# num-methods

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```					      NUMERICAL METHODS - CHEAT SHEET

Compiled from lecture notes of Mrs.V.Valliammal

by Sarath Chandar A P

1     SOLUTION OF EQUATIONS AND EIGENVALUE
PROBLEMS
1.1   Solution of equation
Fixed point iteration: x=g(x) method

Newton’s method

1.2   Solution of linear system
Gaussian elimination method

Gauss-Jordan method

Iterative method

Gauss-Seidel method

1.3   Inverse of a matrix by Gauss Jordan method
1.4   Eigen value of a matrix by power method
1.5   Eigen value of a matrix by Jacobi method for symmetric matrix

2     INTERPOLATION AND APPROXIMATION
Interpolation The process of computing the value of a function inside the given range
is called interpolation.

Extrapolation The process of computing the value of a function outside the given
range is called extrapolation.

2.1   Newton’s forward and backward diﬀerence formulas
E[f(x)] = f(x+h)

∆[f(x)] = f(x+h) - f(x)

E=1+∆

Here ∆ is called Newton’s forward operator and E is called shift operator.
2

Newton’s backward diﬀerence operator is

=1-E

Newton’s forward interpolation formula is

u(u−1) 2                u(u−1)(u−2) 3
y = y0 + u∆y0 +          2!  ∆ y0          +       3!     ∆ y0     + ...

where u = (x - x0 ) / h

The above formula is used mainly for interpolating the values of y near the begin-
ning of a set of tabular values.

Newton’s backward interpolation formula is

u(u+1)     2           u(u+1)(u+2)   3
y = yn + u yn +           2!           yn +        3!            yn + ...

where u = (x - xn ) / h

The above formula is used mainly for interpolating the values of y near the end of
a set of tabular values.

2.2      Lagrangian Polynomials

Newton’s forward and backward interpolation formulae can be used only when the val-
ues of dependent variables are equally spaced; where the values of independent variables
are not equally spaced, then we can use following Lagrangian’s Interpolation formula
and inverse interpolation formula.

Lagrangian interpolation formula for unequal spaces is

(x−x1 )(x−x2 )...(x−xn )
y = f(x) =     (x0 −x1 )(x0 −x2 )...(x0 −xn ) y0   +
(x−x0 )(x−x2 )(x−x3 )...(x−xn )
(x1 −x0 )(x1 −x2 )(x1 −x3 )...(x1 −xn ) y1   +
(x−x0 )(x−x1 )(x−x3 )...(x−xn )
(x2 −x0 )(x2 −x1 )(x2 −x3 )...(x2 −xn ) y2   +
(x−x0 )(x−x1 )(x−x2 )...(x−xn−1 )
... +   (xn −x0 )(xn −x1 )(xn −x2 )...(xn −xn−1 ) yn

Lagrangian inverse interpolation formula for unequal spaces is

(y−y1 )(y−y2 )...(y−yn )
x = f(y) =     (y0 −y1 )(y0 −y2 )...(y0 −yn ) x0   +
(y−y0 )(y−y2 )(y−y3 )...(y−yn )
(y1 −y0 )(y1 −y2 )(y1 −y3 )...(y1 −yn ) x1   +
(y−y0 )(y−y1 )(y−y3 )...(y−yn )
(y2 −y0 )(y2 −y1 )(y2 −y3 )...(y2 −yn ) x2   +
(y−y0 )(y−y1 )(y−y2 )...(y−yn−1 )
... +   (yn −y0 )(yn −y1 )(yn −y2 )...(yn −yn−1 ) xn
3

2.3      Divided diﬀerences

If the values of x are given at unequal intervals then we use divided diﬀerences. Divided
diﬀerences take into consideration the change of the values of the function f(x) and also
the changes in the values of the arguments of x.
Suppose the function y = f(x) assume the values of f(x0 ), f(x1 ), f(x2 ),....f(xn ) re-
spectively, where the intervals x1 − x0 , x2 − x1 ,..... xn − xn−1 need not be equal.
First divided diﬀerence of f(x)

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

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

f (xn )−f (xn−1 )
f(xn−1 , xn ) =          xn −xn−1

Second divided diﬀerence is

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

Third divided diﬀerence is

f (x1 ,x2 ,x3 )−f (x0 ,x1 ,x2 )
f(x0 , x1 , x2 , x3 ) =               x3 −x0

The above steps are called divided diﬀerences of order 1,2,3 respectively.

Property: The divided diﬀerences are symmetrical in all the arguments. i.e. the value
of any diﬀerence is independent of order of arguments.

Newton’s divided diﬀerence formula is

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 )

2.4      Interpolating with a cubic spline

Suppose n+1 data points are given we have to ﬁnd the value of y corresponding to x
where xi  x xi+1 where i = 0,1,2,.. n+1. In this case we have to use cubic spline
method.

1. f(x) is a cubic polynomial
2. f (x) is second degree polynomial.
4

3. f (x) is a ﬁrst degree polynomial.
4. f(x), f (x), f (x) are continuous at each point (xi , yi ) where i = 0,1,2,..n.

If the above all conditions are satisﬁed, then f(x) follows cubic spline.

cubic spline formula The cubic spline in xi−1                  x   xi is

h2 Mi−1                   h2 Mi
f(x) =    1
6h
−x
(xi − x)3 Mi−1 + (x − xi−1 )3 Mi + xih yi−1 −             6      + x−xi−1 yi −
h             6

where h = xi − xi−1 for all i

6
Mi−1 + 4Mi + Mi+1 =       h2   [yi−1 − 2yi + yi+1 ] ∀i

here i = 1,2,3,.. n-1
M0 = 0
Mn = 0
f (xi ) = Mi

3      NUMERICAL DIFFERENTIATION AND INTEGRATION

Numerical diﬀerentiation is a process of calculating the derivative of the given function
by means of a table of given values of that function. In this process, we have to calculate
dy d2 y
dx , dx2

at a given point is called numerical diﬀerentiation. In this case, we have to use Newton’s
forward and backward diﬀerence formula to compute the derivatives.

In the previous units, we are ﬁnding the polynomial curves y = f(x) passing through
n+1 points. But in this unit we have to calculate the derivative of such curves at a par-
ticular point say Xk .

If the values of X are equally spaced, we get the interpolating polynomial using
Newton’s formula.

If the derivative required at a point nearer to starting value in table, use newton’s
forward method and for value at end of the table use newton’s backward method.

In the case of unequal intervals, we can use, Newton’s divided diﬀerence formula or
Lagrangian interpolation formula.

3.1     Diﬀerentiation using interpolation formulae

Newton’s forward diﬀerence formula is
u(u−1) 2         u(u−1)(u−2) 3
y = y0 + u∆y0 +        2!  ∆ y0   +       3!     ∆ y0    + ...
5

where u = (x - x0 ) / h
dy                       1
dx       at(x = x0 ) =   h
1         1
∆y0 − 2 ∆2 y0 + 3 ∆3 y0 − 1 ∆4 y0 + ...
4

d2 y                     1                                        11 4                 13 5
dx2      at(x = x0 ) =   h2       ∆2 y 0 − ∆3 y 0 +               12 ∆ y0        −     16 ∆ y0   + ...
d3 y                     1
dx3      at(x = x0 ) =   h3       ∆3 y0 − 3 ∆4 y0 + ...
2

Newton’s backward diﬀerence formula is
u(u+1)          2            u(u+1)(u+2)               3
y = yn + u yn +          2!                yn +         3!                        yn + ...

where u = (x - xn ) / h
dy                       1                     1   2          1     3              1    4
dx       at(x = xn ) =   h        yn +         2       yn +   3         yn +       4        yn + ...
d2 y                         1        2                3          11     4
dx2      at(x = xn ) =       h2           yn +             yn +   12         yn + ...
d3 y                         1    3       4
dx3      at(x = xn ) =       h3   2           yn + ...

3.2     Numerical integration by trapezoidal and Simpson 1/3 and 3/8 rules
The term numerical integration is the numerical evaluation of a deﬁnite integral say
b
a  f(x) dx

where a and b are given constants and f(x) be a given function.
When we apply numerical integration to any function of two independent variables,
using double integration is called mechanical cubature.

Trapezoidal rule The follwoing rule is said to be trapezoidal rule.

h
I=    2   [(y0 + yn ) + 2(y1 + y2 + ... + yn−1 )]

h is called length of the interval.
U
h= no.ofL−LL
intervals

Simpson’s 1/3 rule is

h
I=    3   [(y0 + yn ) + 2(y2 + y4 + ...) + 4(y1 + y3 + ...)]

Note: In simpson’s 1/3 rule, y(x) is a polynomial of degree 2. To apply this rule
the number of intervals should be even or the number of ordinates must be odd. In
simpson’s 1/3 rule, the truncation error is of the order h4 . i.e.

(b−a)h4 M
|E|          180

iv iv iv
where M = max(y0 , y1 , y2 , ..)
6

Simpson’s 3/8 rule is

3h
I=     8   [(y0 + yn ) + 3(y1 + y2 + y4 + y5 + ... + yn−3 ) + 2(y3 + y6 + ...)]

Note: In simpson’s 3/8 rule, y(x) is a polynomial of degree 3. To apply this rule the
number of intervals must be a multiple of 3.

3.3        Romberg’s method
b
Let I =     a      f(x) dx

Using trapezoidal or simpson’s rule, let us ﬁnd the value of the given deﬁnite inte-
gral by taking the diﬀerent values of h or ∆x.

Let us apply trapezoidal rule for several times by the values as h, h/2, h/4, h/8,
h/16...

In this type of problems, let

I2 −I1
I = I2 +         3

where I1 = the value of the deﬁnite integral, dividing h into two parts.
Similarly I2 = the value of the deﬁnite integral, dividing h into four parts.
Similarly we can calculate the values of
I3 −I2
I = I3 +           3

This method can continue till we get two successive values are equal.

3.4        Two and Three point Gaussian quadrature formulae
Two point formula In this case change the given interval a to b into -1 to 1 using
the following procedure. Transformation formula is

a+b          b−a
x=     2       +    2    t

where b is the upper limit and a is the lower limit.

1
√          √
−1     f(t) dt = f(1/ 3) + f(-1/ 3)

Three point formula In this case change the given interval a to b into -1 to 1 using
the following procedure. Transformation formula is

a+b          b−a
x=     2       +    2    t
7

where b is the upper limit and a is the lower limit.

1
−1     f(t) dt = A1 f(t1 ) + A2 f(t2 ) + A3 f(t3 )

where A1 = A3 = 0.5555
A2 = 0.88888
t1 = −0.7745
t2 = 0
t3 = +0.7745

3.5        Double integral using trapezoidal and Simpson’s rules
trapezoidal rule is

I = hk [sumof thef ourcornervalues(circle)+
4
2(sum of values available in rectangular box) + 4(sum of remaining values) ]

simpson rule is

I = hk [sumof thef ourcornervalues(circle)+
9
4(sum of values available in rectangular box) + 16(sum of remaining values) ]

4      INITIAL VALUE PROBLEMS FOR ODE
There are several methods for solving diﬀerential equations numerically.
1.    Taylor’s series method
2.    Euler’s method
3.    Rungi-kutta method
4.    Milne’s predictor and corrector method
5.    Adam’s predictor and corrector method

4.1        Single step methods
Taylor’s series method Taylor’s series about x = x0 is given by
(x−x0 )2          (x−x0 )3          (x−x0 )4
y(x) = y0 + (x - x0 ) y0 +       2!      y0 +      3!      y0 +      4!      y0 v

where n0 , y0 denotes the initial values of x and y

Euler method for ﬁrst order equation eulers method formula or eulers algorithm.
yn+1 = yn + h f(xn , yn ) where n=0,1,2...

modiﬁed eulers method formula :

yn+1 = yn + h f xn + h ; yn + h f (xn , yn )
2        2
8

Fourth order Runge-Kutta method for solving ﬁrst and second order equa-
tions in this method , let us assume that
dy
dx   = f(x,y) be given

diﬀ eqn, the above eqn can be solved under the condition

y(xo )= y0

let the increment h = length of the interval between two values

fourth order rangi kutta method

K1 = h f(x0 , y0 )
k1
K2 = h f(x0 + h , y0 +
2            2 )

k2
K3 = h f(x0 + h , y0 +
2            2 )

K4 = h f(x0 + h, y0 + k3 )
(K1 +2K2 +2K3 +K4 )
∆y =              6

y1 = y0 + ∆y

x1 = x0 + h

The increments in y for the second interval is computed in a similar manner by the
following formulae.

K1 = h f(x1 , y1 )
k1
K2 = h f(x1 + h , y1 +
2            2 )

k2
K3 = h f(x1 + h , y1 +
2            2 )

K4 = h f(x1 + h, y1 + k3 )
(K1 +2K2 +2K3 +K4 )
∆y =              6

y2 = y1 + ∆y

x2 = x1 + h

4.2     Multistep methods

Milne’s predictor and corrector methods

1. To use Milne’s predictor and corrector formula, we need atleast 4 values.
9

2. If the initial values are not given, we can obtain these values either by using Taylor’s
series method or by R-K method.

3. We can apply milne’s corrector formula only after applying milne’s predictor for-
mula.

Milne’s predictor formula is

4h
yn+1,p = yn−3 +       3    2yn−2 − yn−1 + 2yn

Milne’s corrector formula is

h
yn+1,c = yn−1 +      3    yn−1 + 4yn + yn+1

h
yn+1,p = yn +   24       55yn − 59yn−1 + 37yn−2 − 9yn−3

h
yn+1,c = yn +   24       9yn+1 + 19yn − 5yn−1 + yn−2

5    BOUNDARY VALUE PROBLEMS IN ODE AND PDE
to be included later.

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