100% satisfaction guarantee Immediately available after payment Both online and in PDF No strings attached
logo-home
MAT2613EXAM PACK 2023 R50,00   Add to cart

Exam (elaborations)

MAT2613EXAM PACK 2023

1 review
 13 views  1 purchase

QUESTIONS AND ANSWERS

Preview 4 out of 106  pages

  • September 9, 2023
  • 106
  • 2023/2024
  • Exam (elaborations)
  • Questions & answers
All documents for this subject (7)

1  review

review-writer-avatar

By: ntlantla594 • 1 year ago

avatar-seller
mpumeleloh
MAT2613
EXAM
PACK 2023
QUESTIONS WITH ANSWERS

, 3



I OCTOBER/NOVEMBER )0/7EXAMINATION PAPER AND MEMORANDUM I

QUESTION l

1.1 Use a proof by contradiction to prove that the following statement is true.

2n ;::: 2n for all positive integers n.

[Hint: You may assume the well ordering axiom: Every non-empty set of positive integers has a least
Open Rubric




~~] 00

SOLUTION
Contradiction: There exist at least one positive integer m such that 2m < 2m.
Assumption false for m =
1 and m =
2. The statement must then be: There exist at least one positive integer
m > 2 such that 2m <2m.
Let M = (m lm > 2, m EN, 2m < 2m}. This set M bas a least element by the well·ordening axiom.
Let mo be this element. Then mo > 2 and 2mo < 2m 0 (1)
However, mo - 1 < mo and mo- 1 ¢ M, so 2<mo-l) ;::: 2 (mo- 1) (2)
and so from (1) and (2) we have since (2) is 2mo ;::: 4m0 - 4 that 4m0 - 4 ~ 2mo < 2m0 , i.e 2m0 < 4 or m0 < 2
which is a contradiction.

1.2 Give the contrapositive of the following statement:
00

If L, Or is convergent then (an) is a null sequence. (2)
rei


[10]
SOLUTION
00
If (an) is not a null sequence then L:ar is divergent.
r•l



QUESTION%

Let (an) be the sequence of real numbers defined by a 1 = I and an+l = ,JiCi;,for n EN.
Show that (an) converges and find the limit.
[Hint: Show that 1 ~an < an+l < 2 for all n EN using mathematical induction.] f81

SOLUTION
a1 = 1 and On+1 = -J24,'if n .
Following the hint we have to prove that 1 < an+2 < 2 'if n. (*)
~ an
For n = 1 we have a 1 = 1 ami a 2 = ,J2 thus (*) is true for n = I.
Suppose(*) is true for n = k, i.e 1 ~ at < ak+l < 2 (**)
Then we have from(**) that 2 ~ 2ak < 2aA:+1 < 4 so that ,J2 ~ ,J2iii < ~ < 2.

t
Open Rubric

, 4


But

A - ak+l and J2ak+l = ak+2

so 1 < .J2 ~ ak+l < ak+2 < 2 and the equation (**)is true.

We thus have an increasing sequence which is bounded above by 2.

Suppose
lim an
n-too
= L. Then also lim an+I
11--tOO
= L
We have
lim an+ 1 lim .J2ci:, = Jlim 2an
= n-too
11--too n-too




L = .fi-JI i.e -Jl = v'2 or L =2.
QUESTION3
Prove from first principles that the sequence (an) with

2n 2 +5
a1 = 0, an = ., when n ?: 2
n-- 1
converges. (7)
SOLUTION
2n 2 + 5 2 + 2..
.
We suspect that lrm an
n-too
= .
lun
11--tOO n 2 - 1
= lim ~ =2
n-too ( - ~
'-"
/



Let c > 0 be given. For n :;::: 2 we have




Since

> n when n :;::: 2 we have


lan- 21
7 7/
- -- < - for n > 2
n -l-n
2 -
Clearly
7 7
-<e~n>
n e
By the Archimedean principle there exists ;:: N with N > ~.
f:

For such an N e N we have
• 7 7
n 2: N => n > - => lan - 21 < - < c
e n

, 5


Since c > 0 was arbitrary we have lim a, = 2.
n~oc




QUESTION 4

4.1 Test each of the following series for absolute convergence, conditional convergence or divergence:
r'
4.1.1
oo
I:
r=l
<- 1r ·
3r + 3r1
(4)


SOLUTION

. Ia, I
1un
11--HXl
= I'lm
ri~OO-
3"
3n!
rl
+V=
.3 '
~
1·un
IJ~OO-
-3,--
3JJ!
~+ 1
= 3- V/
1



3
since (
3nJ
; \is a null sequence


lim a, '::f= 0 and fr~e f (-1) a, is divergent.
Since lim
-00 lanl '::f= 0, -00 contrapositive of the vanishing condition
~


1
<- 1y ~~Y'2r
00
4. t. 2 L r::;;===== (7)
r=l 1 2 -



SOLUTION
1
Let lar I = ~~r:==:<===
.:/2r2- 1
1 1
We have ~ > 3r::;-;; =-1 --
1 2


/
2
Y'2r - 1 -v2r·/' rJ
/ 1 1
= 2- < 1
00
By the p- test p 1 the series L diverges and hence 1 L
00
diverges so that ~~.l
~

diverges.
3 r=lr3
2
23 r=ol r' 2 3
r•l 2
2r - 1


Forconditionalllyconvergence: We have lim
r---+00 v
lari.=.P(A1so2(r + 1)2 - 1 > 2r 2 -1 and thus




so that the series f
r=l
larl is decreasing. We thus have that the given series f
r=l
(-1Y
1
J2r2 - 1
is conditiondtt{
convergent:


[Iff (x} = (2x2 - 1r 3
I
then
}
f 1 (x) = -- (2x 2 -
3
1r'.
4
4x = --
4
3
(2x 2 - 1r 4
3
< owhich shows that the series
L larl is decreasing]
4.1.3 L (-1Y
00

r=2
(1
2
r
sin-
r
7r) (7)

The benefits of buying summaries with Stuvia:

Guaranteed quality through customer reviews

Guaranteed quality through customer reviews

Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.

Quick and easy check-out

Quick and easy check-out

You can quickly pay through EFT, credit card or Stuvia-credit for the summaries. There is no membership needed.

Focus on what matters

Focus on what matters

Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!

Frequently asked questions

What do I get when I buy this document?

You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.

Satisfaction guarantee: how does it work?

Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.

Who am I buying this summary from?

Stuvia is a marketplace, so you are not buying this document from us, but from seller mpumeleloh. Stuvia facilitates payment to the seller.

Will I be stuck with a subscription?

No, you only buy this summary for R50,00. You're not tied to anything after your purchase.

Can Stuvia be trusted?

4.6 stars on Google & Trustpilot (+1000 reviews)

79373 documents were sold in the last 30 days

Founded in 2010, the go-to place to buy summaries for 14 years now

Start selling
R50,00  1x  sold
  • (1)
  Buy now