BC Fall Review
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BC Fall Review
Give the first 4 non-zero terms of the MacClaurin Series for the
function:
21
1)(
xxf
Hint: You don’t need to generate each
term. Check your notes/book.
6421)( xxxxP
Use P(x) to find the first 4 non-zero terms for arctan(x)
21)arctan(
x
dxx
Since we already know that arctan(0) = 0, we know that C = 0
dxxxx 6421
Cxxx
xxR 753
)(753
753)(
753 xxxxxR
Use this result to approximate
4
Find the maximum error involved in this approximation.
Use this result to approximate
4
Find the maximum error involved in this approximation.
)1arctan(4
7
1
5
1
3
11)1( R 72381.0
1.09
1error
Because this is an alternating series, we use the next term to find the error.
Use this result to approximate
4
Find the maximum error involved in
this approximation.
)1arctan(4
7
1
5
1
3
11)1( R 72381.0
1.09
1error
Because this is an alternating series, we use the next term to find the error.
Generate a 2nd degree polynomial to approximate ln(0.9) and ln(1.1). Find the error bound involved in each approximation.
0)1( f
xxf ln)(
1)1( f
1)1( f
2)1( f
2
)1()1(ln
2x
xx
2
)19.0()19.0(9.0ln
2
2
)1.0(1.0
2
105.0
2
)11.1()11.1(1.1ln
2
2
)1.0(1.0
2
095.0
Generate a 2nd degree polynomial to approximate ln(0.9) and ln(1.1). Find the error bound involved in each approximation.
2
)1()1(ln
2x
xx
2
)19.0()19.0(9.0ln
2
2
)1.0(1.0
2 105.0
2
)11.1()11.1(1.1ln
2
2
)1.0(1.0
2
095.0
Because this is an alternating series, we can just use the next term…
3 3( 1) (1.1 1).0003
3 3
xerror
Because this is no longer an alternating series, we need the Lagrange error bound for the 3rd term…
Generate a 2nd degree polynomial to approximate ln(0.9) and ln(1.1). Find the error bound involved in each approximation.
2
)19.0()19.0(9.0ln
2
2
)1.0(1.0
2 105.0
Because this is no longer an alternating series, we need the Lagrange error bound for the 3rd term…
3
2( )f c
c And the number between 0.9 and 1 that maximizes
this value is 0.9
3 3
3
(0.9 1) 2(0.9 1)(0.9) .000457
3! 6(0.9)error f
Find the radius and interval of convergence:
1
1
5
)2()1(
n
nn
n
x
n
n
n x
n
n
x
)2(
5
15
)2(lim
1
1)2(lim
xn
1)2(1 x 31 x
1
1
5
)12()1(
n
nn
n
1
1
5
)32()1(
n
nn
n
1
1
5
1)1(
n
n
n
1
1
5
)1()1(
n
nn
n
31 x
Use the integral test to determine if the given series converges:
12 1
1
n n
b
b x
dx1 2 1
limb
bx
1
1tanlim
1tantanlim 11
b
b
42
4
…and therefore it converges
22 1
1
n nUse the integral test to show that the given series converges
1)
Find the sum to which it converges.2)
2 2 1x
dxdx
x
B
x
Ab
b 11lim
2
1
1
11 2
xx
B
x
A 1 BBxAAx
0 BA
1 BA
2
1A
2
1B
dxxx
b
b 1
1
1
1lim
2
12
12
12ln
1
1lnlim
2
1
b
bb
1)
3ln2
1
22 1
1
n nUse the integral test to show that the given series converges
1)
Find the sum to which it converges.2)
2 2 1x
dxdx
x
B
x
Ab
b 11lim
2
12
12ln
1
1lnlim
2
1
b
bb
1)
3
1ln1ln
2
1
11
1lim
b
bb
…or more importantly, since we have a finite limit, the series converges.
22 1
1
n nUse the integral test to show that the given series converges
1)
Find the sum to which it converges.2)
2)
22 1
1
n n 1
1
1
1
2
1
2
nnn
...
6
1
4
1
5
1
3
1
4
1
2
1
3
1
1
1
2
1
4
3
dxx
e x
4/
0 2
tan
cos
Evaluate the integral: dxxe x
4/
0
2tan sec
xu tan
dxxdu 2sec
dueu1
0
0tanu 0u
)4/tan(u 1u
1
0
ue 1e
422 xx
dxEvaluate the integral:
2
x
24 xtan2x
ddx 2sec2
4tan4tan4
sec222
2
d
sec2tan4
sec22
2 d
dcos
1
sin
cos
4
12
2
d
2sin
cos
4
1
422 xx
dxEvaluate the integral:
2
x
24 xtan2x
ddx 2sec2
d 2sin
cos
4
1 sinu
ddu cos
24
1
u
du C
u4
1 C
sin4
1C csc
4
1
Cx
x
4
4 2
xxx
lnlim0
0 Indeterminate
x
xx 1
lnlim
0
L’Hopital’s Rule
2
0 1
1
lim
x
xx
x
x
xx
2
0lim 0
yxdx
dy 2 20 y
dxxy
dy 2 dxxy
dy 2C
xy
3ln
3
Cx
ey
3
3
3
3x
Aey 3
03
2 Ae
2A
3
3
2x
ey
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