Taylor Series Expansion

The 

If 

the 

Taylor Series Expansion 

Taylor’s Theorem : Suppose f is continuous on the closed interval [a, b] and has 

n + 1 continuous derivatives on the open interval (a, b). If x and c are points in 

(a, b), then 

f ( c) 

 

or 

Taylor 

f ( x) 

 


f 

' 

series 

∞ 

∑ 

k 0 

( c)( 

x 

1 

k! 

series 

Series 

c) 

 

 

 

k 0 

converge, 

f 

( k ) 

( c) 

expansion of 

1 

k! 

f 

f 

( x 

( c) 

2! 

(2) 

( k ) 

we 

c) 

( c) 

k 

( x 

( x 

c) 

f ( x) 

2 

c) 

 

k 

can write : 

about c: 

f 

( c) 

3! 

(3) 

( x 

c) 

3 

...

Maclaurin Series 

Maclaurin series is a special case of Taylor series with the 

center of expansion c = 0. 

The 

f 

If 

(0) 

the 

Maclaurin 

 

series 

(2) 

' f (0) 

f (0) x 

2! 

series converge, 

expansion of 

(3) 

( x) : 

2 f (0) 3 

x x ... 

3! 

we can write : 

f 

f 

( x) 

 

∞ ∑ 

k0 

1 

k! 

f 

( k) 

(0) 

x 

k

What If we change the interval ? 

If we change the interval so x=x+h and c=x 

The 


series 

expansion of 

f 

( x 

 

h) 

about 

c 

 

x 

f 

( x 

 

h) 

 

f 

( x) 

 

f 

' 

( x) 

h 

 

f 

( x) 

2! 

(2) 

h 

2 

 

f 

( x) 

3! 

(3) 

h 

3 

...

Finite Differences for Derivation 

Taylor expansion for F( x h) 

F( 

x 

h) 

 

F( 

x) 

 

hF'( 

x) 

 

2 

h 

2! 

F''( 

x) 

..... 

F( 

x 

 

h) 

 

F( 

x) 

 

hF'( 

x) 

O( 

h 

2 

) 

Use only first two terms of the 

expansion 

2 

F ( x h) 

F( 

x) 

O( 

h ) 

F'( 

x) 

 

h 

h 

F'( 

x) 

 

F( 

x 

 

h) 

 

h 

F( 

x) 

Forward Difference Formula


Taylor expansion for F( x h) 

F( 

x 

 

h) 

 

F( 

x) 

 

hF'( 

x) 

 

2 

h 

2! 

F''( 

x) 

..... 

F( 

x) 

 

F( 

x 

 

h) 

 

hF'( 

x) 

 

O( 

h 

2 

) 

F'( 

x) 

 

F( 

x) 

 

F( 

x 

h 

 

h) 

Backward Difference Formula


..... 

) 

'''( 

3! 

) 

''( 

2! 

) 

'( 

) 

( 

) 

( 

3 

2 

 

 

 

 

 

 

x 

F 

h 

x 

F 

h 

x 

hF 

x 

F 

h 

x 

F 

..... 

) 

'''( 

3! 

2 

) 

'( 

2 

) 

( 

) 

( 

3 

 

 

 

 

 

 

x 

F 

h 

x 

hF 

h 

x 

F 

h 

x 

F 

) 

( 

) 

( 

' 

2 

) 

( 

) 

( 3 

h 

O 

x 

F 

h 

h 

x 

F 

h 

x 

F 

 

 

 

 

 

Central Difference Formula 

h 

h 

x 

F 

h 

x 

F 

x 

F 

2 

) 

( 

) 

( 

) 

'( 

 

 

 

 

..... 

) 

'''( 

3! 

) 

''( 

2! 

) 

'( 

) 

( 

) 

( 

3 

2 

 

 

 

 

 

 

x 

F 

h 

x 

F 

h 

x 

hF 

x 

F 

h 

x 

F

Example : Derivation of 

Forward Difference Formula 

) 

( 

) 

( 

) 

( 

2 

) 

( 

) 

''( 

2 

2 h 

O 

h 

h 

x 

F 

x 

F 

h 

x 

F 

x 

F 

 

 

 

 

 

 

..... 

) 

''( 

2! 

) 

'( 

) 

( 

) 

( 

2 

 

 

 

 

 

x 

F 

h 

x 

hF 

x 

F 

h 

x 

F 

..... 

) 

''( 

2! 

) 

'( 

) 

( 

) 

( 

2 

 

 

 

 

 

x 

F 

h 

x 

hF 

x 

F 

h 

x 

F 

+ 

) 

''( 

2! 

) 

''( 

2! 

) 

( 

) 

'( 

) 

( 

2 

) 

( 

) 

( 

2 

2 

x 

F 

h 

x 

F 

h 

x 

hF 

x 

hF 

x 

F 

h 

x 

F 

h 

x 

F 

 

 

 

 

 

 

 

 

 

) 

''( 

2! 

2 

) 

( 

2 

) 

( 

) 

( 

2 

x 

F 

h 

x 

F 

h 

x 

F 

h 

x 

F

Better approximations 

By using Taylor expansions for F( x 2h) 

and F( x 2h) 

, better approximations 

can be obtained: 

F'( 

x) 

 

 

F( 

x 2h) 

4F( 

x h) 

3F( 

x) 

2h 

O( 

h 

2 

) 


F'( 

x) 

 

3F( 

x) 

4F( 

x h) 

 

2h 

F( 

x 2h) 

O( 

h 

2 

) 

Backward Difference Formula 

F'( 

x) 

 

 

F( 

x 2h) 

8F( 

x h) 

8F( 

x h) 

 

12h 

F( 

x 2h) 

O( 

h 

4 

) 

Central Difference Formula

Example : Derivation of 


..... 

) 

''( 

2! 

) 

'( 

) 

( 

) 

( 

2 

 

 

 

 

 

x 

F 

h 

x 

hF 

x 

F 

h 

x 

F 

..... 

) 

''( 

2! 

4 

) 

'( 

2 

) 

( 

) 

2 

( 

2 

 

 

 

 

 

x 

F 

h 

x 

hF 

x 

F 

h 

x 

F 

∗ (−4) 

+ 

) 

''( 

2! 

4 

. 

) 

''( 

2! 

4 

) 

'( 

4 

) 

'( 

2 

) 

( 

4 

) 

( 

) 

( 

4 

) 

2 

( 

2 

2 

x 

F 

h 

x 

F 

h 

x 

hF 

x 

hF 

x 

F 

x 

F 

h 

x 

F 

h 

x 

F 

 

 

 

 

 

 

 

 

 

) 

( 

2 

) 

( 

3 

) 

( 

4 

) 

2 

( x 

hF 

x 

F 

h 

x 

F 

h 

x 

F 

 

 

 

 

 

 

h 

x 

F 

h 

x 

F 

h 

x 

F 

x 

F 

2 

) 

( 

3 

) 

( 

4 

) 

2 

( 

) 

(

Second Derivatives 

Similarly various approximations for second derivatives are possible : 

F( 

x 2h) 

2F( 

x h) 

F( 

x) 

F' '( x) 

 

O( 

h) 

2 

h 

F''( 

x) 

 

F( 

x 

 

h) 

2F( 

x) 

F( 

x h) 

O( 

h 

2 

h 

2 

) 

F''( 

x) 

 

 

F( 

x 2h) 

16F( 

x h) 

30F( 

x) 

16F( 

x h) 

F( 

x 2h) 

O( 

h 

2 

12h 

4 

)

Derivatives of Bivariate & 

Multivariate Functions

First Order Partial Derivatives 

For functions with more variables, the partial derivatives can be approximated 

by grouping together all of the same variables and applying the univariate 

approximation for that group. 

For example, if F(x, y) is our function, then some first order partial derivative 

approximations are: 

F( 

x h, 

y) 

F( 

x 

f x 

( x, 

y) 

 

2h 

 

h, 

y) 

F( 

x, 

y k) 

F( 

x, 

f y 

( x, 

y) 

 

2k 

y 

 

k)

Second Partial Derivatives 

hk 

k 

y 

h 

x 

F 

k 

y 

h 

x 

F 

k 

y 

h 

x 

F 

k 

y 

h 

x 

F 

x 

f 

y 

x 

y 

f 

y 

x 

f xy 4 

) 

, 

( 

) 

, 

( 

) 

, 

( 

) 

, 

( 

) 

, 

( 

2 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2 

2 

2 

) 

, 

( 

) 

, 

( 

2 

) 

, 

( 

) 

, 

( 

h 

y 

h 

x 

F 

y 

x 

f 

y 

h 

x 

F 

x 

f 

x 

x 

f 

y 

x 

f xx 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2 

2 

2 

) 

, 

( 

) 

, 

( 

2 

) 

, 

( 

) 

, 

( 

k 

k 

y 

x 

F 

y 

x 

f 

k 

y 

x 

F 

y 

f 

y 

y 

f 

y 

x 

f yy 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Following formulas can be verified by taking the limits 

and 

0 

h 0 

 

k

The derivatives F x , F y , F xx and F yy just use the univariate approximation formulas. 

The mixed derivative requires slightly more work. The important observation is that the 

approximation for F xy is obtained by applying the x–derivative approximation for F x , 

then applying the y–derivative approximation to the previous approximation. 

1 - the x–derivative approximation for F x : 

F( 

x h, 

y) 

F( 

x 

f x 

( x, 

y) 

 

2h 

h, 

y) 

2- the y–derivative approximation: 

F( 

x, 

y k) 

F( 

x, 

f y 

( x, 

y) 

 

2k 

y 

 

k) 

3- Apply the 2 nd formula to the 1st one: 

F( 

x h, 

y k) 

F( 

x h, 

y k) 

F( 

x 

f xy 

( x, 

y) 

 

4hk 

h, 

y 

 

k) 

 

F( 

x 

 

h, 

y 

k)

Second Partial Derivatives 

Following formulas can be verified by taking the limits 

To find the 

F 

xy 

partial derivative first use the formula for 

h 0 and k 0 

F 

x 

Then apply the approximation for 

( f y 

) 

f xx 

( x, 

y) 

 

F( 

x 

 

h, 

y) 

 

2 

f 

( x, 

h 

2 

y) 

 

F( 

x 

 

h, 

y) 

f yy 

( x, 

y) 

 

F( 

x, 

y 

 

k) 

 

2 

f 

( x, 

k 

2 

y) 

 

F( 

x, 

y 

 

k) 

F( 

x h, 

y k) 

F( 

x h, 

y k) 

F( 

x 

f xy 

( x, 

y) 

 

4hk 

h, 

y 

 

k) 

 

F( 

x 

 

h, 

y 

k)

Classification of the Methods 

Numerical Methods 

for Solving ODE 

Single-Step Methods 

Multiple-Step Methods 

Estimates of the solution 

at a particular step are 

entirely based on 

information on the 

Estimates of the solution 

at a particular step are 

based on information on 

more than one step 

previous step 

17

Taylor Series Method 

The problem to be solved is a first order ODE: 

dy( 

x) 

dx 

 

f 

( x , 

y ), 

y ( x 

0 ) 

 

y 

0 

Estimates of the solution at different base points: 

y 

( 

0 

x0 h), 

y( 

x0 

2h), 

y( 

x 3h), 

.... 

are computed using the truncated Taylor series 

expansions. 

18

Taylor Series Expansion 

Truncated Taylor Series 

Expansion 

y( 

x 

0 

 

h) 

 

n 

 

k 

h 

 

k! 

 

 

k 

d y 

dx 

k 

k 0 xx 

, y 

y 

0 

0 

 

 

 

 

 

y( 

x 

0 

) 

 

h 

dy 

dx 

xx 

y 

y 

0 

0 

, 

 

2 

h 

2! 

2 

d y 

2 

dx 

xx 

y 

y 

0 

0 

, 

... 

 

n 

h 

n! 

n 

d y 

n 

dx 

xx 

y 

y 

0 

0 

, 

The n th order Taylor series method uses the 

n th order Truncated Taylor series expansion. 

19

Euler Method 

• First order Taylor series method is known as 

Euler Method. 

• Only the constant term and linear term are 

used in the Euler method. 

• The error due to the use of the truncated 

Taylor series is of order O(h 2 ). 

20

21 

First Order Taylor Series Method 

(Euler Method) 

) 

, 

( 

) 

, 

( 

), 

( 

, 

: 

) 

( 

) 

( 

) 

( 

1 

, 

0 

2 

, 

0 

0 

0 

0 

i 

i 

i 

i 

i 

i 

y 

y 

x 

x 

n 

n 

n 

y 

y 

x 

x 

y 

x 

f 

h 

y 

y 

Method 

Euler 

y 

x 

f 

dx 

dy 

x 

y 

y 

nh 

x 

x 

Notation 

h 

O 

dx 

dy 

h 

x 

y 

h 

x 

y 

i 

i

22 

Euler Method 

1,2,... 

) 

, 

( 

) 

( 

Method: 

Euler 

1,2,... 

) 

( 

Determine : 

) 

( 

condition : 

initial 

with the 

) 

, 

( 

) 

( 

ODE: 

order 

first 

Given the 

Problem : 

1 

0 

0 

0 

0 

0 

 

 

 

 

 

 

 

 

 

 

i 

for 

y 

x 

f 

h 

y 

y 

x 

y 

y 

i 

for 

ih 

x 

y 

y 

x 

y 

y 

y 

x 

f 

x 

y 

i 

i 

i 

i 

i

Interpretation of Euler Method 

y 2 

y 1 

y 0 

x 0 x 1 x 2 x 

23


y 1 

Slope=f(x 0 ,y 0 ) 

y 1 =y 0 +hf(x 0 ,y 0 ) 

hf(x 0 ,y 0 ) 

y 0 

x 0 x 1 x 2 x 

h 

24


y 2 


hf(x 1 ,y 1 ) 

y 1 


y 2 =y 1 +hf(x 1 ,y 1 ) 

y 1 =y 0 +hf(x 0 ,y 0 ) 

hf(x 0 ,y 0 ) 

y 0 

x 0 x 1 x 2 x 

h 

h 

25

Example 1 

Use Euler method to solve the ODE: 

dy 

dx 

1 

x 

2 

, y(1) 

4 

to determine y(1.01), y(1.02) and y(1.03). 

26

27 

Example 1 

0.01 

4, 

1, 

, 

1 

) 

, 

( 0 

0 

2 

 

 

 

 

 

h 

y 

x 

x 

y 

x 

f 

 

 

 

 

 

 

9394 

3. 

1.02 

0.011 

3.9598 

) 

, 

( 

3: 

3.9598 

1.01 

0.011 

3.98 

) 

, 

( 

2 : 

3.98 

) 

(1) 

0.01(1 

4 

) 

, 

( 

1: 

) 

, 

( 

Method 

Euler 

2 

2 

2 

2 

3 

2 

1 

1 

1 

2 

2 

0 

0 

0 

1 

1 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

y 

x 

f 

h 

y 

y 

Step 

y 

x 

f 

h 

y 

y 

Step 

y 

x 

f 

h 

y 

y 

Step 

y 

x 

f 

h 

y 

y 

i 

i 

i 

i

Example 1 

f 

2 

( x, 

y) 

1 

x , x0 

1, 

y0 

4, 

h 

 

0.01 

Summary of the result: 

i xi yi 

0 1.00 -4.00 

1 1.01 -3.98 

2 1.02 -3.9595 

3 1.03 -3.9394 

28

Example 1 

f 

2 

( x, 

y) 

1 

x , x0 

1, 

y0 

4, 

h 

 

0.01 

Comparison with true value: 

i xi yi True value of yi 

0 1.00 -4.00 -4.00 

1 1.01 -3.98 -3.97990 

2 1.02 -3.9595 -3.95959 

3 1.03 -3.9394 -3.93909 

29

Example 1 

f 

2 

( x, 

y) 

1 

x , x0 

1, 

y0 

4, 

h 

 

0.01 

A graph of the 

solution of the 

ODE for 

1

Types of Errors 

– Local truncation error: 

Error due to the use of truncated Taylor series 

to compute x(t+h) in one step. 

– Global Truncation error: 

Accumulated truncation over many steps. 

– Round off error: 

Error due to finite number of bits used in 

representation of numbers. This error could be 

accumulated and magnified in succeeding 

steps. 

31

Second Order Taylor Series Methods 

Given 

Second 

dy( 

x) 

dx 

 

f 

( y, 

x), 

order Taylor Series 

y( 

x 

0 

) 

 

y 

method 

0 

y 

d 

i1 

2 

dx 

y 

2 

 

y 

i 

dy 

h 

dx 

h 

2 

2! 

d 

needs to be derived analytically. 

2 

dx 

y 

2 

O( 

h 

3 

) 

32

Third Order Taylor Series Methods 

Given 

dy( 

x) 

dx 

 

( y, 

x), 

Third order Taylor Series 

f 

y( 

x 

0 

) 

method 

y 

0 

y 

d 

i1 

2 

dx 

y 

2 

 

y 

i 

and 

dy 

h 

dx 

d 

3 

dx 

y 

3 

h 

2 

2! 

d 

2 

dx 

y 

2 

 

h 

3 

3! 

need to be derived analytically. 

d 

3 

dx 

y 

3 

O( 

h 

4 

) 

33

34 

High Order Taylor Series Methods 

to be derived analytically. 

need 

,....., 

, 

) 

( 

! 

.... 

2! 

method 

Series 

order Taylor 

n 

) 

( 

), 

, 

( 

) 

( 

3 

3 

2 

2 

1 

2 

2 

2 

1 

th 

0 

0 

n 

n 

n 

n 

n 

n 

i 

i 

dx 

y 

d 

dx 

y 

d 

dx 

y 

d 

h 

O 

dx 

y 

d 

n 

h 

dx 

y 

d 

h 

dx 

dy 

h 

y 

y 

y 

x 

y 

x 

y 

f 

dx 

x 

dy 

Given

Higher Order Taylor Series Methods 

• High order Taylor series methods are more 

accurate than Euler method. 

• But, the 2 nd , 3 rd , and higher order derivatives 

need to be derived analytically which may not 

be easy. 

35

Example 2 

Second order Taylor Series Method 

Use 

Second 

order 

Taylor Series 

method 

to 

solve: 

dx 

dt 

2x 

2 

 

t 

1, 

x(0) 

1, 

use 

h 

 

0.01 

What 

is : 

2 

d x( 

t) 

2 

dt 

? 

36

37 

Example 2 

)) 

2 

(1 

4 

1 

( 

2 

) 

2 

(1 

1 

) 

2 

(1 

4 

1 

4 

0 

) 

( 

2 

1 

0.01 

1, 

(0) 

1, 

2 

solve: 

to 

method 

Taylor Series 

order 

Second 

Use 

2 

2 

2 

1 

2 

2 

2 

2 

2 

i 

i 

i 

i 

i 

i 

i 

t 

x 

x 

h 

t 

x 

h 

x 

x 

t 

x 

x 

dt 

dx 

x 

dt 

t 

x 

d 

t 

x 

dt 

dx 

h 

use 

x 

t 

x 

dt 

dx

38 

Example 2 

)) 

2 

(1 

4 

1 

( 

2 

) 

2 

(1 

0.01 

1, 

0, 

, 

2 

1 

) 

, 

( 

2 

2 

2 

1 

0 

0 

2 

i 

i 

i 

i 

i 

i 

i 

t 

x 

x 

h 

t 

x 

h 

x 

x 

h 

x 

t 

t 

x 

x 

t 

f 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0.9716 

3: 

0.9807 

0.01)) 

2(0.9901) 

4(0.9901)(1 

1 

( 

2 

0.01 

0.01) 

2(0.9901) 

0.01(1 

0.9901 

2 : 

0.9901 

0)) 

2 

4(1)(1 

1 

( 

2 

0.01 

0) 

2(1) 

0.01(1 

1 

1: 

3 

2 

2 

2 

2 

2 

2 

1 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

x 

Step 

x 

Step 

x 

Step

Example 2 

f 

2 

( t, 

x) 

1 

2x 

t 

, t0 

0, x0 

1, 

h 

 

0.01 

Summary of the results: 

i t i x i 

0 0.00 1 

1 0.01 0.9901 

2 0.02 0.9807 

3 0.03 0.9716 

39

Programming Euler Method 

Write a MATLAB program to implement Euler 

method to solve: 

dv 

 

1 

2v 

2 

 

t. 

v(0) 

 

1 

dt 

for 

t 

i 

 

0.01i, 

i 

 

1,2,...,100 

40


f=inline('1-2*v^2-t','t','v') 

h=0.01 

t=0 

v=1 

T(1)=t; 

V(1)=v; 

for i=1:100 

v=v+h*f(t,v) 

t=t+h; 

T(i+1)=t; 

V(i+1)=v; 

end 

41


f=inline('1-2*v^2-t','t','v') 

h=0.01 

t=0 

v=1 

T(1)=t; 

V(1)=v; 

for i=1:100 

v=v+h*f(t,v) 

t=t+h; 

T(i+1)=t; 

V(i+1)=v; 

end 

Definition of the ODE 

Initial condition 

Main loop 

Euler method 

Storing information 

42


Plot of the 

solution 

plot(T,V) 

43

Example 1: 

Find y(0.5) if y is the solution of IVP y' = -2x-y, y(0) = -1 using Euler's method 

with step length 0.1. Also find the error in the approximation. 

Solution: f(x, y) = -2x - y, 

y 1 = y 0 + h f(x 0 , y 0 ) = -1 + 0.1* (-2*0 - (-1)) = -0.8999 

y 2 = y 1 + h f(x 1 , y 1 ) = -0.8999 + 0.1* (-2*0.1 - (-0.8999)) = -0.8299 

y 3 = y 2 + h f(x 2 , y 2 ) = -0.8299 + 0.1* (-2*0.2 - (-0.8299)) = -0.7869 

y 4 = y 3 + h f(x 3 , y 3 ) = -0.7869 + 0.1* (-2*0.3 - (-0.7869)) = -0.7683 

y 5 = y 4 + h f(x 4 , y 4 ) = -0.7683 + 0.1* (-2*0.4 - (-0.7683)) = -0.7715

Example 2: 

Use Eulers method to solve for y[0.1] from y' = x + y + xy, y(0) = 1 with h = 0.01 also 

estimate how small h would need to obtain four-decimal accuracy. 

Solution : f(x, y) = x + y + xy, 

y 1 = y 0 + h f(x 0 , y 0 ) = 1.0 + .01*(0 + 1 + 0*1) = 1.01 

y 2 = y 1 + h f(x 1 , y 1 ) = 1.01 + .01*(0.01 + 1.01 + 0.01*1.01) =1.02 

y 3 = y 2 + h f(x 2 , y 2 ) = 1.02 + .01*(0.02 + 1.02 + 0.02*1.02) =1.031 

y 4 = y 3 + h f(x 3 , y 3 ) = 1.031 + .01*(0.03 + 1.031 + 0.03*1.031) =1.042 

y 5 = y 4 + h f(x 4 , y 4 ) = 1.042 + .01*(0.04 + 1.042 + 0.04*1.042) = 1.053 

y 6 = y 5 + h f(x 5 , y 5 ) = 1.053 + .01*(0.05 + 1.053 + 0.05*1.053) = 1.065 

y 7 = y 6 + h f(x 6 , y 6 ) = 1.065 + .01*(0.06 + 1.065 + 0.06*1.065) = 1.076 

y 8 = y 8 + h f(x 7 , y 7 ) = 1.076 + .01*(0.07 + 1.076 + 0.07*1.076) = 1.089 

y 9 = y 9 + h f(x 8 , y 8 ) = 1.089 + .01*(0.08 + 1.089 + 0.08*1.089) = 1.101 

y 10 = y 10 + h f(x 9 , y 9 ) = 1.101 + .01*(0.09 + 1.101 + 0.09*1.101) = 1.114

Example 3: 

Solve the differential equation y' = x/y, y(0)=1 by Euler's method to get y(1). Use the step 

lengths h = 0.1 and 0.2 and compare the results with the analytical solution (y 2 = 1 + x 2 ) 

Solution: 

f(x, y) = x/y, 

with h = 0.1 

y 1 = y 0 + h f(x 0 , y 0 ) = 1.0 + 0.1*0.0/1.0 = 1.00 

y 2 = y 1 + h f(x 1 , y 1 ) = 1.0 + 0.1*0.1/1.0 = 1.01 

y 3 = y 2 + h f(x 2 , y 2 ) = 1.01 + 0.1*0.2/1.01 = 1.0298 

y 4 = y 3 + h f(x 3 , y 3 ) = 1.0298 + 0.1*0.3/1.0298 = 1.0589 

y 5 = y 4 + h f(x 4 , y 4 ) = 1.0589 + 0.1*0.4/1.0589 = 1.0967 

y 6 = y 5 + h f(x 5 , y 5 ) = 1.0967 + 0.1*0.5/1.0967 = 1.1423 

y 7 = y 6 + h f(x 6 , y 6 ) = 1.1423 + 0.1*0.6/1.1423 = 1.1948 

y 8 = y 7 + h f(x 7 , y 7 ) = 1.1948 + 0.1*0.7/1.1948 = 1.2534 

y 9 = y 8 + h f(x 8 , y 8 ) = 1.2534 + 0.1*0.8/1.2534 = 1.3172 

y 10 = y 9 + h f(x 9 , y 9 ) = 1.3172 + 0.1*0.9/1.3172 = 1.3855

with h = 0.2 

y 1 = y 0 + h f(x 0 , y 0 ) = 1.0 + 0.2*0.0/1.0 = 1.0 

y 2 = y 1 + h f(x 1 , y 1 ) = 1.0 + 0.2*0.2/1.0 = 1.0400 

y 3 = y 2 + h f(x 2 , y 2 ) = 1.0400 + 0.2*0.4/1.0400 = 1.1169 

y 4 = y 3 + h f(x 3 , y 3 ) = 1.1169 + 0.2*0.6/1.1169 = 1.2243 

y 5 = y 4 + h f(x 4 , y 4 ) = 1.2243 + 0.2*0.8/1.2243 = 1.3550

Comparison of numerical and 

analytical solutions 

x 

Numerical Solution 

h = 0.1 h = 0.2 

Analytical solution 

0.0 1.0 1.0 1.0 

0.1 1.0 1.0050 

0.2 1.01 1.0 1.0198 

0.3 1.0298 1.0440 

0.4 1.0589 1.0400 1.0770 

0.5 1.0967 1.1180 

0.6 1.1423 1.1169 1.1662 

0.7 1.1948 1.2207 

0.8 1.2534 1.2243 1.2806 

0.9 1.3172 1.3454 

1.0 1.3855 1.3550 1.4142

Taylor Series Expansion

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