Taylor Series Expansion
2016+Week4
2016+Week4
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The<br />
If<br />
the<br />
<strong>Taylor</strong> <strong>Series</strong> <strong>Expansion</strong><br />
<strong>Taylor</strong>’s Theorem : Suppose f is continuous on the closed interval [a, b] and has<br />
n + 1 continuous derivatives on the open interval (a, b). If x and c are points in<br />
(a, b), then<br />
f ( c)<br />
<br />
or<br />
<strong>Taylor</strong><br />
f ( x)<br />
<br />
<strong>Taylor</strong><br />
f<br />
'<br />
series<br />
∞<br />
∑<br />
k 0<br />
( c)(<br />
x<br />
1<br />
k!<br />
series<br />
<strong>Series</strong> <br />
c)<br />
<br />
<br />
<br />
k 0<br />
converge,<br />
f<br />
( k )<br />
( c)<br />
expansion of<br />
1<br />
k!<br />
f<br />
f<br />
( x<br />
( c)<br />
2!<br />
(2)<br />
( k )<br />
we<br />
c)<br />
( c)<br />
k<br />
( x<br />
( x<br />
c)<br />
f ( x)<br />
2<br />
c)<br />
<br />
k<br />
can write :<br />
about c:<br />
f<br />
( c)<br />
3!<br />
(3)<br />
( x<br />
c)<br />
3<br />
...
Maclaurin <strong>Series</strong><br />
Maclaurin series is a special case of <strong>Taylor</strong> series with the<br />
center of expansion c = 0.<br />
The<br />
f<br />
If<br />
(0)<br />
the<br />
Maclaurin<br />
<br />
series<br />
(2)<br />
' f (0)<br />
f (0) x <br />
2!<br />
series converge,<br />
expansion of<br />
(3)<br />
( x) :<br />
2 f (0) 3<br />
x x ...<br />
3!<br />
we can write :<br />
f<br />
f<br />
( x)<br />
<br />
∞ ∑<br />
k0<br />
1<br />
k!<br />
f<br />
( k)<br />
(0)<br />
x<br />
k
What If we change the interval ?<br />
If we change the interval so x=x+h and c=x<br />
The<br />
<strong>Taylor</strong><br />
series<br />
expansion of<br />
f<br />
( x<br />
<br />
h)<br />
about<br />
c<br />
<br />
x<br />
f<br />
( x<br />
<br />
h)<br />
<br />
f<br />
( x)<br />
<br />
f<br />
'<br />
( x)<br />
h<br />
<br />
f<br />
( x)<br />
2!<br />
(2)<br />
h<br />
2<br />
<br />
f<br />
( x)<br />
3!<br />
(3)<br />
h<br />
3<br />
...
Finite Differences for Derivation<br />
<strong>Taylor</strong> expansion for F( x h)<br />
F(<br />
x <br />
h)<br />
<br />
F(<br />
x)<br />
<br />
hF'(<br />
x)<br />
<br />
2<br />
h<br />
2!<br />
F''(<br />
x)<br />
.....<br />
F(<br />
x<br />
<br />
h)<br />
<br />
F(<br />
x)<br />
<br />
hF'(<br />
x)<br />
O(<br />
h<br />
2<br />
)<br />
Use only first two terms of the<br />
expansion<br />
2<br />
F ( x h)<br />
F(<br />
x)<br />
O(<br />
h )<br />
F'(<br />
x)<br />
<br />
h<br />
h<br />
F'(<br />
x)<br />
<br />
F(<br />
x<br />
<br />
h)<br />
<br />
h<br />
F(<br />
x)<br />
Forward Difference Formula
Finite Differences for Derivation<br />
<strong>Taylor</strong> expansion for F( x h)<br />
F(<br />
x<br />
<br />
h)<br />
<br />
F(<br />
x)<br />
<br />
hF'(<br />
x)<br />
<br />
2<br />
h<br />
2!<br />
F''(<br />
x)<br />
.....<br />
F(<br />
x)<br />
<br />
F(<br />
x<br />
<br />
h)<br />
<br />
hF'(<br />
x)<br />
<br />
O(<br />
h<br />
2<br />
)<br />
F'(<br />
x)<br />
<br />
F(<br />
x)<br />
<br />
F(<br />
x<br />
h<br />
<br />
h)<br />
Backward Difference Formula
Finite Differences for Derivation<br />
.....<br />
)<br />
'''(<br />
3!<br />
)<br />
''(<br />
2!<br />
)<br />
'(<br />
)<br />
(<br />
)<br />
(<br />
3<br />
2<br />
<br />
<br />
<br />
<br />
<br />
<br />
x<br />
F<br />
h<br />
x<br />
F<br />
h<br />
x<br />
hF<br />
x<br />
F<br />
h<br />
x<br />
F<br />
.....<br />
)<br />
'''(<br />
3!<br />
2<br />
)<br />
'(<br />
2<br />
)<br />
(<br />
)<br />
(<br />
3<br />
<br />
<br />
<br />
<br />
<br />
<br />
x<br />
F<br />
h<br />
x<br />
hF<br />
h<br />
x<br />
F<br />
h<br />
x<br />
F<br />
)<br />
(<br />
)<br />
(<br />
'<br />
2<br />
)<br />
(<br />
)<br />
( 3<br />
h<br />
O<br />
x<br />
F<br />
h<br />
h<br />
x<br />
F<br />
h<br />
x<br />
F<br />
<br />
<br />
<br />
<br />
<br />
Central Difference Formula<br />
h<br />
h<br />
x<br />
F<br />
h<br />
x<br />
F<br />
x<br />
F<br />
2<br />
)<br />
(<br />
)<br />
(<br />
)<br />
'(<br />
<br />
<br />
<br />
<br />
.....<br />
)<br />
'''(<br />
3!<br />
)<br />
''(<br />
2!<br />
)<br />
'(<br />
)<br />
(<br />
)<br />
(<br />
3<br />
2<br />
<br />
<br />
<br />
<br />
<br />
<br />
x<br />
F<br />
h<br />
x<br />
F<br />
h<br />
x<br />
hF<br />
x<br />
F<br />
h<br />
x<br />
F
Example : Derivation of<br />
Forward Difference Formula<br />
)<br />
(<br />
)<br />
(<br />
)<br />
(<br />
2<br />
)<br />
(<br />
)<br />
''(<br />
2<br />
2 h<br />
O<br />
h<br />
h<br />
x<br />
F<br />
x<br />
F<br />
h<br />
x<br />
F<br />
x<br />
F<br />
<br />
<br />
<br />
<br />
<br />
<br />
.....<br />
)<br />
''(<br />
2!<br />
)<br />
'(<br />
)<br />
(<br />
)<br />
(<br />
2<br />
<br />
<br />
<br />
<br />
<br />
x<br />
F<br />
h<br />
x<br />
hF<br />
x<br />
F<br />
h<br />
x<br />
F<br />
.....<br />
)<br />
''(<br />
2!<br />
)<br />
'(<br />
)<br />
(<br />
)<br />
(<br />
2<br />
<br />
<br />
<br />
<br />
<br />
x<br />
F<br />
h<br />
x<br />
hF<br />
x<br />
F<br />
h<br />
x<br />
F<br />
+<br />
)<br />
''(<br />
2!<br />
)<br />
''(<br />
2!<br />
)<br />
(<br />
)<br />
'(<br />
)<br />
(<br />
2<br />
)<br />
(<br />
)<br />
(<br />
2<br />
2<br />
x<br />
F<br />
h<br />
x<br />
F<br />
h<br />
x<br />
hF<br />
x<br />
hF<br />
x<br />
F<br />
h<br />
x<br />
F<br />
h<br />
x<br />
F<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
)<br />
''(<br />
2!<br />
2<br />
)<br />
(<br />
2<br />
)<br />
(<br />
)<br />
(<br />
2<br />
x<br />
F<br />
h<br />
x<br />
F<br />
h<br />
x<br />
F<br />
h<br />
x<br />
F
Better approximations<br />
By using <strong>Taylor</strong> expansions for F( x 2h)<br />
and F( x 2h)<br />
, better approximations<br />
can be obtained:<br />
F'(<br />
x)<br />
<br />
<br />
F(<br />
x 2h)<br />
4F(<br />
x h)<br />
3F(<br />
x)<br />
2h<br />
O(<br />
h<br />
2<br />
)<br />
Forward Difference Formula<br />
F'(<br />
x)<br />
<br />
3F(<br />
x)<br />
4F(<br />
x h)<br />
<br />
2h<br />
F(<br />
x 2h)<br />
O(<br />
h<br />
2<br />
)<br />
Backward Difference Formula<br />
F'(<br />
x)<br />
<br />
<br />
F(<br />
x 2h)<br />
8F(<br />
x h)<br />
8F(<br />
x h)<br />
<br />
12h<br />
F(<br />
x 2h)<br />
O(<br />
h<br />
4<br />
)<br />
Central Difference Formula
Example : Derivation of<br />
Forward Difference Formula<br />
.....<br />
)<br />
''(<br />
2!<br />
)<br />
'(<br />
)<br />
(<br />
)<br />
(<br />
2<br />
<br />
<br />
<br />
<br />
<br />
x<br />
F<br />
h<br />
x<br />
hF<br />
x<br />
F<br />
h<br />
x<br />
F<br />
.....<br />
)<br />
''(<br />
2!<br />
4<br />
)<br />
'(<br />
2<br />
)<br />
(<br />
)<br />
2<br />
(<br />
2<br />
<br />
<br />
<br />
<br />
<br />
x<br />
F<br />
h<br />
x<br />
hF<br />
x<br />
F<br />
h<br />
x<br />
F<br />
∗ (−4)<br />
+<br />
)<br />
''(<br />
2!<br />
4<br />
.<br />
)<br />
''(<br />
2!<br />
4<br />
)<br />
'(<br />
4<br />
)<br />
'(<br />
2<br />
)<br />
(<br />
4<br />
)<br />
(<br />
)<br />
(<br />
4<br />
)<br />
2<br />
(<br />
2<br />
2<br />
x<br />
F<br />
h<br />
x<br />
F<br />
h<br />
x<br />
hF<br />
x<br />
hF<br />
x<br />
F<br />
x<br />
F<br />
h<br />
x<br />
F<br />
h<br />
x<br />
F<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
)<br />
(<br />
2<br />
)<br />
(<br />
3<br />
)<br />
(<br />
4<br />
)<br />
2<br />
( x<br />
hF<br />
x<br />
F<br />
h<br />
x<br />
F<br />
h<br />
x<br />
F<br />
<br />
<br />
<br />
<br />
<br />
<br />
h<br />
x<br />
F<br />
h<br />
x<br />
F<br />
h<br />
x<br />
F<br />
x<br />
F<br />
2<br />
)<br />
(<br />
3<br />
)<br />
(<br />
4<br />
)<br />
2<br />
(<br />
)<br />
(
Second Derivatives<br />
Similarly various approximations for second derivatives are possible :<br />
F(<br />
x 2h)<br />
2F(<br />
x h)<br />
F(<br />
x)<br />
F' '( x)<br />
<br />
O(<br />
h)<br />
2<br />
h<br />
F''(<br />
x)<br />
<br />
F(<br />
x<br />
<br />
h)<br />
2F(<br />
x)<br />
F(<br />
x h)<br />
O(<br />
h<br />
2<br />
h<br />
2<br />
)<br />
F''(<br />
x)<br />
<br />
<br />
F(<br />
x 2h)<br />
16F(<br />
x h)<br />
30F(<br />
x)<br />
16F(<br />
x h)<br />
F(<br />
x 2h)<br />
O(<br />
h<br />
2<br />
12h<br />
4<br />
)
Derivatives of Bivariate &<br />
Multivariate Functions
First Order Partial Derivatives<br />
For functions with more variables, the partial derivatives can be approximated<br />
by grouping together all of the same variables and applying the univariate<br />
approximation for that group.<br />
For example, if F(x, y) is our function, then some first order partial derivative<br />
approximations are:<br />
F(<br />
x h,<br />
y)<br />
F(<br />
x<br />
f x<br />
( x,<br />
y)<br />
<br />
2h<br />
<br />
h,<br />
y)<br />
F(<br />
x,<br />
y k)<br />
F(<br />
x,<br />
f y<br />
( x,<br />
y)<br />
<br />
2k<br />
y<br />
<br />
k)
Second Partial Derivatives<br />
hk<br />
k<br />
y<br />
h<br />
x<br />
F<br />
k<br />
y<br />
h<br />
x<br />
F<br />
k<br />
y<br />
h<br />
x<br />
F<br />
k<br />
y<br />
h<br />
x<br />
F<br />
x<br />
f<br />
y<br />
x<br />
y<br />
f<br />
y<br />
x<br />
f xy 4<br />
)<br />
,<br />
(<br />
)<br />
,<br />
(<br />
)<br />
,<br />
(<br />
)<br />
,<br />
(<br />
)<br />
,<br />
(<br />
2<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
2<br />
2<br />
2<br />
)<br />
,<br />
(<br />
)<br />
,<br />
(<br />
2<br />
)<br />
,<br />
(<br />
)<br />
,<br />
(<br />
h<br />
y<br />
h<br />
x<br />
F<br />
y<br />
x<br />
f<br />
y<br />
h<br />
x<br />
F<br />
x<br />
f<br />
x<br />
x<br />
f<br />
y<br />
x<br />
f xx<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
2<br />
2<br />
2<br />
)<br />
,<br />
(<br />
)<br />
,<br />
(<br />
2<br />
)<br />
,<br />
(<br />
)<br />
,<br />
(<br />
k<br />
k<br />
y<br />
x<br />
F<br />
y<br />
x<br />
f<br />
k<br />
y<br />
x<br />
F<br />
y<br />
f<br />
y<br />
y<br />
f<br />
y<br />
x<br />
f yy<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
Following formulas can be verified by taking the limits<br />
and<br />
0<br />
h 0<br />
<br />
k
The derivatives F x , F y , F xx and F yy just use the univariate approximation formulas.<br />
The mixed derivative requires slightly more work. The important observation is that the<br />
approximation for F xy is obtained by applying the x–derivative approximation for F x ,<br />
then applying the y–derivative approximation to the previous approximation.<br />
1 - the x–derivative approximation for F x :<br />
F(<br />
x h,<br />
y)<br />
F(<br />
x<br />
f x<br />
( x,<br />
y)<br />
<br />
2h<br />
h,<br />
y)<br />
2- the y–derivative approximation:<br />
F(<br />
x,<br />
y k)<br />
F(<br />
x,<br />
f y<br />
( x,<br />
y)<br />
<br />
2k<br />
y<br />
<br />
k)<br />
3- Apply the 2 nd formula to the 1st one:<br />
F(<br />
x h,<br />
y k)<br />
F(<br />
x h,<br />
y k)<br />
F(<br />
x<br />
f xy<br />
( x,<br />
y)<br />
<br />
4hk<br />
h,<br />
y<br />
<br />
k)<br />
<br />
F(<br />
x<br />
<br />
h,<br />
y<br />
k)
Second Partial Derivatives<br />
Following formulas can be verified by taking the limits<br />
To find the<br />
F<br />
xy<br />
partial derivative first use the formula for<br />
h 0 and k 0<br />
F<br />
x<br />
Then apply the approximation for<br />
( f y<br />
)<br />
f xx<br />
( x,<br />
y)<br />
<br />
F(<br />
x<br />
<br />
h,<br />
y)<br />
<br />
2<br />
f<br />
( x,<br />
h<br />
2<br />
y)<br />
<br />
F(<br />
x<br />
<br />
h,<br />
y)<br />
f yy<br />
( x,<br />
y)<br />
<br />
F(<br />
x,<br />
y<br />
<br />
k)<br />
<br />
2<br />
f<br />
( x,<br />
k<br />
2<br />
y)<br />
<br />
F(<br />
x,<br />
y<br />
<br />
k)<br />
F(<br />
x h,<br />
y k)<br />
F(<br />
x h,<br />
y k)<br />
F(<br />
x<br />
f xy<br />
( x,<br />
y)<br />
<br />
4hk<br />
h,<br />
y<br />
<br />
k)<br />
<br />
F(<br />
x<br />
<br />
h,<br />
y<br />
k)
Classification of the Methods<br />
Numerical Methods<br />
for Solving ODE<br />
Single-Step Methods<br />
Multiple-Step Methods<br />
Estimates of the solution<br />
at a particular step are<br />
entirely based on<br />
information on the<br />
Estimates of the solution<br />
at a particular step are<br />
based on information on<br />
more than one step<br />
previous step<br />
17
<strong>Taylor</strong> <strong>Series</strong> Method<br />
The problem to be solved is a first order ODE:<br />
dy(<br />
x)<br />
dx<br />
<br />
f<br />
( x ,<br />
y ),<br />
y ( x<br />
0 )<br />
<br />
y<br />
0<br />
Estimates of the solution at different base points:<br />
y<br />
(<br />
0<br />
x0 h),<br />
y(<br />
x0<br />
2h),<br />
y(<br />
x 3h),<br />
....<br />
are computed using the truncated <strong>Taylor</strong> series<br />
expansions.<br />
18
<strong>Taylor</strong> <strong>Series</strong> <strong>Expansion</strong><br />
Truncated <strong>Taylor</strong> <strong>Series</strong><br />
<strong>Expansion</strong><br />
y(<br />
x<br />
0<br />
<br />
h)<br />
<br />
n<br />
<br />
k<br />
h <br />
<br />
k!<br />
<br />
<br />
k<br />
d y<br />
dx<br />
k<br />
k 0 xx<br />
, y<br />
y<br />
0<br />
0<br />
<br />
<br />
<br />
<br />
<br />
y(<br />
x<br />
0<br />
)<br />
<br />
h<br />
dy<br />
dx<br />
xx<br />
y<br />
y<br />
0<br />
0<br />
,<br />
<br />
2<br />
h<br />
2!<br />
2<br />
d y<br />
2<br />
dx<br />
xx<br />
y<br />
y<br />
0<br />
0<br />
,<br />
...<br />
<br />
n<br />
h<br />
n!<br />
n<br />
d y<br />
n<br />
dx<br />
xx<br />
y<br />
y<br />
0<br />
0<br />
,<br />
The n th order <strong>Taylor</strong> series method uses the<br />
n th order Truncated <strong>Taylor</strong> series expansion.<br />
19
Euler Method<br />
• First order <strong>Taylor</strong> series method is known as<br />
Euler Method.<br />
• Only the constant term and linear term are<br />
used in the Euler method.<br />
• The error due to the use of the truncated<br />
<strong>Taylor</strong> series is of order O(h 2 ).<br />
20
21<br />
First Order <strong>Taylor</strong> <strong>Series</strong> Method<br />
(Euler Method)<br />
)<br />
,<br />
(<br />
)<br />
,<br />
(<br />
),<br />
(<br />
,<br />
:<br />
)<br />
(<br />
)<br />
(<br />
)<br />
(<br />
1<br />
,<br />
0<br />
2<br />
,<br />
0<br />
0<br />
0<br />
0<br />
i<br />
i<br />
i<br />
i<br />
i<br />
i<br />
y<br />
y<br />
x<br />
x<br />
n<br />
n<br />
n<br />
y<br />
y<br />
x<br />
x<br />
y<br />
x<br />
f<br />
h<br />
y<br />
y<br />
Method<br />
Euler<br />
y<br />
x<br />
f<br />
dx<br />
dy<br />
x<br />
y<br />
y<br />
nh<br />
x<br />
x<br />
Notation<br />
h<br />
O<br />
dx<br />
dy<br />
h<br />
x<br />
y<br />
h<br />
x<br />
y<br />
i<br />
i
22<br />
Euler Method<br />
1,2,...<br />
)<br />
,<br />
(<br />
)<br />
(<br />
Method:<br />
Euler<br />
1,2,...<br />
)<br />
(<br />
Determine :<br />
)<br />
(<br />
condition :<br />
initial<br />
with the<br />
)<br />
,<br />
(<br />
)<br />
(<br />
ODE:<br />
order<br />
first<br />
Given the<br />
Problem :<br />
1<br />
0<br />
0<br />
0<br />
0<br />
0<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
i<br />
for<br />
y<br />
x<br />
f<br />
h<br />
y<br />
y<br />
x<br />
y<br />
y<br />
i<br />
for<br />
ih<br />
x<br />
y<br />
y<br />
x<br />
y<br />
y<br />
y<br />
x<br />
f<br />
x<br />
y<br />
i<br />
i<br />
i<br />
i<br />
i
Interpretation of Euler Method<br />
y 2<br />
y 1<br />
y 0<br />
x 0 x 1 x 2 x<br />
23
Interpretation of Euler Method<br />
y 1<br />
Slope=f(x 0 ,y 0 )<br />
y 1 =y 0 +hf(x 0 ,y 0 )<br />
hf(x 0 ,y 0 )<br />
y 0<br />
x 0 x 1 x 2 x<br />
h<br />
24
Interpretation of Euler Method<br />
y 2<br />
Slope=f(x 1 ,y 1 )<br />
hf(x 1 ,y 1 )<br />
y 1<br />
Slope=f(x 0 ,y 0 )<br />
y 2 =y 1 +hf(x 1 ,y 1 )<br />
y 1 =y 0 +hf(x 0 ,y 0 )<br />
hf(x 0 ,y 0 )<br />
y 0<br />
x 0 x 1 x 2 x<br />
h<br />
h<br />
25
Example 1<br />
Use Euler method to solve the ODE:<br />
dy<br />
dx<br />
1<br />
x<br />
2<br />
, y(1)<br />
4<br />
to determine y(1.01), y(1.02) and y(1.03).<br />
26
27<br />
Example 1<br />
0.01<br />
4,<br />
1,<br />
,<br />
1<br />
)<br />
,<br />
( 0<br />
0<br />
2<br />
<br />
<br />
<br />
<br />
<br />
h<br />
y<br />
x<br />
x<br />
y<br />
x<br />
f<br />
<br />
<br />
<br />
<br />
<br />
<br />
9394<br />
3.<br />
1.02<br />
0.011<br />
3.9598<br />
)<br />
,<br />
(<br />
3:<br />
3.9598<br />
1.01<br />
0.011<br />
3.98<br />
)<br />
,<br />
(<br />
2 :<br />
3.98<br />
)<br />
(1)<br />
0.01(1<br />
4<br />
)<br />
,<br />
(<br />
1:<br />
)<br />
,<br />
(<br />
Method<br />
Euler<br />
2<br />
2<br />
2<br />
2<br />
3<br />
2<br />
1<br />
1<br />
1<br />
2<br />
2<br />
0<br />
0<br />
0<br />
1<br />
1<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
y<br />
x<br />
f<br />
h<br />
y<br />
y<br />
Step<br />
y<br />
x<br />
f<br />
h<br />
y<br />
y<br />
Step<br />
y<br />
x<br />
f<br />
h<br />
y<br />
y<br />
Step<br />
y<br />
x<br />
f<br />
h<br />
y<br />
y<br />
i<br />
i<br />
i<br />
i
Example 1<br />
f<br />
2<br />
( x,<br />
y)<br />
1<br />
x , x0<br />
1,<br />
y0<br />
4,<br />
h<br />
<br />
0.01<br />
Summary of the result:<br />
i xi yi<br />
0 1.00 -4.00<br />
1 1.01 -3.98<br />
2 1.02 -3.9595<br />
3 1.03 -3.9394<br />
28
Example 1<br />
f<br />
2<br />
( x,<br />
y)<br />
1<br />
x , x0<br />
1,<br />
y0<br />
4,<br />
h<br />
<br />
0.01<br />
Comparison with true value:<br />
i xi yi True value of yi<br />
0 1.00 -4.00 -4.00<br />
1 1.01 -3.98 -3.97990<br />
2 1.02 -3.9595 -3.95959<br />
3 1.03 -3.9394 -3.93909<br />
29
Example 1<br />
f<br />
2<br />
( x,<br />
y)<br />
1<br />
x , x0<br />
1,<br />
y0<br />
4,<br />
h<br />
<br />
0.01<br />
A graph of the<br />
solution of the<br />
ODE for<br />
1
Types of Errors<br />
– Local truncation error:<br />
Error due to the use of truncated <strong>Taylor</strong> series<br />
to compute x(t+h) in one step.<br />
– Global Truncation error:<br />
Accumulated truncation over many steps.<br />
– Round off error:<br />
Error due to finite number of bits used in<br />
representation of numbers. This error could be<br />
accumulated and magnified in succeeding<br />
steps.<br />
31
Second Order <strong>Taylor</strong> <strong>Series</strong> Methods<br />
Given<br />
Second<br />
dy(<br />
x)<br />
dx<br />
<br />
f<br />
( y,<br />
x),<br />
order <strong>Taylor</strong> <strong>Series</strong><br />
y(<br />
x<br />
0<br />
)<br />
<br />
y<br />
method<br />
0<br />
y<br />
d<br />
i1<br />
2<br />
dx<br />
y<br />
2<br />
<br />
y<br />
i<br />
dy<br />
h <br />
dx<br />
h<br />
2<br />
2!<br />
d<br />
needs to be derived analytically.<br />
2<br />
dx<br />
y<br />
2<br />
O(<br />
h<br />
3<br />
)<br />
32
Third Order <strong>Taylor</strong> <strong>Series</strong> Methods<br />
Given<br />
dy(<br />
x)<br />
dx<br />
<br />
( y,<br />
x),<br />
Third order <strong>Taylor</strong> <strong>Series</strong><br />
f<br />
y(<br />
x<br />
0<br />
) <br />
method<br />
y<br />
0<br />
y<br />
d<br />
i1<br />
2<br />
dx<br />
y<br />
2<br />
<br />
y<br />
i<br />
and<br />
dy<br />
h <br />
dx<br />
d<br />
3<br />
dx<br />
y<br />
3<br />
h<br />
2<br />
2!<br />
d<br />
2<br />
dx<br />
y<br />
2<br />
<br />
h<br />
3<br />
3!<br />
need to be derived analytically.<br />
d<br />
3<br />
dx<br />
y<br />
3<br />
O(<br />
h<br />
4<br />
)<br />
33
34<br />
High Order <strong>Taylor</strong> <strong>Series</strong> Methods<br />
to be derived analytically.<br />
need<br />
,.....,<br />
,<br />
)<br />
(<br />
!<br />
....<br />
2!<br />
method<br />
<strong>Series</strong><br />
order <strong>Taylor</strong><br />
n<br />
)<br />
(<br />
),<br />
,<br />
(<br />
)<br />
(<br />
3<br />
3<br />
2<br />
2<br />
1<br />
2<br />
2<br />
2<br />
1<br />
th<br />
0<br />
0<br />
n<br />
n<br />
n<br />
n<br />
n<br />
n<br />
i<br />
i<br />
dx<br />
y<br />
d<br />
dx<br />
y<br />
d<br />
dx<br />
y<br />
d<br />
h<br />
O<br />
dx<br />
y<br />
d<br />
n<br />
h<br />
dx<br />
y<br />
d<br />
h<br />
dx<br />
dy<br />
h<br />
y<br />
y<br />
y<br />
x<br />
y<br />
x<br />
y<br />
f<br />
dx<br />
x<br />
dy<br />
Given
Higher Order <strong>Taylor</strong> <strong>Series</strong> Methods<br />
• High order <strong>Taylor</strong> series methods are more<br />
accurate than Euler method.<br />
• But, the 2 nd , 3 rd , and higher order derivatives<br />
need to be derived analytically which may not<br />
be easy.<br />
35
Example 2<br />
Second order <strong>Taylor</strong> <strong>Series</strong> Method<br />
Use<br />
Second<br />
order<br />
<strong>Taylor</strong> <strong>Series</strong><br />
method<br />
to<br />
solve:<br />
dx<br />
dt<br />
2x<br />
2<br />
<br />
t<br />
1,<br />
x(0)<br />
1,<br />
use<br />
h<br />
<br />
0.01<br />
What<br />
is :<br />
2<br />
d x(<br />
t)<br />
2<br />
dt<br />
?<br />
36
37<br />
Example 2<br />
))<br />
2<br />
(1<br />
4<br />
1<br />
(<br />
2<br />
)<br />
2<br />
(1<br />
1<br />
)<br />
2<br />
(1<br />
4<br />
1<br />
4<br />
0<br />
)<br />
(<br />
2<br />
1<br />
0.01<br />
1,<br />
(0)<br />
1,<br />
2<br />
solve:<br />
to<br />
method<br />
<strong>Taylor</strong> <strong>Series</strong><br />
order<br />
Second<br />
Use<br />
2<br />
2<br />
2<br />
1<br />
2<br />
2<br />
2<br />
2<br />
2<br />
i<br />
i<br />
i<br />
i<br />
i<br />
i<br />
i<br />
t<br />
x<br />
x<br />
h<br />
t<br />
x<br />
h<br />
x<br />
x<br />
t<br />
x<br />
x<br />
dt<br />
dx<br />
x<br />
dt<br />
t<br />
x<br />
d<br />
t<br />
x<br />
dt<br />
dx<br />
h<br />
use<br />
x<br />
t<br />
x<br />
dt<br />
dx
38<br />
Example 2<br />
))<br />
2<br />
(1<br />
4<br />
1<br />
(<br />
2<br />
)<br />
2<br />
(1<br />
0.01<br />
1,<br />
0,<br />
,<br />
2<br />
1<br />
)<br />
,<br />
(<br />
2<br />
2<br />
2<br />
1<br />
0<br />
0<br />
2<br />
i<br />
i<br />
i<br />
i<br />
i<br />
i<br />
i<br />
t<br />
x<br />
x<br />
h<br />
t<br />
x<br />
h<br />
x<br />
x<br />
h<br />
x<br />
t<br />
t<br />
x<br />
x<br />
t<br />
f<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
0.9716<br />
3:<br />
0.9807<br />
0.01))<br />
2(0.9901)<br />
4(0.9901)(1<br />
1<br />
(<br />
2<br />
0.01<br />
0.01)<br />
2(0.9901)<br />
0.01(1<br />
0.9901<br />
2 :<br />
0.9901<br />
0))<br />
2<br />
4(1)(1<br />
1<br />
(<br />
2<br />
0.01<br />
0)<br />
2(1)<br />
0.01(1<br />
1<br />
1:<br />
3<br />
2<br />
2<br />
2<br />
2<br />
2<br />
2<br />
1<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
x<br />
Step<br />
x<br />
Step<br />
x<br />
Step
Example 2<br />
f<br />
2<br />
( t,<br />
x)<br />
1<br />
2x<br />
t<br />
, t0<br />
0, x0<br />
1,<br />
h<br />
<br />
0.01<br />
Summary of the results:<br />
i t i x i<br />
0 0.00 1<br />
1 0.01 0.9901<br />
2 0.02 0.9807<br />
3 0.03 0.9716<br />
39
Programming Euler Method<br />
Write a MATLAB program to implement Euler<br />
method to solve:<br />
dv<br />
<br />
1<br />
2v<br />
2<br />
<br />
t.<br />
v(0)<br />
<br />
1<br />
dt<br />
for<br />
t<br />
i<br />
<br />
0.01i,<br />
i<br />
<br />
1,2,...,100<br />
40
Programming Euler Method<br />
f=inline('1-2*v^2-t','t','v')<br />
h=0.01<br />
t=0<br />
v=1<br />
T(1)=t;<br />
V(1)=v;<br />
for i=1:100<br />
v=v+h*f(t,v)<br />
t=t+h;<br />
T(i+1)=t;<br />
V(i+1)=v;<br />
end<br />
41
Programming Euler Method<br />
f=inline('1-2*v^2-t','t','v')<br />
h=0.01<br />
t=0<br />
v=1<br />
T(1)=t;<br />
V(1)=v;<br />
for i=1:100<br />
v=v+h*f(t,v)<br />
t=t+h;<br />
T(i+1)=t;<br />
V(i+1)=v;<br />
end<br />
Definition of the ODE<br />
Initial condition<br />
Main loop<br />
Euler method<br />
Storing information<br />
42
Programming Euler Method<br />
Plot of the<br />
solution<br />
plot(T,V)<br />
43
Example 1:<br />
Find y(0.5) if y is the solution of IVP y' = -2x-y, y(0) = -1 using Euler's method<br />
with step length 0.1. Also find the error in the approximation.<br />
Solution: f(x, y) = -2x - y,<br />
y 1 = y 0 + h f(x 0 , y 0 ) = -1 + 0.1* (-2*0 - (-1)) = -0.8999<br />
y 2 = y 1 + h f(x 1 , y 1 ) = -0.8999 + 0.1* (-2*0.1 - (-0.8999)) = -0.8299<br />
y 3 = y 2 + h f(x 2 , y 2 ) = -0.8299 + 0.1* (-2*0.2 - (-0.8299)) = -0.7869<br />
y 4 = y 3 + h f(x 3 , y 3 ) = -0.7869 + 0.1* (-2*0.3 - (-0.7869)) = -0.7683<br />
y 5 = y 4 + h f(x 4 , y 4 ) = -0.7683 + 0.1* (-2*0.4 - (-0.7683)) = -0.7715
Example 2:<br />
Use Eulers method to solve for y[0.1] from y' = x + y + xy, y(0) = 1 with h = 0.01 also<br />
estimate how small h would need to obtain four-decimal accuracy.<br />
Solution : f(x, y) = x + y + xy,<br />
y 1 = y 0 + h f(x 0 , y 0 ) = 1.0 + .01*(0 + 1 + 0*1) = 1.01<br />
y 2 = y 1 + h f(x 1 , y 1 ) = 1.01 + .01*(0.01 + 1.01 + 0.01*1.01) =1.02<br />
y 3 = y 2 + h f(x 2 , y 2 ) = 1.02 + .01*(0.02 + 1.02 + 0.02*1.02) =1.031<br />
y 4 = y 3 + h f(x 3 , y 3 ) = 1.031 + .01*(0.03 + 1.031 + 0.03*1.031) =1.042<br />
y 5 = y 4 + h f(x 4 , y 4 ) = 1.042 + .01*(0.04 + 1.042 + 0.04*1.042) = 1.053<br />
y 6 = y 5 + h f(x 5 , y 5 ) = 1.053 + .01*(0.05 + 1.053 + 0.05*1.053) = 1.065<br />
y 7 = y 6 + h f(x 6 , y 6 ) = 1.065 + .01*(0.06 + 1.065 + 0.06*1.065) = 1.076<br />
y 8 = y 8 + h f(x 7 , y 7 ) = 1.076 + .01*(0.07 + 1.076 + 0.07*1.076) = 1.089<br />
y 9 = y 9 + h f(x 8 , y 8 ) = 1.089 + .01*(0.08 + 1.089 + 0.08*1.089) = 1.101<br />
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:<br />
Solve the differential equation y' = x/y, y(0)=1 by Euler's method to get y(1). Use the step<br />
lengths h = 0.1 and 0.2 and compare the results with the analytical solution (y 2 = 1 + x 2 )<br />
Solution:<br />
f(x, y) = x/y,<br />
with h = 0.1<br />
y 1 = y 0 + h f(x 0 , y 0 ) = 1.0 + 0.1*0.0/1.0 = 1.00<br />
y 2 = y 1 + h f(x 1 , y 1 ) = 1.0 + 0.1*0.1/1.0 = 1.01<br />
y 3 = y 2 + h f(x 2 , y 2 ) = 1.01 + 0.1*0.2/1.01 = 1.0298<br />
y 4 = y 3 + h f(x 3 , y 3 ) = 1.0298 + 0.1*0.3/1.0298 = 1.0589<br />
y 5 = y 4 + h f(x 4 , y 4 ) = 1.0589 + 0.1*0.4/1.0589 = 1.0967<br />
y 6 = y 5 + h f(x 5 , y 5 ) = 1.0967 + 0.1*0.5/1.0967 = 1.1423<br />
y 7 = y 6 + h f(x 6 , y 6 ) = 1.1423 + 0.1*0.6/1.1423 = 1.1948<br />
y 8 = y 7 + h f(x 7 , y 7 ) = 1.1948 + 0.1*0.7/1.1948 = 1.2534<br />
y 9 = y 8 + h f(x 8 , y 8 ) = 1.2534 + 0.1*0.8/1.2534 = 1.3172<br />
y 10 = y 9 + h f(x 9 , y 9 ) = 1.3172 + 0.1*0.9/1.3172 = 1.3855
with h = 0.2<br />
y 1 = y 0 + h f(x 0 , y 0 ) = 1.0 + 0.2*0.0/1.0 = 1.0<br />
y 2 = y 1 + h f(x 1 , y 1 ) = 1.0 + 0.2*0.2/1.0 = 1.0400<br />
y 3 = y 2 + h f(x 2 , y 2 ) = 1.0400 + 0.2*0.4/1.0400 = 1.1169<br />
y 4 = y 3 + h f(x 3 , y 3 ) = 1.1169 + 0.2*0.6/1.1169 = 1.2243<br />
y 5 = y 4 + h f(x 4 , y 4 ) = 1.2243 + 0.2*0.8/1.2243 = 1.3550
Comparison of numerical and<br />
analytical solutions<br />
x<br />
Numerical Solution<br />
h = 0.1 h = 0.2<br />
Analytical solution<br />
0.0 1.0 1.0 1.0<br />
0.1 1.0 1.0050<br />
0.2 1.01 1.0 1.0198<br />
0.3 1.0298 1.0440<br />
0.4 1.0589 1.0400 1.0770<br />
0.5 1.0967 1.1180<br />
0.6 1.1423 1.1169 1.1662<br />
0.7 1.1948 1.2207<br />
0.8 1.2534 1.2243 1.2806<br />
0.9 1.3172 1.3454<br />
1.0 1.3855 1.3550 1.4142