SME
Calculus II
Taylor and power series
Use this for “Estimate e0.5 with a third-degree Taylor polynomial and bound the error.”
Chart
f(0.5) ≈ 1.649
Dashed: the degree-3 Taylor polynomial of ex about x = 0
At x = 0.5: P = 1.646, f = 1.649, error |f − P| = 0.002888
This is an example. Type your own values to start a new problem.
e
≈
P
(
x
)
=
Pₙ(x)
eˣ
True value eˣ
worked out (not typed)
E
Actual error |eˣ − Pₙ(x)|
worked out (not typed)
M
Largest eᵗ between 0 and x
worked out (not typed)
B
Error bound (Lagrange)
worked out (not typed)
Formulas
Pₙ(
x
)
=
n
Σ
k
=0
x
k
÷
k!
1.6458
=
3
Σ
k
=0
0.5
k
÷
k!
eˣ
=
e
x
1.6487
=
e
0.5
E
=
|eˣ
−
Pₙ(
x
)|
0.002888 = |1.6487 − 1.6458|
M
=
e
max(
x
, 0)
1.6487
=
e
max(0.5, 0)
B
=
M
×
|
x
n
+ 1
|
÷
(
n
+
1)!
0.004294
=
1.6487
×
|
0.5
3 + 1
|
÷
(3
+
1)!
Assumptions
•
Every derivative of eˣ is eˣ, which is 1 at x = 0, so the Maclaurin polynomial is Pₙ(x) = 1 + x + x²/2! + … + xⁿ/n!.
•
Lagrange: the error is at most M|x|ⁿ⁺¹ ÷ (n + 1)!, where M is the largest value of eᵗ for t between 0 and x.
•
eᵗ grows with t, so M = eˣ when x > 0 and M = e⁰ = 1 when x ≤ 0.
Step-by-step
WE KNOW
x = 0.5, n = 3
FIND
polynomial estimate (Taylor) (Pₙ(x)), true value eˣ (eˣ), actual error |eˣ − Pₙ(x)| (E), largest eᵗ between 0 and x (M), error bound (Lagrange) (B)
Step 1 · Find polynomial estimate (Taylor): Pₙ(x)
Use Pₙ(x) = Σ from k = 0 to n of xᵏ ÷ k!
Add the terms from k = 0 up to the degree, each xᵏ divided by k!, since every derivative of eˣ at 0 is 1.
Pₙ(
x
)
=
1
+
x
+
x
2
2!
+
x
3
3!
Pₙ(
x
)
=
1
+
0.5
+
0.5
2
2!
+
0.5
3
3!
Pₙ(
x
)
=
1
+
0.5
+
0.125
+
0.02083
Pₙ(
x
)
=
1.645833
Step 2 · Find true value eˣ (eˣ)
Use eˣ = eˣ
The true value, to compare with the estimate.
eˣ
=
e
0.5
eˣ = 1.648721
Step 3 · Find actual error |eˣ − Pₙ(x)| (E)
Use E = |eˣ − Pₙ(x)|
How
far
the
estimate
really
is
from
eˣ:
the
terms
left
out
after
degree
n
.
E
=
|1.648721
−
1.645833|
E
=
|0.002888|
E
=
0.002888
Step 4 · Find largest eᵗ between 0 and x (M)
Use M = emax(x, 0)
eᵗ
grows,
so
between
0
and
x
it
is
largest
at
t
=
x
.
M
=
e
x
M
=
e
0.5
M
=
1.64872
Step 5 · Find error bound (Lagrange): B
Use B = M × |xⁿ⁺¹| ÷ (n + 1)!
Lagrange’s
form
of
the
remainder:
the
next
derivative,
at
most
M
,
times
|
x
|ⁿ⁺¹
÷
(
n
+
1)!.
The
actual
error
E
is
never
more
than
B
.
B
=
M
×
|
x
|⁴
÷
24
B
=
1.64872
×
|0.5|⁴
÷
24
B
=
1.6487
×
0.5
4
24
B
=
1.6487
×
0.0625
24
B
=
0.10304
24
B
=
0.004294
Check
Put the numbers back in. Both sides should match.
1.6458
=
1
+
0.5
+
0.5
2
2!
+
0.5
3
3!
✓
1.648721
=
e
0.5
✓
0.002888 = |1.648721 − 1.645833| ✓
1.64872
=
e
max(0.5, 0)
✓
0.004294 = 1.6487 × |0.5|⁴ ÷ 24 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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