This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: TUTURIAL LETTER 202.  ' Question 1 1. Use a proof by contradiction to prove that if p is a prime number then ﬂ is an irrational number.
Solution
Suppose ﬂ is a rational number.
Then as observed on p. 43(H), ﬂ can be expressed as a rational in its lowest form. Hence f = E: where m and n are integers n # 0 and n have no common factor greater than one. Now m2 = pn2 and so n22 is divisible by p. As p is a prime number, it follows that m is divisible by p. Thus there exists a k e Z such that m = kp.
Then m2 = [€2,1522 and consequently ,arm2 = kzpz. It now follows that n2 = k2 p and we again deduce that
n is divisible by p. Thus p is a common divisor of both m and n. This contradicts our assumption that the
greatest common divisor of m and n is one. Question 2 2. Investigate the following for absolute and conditional convergence or divergence (a)
few (m —r)
r—l Solution Let Ian! = «M in + l) —n _ ~/nln+l —n t/n(n+1)+n
__._...—.___
rr>oo 1 t/n(n+1)+n n(n+1)—rl2 l'm n
= —————— = l ——_—
"moot/nin+l)+n n><>0t/Jv(n+l)+n' l
= lim———= l
n>oo 1+#+] 2 Since lim [anl # 0 we also have lim an gé 0 and it follows from the contrapositive of the vanishing . {7—)CO . ' ‘ n—mo
condition that the serles IS divergent. 3 MAT213T/202/2 00 .3
(—1 "I
(b) I; r l '
Solution
Let
ng n; 1
Ianl= n2 1 andbn=ﬁzg
Then
3
. lanl . n: l
:: E
El? b” Jl£;f12— ‘n
. n2
= lim 2
n—Hmn —
, l
= 11m i
nDOO] _. :5
= l > 0.
Since
00 CD 1
Zn = Z—l + 1
r=l r=2ri diverges (see 4.1.5H), 2—]; diverges and we have 2 Ia,. diverges (use the second comparison test
r—Z" (4.2.2(H)). Using the diagram on p.l [9(H) we now apply the alternating series test. First we see that lanl is monotone decreasing: Let 3
x5
forx> ]. f(x)= xz—l Then f’ (X) < 0 forx>1. Then f is decreasing on the interval (1, 00) and thus f (n + l) < f (17) Le. la” + ll < lanl for all n > 1.
Also
i
_ , :1
11m a,, = 11m
n—aoo IP9OCn2 —]
, 1
2 11m I I
n—)oo
n?—
E
= 0. 00
It follows from (4.2.4H) that Ear is conditionally convergent.
r.2 4
m 1 l'.
(C) 2%
)‘=1
Solution
We use D’Alembert’s ratio test
,, l . l ”
lim la + I = hm 01+ )‘e_
n—wo fanl n—aoo em“ :1
l , (n+1)
= —lIm
en—mo n
l 1
= — lim (1 + —)
era—mo n
_ l
_ e
< l sincee > 2.
The given series is absolutely convergent.
Question 3
Consider the function f given by
x2—6x+5 '
if x > 1
f(x) = ;' .
x —5 If x 5 1 Solution
We have
f(1) = (I)2 — 5 = —4 and for x > 1 we have
__ (xsxx—n
f(x) — T =x—5. Now I __  f(x)—f(1)
L“) 31.1,"? x—(l)
2— .. _
___ “mix—M
x—H‘ x—l
_ 1‘ x2—]
—.r—l)nlll—x—l
. (x—1)(x+l)
= 11m—
x—>l‘ x—l
= linl]_(x+l)
=2.
Also f() 10)
I _ ' x _
f+(l) —xll»1111+ x—l
= “mtw
x—>I+ x—l
= Iimx_1 x—>l+x 1
1. Consequently ff (1) 75 f: (l) and thus f’ (1) does not exist. [See p. 169(H).] 3.2 Is f continuous at x = 1? Give a reason for your answer. Answer
limf(x) = 1imx2—5=—4=f(l)
x—)— x—)_
limf(x) = limx—5=—4=f(l)
.\‘—)+ x—>l+ It follows from 5.2.3(I—l) that f is continuous in the point x = 1. Question 4 4.1 Find the fourth degree Taylor polynomial of f (x) about a : 1 when f (x) = ﬂ. MAT2 l3T/202/2 Solution
We need to ﬁnd
—]
T4.1f(x) ._ —_2!f(1)+f’(1)(x—l)+f"(1)(x )2
_ 3 “.1
+fm (1) (x3 '1) +fm (1) (X4 '4)
Now
f(x)=xi so f(1)=1
f’(x)=%x%=2—J/§ fr(l)=%
f”(x)= “5—711: —% f”(])=—%
fm (x): %X_§ 7 fm (1)_ _%
flu (x): _%x—§ fiu(1)= __[%
f” (x): '—°~5x % f" (c)_ EC— 3
Substitution give
T4,f(x) = %(x _1)_ Mr; n2 + it: n3 _ :_g(x_4!;ﬁ i+:(x—l)—:(x2!2—l) +:—6‘(x—1)3— .3—3(x—1)4 4.2 Determine R4,, f (x) for the ﬁinctioin in (i). In general we have RM f (x) = (3—1)]??— f "+1 (c) where c lies between a and x.
Solution From the above we have (x  U5
5! _<x1)5 E L
_ 5! 32 c; 7 (x — 1)5
=256 c: ' f 5 (c) where c is between 1 and x. R4,1f(x) = Question 5 Let
f(x)=—x+2 03x5]. ) Let P,r be the partition I O, :IN ,,l}of[0,l]. l
n
5.] Calculate L (P,,) and U (Pn) Solution Pn=0, l
n 2 1
, Z,    , 1]subdivides [0, 1] into n subintervals each of length g? 2
ﬂx)=—x+2
_—%
0% 5 i % 3
u n 1 n n n
J"0 I} 3‘2 x3 *4 In
Since m; = inf[f(x) :xj_. 5 x 5 x;} we have m, = f(x,~) = f (:7) = —n—1 + 2 (the minimum value is at the right end point of the interval) and since M. =Sllplf(x) :xH 5x sxil we have (the maximum value is at the left end point of the interval). Then 2:12
_ 3112 — n
— Zn2
311 — l MAT213T/202/2 =é(—(":')+2)% l” . 2
=;I§(—I+l)+;(n)
1». I =__I(£.(Ll))+.l_+2
2 n —n2 —n+2n+4n2
2H2 _ 3112 + 1n
_ 2.712 _3n+l
_ 2n 5.2 Hence prove that f is Riemann integrable on [0, 1] and evaluate f0] f (x) dx.
Solution Wehave 3 1 3 +1
n— n < < < 211 _£_L[_ 2n and by letting n —> 00 it can be deduced that L S U : %. Hence f is Riemannintegrable on [0, l] with fo‘ f(x)dx =f0‘(—x+2)dx ...
View
Full Document
 Summer '17
 Physics, pH, Prime number, Greatest common divisor, Divisor, lim la

Click to edit the document details