)(x

−x1) . . . (x

−xn)

(n + 1)! f (n+1)

(ξ )


since, ξ is an unknown, it is difficult to find the value of error.However, we can find a bound of the error. The bound of the

error is obtained as

|E (f, x)| ≤ |(x − x0)(x − x1) . . . (x − xn)|(n + 1)!

maxa≤ξ≤b

|f (n+1)(ξ )|


Example


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Ex. Using the data sin(0.1) = 0.09983 and sin(0.2) = 0.19867, find

an approximate value of sin(0.15) by Lagrange interpolation.Obtain a bound on the error at x = 0.15.




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Disadvantages:In practice, we often do not know the degree of the interpolationpolynomial that will give the required accuracy, so we should beprepared to increase the degree if necessary.




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To increase the degree the addition of another interpolation pointleads to re-computation.

i.e. no previous work is useful.




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E.g: In calculating P k(x), no obvious advantage can be taken of the fact that one already has calculated P k−1(x).




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E.g: In calculating P k(x), no obvious advantage can be taken of the fact that one already has calculated P k−1(x).

That means we need to calculate entirely new polynomial.


Divided Difference


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Let f

(x

0), f

(x

1), . . . , f

(xn) be the values of a function

f

corresponding to the arguments x0, x1, . . . , xn where the intervalsx1 − x0, x2 − x1, . . . , xn − xn−1 are not necessarily equally spaced.


Divided Difference


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Let f

(x

0), f

(x

1), . . . , f


f

corresponding to the arguments x0, x1, . . . , xn where the intervalsx1 − x0, x2 − x1, . . . , xn − xn−1 are not necessarily equally spaced.Then the first divided difference of f for the argumentsx0, x1, . . . , xn are defined by ,


Divided Difference


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Let f

(x

0), f

(x

1), . . . , f


f

corresponding to the arguments x0, x1, . . . , xn where the intervalsx1 − x0, x2 − x1, . . . , xn − xn−1 are not necessarily equally spaced.Then the first divided difference of f for the argumentsx0, x1, . . . , xn are defined by ,

f (x0, x1) = f (x1) − f (x0)

x1 − x0


Divided Difference


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Let f (x0

), f (x1

), . . . , f (xn

) be the values of a function f corresponding to the arguments x0, x1, . . . , xn where the intervalsx1 − x0, x2 − x1, . . . , xn − xn−1 are not necessarily equally spaced.Then the first divided difference of f for the argumentsx0, x1, . . . , xn are defined by ,

f (x0, x1) = f (x1) − f (x0)

x1 − x0

f (x1, x2) = f (x2) − f (x1)

x2 − x1


Divided Difference


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The second divided difference of f for three argumentsx0, x1, x2 is defined by

f (x0, x1, x2) = f (x1, x2) − f (x0, x1)

x2

−x0


Divided Difference


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The second divided difference of f for three argumentsx0, x1, x2 is defined by

f (x0, x1, x2) = f (x1, x2) − f (x0, x1)

x2

−x0

and similarly the divided difference of order n is defined by

f (x0, x1, . . . , xn) =

f (x1, x2, . . . , xn)

−f (x0, x1, . . . , xn−1)

xn − x0


Divided Difference


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Properties:

The divided differences are symmetrical in all their arguments;that is, the value of any divided difference is independent of the

order of the arguments.


Divided Difference


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Properties:


order of the arguments.The divided difference operator is linear.


Divided Difference


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Properties:


order of the arguments.The divided difference operator is linear.

The nth order divided differences of a polynomial of degree n areconstant, equal to the coefficient of xn.


Newton’s Divided Difference Interpolation

A i t l ti f l hi h h th t th t


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An interpolation formula which has the property that apolynomial of higher degree may be derived from it by simply

adding new terms.





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adding new terms.Newton’s general interpolation formula is one such formula andterms in it are called divided differences.





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Let f (x0), f (x1), . . . , f (xn) be the values of a function f

corresponding to the arguments x0, x1, . . . , xn where the intervalsx1 − x0, x2 − x1, . . . , xn − xn−1 are not necessarily equally spaced.By the definition of divided difference,



An interpolation formula which has the property that a


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f (x, x0) =

f (x)

−f (x0)

x − x0



An interpolation formula which has the property that a


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f (x, x0) =

f (x)

−f (x0)

x − x0

∴

f (x) = f (x0) + (x − x0)f (x, x0) −−(1)Dr. N. B. Vyas Numerical Methods - Interpolation Unequal Intervals



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Further

f (x, x0, x1) =

f (x, x0)

−f (x0, x1)

x − x1




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Further

f (x, x0, x1) =

f (x, x0)

−f (x0, x1)

x − x1which yields

f (x, x0) = f (x0, x1) + (x − x1)f (x, x0, x1) −−(2)




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Further

f (x, x0, x1) =

f (x, x0)

−f (x0, x1)


f (x, x0) = f (x0, x1) + (x − x1)f (x, x0, x1) −−(2)

Similarly

f (x, x0, x1) = f (x0, x1, x2) + (x − x2)f (x, x0, x1, x2) −−(3)




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Further

f (x, x0, x1) =

f (x, x0)

−f (x0, x1)


f (x, x0) = f (x0, x1) + (x − x1)f (x, x0, x1) −−(2)

Similarly

f (x, x0, x1) = f (x0, x1, x2) + (x − x2)f (x, x0, x1, x2) −−(3)

and in general

f (x, x0,...,xn−1) = f (x0, x1,...,xn) + (x−xn)f (x, x0, x1,...,xn)−−(4)




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multiplying equation (2) by (x − x0)




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multiplying equation (2) by (x − x0) , (3) by (x − x0) (x − x1)and so on,




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multiplying equation (2) by (x − x0) , (3) by (x − x0) (x − x1)and so on, and finally the last term (4) by(x − x0) (x − x1) ... (x − xn−1) and




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multiplying equation (2) by (x − x0) , (3) by (x − x0) (x − x1)and so on, and finally the last term (4) by(x − x0) (x − x1) ... (x − xn−1) and adding (1), (2) , (3) up to (4)we obtain




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multiplying equation (2) by (x − x0) , (3) by (x − x0) (x − x1)and so on, and finally the last term (4) by(x − x0) (x − x1) ... (x − xn−1) and adding (1), (2) , (3) up to (4)we obtain

f (x) =f (x0) + (x − x0) f (x0, x1) + (x − x0) (x − x1) f (x0, x1, x2) + ... +(x − x0) (x − x1) ... (x − xn−1) f (x0, x1,...,xn)This formula is called Newton’s divided difference formula.




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The divided difference upto third order

x y 1stdiv.diff. 2nddiv.diff. 3rddiv.diff.

x0 y0

[x0, x1]x1 y1 [x0, x1, x2][x1, x2] [x0, x1, x2, x3]

x2 y2 [x1, x2, x3][x2, x3]

x3 y3


Example


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Ex. Obtain the divided difference table for the data:

x -1 0 2 3y -8 3 1 12


Example

S l W h th f ll i di id d diff t bl f th d t


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Sol. We have the following divided difference table for the data:

x y First d.d Second d.d Third d.d

-1 -83 + 8

0 + 1 = 11

0 3 −1 − 112 + 1

= −41 − 32 − 0 = −1

4 + 4

3 + 1 = 2

2 1 11 + 1

3−

0 = 4

12 − 13 − 2 = 11

3 12


Example


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Ex. Find f (x) as a polynomial in x for the following data by Newtonsdivided difference formula:

x -4 -1 0 2 5

f (x) 1245 33 5 9 1335


Example


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Sol. We have the following divided difference table for the data:

x y 1st d.d 2nd d.d 3rd d.d 4th d.d

-4 1245−404

-1 33 94−28 −140 5 10 3

2 132 9 88

4425 1335


Example


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The Newtons divided difference formula gives:


Example


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f (x) = f (x0) +


Example


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f (x) = f (x0) + (x − x0)f [x0, x1] + (x − x0)(x − x1)f [x0, x1, x2]


Example


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+


Example


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+ (x − x0)(x − x1)(x − x2)f [x0, x1, x2, x3]


Example


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+ (x − x0)(x − x1)(x − x2)f [x0, x1, x2, x3]+

Dr N B Vyas Numerical Methods - Interpolation Unequal Intervals

Example


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+ (x − x0)(x − x1)(x − x2)f [x0, x1, x2, x3]+ (x − x0)(x − x1)(x − x2)(x − x3)f [x0, x1, x2, x3, x4]


Example


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+ (x − x0)(x − x1)(x − x2)f [x0, x1, x2, x3]+ (x − x0)(x − x1)(x − x2)(x − x3)f [x0, x1, x2, x3, x4]= ...

= 3x4 − 5x3 + 6x2 − 14x + 5

Dr N B Vyas Numerical Methods Interpolation Unequal Intervals

Example


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Ex. Find f (x) as a polynomial in x for the following data by Newtonsdivided difference formula:

x -2 -1 0 1 3 4f (x) 9 16 17 18 44 81

Hence, interpolate at x = 0.5 and x = 3.1.


Example

Sol. We form the divided difference table for the given data.


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x f (x) 1st d.d 2nd d.d 3rd d.d 4th d.d

−2 97

−1 16 −31 1

0 17 0 01 1

1 18 4 013 1

3 44 837

4 81


Example


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Since, the fourth order differences are zeros, the data represents athird degree polynomial. Newtons divided difference formulagives the polynomial as

f (x) = f (x0) +


Example


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Example


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f (x) = f (x0) + (x − x0)f [x0, x1] + (x − x0)(x − x1)f [x0, x1, x2]+


Example


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f (x) = f (x0) + (x − x0)f [x0, x1] + (x − x0)(x − x1)f [x0, x1, x2]+ (x − x0)(x − x1)(x − x2)f [x0, x1, x2, x3]


Example


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f (x) = f (x0) + (x − x0)f [x0, x1] + (x − x0)(x − x1)f [x0, x1, x2]+ (x − x0)(x − x1)(x − x2)f [x0, x1, x2, x3]= ...

= x3 + 17


Example


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Ex. Find the missing term in the following table:

x 0 1 2 3 4y 1 3 9 - 81


Example


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Sol. Divided difference table:


Example


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Sol. Divided difference table:

By Newton’s divided difference formula

f (x) =f (x0) + (x − x0) f (x0, x1) + (x − x0) (x − x1) f (x0, x1, x2) + ...


Spline Interpolation

Spline interpolation is a form of interpolation where the


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Spline interpolation is a form of interpolation where the

interpolant is a special type of piecewise polynomial called aspline

Consider the problem of interpolating between the data points(x0, y0), (x1, y1), . . . , (xn, yn) by means of spline fitting.

Then the cubic spline f (x) is such that(i) f (x) is a linear polynomial outside the interval (x0, xn)

(ii) f (x) is a cubic polynomial in each of the subintervals,

(iii) f (x) and f (x) are continuous at each point.

Since f (x) is cubic in each of the subintervals f

(x) shall belinear.



f (x) = (xi+1 − x)3M i

6h +


103/156



f (x) = (xi+1 − x)3M i

6h +

(x − xi)3M i+16h


104/156



f (x) = (xi+1 − x)3M i

6h +

(x − xi)3M i+16h


105/156

+(xi+1 − x)

h

yi − h

2

6 M i

+



f (x) = (xi+1 − x)3M i

6h +

(x − xi)3M i+16h


106/156

+(xi+1 − x)

h

yi − h

2

6 M i

+

(x − xi)h

yi+1 − h

2

6 M i+1



f (x) = (xi+1 − x)3M i

6h +

(x − xi)3M i+16h


107/156

+(xi+1 − x)

h

yi − h

2

6 M i

+

(x − xi)h

yi+1 − h

2

6 M i+1

where M i−1 + 4M i + M i+1 = 6

h2(yi−1 − 2yi + yi+1),



f (x) = (xi+1 − x)3M i

6h +

(x − xi)3M i+16h


108/156

+(xi+1 − x)

h

yi − h

2

6 M i

+

(x − xi)h

yi+1 − h

2

6 M i+1

where M i−1 + 4M i + M i+1 = 6

h2(yi−1 − 2yi + yi+1),

i = 1, 2, 3, ..., (n − 1)

and M 0 = 0, M n = 0, xi+1

−xi = h.

which gives n + 1 equations in n + 1 unknowns M i(i = 0, 1,...,n)which can be solved. Substituting the value of M i gives theconcerned cubic spline.


Example


109/156

Ex. Obtain cubic spline for the following data:

x 0 1 2 3

y 2 -6 -8 2


Example

Sol. Since points are equispaced with h = 1 and n = 3,


110/156


Example

Sol. Since points are equispaced with h = 1 and n = 3, the cubicspline can be determined from

6


111/156

M i−1 + 4M i + M i+1 = h2 (yi−1 − 2yi + yi+1), i = 1, 2


Example


M M M 6

y y y i


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M i−1 + 4

M i +

M i+1 = h2 (

yi−1 − 2

yi +

yi+1)

, i = 1

,2

also M 0 = 0 , M 3 = 0


Example


M M M 6

y y y i


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M i−1 + 4

M i +

M i+1 = h2 (

yi−1 − 2

yi +

yi+1)

, i = 1

,2

also M 0 = 0 , M 3 = 0

∴ for i = 1,


Example


M M M 6

y y y i


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M i−1 + 4

M i +

M i+1 = h2 (

yi−1 − 2

yi +

yi+1)

, i = 1

,2

also M 0 = 0 , M 3 = 0

∴ for i = 1, M 0 + 4M 1 + M 2 = 6(y0 − 2y1 + y2)therefore, 4M 1 + M 2 = 36; —(1)


Example


M + 4M + M = 6

(y 2y + y ) i = 1 2


115/156

M i−1

+ 4M i + M

i+1

h2(y

i−1 −2y

i + y

i+1), i 1, 2

also M 0 = 0 , M 3 = 0


for i = 2,


Example


M + 4M + M = 6

(y 2y + y ), i = 1, 2


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M i−1

+ 4M i + M

i+1

h2(y

i−1 −2y

i + y

i+1), i 1, 2

also M 0 = 0 , M 3 = 0


for i = 2, M 1 + 4M 2 + M 3 = 6(y1−

2y2 + y3)


Example


M − + 4M + M = 6

(y − 2y + y ), i = 1, 2


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i 1+

i+

i+1 h2(y

i 1 −yi

+ yi+1

), i ,

also M 0 = 0 , M 3 = 0


for i = 2, M 1 + 4M 2 + M 3 = 6(y1−

2y2 + y3)

M 1 + 4M 2 = 72 —(2)


Example


M i− + 4M i + M i = 6

(yi− 2yi + yi ), i = 1, 2


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i1

+ i + i+1 h2

(yi1 −

yi + yi+1

), ,

also M 0 = 0 , M 3 = 0


for i = 2, M 1 + 4M 2 + M 3 = 6(y1−

2y2 + y3)

M 1 + 4M 2 = 72 —(2)

solving these, we get


Example


M i− + 4M i + M i = 6

(yi− 2yi + yi ), i = 1, 2


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i1

i i+1 h2

(yi1 −

yi yi+1

), ,

also M 0 = 0 , M 3 = 0


for i = 2, M 1 + 4M 2 + M 3 = 6(y1−

2y2 + y3)

M 1 + 4M 2 = 72 —(2)

solving these, we get M 1 =


Example


M i− + 4M i + M i = 6

(yi− 2yi + yi ), i = 1, 2


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1 +1 h2(y

1 −y y

+1)

also M 0 = 0 , M 3 = 0


for i = 2, M 1 + 4M 2 + M 3 = 6(y1−

2y2 + y3)

M 1 + 4M 2 = 72 —(2)

solving these, we get M 1 =4.8 and M 2 =


Example


M i−1 + 4M i + M i+1 = 6

(yi−1 2yi + yi+1), i = 1, 2


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h2 −also M 0 = 0 , M 3 = 0∴ for i = 1, M 0 + 4M 1 + M 2 = 6(y0 − 2y1 + y2)

therefore, 4M 1 + M 2 = 36; —(1)

for i = 2, M 1 + 4M 2 + M 3 = 6(y1−

2y2 + y3)

M 1 + 4M 2 = 72 —(2)

solving these, we get M 1 =4.8 and M 2 =16.8


Example


M i−1 + 4M i + M i+1 = 6

(yi−1 2yi + yi+1), i = 1, 2


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therefore, 4M 1 + M 2 = 36; —(1)

for i = 2, M 1 + 4M 2 + M 3 = 6(y1−

2y2 + y3)

M 1 + 4M 2 = 72 —(2)


Now the cubic spline in (xi ≤ x ≤ xi+1) is


Example


M i−1 + 4M i + M i+1 = 6

(yi−1 2yi + yi+1), i = 1, 2


123/156


therefore, 4M 1 + M 2 = 36; —(1)

for i = 2, M 1 + 4M 2 + M 3 = 6(y1−

2y2 + y3)

M 1 + 4M 2 = 72 —(2)


Now the cubic spline in (xi ≤ x ≤ xi+1) isf (x) =

(xi+1 − x)3M i6h

+


Example


M i−1 + 4M i + M i+1 = 6

h(yi−1 2yi + yi+1), i = 1, 2


124/156


therefore, 4M 1 + M 2 = 36; —(1)

for i = 2, M 1 + 4M 2 + M 3 = 6(y1−

2y2 + y3)

M 1 + 4M 2 = 72 —(2)



(xi+1 − x)3M i6h

+(x − xi)3M i+1

6h


Example


M i−1 + 4M i + M i+1 = 6

h2(yi−1 2yi + yi+1), i = 1, 2


125/156


therefore, 4M 1 + M 2 = 36; —(1)

for i = 2, M 1 + 4M 2 + M 3 = 6(y1−

2y2 + y3)

M 1 + 4M 2 = 72 —(2)



(xi+1 − x)3M i6h

+(x − xi)3M i+1

6h

+(xi+1 − x)

h

yi − h

2

6 M i

+


Example


M i−1 + 4M i + M i+1 = 6

h2(yi−1 2yi + yi+1), i = 1, 2


126/156


therefore, 4M 1 + M 2 = 36; —(1)

for i = 2, M 1 + 4M 2 + M 3 = 6(y1

−2y2 + y3)

M 1 + 4M 2 = 72 —(2)



(xi+1 − x)3M i6h

+(x − xi)3M i+1

6h

+(xi+1 − x)

h

yi − h

2

6 M i

+

(x − xi)h

yi+1 − h

2

6 M i+1

—(3)


Example


M i−1 + 4M i + M i+1 = 6

h2(yi−1 2yi + yi+1), i = 1, 2


127/156


therefore, 4M 1 + M 2 = 36; —(1)

for i = 2, M 1 + 4M 2 + M 3 = 6(y1

−2y2 + y3)

M 1 + 4M 2 = 72 —(2)



(xi+1 − x)3M i6h

+(x − xi)3M i+1

6h

+(xi+1 − x)

h

yi − h

2

6 M i

+

(x − xi)h

yi+1 − h

2

6 M i+1

—(3)


Example


128/156

Ex. The following values of x and y are given:

x 1 2 3 4

y 1 2 5 11

Find the cubic splines and evaluate y(1.5) and y(3)


Example

Sol. Since points are equispaced with h = 1 and n = 3,


129/156


Example


M i−1 + 4M i + M i+1 = 6

h2(yi−1 2yi + yi+1), i = 1, 2


130/156

h2 −


Example


M i−1 + 4M i + M i+1 = 6

h2(yi−1 2yi + yi+1), i = 1, 2

l M M


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h2 −also M 0 = 0 , M 3 = 0


Example


M i−1 + 4M i + M i+1 = 6

h2(yi−1 2yi + yi+1), i = 1, 2

l M 0 M 0


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h2 −also M 0 = 0 , M 3 = 0∴ for i = 1,


Example


M i−1 + 4M i + M i+1 = 6

h2(yi−1

−2yi + yi+1), i = 1, 2

l M 0 M 0


133/156


therefore, 4M 1 + M 2 = 12; —(1)


Example


M i−1 + 4M i + M i+1 = 6

h2(yi−1

−2yi + yi+1), i = 1, 2

l M 0 M 0


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h2also M 0 = 0 , M 3 = 0


for i = 2,


Example


M i−1 + 4M i + M i+1 = 6

h2(yi−1

−2yi + yi+1), i = 1, 2

also M 0 M 0


135/156

halso M 0 = 0 , M 3 = 0


for i = 2, M 1 + 4M 2 + M 3 = 6(y1

−2y2 + y3)


Example


M i−1 + 4M i + M i+1 = 6

h2(yi−1

−2yi + yi+1), i = 1, 2

also M = 0 M = 0


136/156

halso M 0 = 0 , M 3 = 0


for i = 2, M 1 + 4M 2 + M 3 = 6(y1

−2y2 + y3)

M 1 + 4M 2 = 18 —(2)


Example


M i−1 + 4M i + M i+1 = 6

h2

(yi−1−

2yi + yi+1), i = 1, 2

also M0 = 0 M3 = 0


137/156

halso M 0 = 0 , M 3 = 0


for i = 2, M 1 + 4M 2 + M 3 = 6(y1

−2y2 + y3)

M 1 + 4M 2 = 18 —(2)solving these, we get


Example


M i−1 + 4M i + M i+1 = 6

h2

(yi−1−

2yi + yi+1), i = 1, 2

also M0 = 0 M3 = 0


138/156

also M 0 = 0 , M 3 = 0


for i = 2, M 1 + 4M 2 + M 3 = 6(y1

−2y2 + y3)

M 1 + 4M 2 = 18 —(2)solving these, we get M 1 =


Example


M i−1 + 4M i + M i+1 = 6

h2

(yi−1−

2yi + yi+1), i = 1, 2

also M0 = 0 M3 = 0


139/156

also M 0 = 0 , M 3 = 0


for i = 2, M 1 + 4M 2 + M 3 = 6(y1

−2y2 + y3)

M 1 + 4M 2 = 18 —(2)solving these, we get M 1 =2 and M 2 =


Example


M i−1 + 4M i + M i+1 = 6

h2

(yi−1−

2yi + yi+1), i = 1, 2

also M0 = 0 M3 = 0


140/156

also M 0 0 , M 3 0


for i = 2, M 1 + 4M 2 + M 3 = 6(y1

−2y2 + y3)

M 1 + 4M 2 = 18 —(2)solving these, we get M 1 =2 and M 2 =4


Example


M i−1 + 4M i + M i+1 = 6

h2

(yi−1−

2yi + yi+1), i = 1, 2

also M0 = 0 , M3 = 0


141/156

also M 0 0 , M 3 0


for i = 2, M 1 + 4M 2 + M 3 = 6(y1

−2y2 + y3)


Now the cubic spline in (xi ≤ x ≤ xi+1) is


Example


M i−1 + 4M i + M i+1 = 6

h2

(yi−1

−2yi + yi+1), i = 1, 2

also M 0 = 0 , M 3 = 0


142/156

0 0 , 3 0


for i = 2, M 1 + 4M 2 + M 3 = 6(y1

−2y2 + y3)



(xi+1 − x)3M i6h

+


Example


M i−1 + 4M i + M i+1 = 6

h2

(yi−1

−2yi + yi+1), i = 1, 2

also M 0 = 0 , M 3 = 0


143/156

0 , 3∴ for i = 1, M 0 + 4M 1 + M 2 = 6(y0 − 2y1 + y2)

therefore, 4M 1 + M 2 = 12; —(1)

for i = 2, M 1 + 4M 2 + M 3 = 6(y1

−2y2 + y3)



(xi+1 − x)3M i6h

+(x − xi)3M i+1

6h


Example


M i−1 + 4M i + M i+1 = 6

h2

(yi−1

−2yi + yi+1), i = 1, 2

also M 0 = 0 , M 3 = 0


144/156

0 , 3∴ for i = 1, M 0 + 4M 1 + M 2 = 6(y0 − 2y1 + y2)

therefore, 4M 1 + M 2 = 12; —(1)

for i = 2, M 1 + 4M 2 + M 3 = 6(y1

−2y2 + y3)



(xi+1 − x)3M i6h

+(x − xi)3M i+1

6h

+(xi+1 − x)

h

yi − h

2

6 M i

+


Example


M i−1 + 4M i + M i+1 = 6

h2

(yi−1

−2yi + yi+1), i = 1, 2

also M 0 = 0 , M 3 = 0


145/156


for i = 2, M 1 + 4M 2 + M 3 = 6(y1

−2y2 + y3)



(xi+1 − x)3M i6h

+(x − xi)3M i+1

6h

+(xi+1 − x)

h

yi − h

2

6 M i

+

(x − xi)h

yi+1 − h

2

6 M i+1

—(3)


Example


M i−1 + 4M i + M i+1 = 6

h2

(yi−1

−2yi + yi+1), i = 1, 2

also M 0 = 0 , M 3 = 0


146/156


for i = 2, M 1 + 4M 2 + M 3 = 6(y1

−2y2 + y3)



(xi+1 − x)3M i6h

+(x − xi)3M i+1

6h

+(xi+1 − x)

h

yi − h

2

6 M i

+

(x − xi)h

yi+1 − h

2

6 M i+1

—(3)


Example

Ex Find whether the following functions are cubic splines ?


147/156

Ex. Find whether the following functions are cubic splines ?

1. f (x) = 5x3 − 3x2, −1 ≤ x ≤ 0

= −5x3 − 3x2, 0 ≤ x ≤ 1

2. f (x) = −2x3 − x2, −1 ≤ x ≤ 0= 2x3 + 3x2, 0 ≤ x ≤ 1


Example

Sol. In both the examples, f (x) is a cubic polynomial in bothintervals (1, 0) and (0, 1).

1. We have

limx→0+

f (x)


148/156

x→0


Example


1. We have

limx→0+

f (x) = 0


149/156

x→0


Example


1. We have

limx→0+

f (x) = 0 = limx→0−

f (x)


150/156

x→0 x→0

therefore, given function f (x) is continuous on (−1, 1).

Nowf (x)


Example


1. We have

limx→0+

f (x) = 0 = limx→0−

f (x)


151/156


Nowf (x) = 15x2 − 6x, −1 ≤ x ≤ 0

=

−15x2

−6x, 0

≤x

≤1


Example


1. We have

limx→0+

f (x) = 0 = limx→0−

f (x)


152/156


Nowf (x) = 15x2 − 6x, −1 ≤ x ≤ 0

=

−15x2

−6x, 0

≤x

≤1

we have,

limx→0+

f (x)


Example


1. We have

limx→0+

f (x) = 0 = limx→0−

f (x)


153/156


Nowf (x) = 15x2 − 6x, −1 ≤ x ≤ 0

=

−15x2

−6x, 0

≤x

≤1

we have,

limx→0+

f (x) = 0


Example


1. We have

limx→0+

f (x) = 0 = limx→0−

f (x)


154/156


Nowf (x) = 15x2 − 6x, −1 ≤ x ≤ 0

=

−15x2

−6x, 0

≤x

≤1

we have,

limx→0+

f (x) = 0 = limx→0−

f (x)

therefore, the function f (x) is continuous on (

−1, 1).

f (x)


Example


1. We have

limx→0+

f (x) = 0 = limx→0−

f (x)


155/156


Nowf (x) = 15x2 − 6x, −1 ≤ x ≤ 0

=

−15x2

−6x, 0

≤x

≤1

we have,

limx→0+

f (x) = 0 = limx→0−

f (x)


−1, 1).

f (x) = 30x−6, −1 ≤ x ≤ 0= −30x − 6, 0 ≤ x ≤ 1


Example


1. We have

limx→0+

f (x) = 0 = limx→0−

f (x)


156/156


Nowf (x) = 15x2 − 6x, −1 ≤ x ≤ 0

=

−15x2

−6x, 0

≤x

≤1

we have,

limx→0+

f (x) = 0 = limx→0−

f (x)


−1, 1).

f (x) = 30x−6, −1 ≤ x ≤ 0= −30x − 6, 0 ≤ x ≤ 1


Dr. Nirav Vyas numerical method 4.pdf

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Transcript of Dr. Nirav Vyas numerical method 4.pdf