SSE2393/SSM3393
1. Thomas Algorithm
Tridiagonal
matrix A for 
Ax = b can be written as A = LU

d1 e1 0
···
0

.. 
.
..
c d
. 
 2 2 e2



.. ..
..

.
.
.
0 
 0

 .

 . ..
. cn−1 dn−1 en−1 
 .

0 ···
0
cn
α1 0
 c
 2 α2

..

.
= 0

 ..
.
..
 .
0 ···
where
dn
0
···
0
···
..
..
.
.

0
0
..
.
cn−1 αn−1 0
0
cn α n
0
···
...

1 β1
0
..
.








0 1 β2
.. . . . . . .
.
.
.
.
0
0 ··· 0
1 βn−1
0 ··· 0
0
1
α 1 = d1
αi = di − ci βi−1 , i = 2, 3, . . . , n
ei
βi =
, i = 1, 2, 3, . . . , n − 1
αi
Solve Lw = b using forward substitution algorithm,
wi = (bi − ci wi−1 )/αi ,
w1 = b1 /α1 ,
i = 2, 3, ..., n
Solve U x = w using backward substitution algorithm,
xn = w n ,
xi = wi − βi xi+1 ,
i = n − 1, n − 2, . . . , 1
2. Lagrange interpolation formula
P (x) =
n
X
Li (x)fi dengan Li (x) =
i=0
n
Y
(x − xj )
j=0 (xi − xj )
j6=i
3. Newton’s divided difference formula
[0]
[1]
[2]
P (x) = f0 + f0 (x − x0 ) + f0 (x − x0 )(x − x1 ) + · · ·
[n]
+ f0 ](x − x0 )(x − x1 ) · · · (x − xn−1 );
[j−1]
[0]
fi
= fi ,
[j]
fi
[j−1]
f
− fi
= i+1
xi+j − xi
8









4. Newton’s forward difference formula
r(r − 1) 2
r(r − 1)(r − 2) 3
∆ f0 +
∆ f0
2!
3!
r(r − 1)(r − 2) · · · (r − n + 1) n
∆ f0 ; h = xi+1 − xi .
+··· +
n!
P (x) = P (x0 + rh) = f0 + r∆f0 +
5. Newton’s backward difference formula
r(r + 1) 2
r(r + 1)(r + 2) 3
∇ fn +
∇ fn
2!
3!
r(r + 1) · · · (r + n − 1) n
∇ fn ; h = xi+1 − xi .
+··· +
n!
P (x) = P (xn + rh) = fn + r∇fn +
6. 2–points centered difference formula
f 0 (x) =
1
[f (x + h) − f (x − h)]
2h
7. 3–points forward difference formula
f 0 (x) =
1
[−f (x + 2h) + 4f (x + h) − 3f (x)]
2h
8. 3–points backward difference formula
f 0 (x) =
1
[3f (x) − 4f (x − h) + f (x − 2h)]
2h
9. 5–points difference formula
f 0 (x) =
1
[−f (x + 2h) + 8f (x + h) − 8f (x − h) + f (x − 2h)]
12h
10. 3–points centered difference formula
f 00 (x) =
1
[f (x + h) − 2f (x) + f (x − h)]
h2
11. 5–points difference formula
f 00 (x) =
1
[−f (x + 2h) + 16f (x + h) − 30f (x) + 16f (x − h) − f (x − 2h)]
12h2
12. 5–points fourth order centered difference formula
f 0000 (x) =
1
[f (x − 2h) − 4f (x − h) + 6f (x) − 4f (x + h) + f (x + 2h)]
h4
9
13. Trapezoidal rule
Z xi+1
f (x)dx =
xi
h
[f (xi ) + f (xi+1 )]
2
14. Simpson’s rule
Z xi+2
xi
f (x)dx =
h
[f (xi ) + 4f (xi+1 ) + f (xi+2 )]
3
15. 3/8 Simpson’s rule
Z xi+3
xi
3
f (x)dx = h[f (xi ) + 3f (xi+1 ) + 3f (xi+2 ) + f (xi+3 )]
8
16. 2-points Gaussian quadrature
Z 1
1
1
f (x) dx = f (− √ ) + f ( √ )
−1
3
3
17. 3–points Gaussian quadrature
s
Z 1
s
5
3
8
5
3
f (x) dx = f (−
) + f (0) + f (
)
9
5
9
9
5
−1
18. Power method
1
Av(k)
v(k+1) =
mk+1
19. Taylor’s series of order n
yi+1 = yi + hyi0 +
h2 00
hn (n)
yi + · · · + yi
2!
n!
20. Euler’s method
yi+1 = yi + hf (xi , yi )
21. Second order Runge–Kutta methods
(a) Midpoint method
h
k1
yi+1 = yi + k2 ; k1 = hf (xi , yi ), k2 = hf (xi + , yi + ).
2
2
(b) Improved Euler’s method
1
yi+1 = yi + (k1 + k2 ); k1 = hf (xi , yi ), k2 = hf (xi + h, yi + k1 ).
2
10
(c) Heun’s method
2
2
1
yi+1 = yi + (k1 + 3k2 ); k1 = hf (xi , yi ), k2 = hf (xi + h, yi + k1 ).
4
3
3
22. Fourth order Runge–Kutta method
1
yi+1 = yi + (k1 + 2k2 + 2k3 + k4 )
6
h
k1
k1 = hf (xi , yi ),
k2 = hf (xi + , yi + ),
2
2
h
k2
k3 = hf (xi + , yi + ), k4 = hf (xi + h, yi + k3 )
2
2
23. Finite difference formulae for partial differential equation
∂u
∂x
!
∂u
∂x
!
∂u
∂x
!
=
ui+1,j − ui,j
h
(forward difference)
=
ui,j − ui−1,j
h
(backward difference)
=
ui+1,j − ui−1,j
2h
(centered difference)
i,j
i,j
∂ 2u
∂x2
i,j
!
=
i,j
ui+1,j − 2ui,j + ui−1,j
h2
(centered difference)
11