MATH 104: INTRODUCTORY ANALYSIS SPRING 2009/10 PROBLEM SET 8 SOLUTIONS. and x i = a + i. i + n(n + 1)(2n + 1) + 2a. (b a)3 6n 2

Size: px
Start display at page:

Download "MATH 104: INTRODUCTORY ANALYSIS SPRING 2009/10 PROBLEM SET 8 SOLUTIONS. and x i = a + i. i + n(n + 1)(2n + 1) + 2a. (b a)3 6n 2"

Transcription

1 MATH 104: INTRODUCTORY ANALYSIS SPRING 2009/10 PROBLEM SET 8 SOLUTIONS 6.9: Let f(x) { x 2 if x Q [, b], 0 if x (R \ Q) [, b], where > 0. Prove tht b. Solutio. Let P { x 0 < x 1 < < x b} be regulr prtitio of [, b] ito subitervls, i.e. x i b ( ) b d x i i for i 1,...,. Hee ( ) b U(P, f) M i x i x 2 i b [ ( )] b 2 i [ b ( ) ( ) ] b b 2 2 2i i 2 b ( ) b 2 ( ) b 2 2 i i 2 ( ) b 2 ( ) (b ) 2 ( 1) b ( 1)(2 1) b (b2 2 )(b ) 2 (b ) 6 2 b Note tht we hve used Exerises 1.1 d 1.14 for the sums. Sie f U(P, f) for every prtitio P, it follows tht f b for ll N. Therefore (b2 2 )(b ) 2 f b (b ) 6 2. (1) By Theorem 6.4, give y > 0, there exists δ > 0 suh tht U(P, f) f wheever P < δ. Sie for lrge eough, we will hve P (b )/ < δ, so Dte: My 6, 2010 (Versio 1.0). U(P, f) f 1

2 d so b (b2 2 )(b ) (b ) Therefore b f d by the rbitrriess of, it follows tht Hee by (1) d (2), b b f. f. (2) 6.12: Prove tht the followig futio is Riem-itegrble o [0, 1] eve though it hs ifiitely my disotiuities o [0, 1]: 1 if x 1 where 1, 2,,..., f(x) 0 otherwise. Solutio. Let > 0. There exists N N suh tht 1/ < /2 wheever > N. Let P be prtitio of [0, 1] where d suh tht So U(P, f) 0 x 0 < x 1 1 N 1 < x 2 < < x k 1 x i x i x i 1 < k 4N. for i 2,,..., k. M i x i 1 N 1 N 4N < 2 2 sie t most 2N of the umbers M 2,..., M k ould be 1 d the rest hve to be 0 (beuse [x 1, 1] [1/(N 1), 1] otis oly 1/N, 1/(N 1),..., 1/2, 1 where f tkes the vlue 1 the worst tht ould hppe is if ll these our mog x 2,..., x k ). Note tht L(P, f) m i x i 0 sie m i 0 for ll i 1, 2,..., k. Hee U(P, f) L(P, f) < d so f is Riem-itegrble by Theorem : Usig Defiitio 6. d Theorem 6.8, prove tht if is y rel umber the f provided f is Riem-itegrble o the lrgest of the itervls [, b], [, b], d [, ]. Solutio. If < < b, the proof of this is give i Theorem 6.8. For the se < b, Theorem 6.8 gi yields f b f f 2

3 d so f b by Defiitio 6.. Likewise for the se < b. 6.20: Prove Theorem 6.1. Hit: First prove tht the result holds whe f d g re oegtive itegble futios o [, b] d the use the idetity f fg (f f )(g g ) f g f g f g f g to estblish Theorem 6.1. Solutio. Suppose f d g re oegtive d Riem itegrble o [, b]. Let P { x 0 < x 1 < < x b} be prtitio. For y bouded futio F : [, b] R, deote m i (F ) For oegtive F, m(f ) Now sie f d g re oegtive, So U(P, fg) L(P, fg) if F (x) d M i(f ) sup F (x), x [x i 1,x i ] x [x i 1,x i ] if F (x) d M(F ) sup F (x). x [,b] x [,b] 0 m(f ) m i (F ) M i (F ) M(F ). M i (fg) M i (f)m i (g) d m i (f)m i (g) m i (fg). [M i (fg) m i (fg)] x i [M i (f)m i (g) m i (f)m i (g)] x i [M i (f)m i (g) M i (f)m i (g) M i (f)m i (g) m i (f)m i (g)] x i M i (f)[m i (g) m i (g)] x i M(f) f m i (g)[m i (f) m i (f)] x i [M i (g) m i (g)] x i M(g) [M i (f) m i (f)] x i M(f)[U(P, g) L(P, g)] M(g)[U(P, f) L(P, f)]. () Note tht if M(f) or M(g) 0, the f or g is the zero futio d therefore fg is the zero futio d therefore Riem itegrble. So we will ssume tht both M(f) > 0 d M(g) > 0. Let > 0 be give. Sie f d g re Riem itegrble, there exists P 1 suh tht U(P 1, g) L(P 1, g) < 2M(f) d there exists P 2 suh tht U(P 2, f) L(P 2, f) < 2M(g). Let P be ommo refiemet of P 1 d P 2. The by Lemm 2 o pp. 15, U(P, g) L(P, g) < U(P 1, g) L(P 1, g) < 2M(f)

4 d Hee by (), U(P, f) L(P, f) < U(P 2, f) L(P 2, f) < 2M(g). U(P, fg) L(P, fg) M(f)[U(P, g) L(P, g)] M(g)[U(P, f) L(P, f)] < 2 2 d so fg is Riem-itegrble by Theorem 6.1. Now for geerl f d g, we pply wht we hve just proved to dedue tht f g, f g, f g, f g (ote tht they re ll produts of two oegtive futios) re Riem-itegrble. The we pply Theorem 6.6 to dedue tht f g f g f g f g is lso Riem-itegrble. But by the hit, this is just fg. 6.22: Prove tht if f d g re otiuous o [, b] d if g does ot hge sig i [, b] the there is poit (, b) suh tht f(x)g(x) dx f() g(x) dx (4) Solutio. We will ssume wlog tht g(x) 0 for ll x [, b] (if ot, osider g). Sie f is otiuous o [, b], so f is bouded o [, b]. Let d so Sie g(x) 0, Hee by Theorem 6.9, m m if f(x) d M sup f(x) x [,b] x [,b] m f(x) M. mg(x) f(x)g(x) Mg(x). g(x) dx f(x)g(x) dx M g(x) dx. If g(x) dx 0, the f(x)g(x) dx 0 d so (4) is trivilly stisfied by y. Otherwise, set f(x)g(x) dx k : g(x) dx d ote tht k [m, M]. By Theorem 4.6 (IVT), there exists (, b) suh tht f() k. 6.2: If f R[, b] defie F (x) x f for every x [, b]. Prove tht F is otiuous o [, b]. Solutio. By Theorem 6.7, sie f R[, b], therefore f R[, ] for y [, b]. Hee F () exists for ll [, b]. We lim tht for ll (, b), d lso lim F (x) F () lim F (x) x x lim F (x) F (), lim x F (x) F (b), x b 4

5 d therefore F is otiuous o [, b] by Theorem.7. Sie f B[, b], there exists M > 0 suh tht f(x) M for ll x [, b]. Let > 0 d set δ /M. If x < δ, the by Problem 6.15 d Theorem 6.12, x F (x) F () f(t) dt f(t) dt x f(t) dt f(t) dt x f(t) dt x f(t) dt M(x ) < Mδ. This shows tht lim x F (x) F () for ll [, b). A essetilly idetil rgumet shows tht lim x F (x) F () for ll (, b]. 6.25: Prove: (1) If f R[0, b] d f is eve futio the f R[, b] d 2 (2) If f R[0, b] d f is odd futio the f R[, b] d 0 0. f. Solutio. (sketh) We defie symmetri prtitio Q of [, b] s prtitio with the property tht (i) 0 Q; d the property tht (ii) x Q if d oly x Q. For y bouded futio f o [, b] (ot eessrily eve or odd), we show tht sup{l(q, f) Q symmetri prtitio of [, b]}, if{u(q, f) Q symmetri prtitio of [, b]}, sie give y osymmetri prtitio P of [, b], we refie it ito symmetri prtitio by simply tkig Q P ( P ). Now it remis to observe tht for eve futio f d symmetri prtitio Q of [, b], L(Q, f) 2L(P, f), U(Q, f) 2U(P, f) where P Q [0, b] is prtitio of [0, b]. symmetri prtitio Q of [, b], L(Q, f) 0, U(Q, f) 0. Riem itegrbility d the required equtios the follows. Wheres for odd futio f d 6.27: Prove the itegrtio-by-prts formul: if f, g re differetible o [, b] d if f, g re itegrble o [, b] the Solutio. fg f(b)g(b) f()g() By the produt rule, f g. (fg) f g fg. (5) 5

6 Sie f, g re differetible d therefore otious d therefore Riem itegrble o [, b] d sie f, g re ssumed to be Riem itegrble o [, b], it follows from Theorem 6.6 d Problem 6.20 tht (fg) is Riem itegrble o [, b]. Hee by Theorem 6.17, (fg) f(b)g(b) f()g(). (6) Now subsitute (5) ito (6) d pply Theorem 6.6 to get the required formul. 6

MATH 104: INTRODUCTORY ANALYSIS SPRING 2008/09 PROBLEM SET 10 SOLUTIONS. f m. and. f m = 0. and x i = a + i. a + i. a + n 2. n(n + 1) = a(b a) +

MATH 104: INTRODUCTORY ANALYSIS SPRING 2008/09 PROBLEM SET 10 SOLUTIONS. f m. and. f m = 0. and x i = a + i. a + i. a + n 2. n(n + 1) = a(b a) + MATH 04: INTRODUCTORY ANALYSIS SPRING 008/09 PROBLEM SET 0 SOLUTIONS Throughout this problem set, B[, b] will deote the set of ll rel-vlued futios bouded o [, b], C[, b] the set of ll rel-vlued futios

More information

Review of the Riemann Integral

Review of the Riemann Integral Chpter 1 Review of the Riem Itegrl This chpter provides quick review of the bsic properties of the Riem itegrl. 1.0 Itegrls d Riem Sums Defiitio 1.0.1. Let [, b] be fiite, closed itervl. A prtitio P of

More information

Riemann Integral Oct 31, such that

Riemann Integral Oct 31, such that Riem Itegrl Ot 31, 2007 Itegrtio of Step Futios A prtitio P of [, ] is olletio {x k } k=0 suh tht = x 0 < x 1 < < x 1 < x =. More suitly, prtitio is fiite suset of [, ] otiig d. It is helpful to thik of

More information

Course 121, , Test III (JF Hilary Term)

Course 121, , Test III (JF Hilary Term) Course 2, 989 9, Test III (JF Hilry Term) Fridy 2d Februry 99, 3. 4.3pm Aswer y THREE questios. Let f: R R d g: R R be differetible fuctios o R. Stte the Product Rule d the Quotiet Rule for differetitig

More information

The Definite Integral

The Definite Integral The Defiite Itegrl A Riem sum R S (f) is pproximtio to the re uder fuctio f. The true re uder the fuctio is obtied by tkig the it of better d better pproximtios to the re uder f. Here is the forml defiitio,

More information

MAS221 Analysis, Semester 2 Exercises

MAS221 Analysis, Semester 2 Exercises MAS22 Alysis, Semester 2 Exercises Srh Whitehouse (Exercises lbelled * my be more demdig.) Chpter Problems: Revisio Questio () Stte the defiitio of covergece of sequece of rel umbers, ( ), to limit. (b)

More information

Chapter 5. The Riemann Integral. 5.1 The Riemann integral Partitions and lower and upper integrals. Note: 1.5 lectures

Chapter 5. The Riemann Integral. 5.1 The Riemann integral Partitions and lower and upper integrals. Note: 1.5 lectures Chpter 5 The Riem Itegrl 5.1 The Riem itegrl Note: 1.5 lectures We ow get to the fudmetl cocept of itegrtio. There is ofte cofusio mog studets of clculus betwee itegrl d tiderivtive. The itegrl is (iformlly)

More information

Sequence and Series of Functions

Sequence and Series of Functions 6 Sequece d Series of Fuctios 6. Sequece of Fuctios 6.. Poitwise Covergece d Uiform Covergece Let J be itervl i R. Defiitio 6. For ech N, suppose fuctio f : J R is give. The we sy tht sequece (f ) of fuctios

More information

MATH 104 FINAL SOLUTIONS. 1. (2 points each) Mark each of the following as True or False. No justification is required. y n = x 1 + x x n n

MATH 104 FINAL SOLUTIONS. 1. (2 points each) Mark each of the following as True or False. No justification is required. y n = x 1 + x x n n MATH 04 FINAL SOLUTIONS. ( poits ech) Mrk ech of the followig s True or Flse. No justifictio is required. ) A ubouded sequece c hve o Cuchy subsequece. Flse b) A ifiite uio of Dedekid cuts is Dedekid cut.

More information

1.3 Continuous Functions and Riemann Sums

1.3 Continuous Functions and Riemann Sums mth riem sums, prt 0 Cotiuous Fuctios d Riem Sums I Exmple we sw tht lim Lower() = lim Upper() for the fuctio!! f (x) = + x o [0, ] This is o ccidet It is exmple of the followig theorem THEOREM Let f be

More information

Convergence rates of approximate sums of Riemann integrals

Convergence rates of approximate sums of Riemann integrals Jourl of Approximtio Theory 6 (9 477 49 www.elsevier.com/locte/jt Covergece rtes of pproximte sums of Riem itegrls Hiroyuki Tski Grdute School of Pure d Applied Sciece, Uiversity of Tsukub, Tsukub Ibrki

More information

Week 13 Notes: 1) Riemann Sum. Aim: Compute Area Under a Graph. Suppose we want to find out the area of a graph, like the one on the right:

Week 13 Notes: 1) Riemann Sum. Aim: Compute Area Under a Graph. Suppose we want to find out the area of a graph, like the one on the right: Week 1 Notes: 1) Riem Sum Aim: Compute Are Uder Grph Suppose we wt to fid out the re of grph, like the oe o the right: We wt to kow the re of the red re. Here re some wys to pproximte the re: We cut the

More information

1. (25 points) Use the limit definition of the definite integral and the sum formulas to compute. [1 x + x2

1. (25 points) Use the limit definition of the definite integral and the sum formulas to compute. [1 x + x2 Mth 3, Clculus II Fil Exm Solutios. (5 poits) Use the limit defiitio of the defiite itegrl d the sum formuls to compute 3 x + x. Check your swer by usig the Fudmetl Theorem of Clculus. Solutio: The limit

More information

1 Tangent Line Problem

1 Tangent Line Problem October 9, 018 MAT18 Week Justi Ko 1 Tget Lie Problem Questio: Give the grph of fuctio f, wht is the slope of the curve t the poit, f? Our strteg is to pproimte the slope b limit of sect lies betwee poits,

More information

MA123, Chapter 9: Computing some integrals (pp )

MA123, Chapter 9: Computing some integrals (pp ) MA13, Chpter 9: Computig some itegrls (pp. 189-05) Dte: Chpter Gols: Uderstd how to use bsic summtio formuls to evlute more complex sums. Uderstd how to compute its of rtiol fuctios t ifiity. Uderstd how

More information

The Basic Properties of the Integral

The Basic Properties of the Integral The Bsic Properties of the Itegrl Whe we compute the derivtive of complicted fuctio, like x + six, we usully use differetitio rules, like d [f(x)+g(x)] d f(x)+ d g(x), to reduce the computtio dx dx dx

More information

12.2 The Definite Integrals (5.2)

12.2 The Definite Integrals (5.2) Course: Aelerted Egieerig Clulus I Istrutor: Mihel Medvisky. The Defiite Itegrls 5. Def: Let fx e defied o itervl [,]. Divide [,] ito suitervls of equl width Δx, so x, x + Δx, x + jδx, x. Let x j j e ritrry

More information

POWER SERIES R. E. SHOWALTER

POWER SERIES R. E. SHOWALTER POWER SERIES R. E. SHOWALTER. sequeces We deote by lim = tht the limit of the sequece { } is the umber. By this we me tht for y ε > 0 there is iteger N such tht < ε for ll itegers N. This mkes precise

More information

UNIVERSITY OF BRISTOL. Examination for the Degrees of B.Sc. and M.Sci. (Level C/4) ANALYSIS 1B, SOLUTIONS MATH (Paper Code MATH-10006)

UNIVERSITY OF BRISTOL. Examination for the Degrees of B.Sc. and M.Sci. (Level C/4) ANALYSIS 1B, SOLUTIONS MATH (Paper Code MATH-10006) UNIVERSITY OF BRISTOL Exmitio for the Degrees of B.Sc. d M.Sci. (Level C/4) ANALYSIS B, SOLUTIONS MATH 6 (Pper Code MATH-6) My/Jue 25, hours 3 miutes This pper cotis two sectios, A d B. Plese use seprte

More information

f(t)dt 2δ f(x) f(t)dt 0 and b f(t)dt = 0 gives F (b) = 0. Since F is increasing, this means that

f(t)dt 2δ f(x) f(t)dt 0 and b f(t)dt = 0 gives F (b) = 0. Since F is increasing, this means that Uiversity of Illiois t Ur-Chmpig Fll 6 Mth 444 Group E3 Itegrtio : correctio of the exercises.. ( Assume tht f : [, ] R is cotiuous fuctio such tht f(x for ll x (,, d f(tdt =. Show tht f(x = for ll x [,

More information

Convergence rates of approximate sums of Riemann integrals

Convergence rates of approximate sums of Riemann integrals Covergece rtes of pproximte sums of Riem itegrls Hiroyuki Tski Grdute School of Pure d Applied Sciece, Uiversity of Tsuku Tsuku Irki 5-857 Jp tski@mth.tsuku.c.jp Keywords : covergece rte; Riem sum; Riem

More information

Linford 1. Kyle Linford. Math 211. Honors Project. Theorems to Analyze: Theorem 2.4 The Limit of a Function Involving a Radical (A4)

Linford 1. Kyle Linford. Math 211. Honors Project. Theorems to Analyze: Theorem 2.4 The Limit of a Function Involving a Radical (A4) Liford 1 Kyle Liford Mth 211 Hoors Project Theorems to Alyze: Theorem 2.4 The Limit of Fuctio Ivolvig Rdicl (A4) Theorem 2.8 The Squeeze Theorem (A5) Theorem 2.9 The Limit of Si(x)/x = 1 (p. 85) Theorem

More information

FOURIER SERIES PART I: DEFINITIONS AND EXAMPLES. To a 2π-periodic function f(x) we will associate a trigonometric series. a n cos(nx) + b n sin(nx),

FOURIER SERIES PART I: DEFINITIONS AND EXAMPLES. To a 2π-periodic function f(x) we will associate a trigonometric series. a n cos(nx) + b n sin(nx), FOURIER SERIES PART I: DEFINITIONS AND EXAMPLES To -periodic fuctio f() we will ssocite trigoometric series + cos() + b si(), or i terms of the epoetil e i, series of the form c e i. Z For most of the

More information

The Reimann Integral is a formal limit definition of a definite integral

The Reimann Integral is a formal limit definition of a definite integral MATH 136 The Reim Itegrl The Reim Itegrl is forml limit defiitio of defiite itegrl cotiuous fuctio f. The costructio is s follows: f ( x) dx for Reim Itegrl: Prtitio [, ] ito suitervls ech hvig the equl

More information

The Weierstrass Approximation Theorem

The Weierstrass Approximation Theorem The Weierstrss Approximtio Theorem Jmes K. Peterso Deprtmet of Biologicl Scieces d Deprtmet of Mthemticl Scieces Clemso Uiversity Februry 26, 2018 Outlie The Wierstrss Approximtio Theorem MtLb Implemettio

More information

Second Mean Value Theorem for Integrals By Ng Tze Beng. The Second Mean Value Theorem for Integrals (SMVT) Statement of the Theorem

Second Mean Value Theorem for Integrals By Ng Tze Beng. The Second Mean Value Theorem for Integrals (SMVT) Statement of the Theorem Secod Me Vlue Theorem for Itegrls By Ng Tze Beg This rticle is out the Secod Me Vlue Theorem for Itegrls. This theorem, first proved y Hoso i its most geerlity d with extesio y ixo, is very useful d lmost

More information

Definition Integral. over[ ab, ] the sum of the form. 2. Definite Integral

Definition Integral. over[ ab, ] the sum of the form. 2. Definite Integral Defiite Itegrl Defiitio Itegrl. Riem Sum Let f e futio efie over the lose itervl with = < < < = e ritrr prtitio i suitervl. We lle the Riem Sum of the futio f over[, ] the sum of the form ( ξ ) S = f Δ

More information

Math 104: Final exam solutions

Math 104: Final exam solutions Mth 14: Fil exm solutios 1. Suppose tht (s ) is icresig sequece with coverget subsequece. Prove tht (s ) is coverget sequece. Aswer: Let the coverget subsequece be (s k ) tht coverges to limit s. The there

More information

f(bx) dx = f dx = dx l dx f(0) log b x a + l log b a 2ɛ log b a.

f(bx) dx = f dx = dx l dx f(0) log b x a + l log b a 2ɛ log b a. Eercise 5 For y < A < B, we hve B A f fb B d = = A B A f d f d For y ɛ >, there re N > δ >, such tht d The for y < A < δ d B > N, we hve ba f d f A bb f d l By ba A A B A bb ba fb d f d = ba < m{, b}δ

More information

Limit of a function:

Limit of a function: - Limit of fuctio: We sy tht f ( ) eists d is equl with (rel) umer L if f( ) gets s close s we wt to L if is close eough to (This defiitio c e geerlized for L y syig tht f( ) ecomes s lrge (or s lrge egtive

More information

Riemann Integral and Bounded function. Ng Tze Beng

Riemann Integral and Bounded function. Ng Tze Beng Riem Itegrl d Bouded fuctio. Ng Tze Beg I geerlistio of re uder grph of fuctio, it is ormlly ssumed tht the fuctio uder cosidertio e ouded. For ouded fuctio, the rge of the fuctio is ouded d hece y suset

More information

a f(x)dx is divergent.

a f(x)dx is divergent. Mth 250 Exm 2 review. Thursdy Mrh 5. Brig TI 30 lultor but NO NOTES. Emphsis o setios 5.5, 6., 6.2, 6.3, 3.7, 6.6, 8., 8.2, 8.3, prt of 8.4; HW- 2; Q-. Kow for trig futios tht 0.707 2/2 d 0.866 3/2. From

More information

0 otherwise. sin( nx)sin( kx) 0 otherwise. cos( nx) sin( kx) dx 0 for all integers n, k.

0 otherwise. sin( nx)sin( kx) 0 otherwise. cos( nx) sin( kx) dx 0 for all integers n, k. . Computtio of Fourier Series I this sectio, we compute the Fourier coefficiets, f ( x) cos( x) b si( x) d b, i the Fourier series To do this, we eed the followig result o the orthogolity of the trigoometric

More information

Numerical Methods. Lecture 5. Numerical integration. dr hab. inż. Katarzyna Zakrzewska, prof. AGH. Numerical Methods lecture 5 1

Numerical Methods. Lecture 5. Numerical integration. dr hab. inż. Katarzyna Zakrzewska, prof. AGH. Numerical Methods lecture 5 1 Numeril Methods Leture 5. Numeril itegrtio dr h. iż. Ktrzy Zkrzewsk, pro. AGH Numeril Methods leture 5 Outlie Trpezoidl rule Multi-segmet trpezoidl rule Rihrdso etrpoltio Romerg's method Simpso's rule

More information

Project 3: Using Identities to Rewrite Expressions

Project 3: Using Identities to Rewrite Expressions MAT 5 Projet 3: Usig Idetities to Rewrite Expressios Wldis I lger, equtios tht desrie properties or ptters re ofte lled idetities. Idetities desrie expressio e repled with equl or equivlet expressio tht

More information

We will begin by supplying the proof to (a).

We will begin by supplying the proof to (a). (The solutios of problem re mostly from Jeffrey Mudrock s HWs) Problem 1. There re three sttemet from Exmple 5.4 i the textbook for which we will supply proofs. The sttemets re the followig: () The spce

More information

Riemann Integration. Chapter 1

Riemann Integration. Chapter 1 Mesure, Itegrtio & Rel Alysis. Prelimiry editio. 8 July 2018. 2018 Sheldo Axler 1 Chpter 1 Riem Itegrtio This chpter reviews Riem itegrtio. Riem itegrtio uses rectgles to pproximte res uder grphs. This

More information

Review of Sections

Review of Sections Review of Sectios.-.6 Mrch 24, 204 Abstrct This is the set of otes tht reviews the mi ides from Chpter coverig sequeces d series. The specific sectios tht we covered re s follows:.: Sequces..2: Series,

More information

( ) dx ; f ( x ) is height and Δx is

( ) dx ; f ( x ) is height and Δx is Mth : 6.3 Defiite Itegrls from Riem Sums We just sw tht the exct re ouded y cotiuous fuctio f d the x xis o the itervl x, ws give s A = lim A exct RAM, where is the umer of rectgles i the Rectgulr Approximtio

More information

B. Examples 1. Finite Sums finite sums are an example of Riemann Sums in which each subinterval has the same length and the same x i

B. Examples 1. Finite Sums finite sums are an example of Riemann Sums in which each subinterval has the same length and the same x i Mth 06 Clculus Sec. 5.: The Defiite Itegrl I. Riem Sums A. Def : Give y=f(x):. Let f e defied o closed itervl[,].. Prtitio [,] ito suitervls[x (i-),x i ] of legth Δx i = x i -x (i-). Let P deote the prtitio

More information

Definite Integral. The Left and Right Sums

Definite Integral. The Left and Right Sums Clculus Li Vs Defiite Itegrl. The Left d Right Sums The defiite itegrl rises from the questio of fidig the re betwee give curve d x-xis o itervl. The re uder curve c be esily clculted if the curve is give

More information

Content: Essential Calculus, Early Transcendentals, James Stewart, 2007 Chapter 1: Functions and Limits., in a set B.

Content: Essential Calculus, Early Transcendentals, James Stewart, 2007 Chapter 1: Functions and Limits., in a set B. Review Sheet: Chpter Cotet: Essetil Clculus, Erly Trscedetls, Jmes Stewrt, 007 Chpter : Fuctios d Limits Cocepts, Defiitios, Lws, Theorems: A fuctio, f, is rule tht ssigs to ech elemet i set A ectly oe

More information

Section IV.6: The Master Method and Applications

Section IV.6: The Master Method and Applications Sectio IV.6: The Mster Method d Applictios Defiitio IV.6.1: A fuctio f is symptoticlly positive if d oly if there exists rel umer such tht f(x) > for ll x >. A cosequece of this defiitio is tht fuctio

More information

SUTCLIFFE S NOTES: CALCULUS 2 SWOKOWSKI S CHAPTER 11

SUTCLIFFE S NOTES: CALCULUS 2 SWOKOWSKI S CHAPTER 11 SUTCLIFFE S NOTES: CALCULUS SWOKOWSKI S CHAPTER Ifiite Series.5 Altertig Series d Absolute Covergece Next, let us cosider series with both positive d egtive terms. The simplest d most useful is ltertig

More information

ALGEBRA. Set of Equations. have no solution 1 b1. Dependent system has infinitely many solutions

ALGEBRA. Set of Equations. have no solution 1 b1. Dependent system has infinitely many solutions Qudrtic Equtios ALGEBRA Remider theorem: If f() is divided b( ), the remider is f(). Fctor theorem: If ( ) is fctor of f(), the f() = 0. Ivolutio d Evlutio ( + b) = + b + b ( b) = + b b ( + b) 3 = 3 +

More information

lecture 16: Introduction to Least Squares Approximation

lecture 16: Introduction to Least Squares Approximation 97 lecture 16: Itroductio to Lest Squres Approximtio.4 Lest squres pproximtio The miimx criterio is ituitive objective for pproximtig fuctio. However, i my cses it is more ppelig (for both computtio d

More information

Chapter 7 Infinite Series

Chapter 7 Infinite Series MA Ifiite Series Asst.Prof.Dr.Supree Liswdi Chpter 7 Ifiite Series Sectio 7. Sequece A sequece c be thought of s list of umbers writte i defiite order:,,...,,... 2 The umber is clled the first term, 2

More information

Remarks: (a) The Dirac delta is the function zero on the domain R {0}.

Remarks: (a) The Dirac delta is the function zero on the domain R {0}. Sectio Objective(s): The Dirc s Delt. Mi Properties. Applictios. The Impulse Respose Fuctio. 4.4.. The Dirc Delt. 4.4. Geerlized Sources Defiitio 4.4.. The Dirc delt geerlized fuctio is the limit δ(t)

More information

3.7 The Lebesgue integral

3.7 The Lebesgue integral 3 Mesure d Itegrtio The f is simple fuctio d positive wheever f is positive (the ltter follows from the fct tht i this cse f 1 [B,k ] = for ll k, ). Moreover, f (x) f (x). Ideed, if x, the there exists

More information

Numbers (Part I) -- Solutions

Numbers (Part I) -- Solutions Ley College -- For AMATYC SML Mth Competitio Cochig Sessios v.., [/7/00] sme s /6/009 versio, with presettio improvemets Numbers Prt I) -- Solutios. The equtio b c 008 hs solutio i which, b, c re distict

More information

BRAIN TEASURES INDEFINITE INTEGRATION+DEFINITE INTEGRATION EXERCISE I

BRAIN TEASURES INDEFINITE INTEGRATION+DEFINITE INTEGRATION EXERCISE I EXERCISE I t Q. d Q. 6 6 cos si Q. Q.6 d d Q. d Q. Itegrte cos t d by the substitutio z = + e d e Q.7 cos. l cos si d d Q. cos si si si si b cos Q.9 d Q. si b cos Q. si( ) si( ) d ( ) Q. d cot d d Q. (si

More information

INSTRUCTOR: CEZAR LUPU. Problem 1. a) Let f(x) be a continuous function on [1, 2]. Prove that. nx 2 lim

INSTRUCTOR: CEZAR LUPU. Problem 1. a) Let f(x) be a continuous function on [1, 2]. Prove that. nx 2 lim WORKSHEET FOR THE PRELIMINARY EXAMINATION-REAL ANALYSIS (RIEMANN AND RIEMANN-STIELTJES INTEGRAL OF A SINGLE VARIABLE, INTEGRAL CALCULUS, FOURIER SERIES AND SPECIAL FUNCTIONS INSTRUCTOR: CEZAR LUPU Problem.

More information

18.01 Calculus Jason Starr Fall 2005

18.01 Calculus Jason Starr Fall 2005 18.01 Clculus Jso Strr Lecture 14. October 14, 005 Homework. Problem Set 4 Prt II: Problem. Prctice Problems. Course Reder: 3B 1, 3B 3, 3B 4, 3B 5. 1. The problem of res. The ciet Greeks computed the res

More information

Approximations of Definite Integrals

Approximations of Definite Integrals Approximtios of Defiite Itegrls So fr we hve relied o tiderivtives to evlute res uder curves, work doe by vrible force, volumes of revolutio, etc. More precisely, wheever we hve hd to evlute defiite itegrl

More information

Exponential and Logarithmic Functions (4.1, 4.2, 4.4, 4.6)

Exponential and Logarithmic Functions (4.1, 4.2, 4.4, 4.6) WQ017 MAT16B Lecture : Mrch 8, 017 Aoucemets W -4p Wellm 115-4p Wellm 115 Q4 ue F T 3/1 10:30-1:30 FINAL Expoetil Logrithmic Fuctios (4.1, 4., 4.4, 4.6) Properties of Expoets Let b be positive rel umbers.

More information

n 2 + 3n + 1 4n = n2 + 3n + 1 n n 2 = n + 1

n 2 + 3n + 1 4n = n2 + 3n + 1 n n 2 = n + 1 Ifiite Series Some Tests for Divergece d Covergece Divergece Test: If lim u or if the limit does ot exist, the series diverget. + 3 + 4 + 3 EXAMPLE: Show tht the series diverges. = u = + 3 + 4 + 3 + 3

More information

Calculus II Homework: The Integral Test and Estimation of Sums Page 1

Calculus II Homework: The Integral Test and Estimation of Sums Page 1 Clculus II Homework: The Itegrl Test d Estimtio of Sums Pge Questios Emple (The p series) Get upper d lower bouds o the sum for the p series i= /ip with p = 2 if the th prtil sum is used to estimte the

More information

Chapter 2. LOGARITHMS

Chapter 2. LOGARITHMS Chpter. LOGARITHMS Dte: - 009 A. INTRODUCTION At the lst hpter, you hve studied bout Idies d Surds. Now you re omig to Logrithms. Logrithm is ivers of idies form. So Logrithms, Idies, d Surds hve strog

More information

Approximate Integration

Approximate Integration Study Sheet (7.7) Approimte Itegrtio I this sectio, we will ler: How to fid pproimte vlues of defiite itegrls. There re two situtios i which it is impossile to fid the ect vlue of defiite itegrl. Situtio:

More information

Math 113, Calculus II Winter 2007 Final Exam Solutions

Math 113, Calculus II Winter 2007 Final Exam Solutions Math, Calculus II Witer 7 Fial Exam Solutios (5 poits) Use the limit defiitio of the defiite itegral ad the sum formulas to compute x x + dx The check your aswer usig the Evaluatio Theorem Solutio: I this

More information

Data Compression Techniques (Spring 2012) Model Solutions for Exercise 4

Data Compression Techniques (Spring 2012) Model Solutions for Exercise 4 58487 Dt Compressio Tehiques (Sprig 0) Moel Solutios for Exerise 4 If you hve y fee or orretios, plese ott jro.lo t s.helsii.fi.. Prolem: Let T = Σ = {,,, }. Eoe T usig ptive Huffm oig. Solutio: R 4 U

More information

8.3 Sequences & Series: Convergence & Divergence

8.3 Sequences & Series: Convergence & Divergence 8.3 Sequeces & Series: Covergece & Divergece A sequece is simply list of thigs geerted by rule More formlly, sequece is fuctio whose domi is the set of positive itegers, or turl umbers,,,3,. The rge of

More information

10.5 Power Series. In this section, we are going to start talking about power series. A power series is a series of the form

10.5 Power Series. In this section, we are going to start talking about power series. A power series is a series of the form 0.5 Power Series I the lst three sectios, we ve spet most of tht time tlkig bout how to determie if series is coverget or ot. Now it is time to strt lookig t some specific kids of series d we will evetully

More information

THE NATIONAL UNIVERSITY OF IRELAND, CORK COLÁISTE NA hollscoile, CORCAIGH UNIVERSITY COLLEGE, CORK SUMMER EXAMINATION 2005 FIRST ENGINEERING

THE NATIONAL UNIVERSITY OF IRELAND, CORK COLÁISTE NA hollscoile, CORCAIGH UNIVERSITY COLLEGE, CORK SUMMER EXAMINATION 2005 FIRST ENGINEERING OLLSCOIL NA héireann, CORCAIGH THE NATIONAL UNIVERSITY OF IRELAND, CORK COLÁISTE NA hollscoile, CORCAIGH UNIVERSITY COLLEGE, CORK SUMMER EXAMINATION 2005 FIRST ENGINEERING MATHEMATICS MA008 Clculus d Lier

More information

Theorem 5.3 (Continued) The Fundamental Theorem of Calculus, Part 2: ab,, then. f x dx F x F b F a. a a. f x dx= f x x

Theorem 5.3 (Continued) The Fundamental Theorem of Calculus, Part 2: ab,, then. f x dx F x F b F a. a a. f x dx= f x x Chpter 6 Applictios Itegrtio Sectio 6. Regio Betwee Curves Recll: Theorem 5.3 (Cotiued) The Fudmetl Theorem of Clculus, Prt :,,, the If f is cotiuous t ever poit of [ ] d F is tiderivtive of f o [ ] (

More information

(200 terms) equals Let f (x) = 1 + x + x 2 + +x 100 = x101 1

(200 terms) equals Let f (x) = 1 + x + x 2 + +x 100 = x101 1 SECTION 5. PGE 78.. DMS: CLCULUS.... 5. 6. CHPTE 5. Sectio 5. pge 78 i + + + INTEGTION Sums d Sigm Nottio j j + + + + + i + + + + i j i i + + + j j + 5 + + j + + 9 + + 7. 5 + 6 + 7 + 8 + 9 9 i i5 8. +

More information

Chapter 5. Integration

Chapter 5. Integration Chpter 5 Itegrtio Itrodutio The term "itegrtio" hs severl meigs It is usully met s the reverse proess to differetitio, ie fidig ti-derivtive to futio A ti-derivtive of futio f is futio F suh tht its derivtive

More information

{ } { S n } is monotonically decreasing if Sn

{ } { S n } is monotonically decreasing if Sn Sequece A sequece is fuctio whose domi of defiitio is the set of turl umers. Or it c lso e defied s ordered set. Nottio: A ifiite sequece is deoted s { } S or { S : N } or { S, S, S,...} or simply s {

More information

Solutions to RSPL/1. log 3. When x = 1, t = 0 and when x = 3, t = log 3 = sin(log 3) 4. Given planes are 2x + y + 2z 8 = 0, i.e.

Solutions to RSPL/1. log 3. When x = 1, t = 0 and when x = 3, t = log 3 = sin(log 3) 4. Given planes are 2x + y + 2z 8 = 0, i.e. olutios to RPL/. < F < F< Applig C C + C, we get F < 5 F < F< F, $. f() *, < f( h) f( ) h Lf () lim lim lim h h " h h " h h " f( + h) f( ) h Rf () lim lim lim h h " h h " h h " Lf () Rf (). Hee, differetile

More information

Math 3B Midterm Review

Math 3B Midterm Review Mth 3B Midterm Review Writte by Victori Kl vtkl@mth.ucsb.edu SH 643u Office Hours: R 11:00 m - 1:00 pm Lst updted /15/015 Here re some short otes o Sectios 7.1-7.8 i your ebook. The best idictio of wht

More information

( a n ) converges or diverges.

( a n ) converges or diverges. Chpter Ifiite Series Pge of Sectio E Rtio Test Chpter : Ifiite Series By the ed of this sectio you will be ble to uderstd the proof of the rtio test test series for covergece by pplyig the rtio test pprecite

More information

SUTCLIFFE S NOTES: CALCULUS 2 SWOKOWSKI S CHAPTER 11

SUTCLIFFE S NOTES: CALCULUS 2 SWOKOWSKI S CHAPTER 11 UTCLIFFE NOTE: CALCULU WOKOWKI CHAPTER Ifiite eries Coverget or Diverget eries Cosider the sequece If we form the ifiite sum 0, 00, 000, 0 00 000, we hve wht is clled ifiite series We wt to fid the sum

More information

The Definite Riemann Integral

The Definite Riemann Integral These otes closely follow the presettio of the mteril give i Jmes Stewrt s textook Clculus, Cocepts d Cotexts (d editio). These otes re iteded primrily for i-clss presettio d should ot e regrded s sustitute

More information

Infinite Sequences and Series. Sequences. Sequences { } { } A sequence is a list of number in a definite order: a 1, a 2, a 3,, a n, or {a n } or

Infinite Sequences and Series. Sequences. Sequences { } { } A sequence is a list of number in a definite order: a 1, a 2, a 3,, a n, or {a n } or Mth 0 Clculus II Ifiite Sequeces d Series -- Chpter Ifiite Sequeces d Series Mth 0 Clculus II Ifiite Sequeces d Series: Sequeces -- Chpter. Sequeces Mth 0 Clculus II Ifiite Sequeces d Series: Sequeces

More information

SPH3UW Unit 7.5 Snell s Law Page 1 of Total Internal Reflection occurs when the incoming refraction angle is

SPH3UW Unit 7.5 Snell s Law Page 1 of Total Internal Reflection occurs when the incoming refraction angle is SPH3UW Uit 7.5 Sell s Lw Pge 1 of 7 Notes Physis Tool ox Refrtio is the hge i diretio of wve due to hge i its speed. This is most ommoly see whe wve psses from oe medium to other. Idex of refrtio lso lled

More information

EVALUATING DEFINITE INTEGRALS

EVALUATING DEFINITE INTEGRALS Chpter 4 EVALUATING DEFINITE INTEGRALS If the defiite itegrl represets re betwee curve d the x-xis, d if you c fid the re by recogizig the shpe of the regio, the you c evlute the defiite itegrl. Those

More information

General properties of definite integrals

General properties of definite integrals Roerto s Notes o Itegrl Clculus Chpter 4: Defiite itegrls d the FTC Sectio Geerl properties of defiite itegrls Wht you eed to kow lredy: Wht defiite Riem itegrl is. Wht you c ler here: Some key properties

More information

 n. A Very Interesting Example + + = d. + x3. + 5x4. math 131 power series, part ii 7. One of the first power series we examined was. 2!

 n. A Very Interesting Example + + = d. + x3. + 5x4. math 131 power series, part ii 7. One of the first power series we examined was. 2! mth power series, prt ii 7 A Very Iterestig Emple Oe of the first power series we emied ws! + +! + + +!! + I Emple 58 we used the rtio test to show tht the itervl of covergece ws (, ) Sice the series coverges

More information

ANALYSIS HW 3. f(x + y) = f(x) + f(y) for all real x, y. Demonstration: Let f be such a function. Since f is smooth, f exists.

ANALYSIS HW 3. f(x + y) = f(x) + f(y) for all real x, y. Demonstration: Let f be such a function. Since f is smooth, f exists. ANALYSIS HW 3 CLAY SHONKWILER () Fid ll smooth fuctios f : R R with the property f(x + y) = f(x) + f(y) for ll rel x, y. Demostrtio: Let f be such fuctio. Sice f is smooth, f exists. The The f f(x + h)

More information

Probability for mathematicians INDEPENDENCE TAU

Probability for mathematicians INDEPENDENCE TAU Probbility for mthemticis INDEPENDENCE TAU 2013 21 Cotets 2 Cetrl limit theorem 21 2 Itroductio............................ 21 2b Covolutio............................ 22 2c The iitil distributio does

More information

Topics. Homework Problems. MATH 301 Introduction to Analysis Chapter Four Sequences. 1. Definition of convergence of sequences.

Topics. Homework Problems. MATH 301 Introduction to Analysis Chapter Four Sequences. 1. Definition of convergence of sequences. MATH 301 Itroductio to Aalysis Chapter Four Sequeces Topics 1. Defiitio of covergece of sequeces. 2. Fidig ad provig the limit of sequeces. 3. Bouded covergece theorem: Theorem 4.1.8. 4. Theorems 4.1.13

More information

Chapter System of Equations

Chapter System of Equations hpter 4.5 System of Equtios After redig th chpter, you should be ble to:. setup simulteous lier equtios i mtrix form d vice-vers,. uderstd the cocept of the iverse of mtrix, 3. kow the differece betwee

More information

INTEGRATION IN THEORY

INTEGRATION IN THEORY CHATER 5 INTEGRATION IN THEORY 5.1 AREA AROXIMATION 5.1.1 SUMMATION NOTATION Fibocci Sequece First, exmple of fmous sequece of umbers. This is commoly ttributed to the mthemtici Fibocci of is, lthough

More information

82A Engineering Mathematics

82A Engineering Mathematics Clss Notes 9: Power Series /) 8A Egieerig Mthetics Secod Order Differetil Equtios Series Solutio Solutio Ato Differetil Equtio =, Hoogeeous =gt), No-hoogeeous Solutio: = c + p Hoogeeous No-hoogeeous Fudetl

More information

AP Calculus AB AP Review

AP Calculus AB AP Review AP Clulus AB Chpters. Re limit vlues from grphsleft-h Limits Right H Limits Uerst tht f() vlues eist ut tht the limit t oes ot hve to.. Be le to ietify lel isotiuities from grphs. Do t forget out the 3-step

More information

The Real Numbers. RATIONAL FIELD Take rationals as given. is a field with addition and multiplication defined. BOUNDS. Addition: xy= yx, xy z=x yz,

The Real Numbers. RATIONAL FIELD Take rationals as given. is a field with addition and multiplication defined. BOUNDS. Addition: xy= yx, xy z=x yz, MAT337H Itrodutio to Rel Alysis The Rel Numers RATIONAL FIELD Tke rtiols s give { m, m, R, } is field with dditio d multiplitio defied Additio: y= y, y z= yz, There eists Q suh tht = For eh Q there eists

More information

HOMEWORK 1 1. P 229. # Then ; then we have. goes to 1 uniformly as n goes to infinity. Therefore. e x2 /n dx = = sin x.

HOMEWORK 1 1. P 229. # Then ; then we have. goes to 1 uniformly as n goes to infinity. Therefore. e x2 /n dx = = sin x. HOMEWORK 1 SHUANGLIN SHAO 1. P 229. # 7.1.2. Proof. (). Let f (x) x99 + 5. The x 66 + x3 f x 33 s goes to ifiity. We estimte the differece, f (x) x 33 5 x 66 + 3 5 x 66 5, for ll x [1, 3], which goes to

More information

Lecture 8. Dirac and Weierstrass

Lecture 8. Dirac and Weierstrass Leture 8. Dira ad Weierstrass Audrey Terras May 5, 9 A New Kid of Produt of Futios You are familiar with the poitwise produt of futios de ed by f g(x) f(x) g(x): You just tae the produt of the real umbers

More information

Integration. Table of contents

Integration. Table of contents Itegrtio Tle of cotets. Defiitio of Riem Itegrtio.................................. Prtitio of itervls........................................... Upper/lower Riem sum......................................3.

More information

Thomas J. Osler Mathematics Department Rowan University Glassboro NJ Introduction

Thomas J. Osler Mathematics Department Rowan University Glassboro NJ Introduction Ot 0 006 Euler s little summtio formul d speil vlues of te zet futio Toms J Osler temtis Deprtmet Row Uiversity Glssboro J 0608 Osler@rowedu Itrodutio I tis ote we preset elemetry metod of determiig vlues

More information

MATH 10550, EXAM 3 SOLUTIONS

MATH 10550, EXAM 3 SOLUTIONS MATH 155, EXAM 3 SOLUTIONS 1. I fidig a approximate solutio to the equatio x 3 +x 4 = usig Newto s method with iitial approximatio x 1 = 1, what is x? Solutio. Recall that x +1 = x f(x ) f (x ). Hece,

More information

Exploring the Rate of Convergence of Approximations to the Riemann Integral

Exploring the Rate of Convergence of Approximations to the Riemann Integral Explorig the Rte of Covergece of Approximtios to the Riem Itegrl Lus Owes My 7, 24 Cotets Itroductio 2. Prelimiries............................................. 2.2 Motivtio..............................................

More information

Advanced Calculus Test File Spring Test 1

Advanced Calculus Test File Spring Test 1 Advced Clculus Test File Sprig 009 Test Defiitios - Defie the followig terms.) Crtesi product of A d B.) The set, A, is coutble.) The set, A, is ucoutble 4.) The set, A, is ifiite 5.) The sets A d B re

More information

5.1 - Areas and Distances

5.1 - Areas and Distances Mth 3B Midterm Review Writte by Victori Kl vtkl@mth.ucsb.edu SH 63u Office Hours: R 9:5 - :5m The midterm will cover the sectios for which you hve received homework d feedbck Sectios.9-6.5 i your book.

More information

Exponents and Radical

Exponents and Radical Expoets d Rdil Rule : If the root is eve d iside the rdil is egtive, the the swer is o rel umber, meig tht If is eve d is egtive, the Beuse rel umber multiplied eve times by itself will be lwys positive.

More information

HOMEWORK #10 SOLUTIONS

HOMEWORK #10 SOLUTIONS Math 33 - Aalysis I Sprig 29 HOMEWORK # SOLUTIONS () Prove that the fuctio f(x) = x 3 is (Riema) itegrable o [, ] ad show that x 3 dx = 4. (Without usig formulae for itegratio that you leart i previous

More information

Calculus in R. Chapter Di erentiation

Calculus in R. Chapter Di erentiation Chpter 3 Clculus in R 3.1 Di erentition Definition 3.1. Suppose U R is open. A function f : U! R is di erentible t x 2 U if there exists number m such tht lim y!0 pple f(x + y) f(x) my y =0. If f is di

More information

T b a(f) [f ] +. P b a(f) = Conclude that if f is in AC then it is the difference of two monotone absolutely continuous functions.

T b a(f) [f ] +. P b a(f) = Conclude that if f is in AC then it is the difference of two monotone absolutely continuous functions. Rel Vribles, Fll 2014 Problem set 5 Solution suggestions Exerise 1. Let f be bsolutely ontinuous on [, b] Show tht nd T b (f) P b (f) f (x) dx [f ] +. Conlude tht if f is in AC then it is the differene

More information

MATH301 Real Analysis (2008 Fall) Tutorial Note #7. k=1 f k (x) converges pointwise to S(x) on E if and

MATH301 Real Analysis (2008 Fall) Tutorial Note #7. k=1 f k (x) converges pointwise to S(x) on E if and MATH01 Real Aalysis (2008 Fall) Tutorial Note #7 Sequece ad Series of fuctio 1: Poitwise Covergece ad Uiform Covergece Part I: Poitwise Covergece Defiitio of poitwise covergece: A sequece of fuctios f

More information

MATH 118 HW 7 KELLY DOUGAN, ANDREW KOMAR, MARIA SIMBIRSKY, BRANDEN LASKE

MATH 118 HW 7 KELLY DOUGAN, ANDREW KOMAR, MARIA SIMBIRSKY, BRANDEN LASKE MATH 118 HW 7 KELLY DOUGAN, ANDREW KOMAR, MARIA SIMBIRSKY, BRANDEN LASKE Prt 1. Let be odd rime d let Z such tht gcd(, 1. Show tht if is qudrtic residue mod, the is qudrtic residue mod for y ositive iteger.

More information