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Use the second order Taylor polynomial for f(x)=√x at x=16 to approximate √17.…
Related Topics
Wize University Calculus 1 Textbook > Applications of Differentiation
Taylor Polynomial Remainder
3 Activities
Use the second order Taylor polynomial for
f
(
x
)
=
x
f(x)=\sqrt{x}
f
(
x
)
=
x
at
x
=
16
x=16
x
=
16
to approximate
17
\sqrt{17}
17
. What is a good bound on the error in this approximation?
4
−
1
8
−
1
2
⋅
1
4
4
4-\frac{1}{8}-\frac{1}{2}\cdot\frac{1}{4^4}
4
−
8
1
−
2
1
⋅
4
4
1
,
≥
3
(
2
!
)
(
8
)
(
16
5
2
)
\ge\frac{3}{(2!)(8)(16^{\frac{5}{2}})}
≥
(
2
!)
(
8
)
(
1
6
2
5
)
3
4
+
1
8
+
1
2
⋅
1
4
4
4+\frac{1}{8}+\frac{1}{2}\cdot\frac{1}{4^4}
4
+
8
1
+
2
1
⋅
4
4
1
,
≥
3
(
3
!
)
(
8
)
(
16
5
2
)
\ge\frac{3}{(3!)(8)(16^{\frac{5}{2}})}
≥
(
3
!)
(
8
)
(
1
6
2
5
)
3
4
+
1
8
−
1
2
⋅
1
4
4
4+\frac{1}{8}-\frac{1}{2}\cdot\frac{1}{4^4}
4
+
8
1
−
2
1
⋅
4
4
1
,
≤
3
(
3
!
)
(
8
)
(
16
5
2
)
\le\frac{3}{(3!)(8)(16^{\frac{5}{2}})}
≤
(
3
!)
(
8
)
(
1
6
2
5
)
3
4
−
1
8
+
1
2
⋅
1
4
4
4-\frac{1}{8}+\frac{1}{2}\cdot\frac{1}{4^4}
4
−
8
1
+
2
1
⋅
4
4
1
,
≤
3
(
2
!
)
(
8
)
(
16
5
2
)
\le\frac{3}{(2!)(8)(16^{\frac{5}{2}})}
≤
(
2
!)
(
8
)
(
1
6
2
5
)
3
I don't know
Check Submission
More Taylor Polynomial Remainder Questions:
Let
f
(
x
)
=
cos
(
3
x
)
+
x
−
1
f(x)=\cos{(3x)}+x-1
f
(
x
)
=
cos
(
3
x
)
+
x
−
1
. Approximate
f
(
π
3
+
1
5
)
f(\frac{\pi}{3}+\frac{1}{5})
f
(
3
π
+
5
1
)
using linear approximation. Is this overestimation or underestimation of the actual value? Estimate the max error bound for this approximation.
Estimate the size of the error made in the linear approximation to estimate
99.9
\sqrt{99.9}
99.9
.
Let
f
(
x
)
=
cos
(
3
x
)
+
x
−
1
f(x)=\cos{(3x)}+x-1
f
(
x
)
=
cos
(
3
x
)
+
x
−
1
. Approximate
f
(
π
3
+
1
5
)
f(\frac{\pi}{3}+\frac{1}{5})
f
(
3
π
+
5
1
)
using linear approximation. Is this overestimation or underestimation of the actual value?
Find the second order Taylor polynomial
T
2
(
x
)
T_2(x)
T
2
(
x
)
for
f
(
x
)
=
x
f(x)=\sqrt{x}
f
(
x
)
=
x
at
x
=
16
x=16
x
=
16
. Use this to approximate
17
\sqrt{17}
17
. What is a good bound on the error in this approximation?
Consider
f
(
x
)
=
1
+
x
f(x) = \sqrt{1+x}
f
(
x
)
=
1
+
x
.
Taylor polynomials
Find the degree 4 Taylor polynomial for
f
(x) = sin(3
x
) centered at
x
=
π
/
6.
x=\pi/6.
x
=
π
/6.
What is the lowest error bound within
[
0
,
π
/
6
]
[0,\pi/6]
[
0
,
π
/6
]
?
Concept Clarifier
Find the fifth order Taylor polynomial for
y
=
sin
x
y=\sin\ x
y
=
sin
x
centred about
x
0
=
π
x_0=\pi
x
0
=
π
. What is the minimum error bound for such Taylor polynomial in the interval
x
∈
[
0
,
3
π
]
x \in [0,3\pi]
x
∈
[
0
,
3
π
]
?
If
f
(
x
)
=
e
−
2
x
+
3
x
2
f(x)=e^{-2x}+3x^2
f
(
x
)
=
e
−
2
x
+
3
x
2
approximate
f
(
0.1
)
f(0.1)
f
(
0.1
)
? Find the maximum error for approximating positive numbers?
Find the second order Taylor polynomial
T
2
(
x
)
T_2(x)
T
2
(
x
)
for
f
(
x
)
=
x
f(x)=\sqrt{x}
f
(
x
)
=
x
at
x
=
16
x=16
x
=
16
. Use this to approximate
17
\sqrt{17}
17
. What is a good bound on the error in this approximation?
Let
f
(
x
)
=
cos
(
3
x
)
+
x
−
1
f(x)=\cos{(3x)}+x-1
f
(
x
)
=
cos
(
3
x
)
+
x
−
1
. Approximate
f
(
π
3
+
1
5
)
f(\frac{\pi}{3}+\frac{1}{5})
f
(
3
π
+
5
1
)
using linear approximation. Is this overestimation or underestimation of the actual value? Estimate the max error bound for this approximation.
Consider
f
(
x
)
=
1
+
x
f(x) = \sqrt{1+x}
f
(
x
)
=
1
+
x
.