Taylor’s Method
Dr N Ndou
March 17, 2026
Content
• Derivation of the Taylor Method 2,3 and 4.
• Local Truncation error. • Examples.
Dr N Ndou
Taylor’s Method
Initial Value Problem
Consider the first-order differential equation
dy
= f (t, y ),
dt
y (t0 ) = y0
where
f (t, y ) is a known function
y (t) is the unknown solution
numerical methods approximate the solution at discrete points
Let
tn = t0 + nh
where h is the step size.
Dr N Ndou
Taylor’s Method
Taylor Method
We first express the solution y (ti+1 ) to the initial-value problem
(*) by its nth Taylor polynomial and corresponding remainder
centered at ti :
y (ti+1 ) = y (ti ) + y 0 (ti ) h +
y (n) (ti ) n
y 00 (ti ) 2
h + ... +
h
2!
n!
y (n+1) (ti∗ ) n+1
h , where ti ≤ ti∗ ≤ ti+1
(n + 1)!
f 0 (ti , y (ti )) 2
h + ...
= y (ti ) + f (ti , y (ti )) h +
2!
f (n−1) (ti , y (ti )) n f (n) (ti∗ , y (ti∗ )) n+1
+
h +
h
n!
(n + 1)!
+
Dr N Ndou
Taylor’s Method
(1)
Taylor Method





f (n−1) (ti , y (ti )) n−1 f (n) (ti∗ , y (ti∗ )) n 
f 0 (ti , y (ti ))
.
h
+
...
+
h
+
h
y (ti ) + h 
f
(t
,
y
(t
))
+
i
i


2!
n!
|
{z
} | (n +{z1)!
}
Tn (ti , y (ti ))
Rn (ti )
Define
Tn (t, y (t)) = f (t, y (t))+
Dr N Ndou
Taylor’s Method
f (n−1) (t, y (t)) n−1
f 0 (t, y (t))
h+...+
h
2!
n!
Taylor Method of order n (nth order of Taylor Method):
y0 = α
yi+1 = yi + hTn (ti , yi ) , i = 0, 1, 2, ..., N − 1,
Notes:
T1 (t, y (t)) = f (t, y (t)) . So, Euler’s Method is the Taylor
Method of order 1.
f (n−1) (t, y (t)) n−1
h . If h is
2 Tn (t, y (t)) = Tn−1 (t, y (t)) +
n!
small, for large n, Tn (t, y (t)) ≈ Tn−1 (t, y (t)), that means
Tn (t, y (t)) will not improve much the approximation.
1
Dr N Ndou
Taylor’s Method
Example Use Taylor Method of order 2, 3 and 4 to approximate
the solution of the initial-value problem.
y 0 (t) = y − t 2 + 1, 0 ≤ t ≤ 2, y (0) = 0.5.
f 0 (ti , yi )
h
2!
f 00 (ti , yi ) 2
T3 (ti , yi ) = T2 (ti , yi ) +
h
3!
f 000 (ti , yi ) 3
h
T4 (t, y (t)) = T3 (ti , yi ) +
4!
T2 (ti , yi ) = f (ti , yi ) +
f (t, y (t)) = y − t 2 + 1,
f 0 (t, y (t)) = y 0 − 2t = y − t 2 + 1 − 2t = y − t 2 − 2t + 1
f 00 (t, y (t)) = y 0 − 2t − 2 = y − t 2 + 1 − 2t − 2 = y − t 2 − 2t − 1
f 000 (t, y (t)) = y 0 − 2t − 2 = y − t 2 + 1 − 2t − 2 = y − t 2 − 2t − 1
Dr N Ndou
Taylor’s Method
Taylor Method of order 2 and 3
Taylor Method of order 2:
y0 = 0.5
yi+1 = yi + h
h
yi − ti2 + 1 +
2
yi − ti2 − 2ti + 1
Taylor Method of order 3:
y0 = 0.5
h
yi+1 = yi + h(yi − ti2 + 1 + (yi − ti2 − 2ti + 1)
2
h2
+ (yi − ti2 − 2ti − 1))
6
Dr N Ndou
Taylor’s Method
(2)
(3)
Taylor Method of order 4
Taylor Method of order 4:
y0 = 0.5
h
yi+1 = yi + h(yi − ti2 + 1 + (yi − ti2 − 2ti + 1)
2
h2
h3
+ (yi − ti2 − 2ti − 1) + (yi − ti2 − 2ti − 1))
6
24
Dr N Ndou
Taylor’s Method
(4)
(5)
Local truncation error
definition (local truncation error) Consider the difference
method:
y0 = α
yi+1 = yi + h φ (ti , yi ) , i = 0, 1, 2, ..., N − 1,
The local truncation error of this difference method is defined as:
τi+1 (h) =
=
y (ti+1 ) − (y (ti ) + h φ (ti , y (ti )))
h
y (ti+1 ) − y (ti )
− φ (ti , y (ti )) , for i = 0, 1, ..., N − 1.
h
Note that τi+1 (h) is a function of h for i = 0, 1, ..., N − 1.
Dr N Ndou
Taylor’s Method
Local truncation error
When φ (ti , y (ti )) = Tn (ti , yi ) , the local truncation error can be
derived as follows.
τi+1 (h)
=
=
=
y (ti+1 ) − y (ti )
− φ (ti , y (ti )) , for i = 0, 1, ..., N − 1.
h
1
y 00 (ti ) 2
y (n) (ti ) n y (n+1) (ti∗ ) n
y (ti ) + y 0 (ti ) h +
h + ... +
h +
h
h
2!
n!
(n + 1)!
y (n+1) (ti∗ ) n
h , where ti < ti∗ < ti+1 .
(n + 1)!
If y (n+1) (t) ≤ M for t in [a, b] , then
|τi+1 (h)| ≤
Dr N Ndou
Taylor’s Method
Mhn
for all i = 0, 1, ..., N.
(n + 1)!
Example
Use Taylor method of order 2,3, and 4 to solve the following
equation
y 0 (t) = y (t) − t 2 + 1,
0 ≤ t ≤ 2,
y (0) = 0.5, h = 0.2
b−a
2−0
=
= 10
h
0.2
t0 = 0, t1 = 0.2 t2 = 0.6, t3 = 0.8, t4 = 1, t5 = 1.2, t6 =
1.4, t7 = 1.6, t8 = 1.8, t9 = 2.
N=
Dr N Ndou
Taylor’s Method
Taylor Method of order 2 and 3
Taylor Method of order 2:
y0 = 0.5
yi+1 = yi + h
h
yi − ti2 + 1 +
2
yi − ti2 − 2ti + 1
Taylor Method of order 3:
y0 = 0.5
h
yi+1 = yi + h(yi − ti2 + 1 + (yi − ti2 − 2ti + 1)
2
h2
+ (yi − ti2 − 2ti − 1))
6
Dr N Ndou
Taylor’s Method
(6)
(7)
Table: Comparison of Taylor methods (orders 2, 3, 4) with h = 0.2
t
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
Dr N Ndou
Taylor’s Method
Exact y (t)
0.50000000
0.82929862
1.21408765
1.64894060
2.12722954
2.64085909
3.17994154
3.73240002
4.28348379
4.81517627
5.30547195
Taylor 2
0.50000000
0.83000000
1.21580000
1.65207600
2.13233272
2.64864592
3.19134802
3.74864458
4.30614639
4.84629860
5.34768429
Error2
0.00e0
7.01e-4
1.71e-3
3.14e-3
5.10e-3
7.79e-3
1.14e-2
1.62e-2
2.27e-2
3.11e-2
4.22e-2
Taylor 3
0.50000000
0.82933333
1.21417244
1.64909595
2.12748251
2.64124531
3.18050761
3.73320662
4.28460969
4.81672330
5.30757139
Error3
0.00e0
3.47e-5
8.48e-5
1.55e-4
2.53e-4
3.86e-4
5.66e-4
8.07e-4
1.13e-3
1.55e-3
2.10e-3
Taylor 4
0.50000000
0.82930000
1.21409102
1.64894677
2.12723959
2.64087443
3.17996403
3.73243207
4.28352853
4.81523774
5.30555538
Error4
0.00e0
1.38e-6
3.37e-6
6.17e-6
1.01e-5
1.53e-5
2.25e-5
3.21e-5
4.47e-5
6.15e-5
8.34e-5
Solutions Comparison
i
0
1
2
3
4
5
6
7
8
9
ti ExactValues Eulers
Error
Heun0 s Error
0
0.5000
0.5000
0.0000 0.5000 0.0000
0.2
0.8293
0.8000X 0.0293X 0.8260 0.0033
0.4
1.2141
1.1520X 0.0621X 1.2069 0.0072
0.6
1.6489
1.5504X 0.0985X 1.6372 0.0117
0.8
2.1272
1.9885X 0.1387X 2.1102 0.0170
1.0
2.6409
2.4582X 0.1827X 2.6177 0.0232
1.2
3.1799
2.9498X 0.2301X 3.1496 0.0304
1.6
3.7324
3.4518X 0.2806X 3.6937 0.0387
1.8
4.2835
3.9501X 0.3334X 4.2351 0.0484
2.0
4.8152
4.4282X 0.3870X 4.7556 0.0596
Dr N Ndou
Taylor’s Method
Solutions Comparison
i
1
2
3
4
5
6
7
8
9
Dr N Ndou
Taylor’s Method
Midpoint
0.5000
0.8280
1.2114
1.6447
2.1213
2.6332
3.1705
3.7212
4.2706
Error
0.0000
0.0013
0.0027
0.0043
0.0059
0.0077
0.0095
0.0112
0.0129
Ralson
0.5000
0.8270
1.2091
1.6410
2.1158
2.6254
3.1600
3.7074
4.2529
Error
0.0000
0.0023
0.0049
0.0080
0.0115
0.0154
0.0199
0.0250
0.0306
Solutions Comparison
i
1
2
3
4
5
6
7
8
9
Dr N Ndou
Taylor’s Method
RKM4
0.5000
0.8293
1.2141
1.6489
2.1272
2.6408
3.1799
3.7323
4.2834
Error
0.0000
0.0053
0.0114
0.0186
0.0269
0.0364
0.0474
0.0599
0.0743
RKM3
0.5000
0.8292 0
1.2139
1.6486
2.1267
2.6402
3.1791
3.7314
4.2822
Error
0.0000
0.0001
0.0002
0.0003
0.0005
0.0006
0.0008
0.0010
0.0013