SME
Calculus II · Problem type
A Taylor polynomial from given derivatives
Use this for “g(2) = 5, g′(2) = 3, g″(2) = −2 and g‴(2) = 12. Use P₃ about x = 2 to estimate g(1.9).”
Chart
P₃(1.2) = 1.888
This is an example. Type your own values to start a new problem.
P₃(x)
=
f(a)
+
(
f′(a)
)
(x
−
a
)
+
f″(a)
2!
(x
−
)
2
+
f‴(a)
3!
(x
−
)
3
P₃(
x
)
=
h
Step from the center, x − a
worked out (not typed)
Formulas
h
=
x
−
a
0.2 = 1.2 − 1
P₃(
x
)
=
f
(
a
)
+
f
′(
a
)
×
h
+
f
″(
a
) ×
h
2
2
+
f
‴(
a
) ×
h
3
6
1.888
=
2
+
(−1)
×
0.2
+
4 ×
0.2
2
2
+
6 ×
0.2
3
6
Assumptions
•
Only the values of f and its first three derivatives at the center a are known.
•
P₃(x) = f(a) + f′(a)(x − a) + f″(a)(x − a)²/2! + f‴(a)(x − a)³/3!: each term matches one more derivative at a.
•
The estimate is best near a: the farther x is from a, the larger the part P₃ leaves out can be.
Step-by-step
WE KNOW
a = 1, f(a) = 2, f′(a) = −1, f″(a) = 4, f‴(a) = 6, x = 1.2
FIND
step from the center, x − a (h), estimate of f(x) (P₃(x))
Step 1 · Find step from the center, x − a (h)
Use h = x − a
Every
term
is
a
power
of
the
step
from
the
center
to
x
.
h
=
1.2
−
1
h
=
0.2
Step 2 · Find estimate of f(x): P₃(x)
Use P₃(x) = f(a) + f′(a) × h + (f″(a) × h²) ÷ 2 + (f‴(a) × h³) ÷ 6
Each
derivative
is
divided
by
its
factorial
(2!
=
2,
3!
=
6)
and
multiplied
by
the
matching
power
of
h
=
x
−
a
.
P₃(
x
)
=
2
+
(−1)
×
0.2
+
4 ×
0.2
2
2
+
6 ×
0.2
3
6
P₃(
x
)
=
2
+
(−0.2)
+
0.08
+
0.008
P₃(
x
)
=
1.888
Check
Put the numbers back in. Both sides should match.
0.2 = 1.2 − 1 ✓
1.888
=
2
+
(−1)
×
0.2
+
4 ×
0.2
2
2
+
6 ×
0.2
3
6
✓
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Radius and interval of convergence
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