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Find the following Taylor polynomials
Related Topics
Wize University Calculus 1 Textbook > Applications of Differentiation
Taylor and MacLaurin and Polynomials
4 Activities
Find the following Taylor polynomials
a) T_2 (x) for f(x)=√x about x =1.
b) T_3 (x) for f(x)=sin(x) about x = 0.
I don't know
Check Submission
More Taylor and MacLaurin and Polynomials Questions:
Taylor and MacLaurin and Polynomials
Find the 2
nd
degree Taylor polynomial of
f
(
x
)
=
e
2
x
f\left(x\right)=e^{2x}
f
(
x
)
=
e
2
x
centered at
a
=
ln
3
a=\ln\sqrt{3}
a
=
ln
3
.
(i.e. find the first 3 terms (i.e.
n
=
0
,
1
,
2
n=0,1,2
n
=
0
,
1
,
2
) of the Taylor polynomial of
f
(
x
)
f\left(x\right)
f
(
x
)
with
a
=
ln
3
a=\ln\sqrt{3}
a
=
ln
3
)
Taylor and MacLaurin and Polynomials
Find the 2
nd
degree Taylor polynomial of
f
(
x
)
=
e
2
x
f\left(x\right)=e^{2x}
f
(
x
)
=
e
2
x
centered at
a
=
ln
3
a=\ln\sqrt{3}
a
=
ln
3
.
(i.e. find the first 3 terms (i.e.
n
=
0
,
1
,
2
n=0,1,2
n
=
0
,
1
,
2
) of the Taylor polynomial of
f
(
x
)
f\left(x\right)
f
(
x
)
with
a
=
ln
3
a=\ln\sqrt{3}
a
=
ln
3
)
Taylor and MacLaurin and Polynomials
Find the 2
nd
degree Taylor polynomial of
f
(
x
)
=
e
2
x
f\left(x\right)=e^{2x}
f
(
x
)
=
e
2
x
centered at
a
=
ln
3
a=\ln\sqrt{3}
a
=
ln
3
.
(i.e. find the first 3 terms (i.e.
n
=
0
,
1
,
2
n=0,1,2
n
=
0
,
1
,
2
) of the Taylor polynomial of
f
(
x
)
f\left(x\right)
f
(
x
)
with
a
=
ln
3
a=\ln\sqrt{3}
a
=
ln
3
)
Determine the second order Taylor approximation of
3
+
x
2
\sqrt{3+x^2}
3
+
x
2
about
x
=
1
x=1
x
=
1
.
(a)
2
+
1
2
x
2
+
3
16
x
4
2+\frac12x^2+\frac{3}{16}x^4
2
+
2
1
x
2
+
16
3
x
4
(b)
2
+
1
2
(
x
−
1
)
+
3
16
(
x
−
1
)
2
2+\frac{1}{2}(x-1)+\frac{3}{16}(x-1)^2
2
+
2
1
(
x
−
1
)
+
16
3
(
x
−
1
)
2
(c)
4
+
1
4
(
x
−
1
)
+
3
8
(
x
−
1
)
2
4+\frac{1}{4}(x-1)+\frac{3}{8}(x-1)^2
4
+
4
1
(
x
−
1
)
+
8
3
(
x
−
1
)
2
Find the 2
nd
degree Taylor polynomial of
f
(
x
)
=
e
2
x
f\left(x\right)=e^{2x}
f
(
x
)
=
e
2
x
centered at
a
=
ln
3
a=\ln\sqrt{3}
a
=
ln
3
.
(a)
3
+
(
x
−
ln
3
)
+
(
x
−
ln
3
)
2
2
!
3+\left(x-\ln\sqrt{3}\right)+\frac{\left(x-\ln\sqrt{3}\right)^2}{2!}
3
+
(
x
−
ln
3
)
+
2
!
(
x
−
l
n
3
)
2
(b)
3
+
6
(
x
−
ln
3
)
+
6
(
x
−
ln
3
)
2
2
!
3+6\left(x-\ln\sqrt{3}\right)+\frac{6\left(x-\ln\sqrt{3}\right)^2}{2!}
3
+
6
(
x
−
ln
3
)
+
2
!
6
(
x
−
l
n
3
)
2
(c)
3
+
6
(
x
−
ln
3
)
+
6
(
x
−
ln
3
)
2
3+6\left(x-\ln\sqrt{3}\right)+6\left(x-\ln\sqrt{3}\right)^2
3
+
6
(
x
−
ln
3
)
+
6
(
x
−
ln
3
)
2
(d)
(
x
−
ln
3
)
+
(
x
−
ln
3
)
2
2
!
+
(
x
−
ln
3
)
3
3
!
\left(x-\ln\sqrt{3}\right)+\frac{\left(x-\ln\sqrt{3}\right)^2}{2!}+\frac{\left(x-\ln\sqrt{3}\right)^3}{3!}
(
x
−
ln
3
)
+
2
!
(
x
−
l
n
3
)
2
+
3
!
(
x
−
l
n
3
)
3
Taylor and MacLaurin: Polynomials
Now let
T
n
(
x
)
T_n(x)
T
n
(
x
)
be the
n
n
n
th degree Taylor polynomial centred at
x
=
1
x = 1
x
=
1
for the function
f
(
x
)
=
ln
(
x
)
f(x) = \ln(x)
f
(
x
)
=
ln
(
x
)
For which value(s) of
n
n
n
will
T
n
(
1.1
)
T_n(1.1)
T
n
(
1.1
)
give an
underestimate
of
ln
(
1.1
)
\ln(1.1)
ln
(
1.1
)
?
Taylor and MacLaurin: Polynomials
Let
T
3
(
x
)
T_3(x)
T
3
(
x
)
be the third degree Taylor polynomial centred at
x
=
0
x = 0
x
=
0
for
f
(
x
)
=
(
x
+
1
)
cos
(
x
)
f(x) = (x + 1)\cos(x)
f
(
x
)
=
(
x
+
1
)
cos
(
x
)
Write down
T
3
(
x
)
T_3(x)
T
3
(
x
)
. Make sure that you simplify the coefficients.
Taylor and MacLaurin and Polynomials
Find the 2
nd
degree Taylor polynomial of
f
(
x
)
=
e
2
x
f\left(x\right)=e^{2x}
f
(
x
)
=
e
2
x
centered at
a
=
ln
3
a=\ln\sqrt{3}
a
=
ln
3
.
(i.e. find the first 3 terms (i.e.
n
=
0
,
1
,
2
n=0,1,2
n
=
0
,
1
,
2
) of the Taylor polynomial of
f
(
x
)
f\left(x\right)
f
(
x
)
with
a
=
ln
3
a=\ln\sqrt{3}
a
=
ln
3
)
Find the 2
nd
degree Taylor polynomial of
f
(
x
)
=
e
2
x
f\left(x\right)=e^{2x}
f
(
x
)
=
e
2
x
centered at
a
=
ln
3
a=\ln\sqrt{3}
a
=
ln
3
.
(i.e. find the first 3 terms (i.e.
n
=
0
,
1
,
2
n=0,1,2
n
=
0
,
1
,
2
) of the Taylor polynomial of
f
(
x
)
f\left(x\right)
f
(
x
)
with
a
=
ln
3
a=\ln\sqrt{3}
a
=
ln
3
)
Find the 2
nd
degree Taylor polynomial of
f
(
x
)
=
e
2
x
f\left(x\right)=e^{2x}
f
(
x
)
=
e
2
x
centered at
a
=
ln
3
a=\ln\sqrt{3}
a
=
ln
3
.
(i.e. find the first 3 terms (i.e.
n
=
0
,
1
,
2
n=0,1,2
n
=
0
,
1
,
2
) of the Taylor polynomial of
f
(
x
)
f\left(x\right)
f
(
x
)
with
a
=
ln
3
a=\ln\sqrt{3}
a
=
ln
3
)
Determine the second order Taylor approximation of
3
+
x
2
\sqrt{3+x^2}
3
+
x
2
about
x
=
1
x=1
x
=
1
.
3
+
x
2
a
b
o
u
t
x
=
1
[
second order approximation]
\sqrt{3+x^2}\ \ about\ x=1\ \ [\text{second\ order\ approximation]}
3
+
x
2
ab
o
u
t
x
=
1
[
second order approximation]
f
(
0
)
=
f
(
x
)
=
3
+
x
2
f^{(0)}=f(x)=\sqrt{3+x^2}
f
(
0
)
=
f
(
x
)
=
3
+
x
2
f
(
1
)
(
x
)
=
1
2
(
3
+
x
2
)
−
1
2
.2
x
f^{(1)}(x)=\frac{1}{2}(3+x^2)^{-\frac{1}{2}}.2x
f
(
1
)
(
x
)
=
2
1
(
3
+
x
2
)
−
2
1
.2
x
Taylor and MacLaurin and Polynomials
If
f
(
x
)
f(x)
f
(
x
)
has the following Taylor series approximation find the 10th derivative of
f
(
x
)
f(x)
f
(
x
)
at
x
=
0.
x=0.
x
=
0.
f
(
x
)
=
1
−
x
2
2
!
+
x
4
4
!
−
x
6
6
!
+
…
f(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\ldots
f
(
x
)
=
1
−
2
!
x
2
+
4
!
x
4
−
6
!
x
6
+
…
Taylor Polynomials
Find quadratic approximation for
x
\sqrt x
x
at
x
=
4
x=4
x
=
4
Taylor and MacLaurin and Polynomials
Find the 2
nd
degree Taylor polynomial of
f
(
x
)
=
e
2
x
f\left(x\right)=e^{2x}
f
(
x
)
=
e
2
x
centered at
a
=
ln
3
a=\ln\sqrt{3}
a
=
ln
3
.
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
.
Find the 2
nd
degree Taylor polynomial of
f
(
x
)
=
e
2
x
f\left(x\right)=e^{2x}
f
(
x
)
=
e
2
x
centered at
a
=
ln
3
a=\ln\sqrt{3}
a
=
ln
3
.
(i.e. find the first 3 terms (i.e.
n
=
0
,
1
,
2
n=0,1,2
n
=
0
,
1
,
2
) of the Taylor polynomial of
f
(
x
)
f\left(x\right)
f
(
x
)
with
a
=
ln
3
a=\ln\sqrt{3}
a
=
ln
3
)
Find that second order Taylor series polynomial used to approximate
f
(
x
)
=
e
2
x
f(x)=e^{2x}
f
(
x
)
=
e
2
x
at x = 0
If
f
(
x
)
f(x)
f
(
x
)
has the following Taylor series approximation find the 10th derivative of
f
(
x
)
f(x)
f
(
x
)
at
x
=
0.
x=0.
x
=
0.
f
(
x
)
=
1
−
x
2
2
!
+
x
4
4
!
−
x
6
6
!
+
…
f(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\ldots
f
(
x
)
=
1
−
2
!
x
2
+
4
!
x
4
−
6
!
x
6
+
…
Find the Maclaurin polynomial of degree 2019 of the function
f
(
x
)
=
20
x
2
+
16
f(x) = 20x^2+16
f
(
x
)
=
20
x
2
+
16
.
Find second order Taylor series approximation for
x
\sqrt{x}
x
about
x
=
2
x=2
x
=
2
.
Find the Maclaurin series for
f
(
x
)
=
cos
3
x
f(x)=\cos{3x}
f
(
x
)
=
cos
3
x
up to and including
x
4
x^4
x
4
terms.
Let
F
3
(
x
)
=
c
0
+
c
1
x
+
c
2
x
2
+
c
3
x
3
F_3(x) = c_0 + c_1x + c_2x^2 + c_3x^3
F
3
(
x
)
=
c
0
+
c
1
x
+
c
2
x
2
+
c
3
x
3
be the third order Taylor polynomial for the function
Taylor polynomials
In the Maclaurin polynomial of degree 30 of the function
f
(
x
)
f(x)
f
(
x
)
, the coefficient of
x
10
x^{10}
x
10
is given by
1
/
72000
1/72000
1/72000
. Find
f
(
10
)
(
0
)
f^{(10)}(0)
f
(
10
)
(
0
)
.
Taylor and MacLaurin and Polynomials
Find the 2
nd
degree Taylor polynomial of
f
(
x
)
=
e
2
x
f\left(x\right)=e^{2x}
f
(
x
)
=
e
2
x
centered at
a
=
ln
3
a=\ln\sqrt{3}
a
=
ln
3
.
(i.e. find the first 3 terms (i.e.
n
=
0
,
1
,
2
n=0,1,2
n
=
0
,
1
,
2
) of the Taylor polynomial of
f
(
x
)
f\left(x\right)
f
(
x
)
with
a
=
ln
3
a=\ln\sqrt{3}
a
=
ln
3
)
Find the following Taylor polynomials
Find the following Taylor polynomials (just type the answer, do NOT type "
T
2
(
x
)
=
T_2\left(x\right)=
T
2
(
x
)
=
" in front of your answer).
Find the following Taylor polynomials
a) 2nd degree for
f
(
x
)
=
x
f\left(x\right)=\sqrt{x}
f
(
x
)
=
x
b) 3rd degree for
g
(
x
)
=
sin
x
g\left(x\right)=\sin x
g
(
x
)
=
sin
x
Taylor Polynomials
Find the following Taylor polynomials
T
4
(
x
)
T_4(x)
T
4
(
x
)
of a function
f
f
f
about point
x
=
−
1
x=-1
x
=
−
1
is given as follows:
T
4
(
x
)
=
2
+
5
(
x
+
1
)
+
12
(
x
+
1
)
2
+
2
5
(
x
+
1
)
4
T_4(x)=2+5(x+1)+12(x+1)^2+\frac{2}{5}(x+1)^4
T
4
(
x
)
=
2
+
5
(
x
+
1
)
+
12
(
x
+
1
)
2
+
5
2
(
x
+
1
)
4
Find
Consider
f
(
x
)
=
1
+
x
f(x) = \sqrt{1+x}
f
(
x
)
=
1
+
x
.