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math 129 exam 1 document solutions, Exams of Mathematics

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2023/2024

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Math 129 Exam 3 Fall 2024
Name (Please Print.):
You need to show all your work to get a full credit.
Problem 1 (5 points each) Determine whether the series converges or diverges. Give a
reason such as a test.
(a) n1
cos 2n
n23n
(b) n1
2n23
n34n1
(c) n1
3n
n5n
1
Diverge
general
term
:
Eo
[
+
3
n
)
n
,
3
m
=
-
3
-
o
since
cos
(
x
)
oscillates
betuwen
-
t
aud
1
,
the
senies
do
o
st
fy
the
necessary
condition
br
con
eg
a
L
is
an
#
0
)
,
thereforer
,
it
'
s
d
irerges
highest
-
degree
ion
the
naoom
&
darom
,
24
~
=
도금
,
which
diverges
.
By
the
limit
Comparion
Tes t
,
the
givem
series
is
divege
onverges
.
Bominant
Jeron
.
났다
.
~
(
)
This
is
aa
gronatra
series
wite
γ=
5
L
1
.
which
Eonwerges
.
By
the
Limit
Comparrts
Tes t
with
도록
)
"
,
the
given
seises
conwerages
_
pf3
pf4
pf5

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Math 129 Exam 3 Fall 2024 Name (Please Print.): You need to show all your work to get a full credit. Problem 1 (5 points each) Determine whether the series converges or diverges. Give a reason such as a test.

(a)  n  1

 cos 2 n n^2  3 n

(b)  n  1

 (^2) n^2  3 n^3  4 n  1

(c)  n  1

 3 n n  5 n Diverge general term (^) :Eo (^) [+ 3 n) n →^ ∞ (^) , 3 m= -^3 - → o since (^) cos (x)oscillates betuwen^ -^ taud^1 ,the senies dost (^) o fythe necessaryconditionbrcon^ eg is^ a^ L an^ #^0 ) (^) , thereforer,^ it d's irerges

highest - degree ionthenaoom^ &darom, 24

~ (^) = ,^ which^ diverges.

By the^ limitComparionTest^ ,thegivemseriesis

divege onverges . Bominant Jeron (^). .~ (^) ( ) This is aagronatra^ series^ wite^ γ=^5 L^1.^ which^ Eonwerges. By the^ Limit^ ComparrtsTestwith^ ) " ,

the given seises^ _conwerages

Problem 2 (5 points) Determine whether the series  n  1

  1  n ^1 2 n 2 n^4  3 n^3  1 converges

absolutely. Also find the maximum possible error if  n  1

  1  n ^1 2 n 2 n^4  3 n^3  1 is approximated by S 4.

Problem 3 (5 points) Find the sum of the series  n  1

 5  ^3  n 4 n ^1 , if it converges. 2 Determine whether^ the^ series^ converges absolutely : an 2 Ʃ )^ n (^4) + (^3) n 3 tt Ʃ (^) converges

because it is a

Dominant P

×

. "□= (^2) as k (^) ts anmjsngm a Termms FInd (^) maximum possible emor if^ S^4 is^ used^ : Eror ≤^ lak+^11 , an =*^3.

s 4 , a ; =^ ≈(^55

) 4

  • 3 ( 5 )+)= → (^25 ). 25 +^37 s^1 =^. 0 ≈ (^0). (^04995). 6

.^ themaximumpossible^ eronwhenusig^ Se^ is^ approximately 0. 04 5 ." π (^3)

= .( ) ( ) ) For (^) a geometric

eies to^

convenge ,^ the^ absolute value^ of^ r^ must fit wth^ r^1 4. =

. Thisseq es c =.

,

, , ( (^) ) (^) , × (^). μ m = . C

Problem 6 (6 points) Find the Taylor series of fx   sin x in x. Write down the four terms of the Taylor series. Do it by direct computation. Problem 7 (a) (3 points) Using the Taylor series of ex^ in x , find the Taylor series of xex 2 in x.

(b) (4 points) Approximate 

0 1 ex 2 dx using the first four nonzero terms of a Taylor series of ex 2 in x.

a =^ π^ f(a)^ =^ sin^ a)^ f(x)^ =^ f^ (π)^ ←^ t^

' (xx- ) +

  • " : (x-x) p … f(x)=^ simx (^) , f(a) =^ sin^ (a)^ =^0 +(←^ ) =^ O

f'^ (c)=^ Cosx^ ,f^ )^ (x )=[^ os(x^ )=-^1

f '^ (a)^ =^ -^1 f" (x) =^ - sin, (^) f(π]^ =^

. (^) sin (π) =^ o f " (π (^) ) -^0 f (^) (x) (β^3 =

  • sosc] , fi"[a )^ ÷ - cos)

f "' ( )a f( 4 ) c)^ : sin (^) x,^ f^ @" (π) = (^) sin (e) = (^) o ∴ f(s)=^0 -^ (x)+ .(x-π^ )^ (x " repeat (^) cylically

fin'(^ x) = sinx,^ cosx^ ,-si^ cosc)

fG) =-c-λ+← . = xe "^ ' = ②. . " = . " e

  • (^) π^ ' = . " " 1 - x+ ③ : - :

S. ee

'bs= f :[ 1 -^ π

  • ] d.. (^) =

[ ] =.

5 : 1. da) -

S!^ π^ '+ /: (^) : '

[ ]!=

. s = (^) = =

Problem 8 (a) (5 points) Find the power series expansion of x 1  3 x in x , and also find the radius of convergence. (b) (6 points) Find the power series expansion of 1 1  2 x and find the power series expansion of ln 1  2 x  in x using term by term integration of power series expansion of 1 1  2 x

  • , lef γ=^3 x (^) ,
  • (^) .( 3 a)" =^. 3 mxu^ , (^13) l 4 or bal<
  • =- ".^ 3 nx μ (^) = ) 3 m + x (^) ) m^ =^ ntl
  • =

)^3 m^ "

x ". O ① = =.

) " 2 μ x " ,^ 2 , = . ② |^ n^ (^1 +^2 )^ = S *u = S . ]uznsaud. = S

]

" (^2) " x "d =. S ) znxndsc =

. " C