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    II/IV B.Tech Mathematical Methods (13BS201) Chapter-2

    Page 1

    Finite Differences and Interpolation

    Suppose we are given the following values of y = f(x) for a set of values of x :

    x : x0 x1 x2 xn

    y : y0 y1 y2 yn

    The process of finding the values of y corresponding to any value of x=x ibetween x0and xn is

    called interpolation.

    The technique of estimating the value of a function for any intermediate value of the

    independent variable is called interpolation.

    The technique of estimating the value of a function outside the given range is called

    extrapolation.

    The study of interpolation is based on the concept of differences of a function.

    Suppose that the function y=f(x) is tabulated for the equally spaced values x = x0,

    x1=x0+h, x2=x0+2h, , xn=x0+nh giving y = y0, y1, y2, , yn. To determine the values off(x) and f '(x) for some intermediate values of x, we use the following three types of

    differences

    1. Forward differences

    2. Backward differences3. Central differences

    Forward differences : The forward differences are defined and denoted by f(x)=f(x+h)-f(x), y0= y1y0

    y1= y2y1y2= y3y2.yr= yr+1yr.

    yn-1 = ynyn-1These are called the first forward differences and is the forward difference operator.Similarly the second forward differences are defined by

    2yr= yr+1yr.In general

    pyr= p-1 yr+1

    p-1yr,

    pthforward differences.The forward differences systematically set out in a table called forward difference table.

    Value of

    x

    Value of

    y

    1s diff.

    2n diff.

    23r diff.

    34 diff.

    45 diff.

    5

    x0 y0

    Chapter-2

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    II/IV B.Tech Mathematical Methods (13BS201) Chapter-2

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    y0

    x1 y1 y0

    y1 y0

    x2 y2 y1 y0y2 y1 y0

    x3 y3 y2 y1

    y3 y2

    x4 y4 y3

    y4

    x5 y5

    Backward Differences:The backward differences are defined and denoted by f(x)= f(x)-f(x-h),y1= y1y0y2= y2y1

    y3= y3y2.yr= yryr-1

    .yn= ynyn-1.

    These are called the first backward differences and is the backward difference operator.

    Similarly the second backward differences are defined by2yr= yr yr-1.

    In generalpyr=

    p-1yr p-1yr-1,pth backward differences. The backward differences systematically set out in a table calledbackward difference table.

    Value of

    x

    Value of

    y

    1stdiff. 2n diff.2

    3r diff.3

    4t diff.4

    5t diff.5

    x0 y0

    y1

    x1 y12y2

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    II/IV B.Tech Mathematical Methods (13BS201) Chapter-2

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    y23 y3

    x2 y2 y3 y4

    y3 3 y4 5 y5

    x3 y32y4

    4 y5

    y43 y5

    x4 y42y5

    y5

    x5 y5

    Example#1. Evaluate(i) tan-1x(ii) (exlog 2x)(iii) 2cos 2x

    Sol. From the definition of forward differences f(x) = f(x+h)f(x).

    (i) Let f(x) = tan-1

    x, thentan-1x = tan-1(x+h) - tan-1x

    = .1

    tan)(1

    tan2

    11

    xhx

    h

    xhx

    xhx

    (ii)

    .2log)1()1log(

    2log)(log

    2log2log2log)(2log

    2log)(2log)2log(

    xex

    hee

    xeex

    hxe

    xexexehxe

    xehxexe

    hhx

    xhxhx

    xhxhxhx

    xhxx

    (iii) 2cos 2x = [cos 2x]= [ cos 2(x+h)cos 2x]= cos 2(x+h)cos 2x= cos 2(x+2h)cos 2(x+h)[cos 2(x+h) )cos 2x]

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    = 2 cos (2x+3h) sin h + 2 sin(2x+h) sin h=2 sin h [sin(2x+3h)sin(2x+h) ]=2 sin h [2 cos(2x+2h)sin h]=2 sin2h cos 2(x+h).

    Example#2.Evaluate the following, with interval of difference being unity

    (i)

    2

    (ab

    x

    ) (ii)

    n

    e

    x

    Sol. From the definition of forward differences f(x) = f(x+h)f(x).

    (i) (abx) = a bx= a(bx+1bx) = abx(b1)2(abx) = [(abx)]

    = abx(b1) = a(b1) (bx)= a(b1) (bx+1bx)= a(b1)2bx.

    (ii) ex= ex+1ex= ex(e1)2ex= [ex] = [ex+1ex]= (e1)ex

    = (e1) ex(e1) = (e1)2ex.

    Similarly 2ex = (e1)2ex, 3ex = (e1)3ex, and nex = (e1)nex.

    Differences of a Polynomial:

    Let f(x) = a0xn+ a1x

    n-1+ a2xn-2+ + an-1x + anbe an nth degree polynomial in x, then

    f(x) = f(x+h)f(x)

    = a0[(x+h)nxn] + a1[(x+h)

    n-1 xn-1]+ a2[(x+h)n-2xn-2 ]+ + an-1 [x+h-x]

    = a0 n h xn-1 + a11x

    n-2+ a21xn-2+ + an-1h,

    where a11, a12, are new constant coefficients. Thus, the first difference of a polynomial of nth

    degree is a polynomial of degree n-1.Similarly

    2f(x) = [f(x)]

    = a0n h [(x+h)n-1 xn] + a11[(x+h)n-2 xn-2 ]+ a21[(x+h)n-3 xn-3]+ + an-2, 1 [x+h-x]

    = a0 n(n1) h2xn-2 + a12x

    n-3+ a22xn-4+ + an-2,1h,

    where a12, a12, are new constant coefficients. Thus, the second difference of a polynomial ofnth degree is a polynomial of degree n-2.

    Continuing this process, for the nth difference, we get a polynomial of degree zero i.e.

    nf(x) = a0n(n1) (n2) 3.2.1. hn= a0n! h

    n

    which is a constant.

    From the above discussion, we have the following results :

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    The differences of a polynomial of the nth degree are constant and all higher orderdifferences are zero.

    If the nth differences of a function tabulated at equally spaced intervals are constant, then

    the function is a polynomial of degree n (It is important in numerical analysis as itenables as to approximate a function by a polynomial).

    Example#3.Evaluate 10[(1-ax)(1-bx2)(1-cx3)(1-dx4)].

    Sol. Taking interval of difference h = 1.

    10[(1-ax)(1-bx2)(1-cx3)(1-dx4)] = 10[abcd x10+ k1x9+ k2x

    8+ + 1]

    = abcd 10(x10) + k110(x9)+ k2

    10(x8) + + 10(1),

    where k1, k2, are constant coefficients. Since 10(xn) = 0 for n < 10, we have

    10[(1-ax)(1-bx2)(1-cx3)(1-dx4)] = abcd 10(x10) = abcd 10!.

    Factorial Notation: A product of the form x(x1) (x2) (x r +1) is denoted by [x]rand iscalled a factorial. In particular,

    [x] = x, [x]2= x(x1) , [x]3= x(x1) (x2), [x]n= x(x1) (x2) (x n +1).

    If the interval of difference is h, then

    [x]n= x(xh) (x2h) (x (n1)h).

    The factorial notation is of special utility in the theory of finite differences. It helps in finding the

    successive differences o f a polynomial directly by simple rule of differentiation ([x]ras xr).

    To express a polynomial of nth degree in the factorial notation, we use the following two steps

    1.Arrange the coefficients of the powers of x in descending order, replacing missing

    powers by zeros.

    2.Using detached coefficients divide by x, x1, x2, x (n1) successively.

    Example#4. Express f(x) = 2x3 3x2 + 3x 10 in a factorial notation and hence find alldifferences.

    Sol. Let f(x) = A[x]3+ B[x]2+ C[x] + D. Then

    x3 x2 x

    1 2 -3 3 -10 = D_ 2 -1

    2 2 -1 2 = C_ 4

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    3 2 3 = B

    _

    2 = A

    Hence f(x) = 2[x]3+ 3[x]2+ 2[x]10. Therefore,f(x) = 6[x]2+ 6[x] + 22f(x) = 12 [x] + 63f(x) = 12.

    Other Difference Operators:(1) Shift operator : Shift operator E is the operation of increasing the argument x by h so that

    E f(x) = f(x+h), E2 f(x) = f(x+2h), Enf(x) = f(x+nh).

    The inverse operator E-1is defined byE-1f(x) = f(x-h).

    Similarly E-nf(x) = f(x-nh).(2) Averaging operator : Averaging operator is defined by the equation

    f(x) = [f(x + h/2) + f(x - h/2)].In the difference calculus, and E are regarded as the fundamental operators and ,

    and can be expressed in terms of these.

    Relations Between the Operators :

    1. = E 12. = 1E-13. = E1/2E-1/24. = [E1/2+ E-1/2 ]5. = E = E =E1/26. E = ehD.

    Example#1. Determine the missing values in the following table:

    x 45 50 55 60 65

    y 3 ? 2 ? -2.4

    Sol. Let p and q be the missing values in the given table, then the difference table is as follows:

    x y y y y45 3

    p350 p 52p

    2p 3p + q955 2 p + q4q2 3.6p3q

    60 q 0.42q2.4q

    65 -2.4

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    Since three entries are given, the function y can be represented by a second degree polynomial.Therefore, 3y0= 0 and 3y1= 0. Thus 3p + q9 = 0 and 3.6p3q = 0. Solving theseequations, we get p = 2.925 and q = 0.225.Example#2. Determine the missing values in the following table without using difference table.

    x 45 50 55 60 65

    y 3 ? 2 ? -2.4Sol. Given that y0= 3, y2= 2 and y4= -2.4 and missing values be taken as y1= p and y3= q.

    Since three entries are given, the function y can be represented by a second degree polynomial.Therefore, 3y0= 0 and

    3y1= 0.

    (E1)3y0= 0 (E1)3y1= 0(E33E2+ 3E1)y0= 0 (E

    33E2+ 3E1)y1= 0y33y2+ 3y1y0= 0 y43y3+ 3y2y1= 0

    q3(2)+ 3p3 = 0 -2.43q + 3(2)p = 03p + q9 = 0 3.6p3q = 0.

    Solving these equations, we get p = 2.925 and q = 0.225.

    Newtons Forward Interpolation Formulae:Let the function y=f(x) take the values y0, y1, y2, corresponding to the values x0, x1, x2, of x. Suppose it is required to evaluate f(x) for x=x0+ph, p is any real number.

    For any real number p, we have defined E such that

    Epf(x) = f(x0+ph)yp= f(x0+ph) = E

    pf(x0)= (1+)py0= [1+p+p(p-1)/2! 2+ p(p-1)(p-2)/3! 3+] y0 = y0+ p y0+ p(p-1)/2!

    2y0+ p(p-1)(p-2)/3! 3y0+

    It is called Newtons forward interpolation formulae.

    Newtons Backward Interpolation Formulae:Suppose it is required to evaluate f(x) for x=xn+ph, where p is any real number.Epf(x) = f(xn+ph)

    yp= f(xn+ph) = Epf(xn)= (1- )

    -p yn

    = [1+p +p(p+1)/2! 2+ p(p+1)(p+2)/3! 3+] yn= yn+ p yn+ p(p+1)/2!

    2yn+ p(p+1)(p+2)/3!3yn+

    It is called Newtons backward interpolation formulae.

    Choice of Newtons Interpolation formulae :

    Newtons forward interpolat ion formulae is used for interpolating the values of y near thebeginning of a set of tabulated values and extrapolating values of y a little backward o f

    y0. Newtons backward interpolation formulae is used for interpolating the values of y near

    the end of a set of tabulated values and also extrapolating values of y a little ahead of yn.

    Example#1. The table gives the distances in nautical miles of the visible horizon for the given

    heights in feet above the earths surface :

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    x=height 100 150 200 250 300 350 400

    y=distance 10.63 13.03 15.04 16.81 18.42 19.90 21.27

    Find the values of y when (i) x= 218 ft. (ii) x= 410 ft.

    Sol. The difference table is

    x y

    100 10.63

    2.4

    150 13.03 -0.39

    2.01 0.15

    x0=200 15.04 -0.24 -0.07

    1.77 0.08

    250 16.81 -0.16 -0.05

    1.61 0.03

    300 18.42 -0.13 -0.01

    1.48 0.02

    350 19.90 -0.11

    1.37

    xn=400 21.27

    (i) If we take x0=200, then y0=15.04, y0=1.77, 2y0=-0.16,

    3y0=0.03, 4y0=-0.01.

    Since x=218, step length h=50 and p=(x-x0)/h =18/50 = 0.36.

    By Newtons forward interpolation formula, we have

    y(218) = y0+ p y0+ p(p-1)/2! 2y0+ p(p-1)(p-2)/3!

    3y0+ p(p-1)(p-2)(p-3)/4! 4y0

    = 15.04 + 0.36 (1.77) + 0.36(0.36-1)/2 (-0.16)+ 0.36(0.36-1)(0.36-2)/6 (0.03)

    + 0.36(0.36-1)(0.36-2)(0.36-3)/24 (-0.01)

    =15.04+0.6372+0.0184+0.0018+ 0.00041 = 15.69741

    15.7 nautical miles.

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    (ii) If we take xn=400, then yn=21.27, yn=1.37,2yn=-0.11,

    3yn=0.02,4yn=-0.01.

    Since x=410, step length h=50 and p=(x-xn)/h =10/50 = 0.2.

    By Newtons backward interpolation formula, we have

    y(410) = yn+ p yn+ p(p+1)/2! 2yn+ p(p+1)(p+2)/3! 3yn+ p(p+1)(p+2)(p+3)/4! 4yn

    = 21.27 + 0.2 (1.37) + 0.2(0.2+1)/2 (-0.11) + 0.2(0.2+1)(0.2+2)/6 (0.02)

    + 0.2(0.2+1)(0.2+2)(0.2+3)/24 (-0.01)

    =21.27+0.274-0.0132+0.0017- 0.0007 = 21.5318

    21.53 nautical miles.

    Interpolation with unequal intervals:

    The disadvantage for the previous interpolation formulas is that, they are used only forequal intervals. The following are the interpolation with unequal intervals;

    1) Lagranges formula for unequal intervals,2) Newtons divided difference formula.

    Lagranges interpolation formula:If y = f(x) takes the values y0, y1, y2, , yncorrespondingto x0, x1, x2, , xn, then

    ,y)x(x)x)(xx(x

    )x(x)x)(xx(x

    y)x(x)x)(xx(x

    )x(x)x)(xx(xy

    )x(x)x)(xx(x

    )x(x)x)(xx(xf(x)

    n

    1nn1n0n

    1n10

    1

    n12101

    n200

    n02010

    n21

    which is known as Lagranges formula.

    Divided Differences: If (x0, y0), (x1, y1), , (xn, yn) are given points, then the first divideddifferences for the argument x0, x1is defined by

    01

    0110

    xx

    yy]x,[x .

    Similarly

    .

    xx

    yy]x,[x,,

    xx

    yy]x,[x,

    xx

    yy]x,[x

    1nn

    1nn

    n1n

    23

    23

    32

    12

    12

    21

    The second divided differences for x0, x1, x2 is

    02

    1021210

    xx

    ]x,[x]x,[x]x,x,[x .

    The third divided differences for x0, x1, x2, x3 is

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    03

    2103213210

    xx

    ]x,x,[x]x,x,[x]x,x,x,[x .

    And so on, the nth divided differences for x0, x1, x2, ,xn is

    .

    xx

    ]x,,x,[x]x,,x,[x]x,,x,x,[x

    0n

    1-n10n21

    n210

    All the divided differences systematically set out in a table called divided difference table.

    Valueof x

    Valueof y

    1stdivided

    difference

    2n divided

    difference

    3r divideddifference

    4t divideddifference

    5t divideddifference

    x0 y0

    [x0,x1]

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

    x2 y2 [x1,x2,x3] [x0,x1,x2,x3,x4]

    [x2,x3] [x1,x2,x3,x4] [x0,x1,x2,x3,x4,x5]

    x3 y3 [x2,x3,x4] [x1,x2,x3,x4,x5]

    [x3,x4] [x2,x3,x4,x5]

    x4 y4 [x3,x4,x5]

    [x4,x5]

    x5 y5

    Newtons divided difference formula: If y = f(x) takes the values y0, y1, y2, , yncorresponding to x0, x1, x2, , xn, then

    f(x) = y0+(x-x0)[x0, x1] + (x-x0)(x-x1)[x0, x1, x2] + +(x-x0)(x-x1)(x-xn-1)[x0, x1, x2, , xn],which is known as Newtons general interpolation formula with divided differences.

    Example#1. Given the values

    x : 5 7 11 13 17

    f(x): 150 392 1452 2366 5202

    Evaluate f(9), using

    (i) Lagranges formula(ii) Newtons divided difference formula.Sol. Let y = f(x), then from the given data, we have x0= 5, x1= 7, x2= 11, x3= 13, x4= 17 and y0= 150, y1= 392, y2= 1452, y3 = 2366, y4= 5202.

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    (i) By Lagrnges interpolation formula

    .y)x(x)()x)(xx(x

    )x(x)()x)(xx(x

    y)x(x)()x)(xx(x

    )x(x)()x)(xx(xy

    )x(x)()x)(xx(x

    )x(x)()x)(xx(x

    y)x(x)()x)(xx(x

    )x(x)()x)(xx(xy

    )x(x)()x)(xx(x

    )x(x)()x)(xx(xf(x)

    4

    34241404

    3210

    3

    43231303

    42102

    42321202

    4310

    1

    n1312101

    4320

    0

    n0302010

    4321

    xx

    xx

    xx

    xx

    xx

    xx

    xx

    xx

    xx

    xx

    .8105

    578

    3

    2366

    3

    3872

    15

    3136

    3

    505202

    13)11)(177)(175)(17(17

    13)11)(97)(95)(9(9

    236617)11)(137)(135)(13(13

    17)11)(97)(95)(9(91452

    17)13)(117)(118)(11(11

    17)13)(97)(95)(9(9

    39217)13)(711)(75)(7(7

    17)13)(911)(95)(9(9150

    17)13)(511)(57)(5(5

    17)13)(911)(97)(9(9f(9)

    (ii) The divided difference table is

    Value

    of x

    Value

    of y

    1st

    divideddifference

    2n

    divideddifference

    3r divided

    difference

    4t divided

    difference

    5 150

    121

    7 392 24

    265 1

    11 1452 32 0

    457 1

    13 2366 42

    709

    17 5202

    By Newton divided difference formulaf(x) = y0+(x-x0)[x0, x1] + (x-x0)(x-x1)[x0, x1, x2] +(x-x0)(x-x1)(x-x2)[x0, x1, x2,x3]

    + (x-x0)(x-x1)(x-x2)(x-x3)[x0, x1, x2,x3,x4].f(9) = 150 + (95)121 + (95) (97)24 + (95)(97)(911)1

    + (95)(97)(911)(913)0= 150 + 484 + 19216 + 0= 810.

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    Numerical Differentiation:

    Mathematically, the derivative represents the rate of change of a dependent variable with respect

    to an independent variable. For example, if we are given a function y(t) that specifies an objects

    position as a function of time, differentiation provides a means to determine its velocity, as in:

    As in following Figure, the derivative can be visualized as the slope of a function.

    Numerical differentiation is used when the function y = f(x) is given in tabular form or it

    is highly complex. The basic idea in numerical differentiation is to replace the given function y =

    f(x) on the interval by an interpolating polynomial P(x) and set f (x)= P(x), f (x) = P(x) etc. Numerical differentiation is less exact than interpolat ion.

    Numerical differentiation using Newtons forward formula:Suppose y = f(x) is specified in

    an interval [a, b] at equally spaced points xi= x0+ ih (i = 0, 1, , n) (x0=a, xn=b) by means ofvalues yi= f(xi). By Newton forward interpolation formula

    ...y4!

    3)2)(p1)(pp(py

    3!

    2)1)(pp(py

    2!

    1)p(pypyf(x)y 0

    4

    0

    3

    0

    2

    00 ,

    whereh

    xxp 0 and h = xi+1xi, for i = 1, 2, , n. Here p is a function of x and

    h

    1

    dx

    dp.

    Rewriting the above equation, we have

    ...y

    24

    3p11p6ppy

    6

    2p3ppy

    2

    ppypyy(x) 0

    4234

    0

    323

    0

    22

    00

    Differentiating the above equation with respect to x, we have

    ....y24

    622p18p4py

    6

    26p3py

    2

    12py

    h

    1

    dp

    dy

    h

    1

    dx

    dp

    dp

    dy

    dx

    dy0

    423

    0

    32

    0

    2

    0

    Again differentiating with respect to x, we get

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    ....y12

    1118p6py)1(y

    h

    1

    dx

    dy

    dp

    d

    h

    1

    dx

    dp

    dx

    dy

    dp

    d

    dx

    dy

    dx

    d

    dx

    yd0

    42

    0

    3

    0

    2

    22

    2

    p

    Special case: If the derivative is required to find at a basic tabulated point x = x i, then choosex0= xiand the formulas become

    050403020xx

    y5

    1y

    4

    1y

    3

    1y

    2

    1y

    h

    1

    dx

    dy

    0

    And

    .y6

    5y

    12

    11yy

    h

    1

    dx

    yd0

    50

    40

    30

    2

    2xx

    2

    2

    0

    Numerical differentiation using Newtons backward formula:

    In this case we replace y(x) by Newtons backward interpolation formula

    ...y4!

    3)2)(p1)(pp(py

    3!

    2)1)(pp(py

    2!

    1)p(pypyy(x) n

    4n

    3n

    2nn

    whereh

    xxp n and h = xi+1xi, for i = 1, 2, , n. Here p is a function of x and

    h

    1

    dx

    dp.

    Rewriting the above equation, we have

    ...y24

    3p11p6ppy

    6

    2p3ppy

    2

    ppypyy(x) n

    4234

    n3

    23

    n2

    2

    0n

    Differentiating the above equation with respect to x, we have

    ....y24

    622p18p4py

    6

    26p3py

    2

    12py

    h

    1

    dp

    dy

    h

    1

    dx

    dp

    dp

    dy

    dx

    dyn

    423

    n3

    2

    n2

    n

    Again differentiating with respect to x, we get

    ....y12

    1118p6py)1(y

    h

    1

    dx

    dy

    dp

    d

    h

    1

    dx

    dp

    dx

    dy

    dp

    d

    dx

    dy

    dx

    d

    dx

    ydn

    42

    n3

    n2

    22

    2

    p

    Special case: If the derivative is required to find at a basic tabulated point x = x i, then choosexn= xiand the formulas become

    n5

    n4

    n3

    n2

    0xx

    y5

    1y

    4

    1y

    3

    1y

    2

    1y

    h

    1

    dx

    dy

    n

    and

    .y

    6

    5y

    12

    11yy

    h

    1

    dx

    ydn

    5n

    4n

    3n

    2

    2xx2

    2

    n

    Example#1. Compute f (x) and f (x) at (i) x = 16 (ii) x = 15 (iii) x = 24 (iv) x = 25 from thefollowing table

    x 15 17 19 21 23 25

    f(x) 3.873 4.123 4.359 4.583 4.796 5.8

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    Sol. Let y = f(x), then from the given table x0= 15, x1= 17, x2= 19, x3= 21, x4= 23, x5= 25and y0= 3.873, y1= 4.123, y2= 4.359, y3= 4.583, y4= 4.796, y5= 5.8. The finite difference table

    is

    x y y y y y y15 3.873

    0.2517 4.123 -0.014

    0.236 0.002

    19 4.359 -0.012 -0.001

    0.224 0.001 0.002

    21 4.583 -0.011 0.001

    0.213 0.002

    23 4.796 -0.009

    0.204

    25 5

    (i) Since x = 16 is nearer to the beginning of the table we use Newton forward formula. Here the

    step size h = 2. Taking x0= 15, then .5.02

    1

    2

    15160

    h

    xxp

    Newton forward formula to compute first derivative of y=f(x) is

    ....y24

    622p18p4py

    6

    26p3py

    2

    12py

    h

    1

    dx

    dy0

    423

    03

    2

    02

    0

    Substituting x = 16, p = 0.5, y0 = 0.25, 2y0=0.014,

    3y0 = 0.002, 4y0= 0.001,

    5y0=0.002,

    we have

    .(-0.001)24

    622(0.5)18(0.5)4(0.5)

    (0.002)6

    26(0.5)3(0.5)(-0.014)

    2

    12(0.5)0.25

    2

    1

    dx

    dy

    23

    2

    16x

    f (16) = 0.1249375.

    Newton forward formula to compute second derivative of y=f(x) is

    ....y12

    1118p6py)1(y

    h

    1

    dx

    yd0

    42

    03

    02

    22

    2

    p

    Substituting the values from the table, we have

    .(-0.001)12

    1118(0.5)6(0.5)(0.002))15.0(0.014-

    2

    1

    dx

    yd2

    216

    2

    2

    x

    f (16) = -0.0038229.

    (ii) Since x = 15 is in the beginning of the table, we use Newton forward formula. Here the step

    size h = 2, x = x0= 15 and p = 0.Newton forward formula to compute first derivative of y=f(x) at x = x0is

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    05

    04

    03

    02

    0xx

    y5

    1y

    4

    1y

    3

    1y

    2

    1y

    h

    1

    dx

    dy

    0

    Substituting the values from the table, we have

    )(0.002

    5

    1(-0.001)

    4

    1(0.002)

    3

    1(-0.014)

    2

    10.25

    2

    1

    dx

    dy

    15x

    f (15) = 0.128958.

    Newton forward formula to compute second derivative of y=f(x) at x = x0is

    .y6

    5y

    12

    11yy

    h

    1

    dx

    yd0

    50

    40

    30

    2

    2xx

    2

    2

    0

    Substituting the values from the table, we have

    .(0.002)6

    5(-0.001)

    12

    110.0020.014-

    2

    1

    dx

    yd

    215x

    2

    2

    f (15) = -0.004229.(iii) Since x = 24 is nearer to the ending of the table, we use Newton backward formula. Here the

    step size h = 2. Taking xn= 25, then .5.02

    1

    2

    2524

    h

    xxp n

    Newton backward formula to compute first derivative of y=f(x) is

    ....y24

    622p18p4py

    6

    26p3py

    2

    12py

    h

    1

    dx

    dyn

    423

    n3

    2

    n2

    n

    Substituting x = 25, p =0.5, yn = 0.204,2yn=0.009,

    3yn = 0.002,4yn= 0.001,

    5yn=0.002, we have

    .)001.0(24

    622(-0.5)18(-0.5)4(-0.5)

    )002.0(6

    26(-0.5)3(-0.5)

    )009.0(2

    12(-0.5)

    0.2042

    1

    dx

    dy

    23

    2

    25x

    f (24) = 0.09727.

    Newton backward formula to compute second derivative of y=f(x) is

    ....y12

    1118p6py)1(y

    h

    1

    dx

    ydn

    42

    n3

    n2

    22

    2

    p

    Substituting the values from the table, we have

    .)001.0(12

    1118(-0.5)6(-0.5))002.0)(15.0(009.02

    1

    dx

    yd 22

    242

    2

    x

    f (24) = -0.00242708.

    (iv) Since x = 25 is in the ending of the table, we use Newton forward formula. Here the step sizeh = 2, x = x0= 15 and p = 0.

    Newton backward formula to compute first derivative of y=f(x) at x = xnis

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    n5

    n4

    n3

    n2

    0xx

    y5

    1y

    4

    1y

    3

    1y

    2

    1y

    h

    1

    dx

    dy

    n

    .

    Substituting the values from the table, we have

    .)002.0(

    5

    1)001.0(

    4

    1)002.0(

    3

    1)009.0(

    2

    10.204

    2

    1

    dx

    dy

    25x

    f (25) = 0.10048Newton backward formula to compute second derivative of y=f(x) at x = xnis

    .y6

    5y

    12

    11yy

    h

    1

    dx

    ydn

    5n

    4n

    3n

    2

    2xx

    2

    2

    n

    Substituting the values from the table, we have

    .)002.0(6

    5)001.0(

    12

    11002.0009.0

    2

    1

    dx

    yd

    225x

    2

    2

    f (25) =-0.001833.

    Numerical Integration

    Integration is the inverse of differentiation. Just as differentiation uses differences to quantify an

    instantaneous process, integration involves summing instantaneous information to give a totalresult over an interval. Thus, if we are provided with velocity as a function of time, integrationcan be used to determine the distance traveled:

    According to the dict ionary definition, to integrate means to bring together, as parts, intoa whole; to unite; to indicate the total amountMathematically, definite integration is

    represented by(1)

    which stands for the integral of the functionf (x) with respect to the independent variablex,evaluated between the limitsx = a tox = b. As suggested by the dictionary definition, the

    meaning of Eq. (1)is the total value, or summation, off (x)dx over the rangex = a to b. In fact,the symbol is actually a stylized capital S that is intended to signify the close connectionbetween integration and summation.

    Geometrically, integration is just finding the area under a curve from one point to another. It is

    represented by

    b

    a

    dxxf )( , where the numbers a and b are the lower and upper limits of

    integration, respectively, the function f is the integrand of the integral, and x is the variable ofintegration. Figure 1 represents a graphical demonstrat ion of the concept.

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    Why are we interested in integration: because most equations in physics are differential

    equations that must be integrated to find the solution(s). Furthermore, some physical quantities

    can be obtained by integration (example: displacement from velocity).

    The problem is that sometimes integrating analytically some functions can easily become

    laborious. For this reason, a wide variety of numerical methods have been developed to find theintegral.

    The process of evaluating a definite integral from a set of tabulated values of the integrand f(x) is

    called numerical integration. This process when applied to a function of single variable, is known

    as quadrature.

    Let dxy(x)Ib

    a

    , where y(x) takes the values y0, y1, y2, , ynfor x = x0, x1, x2, , xn. Let

    us divide the interval (a, b) into n sub-intervals of width h so that x0= a, x1= x0+ h, x2= x0+ 2h,

    , xn= x0+ nh = b.

    Trapezoidal rule: )y...y2(y)y(y2

    hy(x)dx 1-n21n0

    x

    x

    n

    0

    .

    Simpsons 1/3rdrule :

    )y...y2(y)y...y4(y)y(y3

    hy(x)dx

    2-n421-n31n0

    x

    x

    n

    0

    Simpsons 3/8th rule :

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    )y...y2(y)yy...yyy3(y)y(y8

    3hy(x)dx 3-n632-n1-n5421n0

    x

    x

    n

    0

    Problem#1. Compute the integral

    6

    021 x

    dx

    , using (i) Trapezoidal rule (ii) Simpsons 1/3rd

    rule

    (iii) Simpsons 3/8th rule and also determine the relative true error.

    Sol. Let2

    x1

    1y(x) and divide the interval (0, 6) into n = 6 subintervals each of length h = 1.

    Then, we have the following tabular values.

    x 0 1 2 3 4 5 6

    y(x) 1 0.5 0.2 0.1 0.0588 0.0385 0.027

    (i) Trapezoidal rule

    )y...y2(y)y(y

    2

    hy(x)dx 1-n21n0

    x

    x

    n

    0

    .

    1.4108.

    0.0385)0.05880.10.22(0.5)027.0(12

    1

    )yyyy2(y)y(y2

    hy(x)dx 5432160

    6

    0

    (ii) Simpsons 1/3rdrule

    )y...y2(y)y...y4(y)y(y

    3

    hy(x)dx 2-n421-n31n0

    x

    x

    n

    0

    1.3662.

    0.0588)2(0.20.0385)0.14(0.50.027)(13

    1

    )y2(y)yy4(y)y(y3

    hy(x)dx 4253160

    6

    0

    (iii) Simpsons 3/8th rule

    )y...y2(y)yy...yyy3(y)y(y8

    3hy(x)dx 3-n632-n1-n5421n0

    x

    x

    n

    0

    .3571.12(0.1))0.03850.05880.23(0.50.027)(08

    3

    )2(y)yyy3(y)y(y8

    3hy(x)dx 3542160

    6

    0

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    %100valueTrue

    valueeApproximat-valueTrueerrortruerelativeThe .

    True value :

    .4056.104056.1)0(tan)6(tan)(tan

    1

    116

    01

    6

    02

    x

    x

    dx

    Relative true error for (i) Trapezoidal rule = %.3699.0%1001.4056

    1.4108-1.4056

    (ii) Simpsons 1/3rdrule = %.803.2%1001.4056

    1.3662-1.4056

    (iii) Simpsons 3/8th rule = %.4504.3%1001.4056

    1.3571-1.4056

    Problems1. Estimate the missing values in the following table

    x 45 50 55 60 65

    y 3.0 ? 2.0 ? -2.4

    2. Express y=2x3-3x2+3x-10 in factorial notation and hence show that 3y=12.3. Given Sin450= 0.7071, Sin500= 0.7660, Sin550= 0.8192, Sin 600= 0.8660, find Sin520, using

    Newtons forward formula.4. From the following table, estimate the number of students who obtained marks between 40

    and45

    Marks 30-40 40-50 50-60 60-70 70-80

    No. ofstudents

    31 42 51 35 31

    5.

    Construct a cubic polynomial which takes the following values :

    x 0 1 2 3

    f(x) 1 2 1 10

    6. The area of a circle of diameter d is given for the following values:

    d 80 85 90 95 100

    A 5026 5674 6362 7088 7854

    Calculate the area of a circle of diameter 105.

    7.

    Given the valuesx: 5 7 11 13 17

    f(x): 150 392 1452 2366 5202

    Evaluate f(9) by using (a) Lagranges formula (b) Newtons divided difference formula.

    8. Use Lagranges formula to find the form of f(x) when,

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    x: 0 2 3 6

    f(x): 648 704 729 792

    9. Determine f(x) as a polynomial in x for the following data using Newtons divided differenceformula

    x: -4 -1 0 2 5

    f(x): 1245 33 5 9 1335

    10.Apply Lagranges formula inversely to obtain a root of the equation f(x) = 0 f(30)= -30, f(34)= -13,f(38) = 3 and f(42)= 18.

    15. Given that

    x 1.0 1.1 1.2 1.3 1.4 1.5 1.6

    y 7.989 8.403 8.781 9.129 9.451 9.750 10.031

    find the values of dy/dx and d2y/dx2 at x = 1.1and at x 1.6.

    16.Find the first and second derivatives of the function tabulated be low, at the point x = 1.1

    x: 1.0 1.2 1.4 1.6 1.8 2.0

    f(x): 0 0.128 0.544 1.296 2.432 4.00

    17.From the following table, find the values of dy/dx and d2y/dx2 at x = 2.03

    x: 1.96 1.98 2.00 2.02 2.04

    y: 0.7825 0.7739 0.7651 0.7563 0.7473

    18. The following data gives the corresponding values of pressure and specific volume of asuperheated steam.

    v 2 4 6 8 10

    p 105 42.7 25.3 16.7 13

    19.From the table below, for each value of x, y is minimum? Also find the value of y

    x 3 4 5 6 7 8

    y 0.205 0.240 0.259 0.262 0.250 0.224

    20. Given that ,

    x: 4.0 4.2 4.4 4.6 4.8 5.0 5.2

    logx: 1.3863 1.4351 1.4816 1.5261 1.5686 1.6094 1.6484

    Evaluate by (a) Trapezoidal rule(b) Simpsons 1/3 rule (c) Simpsons

    3/8th rule.

    21.Use Simpsons 1/3rdrule to find dx by taking seven ordinates.22.The velocity (km/min) of a moped which starts from rest, is given at fixed intervals of time t

    (min) as follows:

    T 2 4 6 8 10 12 14 16 18 20

    V 10 18 25 29 32 20 11 5 2 0

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    Estimate approximately the distance covered in 20 minutes.23.A river is 80 feet wide. The depth d in feet at a distance x feet from one bank is given by the

    following table:

    x 0 10 20 30 40 50 60 70 80

    d 0 4 7 9 12 15 14 8 3

    Find approximately the area of the cross-section.24.A solid of revolution is formed by rotating about the X-axis, the area between the X-axis, the

    lines x=0 and x=1 and a curve through the points with the following coordinate:

    x 0.00 0.25 0.50 0.75 1.00

    y 1.0000 0.9896 0.9589 0.9089 0.8415

    Short Answer Questions

    1. Evaluate 2(abx), interval of differences being unity.2. Show that log f(x)= log{1+f(x)/f(x)}

    3.

    Evaluate 10

    [(1-x) (1-2x2

    ) (1-3x3

    ) (1-4x4

    )], if the interval of differencing is 2.4. Prove that y3= y2+ y1+

    2y0+ 3y0.

    5. Prove with usual notations that (E1/2+ E-1/2) (1 + ) = 2 + . 6. Prove with usual notations that 3y2=

    3y5.

    7. State Newtons forward interpolative formula.8. Write the relation between and E. 9. Write the relation between , E and10.Prove with usual notations that (1+)(1- ) = 1

    11.State Lagranges Interpolation formula12.State Newtons divided difference formula

    18.By Trapezoidal rule, write the value of dx

    19.State Simpsons 3/8 th rule. 20.If f(x) is given by x = 0 0.5 1 1.5 2

    f(x) = 0 0.25 1 2.25 4.

    Then the value of dx by Simpsons 1/3 rule.