This document discusses numerical methods for solving differential equations and calculating integrals. It introduces Euler's method, modified Euler's method (Heun's method), Taylor's method of order two, Lagrange interpolation, and Simpson's rule for numerical integration. Euler's method approximates the slope at each step to calculate the next value. Modified Euler's method takes the average of the slopes. Taylor's method uses the slope and second derivative. Lagrange interpolation fits a polynomial through given data points. Simpson's rule approximates the integral of a function using a weighted sum over intervals.

1. By:
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2. IINNIITTIIAALL VVAALLUUEE PPRROOBBLLEEMMSS
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3. RREEMMEEMMBBEERR
NNUUMMEERRIICCAALL DDIIFFFFEERREENNTTIIAATTIIOONN
The derivative of y(x) at xis:
0 ( ) ( + ) -
( ) 0 0
Forward Difference
Formula
lim
h
0 0
y x h y x
y x
® h
¢ =
An approximation to this is:
( ) y ( x + h ) -
y ( x
) y x
0 0
0
for small values of h. h
¢ »
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4. ( ) ( ) 0 0
¢ »
0
hy¢ ( x ) » y ( x + h ) - y(x )
0 0 0 y x h y(x )
y x
+ -
h
( ) ( ) 0 0 0 y(x ) + hy¢ x » y x + h
( ) ( ) 0 0 0 y x + h » y(x ) + hy¢ x
Provided h is small.
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5. y ( x0 + h) » y(x0 ) + hy¢ ( x0 )
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6. y ( x0 + h) » y(x0 ) + hy¢ ( x0 )
The value of a function at x + h 0
can be approximated if we know its value
and its slope at an earlier point, 0 x
provided h is small.
A sequence can be written as:
1 0 0 y » y + hy¢
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7. 1 0 0 y » y + hy¢
2 1 1 y » y + hy¢
n 1 n n y y hy + » + ¢
In an initial value problem, n y¢ is given
and often written as: ( ) n n f x , y
( ) n n n n y » y + hf x , y Therefore, +1
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8. Euler’s Method
n n ( n n ) y » y + hf x , y +1
( ) n n y = y x Value of y at n x = x
( ) +1 +1 = n n y y x Value of y at +1 = n x x
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9. ( n n ) ( n ) f x , y = y¢ x
First derivative of y at n x = x
h is a small number by which x is incremented.
x = x + h 1 0
x x h x 2h 2 1 0 = + = +
x x h x 3h 3 2 0 = + = +
x x nh n = + 0
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10. So what are we doing in Euler’s method?
We are approximating the value of the
function y at x = xn+1 based on the
information that we obtained at the
previous point, x = xn .
The previous information are:
The value of y at x = xn
The value of the first derivative of y at x = xn
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11. n n ( n n ) y » y + hf x , y +1
Consider the following initial value problem.
y¢ = y - x , y(0) = 0.5
Find the value of y(0.8).
Let h = 0.1
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12. Given:
( n n ) n n n f x , y = y¢(at x = x ) = y - x
0 5 0 0 y( at x = x ) = y = .
where 0 0 x =
Our starting point.
. .
x x h
= + = + =
0 0 1 0 1
. .
x x h
= + = + ´ =
2 0 2 0 1 0 2
x x h
3 0 3 0 1 0 3
1 0
2 0
3 0
. .
= + = + ´ =
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13. Based on the sequence generated by Euler’s
method we can write:
1 0 ( 0 0 ) y » y + hf x , y
0 5 0 0 5 0 0 0 0 f (x , y ) = y - x = . - = .
0 5 0 1 0 5 0 55 1 y » . + . ´ . = .
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14. 2 1 ( 1 1 ) y » y + hf x , y
0 0 1 0 1 1 0 x = x + h = + . = .
0 55 0 1 0 45 1 1 1 1 f (x , y ) = y - x = . - . = .
0 55 0 1 0 45 0 595 2 y » . + . ´ . = .
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15. 3 2 ( 2 2 ) y » y + hf x , y
2 0 2 0 1 0 2 2 0 x = x + h = + ´ . = .
0 595 0 2 0 395 2 2 2 2 f (x , y ) = y - x = . - . = .
0 595 0 1 0 395 0 6345 3 y » . + . ´ . = .
Approximate value of y at x = x3 is 0.6345.
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16. Therefore, we can write that
y y
( . ) = »
.
0 1 0 55
1
y y
( . ) = »
.
0 2 0 595
2
y y
= »
0 3 0 6345
3
y y
( . ) .
0 8 0 728205
8
.
.
( . ) .
= »
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17. Modified Euler’s Method (Heun’s Method)
In the Modified Euler’s method an average value
of the slope is used.
{ úû
( ) ( ) }ù
» + é + + + +
, , * 1 2 1 1
êë
1
n n n n n n y y h f x y f x y
Where
*
n+1 y is calculated using the Euler’s method.
( ) n n n n y* » y + hf x , y +1
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18. ( ) n n n n y* » y + hf x , y +1
*
n+1 y as calculated using the Euler’s
method shown above is used to
obtain the 2nd slope.
( * )
n n f x , y +1 +1
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19. { úû
( ) ( ) }ù
» + é + + + +
, , * 1 2 1 1
êë
1
n n n n n n y y h f x y f x y
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20. Consider the previous initial value
problem.
y¢ = y - x , y(0) = 0.5
Find the value of y(0.8).
Again, let h = 0.1
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21. 0 5 0 0 5 0 0 0 0 f (x , y ) = y - x = . - = .
0 5 0 1 0 5 0 55 1 y* » . + . ´ . = .
0 55 0 1 0 45 1 1 1 1 f (x , y* ) = y* - x = . - . = .
{ úû
( ) ( ) }ù
y y h 1 f x y f x y
» + é , + , * 1 0 2 0 0 1 1
êë
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22. { úû
( ) ( ) }ù
y y h 1 f x y f x y
» + é , + , * 1 0 2 0 0 1 1
êë
úû
0 . 5 0 . 1 1 . . ù
= . 1 y » + é +
{0 5 0 45} 0 5475
2
êë
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23. f x y = y - x = -
( , ) . .
0 5475 0 1 1 1 1 1
.
0 4475
=
* . . .
»
y » + ´
0 5475 0 1 0 4475 2
.
0 59225
( , * ) * . .
f x y = y - x = -
0 59225 0 2 2 2 2 2
.
0 39225
=
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24. { úû
( ) ( ) }ù
y y h 1 f x y f x y
» + é , + , *2 1 2 1 1 2 2
êë
y » + é +
{ }
0 5475 0 1 1 2
. . . .
0 589487
0 4475 0 39225
2
.
»
ù
úû
êë
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25. Taylor’s Method of Order Two
2
2
2
y x + h = y x + hy¢ x + h y¢¢(x) + R
( ) ( ) ( )
!
Ris called Taylor’s remainder.
2 If h is small then:
2
y(x + h) » y(x) + hy¢(x) + h y¢¢ x
( )
!
2
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26. 2
y(x + h) » y(x) + hy¢(x) + h y¢¢ x
( )
!
2
Taylor’s method of order two can be written
in the form of the following sequence.
2
y » y + hy¢ + h y¢¢ n + 1
n n 2
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27. Solve the following IVP by using Taylor’s method of
order two.
y¢ = - y + x +1 , 0 £ x £ 1 , y(0) = 1
Solution:
y¢¢ = - y¢ +1 = -(- y + x +1) +1 = y - x
2
y » y + hy¢ + h y¢¢ n + 1
n n 2
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28. ¢ = - + +1
y y x
¢¢ = -
n n n
y y x
n n n
2
y » y + hy¢ + h y¢¢ n + 1
n n 2
n y » y + h ( - y + x + ) + h ( y - x ) n + n n n 1
2
n n 2
1
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29. ö
æ
= - + + 2 2
æ
- + ÷ ÷ø
y h h y h h x h n n n + ÷ ÷ø
ç çè
ö
ç çè
1
2 2
1
Let h = 0.1
0 905 0 095 0 1 1 = . + . + . n+ n n y y x
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30. Results with Taylor’s Method of Order Two
n xn yn+1
0 0.0 1.005
1 0.1 1.019025
2 0.2 1.041218
3 0.3 1.070802
4 0.4 1.107076
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31. IINNTTEERRPPOOLLAATTIIOONN
Interpolation produces a
function that matches the given
data exactly. The function then
should provide a good
approximation to the data
values at intermediate points.
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32. Interpolation may also be used
to produce a smooth graph of a
function for which values are
known only at discrete points,
either from measurements or
calculations.
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33. Given data points
Obtain a function, P(x)
P(x) goes through the data points
Use P(x)
To estimate values at intermediate
points
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34. school.edhole.com

35. Assume that a function goes through three points:
( ( ) ) ( ( ) ) ( ( ) ) 0 0 1 1 2 2 x , y x , x , y x and x , y x .
y(x) » P(x)
( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 0 1 1 2 2 P x = L x y x + L x y x + L x y x
Lagrange Interpolating Polynomial
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36. ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 0 1 1 2 2 P x = L x y x + L x y x + L x y x
P x x x x x y x
( - )( -
)
( )( )
( )( )
( )( )
( )( )
( )( )
( ) ( )
x x x x
x x x x y x
x x x x
x x x x y x
x x x x
( )
( )
1 2
0
0 1 0 2
0 2
1
1 0 1 2
0 1
2
2 0 2 1
=
- -
- -
+
- -
- -
+
- -
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37. y¢(x) » P¢(x)
( ) ( )
P ¢ x x - x - x =
y x
x x x x
x x x y x
x x x x
x x x y x
x x x x
( )
( )
2
1 2
( )( )
2
( )( )
2
( )( )
0
- -
- -
0 1 0 2
0 2
1
1 0 1 2
0 1
2
2 0 2 1
+
- -
- -
+
- -
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38. NNUUMMEERRIICCAALL IINNTTEEGGRRAATTIIOONN
ò y ( x )
dx = area under the curve f(x) between
b
a
x = a to x = b.
In many cases a mathematical expression for y(x) is
unknown and in some cases even if y(x) is known its
complex form makes it difficult to perform the integration.
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39. Simpson’s Rule:
x x
x x
ò y x dx » ò P x dx
( ) ( ) 2 2
0 0
x x h x x 2h 1 0 2 0 = + and = +
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40. P x dx x x x x y x dx
( )( )
( )( )
x x
x x
- -
( ) ( )
- -
( )
( )
ò ò
2 2
=
0 0
2
0
2
0
1 2
0
0 1 0 2
( )( )
0 2
( )( )
( )( )
( )( )
1
1 0 1 2
0 1
2
2 0 2 1
x
x
x
x
x x x x
x - x x -
x y x dx
x x x x
x x x x y x dx
x x x x
+
- -
- -
+
- -
ò
ò
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41. x x
x x
( ) ( )
( ) ( ) ( )
ò ò
2 2
y x dx P x dx
h y x y x y x
0 0
0 1 2 4
3
»
= éë + + ùû
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42. Runge-Kutta method uses a
sampling of slopes through an
interval and takes a weighted average
slope to calculate the end point. By
using fundamental theorem as shown
in Figure 1 we can write:
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43. y ( x0 + h) » y(x0 ) + hy¢ ( x0 )
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44. school.edhole.com

45. dy = y¢( x) dx  (1)
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46. Integrating both sides of Eqn. (1) we get
ò dy = ò y¢( x) dx  (2)
Applying appropriate limits to Eqn.
(2) we get
( ) ( ) x
1
( ) (3) n n òx + - = ¢ +
y x 1 y x y x dx  n
n
x h
n ò + + - = ¢
y( x h) y( x ) y ( x)dx  (4)
n n x
n
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47. Let us now concentrate on the right-hand
side of Eqn. (4).
x h
x
n ò ò + +
y ( x) dx P( x) dx  (5)
x h
x
n
n
n
¢ »
P(x) can be generated by utilizing Lagrange
Interpolating Polynomial. Assume that the
only information we have about a function,
f(x) is that it goes through three points:
( ( ) ) ( ( ) ) ( ( ) ) 0 0 1 1 2 2 x , y x , x , y x and x , y x .
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48. P(x) » y(x)
P ( x ) = L ( x ) y ( x ) + L ( x ) y ( x ) + L ( x ) y ( x
) 0 0 1 1 2 2 P ( x ) x - x x -
x y ( x
)
x x x x
x x x x y x
x x x x
x x x x y x
x x x x
( )
( )
( )( )
1 2
( )( )
( )( )
( )( )
( )( )
( )( )
0
0 1 0 2
0 2
1
1 0 1 2
0 1
2
2 0 2 1
=
- -
- -
+
- -
- -
+
- -
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49. Using Simpson’s Integration,
x x
x x
( ) ( ) 2 2
0 0
ò ò
y x dx »
P x dx
= + ¢ = + ¢
x x h x x h
1 0 2 0 and 2
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50. P x dx x x x x y x dx
( )( )
( )( )
x x
x x
- -
( ) ( )
- -
( )
( )
ò =
ò
( )( )
( )( )
( )( )
( )( )
ò
ò
2 y ( x ) dx 2
P ( x ) dx h y ( x ) y ( x ) y ( x
) 0 0
...………………….(6)
2 2
0 0
2
0
2
0
1 2
0
0 1 0 2
0 2
1
1 0 1 2
0 1
2
2 0 2 1
x
x
x
x
x x x x
x - x x -
x y x dx
x x x x
x x x x y x dx
x x x x
+
- -
- -
+
- -
0 1 2 4
3
x x
x x
¢
ò » ò = éë + + ùû
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51. We can apply Simpson’s Integration to Eqn.
(6) with the following substitutions:
x = xn 0 x x h n = + ; 2 ;
h¢ = h/ 2 and the midpoint, 1 2
x x h n = +
ù
ò ¢ » ¢ + ¢æ + + y x dx h y x y x h y x h n n n
é + ¢ + ÷ø
( ) ( ) ( )úû
êë
ö çè
x n
h
x
4
n 6
2
….(7)
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52. Compared to Euler’s Formula, an average of
six slopes is used in Eqn. (7) instead of just
one slope. Actually, the slope at the midpoint
has a weight of 4. The slope at the midpoint
can be estimated in two ways.
ù
ò ¢ » ¢ + ¢æ + + y x dx h y x y x h y x h y x h n n n n
÷ø
é ÷ø
+ ¢ æ + + ¢ + ( ) ( ) ( )úû
êë
ö çè
ö çè
x n
h
x
2
n 2
2
2
6
ù
y x h y x h é
+ - » y¢ x + y¢æ x + h y x h y x h n n n n n 2
n ö çè
ö çè
( ) ( ) ( ) ( )úû
êë
+ ¢ + ÷ø
+ 2
¢ ÷ø
æ + 2
2
6
ù
y x h y x h é
+ » + y¢ x + y¢æ x + h y x h y x h n n n n n 2
n ö çè
ö çè
( ) ( ) ( ) ( )úû
êë
+ ¢ + ÷ø
+ 2
¢ ÷ø
æ + 2
2
6
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53. é ¢ + ÷ø
n n 6 n n n n y x y x h y x y x h y x h y x
» + ¢ + ¢æ + +1 2 +1
ö çè
( ) ( ) ( ) ( )úû
The slopes can be estimated in the
following manner:
( n ) ( n n ) y x k f x , y 1 ¢ » =
y x h k f x h y h k n n n
¢æ + 2 2 1
ö çè
÷ø
ö çè
» = æ + ÷ø
,
+ 2 2
ù
êë
+ 2
çè¢ ÷ø
æ + ö 2
2
........... (8)
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54. y ¢æ x + h k f h h n 3 çè
x y k ö n n
2 2
y ¢ ( x + h ) » k = f ( x + h, y +
hk ) n 4 n n 3 Substituting the estimated slopes into Eqn.
(8) gives the formula for Runge-Kutta
Method of Fourth Order:
÷ø
ö çè
» = æ + ÷ø
,
+ 2 2
y y h k k k k n n = + + + + +
[ ] 1 1 2 3 4 2 2
6
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