Lecture 11 Complex Numbers
|
|
- Jane Greene
- 6 years ago
- Views:
Transcription
1 Lecture 11 Complex Numbers Holde Lee /19/11 1 Itroductio 1.1 Defiitios The set of real umbers R from the familiar itegers 0, 1, 1,,,... to ratioal umbers 1, 3,... to irratioal umbers, π, e are familiar to us ad have a clear place i the world aroud us. But math is ot quite complete without the complex umbers. I the reals there is o solutio to the polyomial equatio x = 1, as the square of ay real umber is positive. However, we ca create a solutio: let s adjoi a umber to R whose square is 1. Call it i, ad let the resultig umber system be C. Sice we wat additio ad multiplicatio to work out, we have to iclude i C all umbers of the form z = a + bi, where a ad b are real. Defiitio 1.1: The set of umbers i the form a + bi, where a ad b are real, is called the complex umbers C. We call a the real part of z ad b the imagiary part, deoted by R(z = Re(z ad I(z = Im(z, respectively. A good way to imagie complex umbers is to thik of them as poits o a plae (the complex plae, by graphig a + bi as the poit (a, b. The x-axis cosists of the real umbers a R, while the y-axis cosists of the imagiary umbers bi, b R. However, the complex umbers have much more structure tha just ordered pairs of real umbers for istace, we ca multiply complex umbers as we show below. Let s see how basic operatios work i C. To add or subtract complex umbers, we simply add the real ad imagiary compoets: (1+i (3 5i = (1 3i+( ( 5i = + 7i. We multiply complex umbers by the distributive law, ad recall that i = 1: for example (1 + i(3 5i = i + i 3 + i 5i = 3 + 6i 10i = i. To defie divisio, we eed the followig: Defiitio 1.: The cojugate of a complex umber z = a + bi, where a, b are real, is z = a bi. Note that the product of a complex umber ad its cojugate is always real: (a + bi(a bi = a (bi = a + b. This allows us to divide complex umbers: to evaluate a+bi we multiply both the umerator ad the deomiator by the complex cojugate of c + di, c di. For c+di example, 1 + i 3 5i = (1 + i(3 + 5i (3 5i(3 + 5i = i + 10i = i.
2 Note that complex cojugatio preserves additio ad multiplicatio, i.e. z 1 + z = z 1 + z ad z 1 z = z 1 z. Ideed, let z 1 = a + bi ad z = c + di. The z 1 + z = (a bi + (c di = (a + c (b + di = z 1 + z. z 1 z = (a bi(c di = (ac bd (ad + bci = (ac bd + (ad + bci = z 1 z. The absolute value of z = a + bi is defied by z = a + b. It is the distace from z to 0 whe we plot z i the complex plae. Now egative umbers have square roots too, sice ( ai = a. With this, we ca solve ay quadratic equatio by completig the square or usig the quadratic formula. But what about geeral polyomials equatios? Do we have to adjoi more elemets to C so that every cubic polyomial ax 3 + bx + cx + d = 0 where a, b, c, d C, has a solutio? Due to the followig miracle, we do ot: Theorem 1.3 (Fudametal Theorem of Algebra: Every ocostat polyomial with coefficiets i C has a zero i C. (See lecture 8 for a proof. This shows that C is a very atural umber system to do algebra over. 1. Motivatios So what are complex umbers good for? Who cares about polyomial equatios that ca t be solved over the reals after all, complex solutios do t make sese i real life, right? Complex umbers were first developed to solve cubic polyomial equatios the formula higed o the use of complex umbers eve whe the solutios of the equatio were real. At first mathematicias were hesitat to adopt complex umbers, but complex umbers soo proved their use i a variety of other ways. A math problem (for example, a differetial equatio might require as a itermediate step to solve a polyomial equatio that may ot have real solutios. But if we kow how to work with complex umbers, we ca proceed just as we would i the real case, ad the simplified aswer may i fact ot have complex umbers at all, but would be hard to obtai otherwise. Calculus over the complex umbers turs be much icer so that it is ofte advatageous to exted the domai of real fuctios to the complex umbers, ad the look at their properties. This is the case with the ifamous zeta fuctio, which has applicatios to umber theory. May problems that do t look like they ivolve complex umbers tur out to be much easier if we use them. For istace, to fid Pythagorea triples we wat to solve a + b = c over the itegers. We may rewrite this as c b = a ad factor as (c b(c + b = a, ad the use umber theory. Suppose we wat to do the same with a 3 + b 3 = c 3.
3 (Thik Fermat s Last Theorem! We rewrite this as c 3 b 3 = a 3 ad factor (c b(c + cb + b = a 3. However, the left side is ot completely factored. If, however, we were workig over the complex umbers, we could factor this further: (c b(c ωb(c ω b = a 3 for some complex umber ω. The we could solve this by usig umber theory but over complex umbers istead (this is part of what is called algebraic umber theory. 1.3 Examples Let s warm up with some problems. Example 1.4: Fid the sum Solutio. We calculate 1 + i + i + i i = 1, i = i, i = 1, i 3 = i, i 4 = 1. Hece the powers of i cycle through 1, i, 1, ad i. So we group the sum ito four terms at a time (1 + i + i + i 3 + (i 4 + i 5 + i 6 + i (i i i i 011 ad ote that the sum i each group is 1 + i 1 i = 0. Hece the aswer is 0. Example 1.5: Show that if f is a polyomial with real coefficiets ad f(z = 0, the f( z = 0. Coclude that the roots of f (with multiplicity ca be grouped ito complex cojugate pairs. Proof. Let f(z = a z + +a 1 z+a 0. Usig the fact that cojugatio preserves additio ad multiplicatio, f( z = a z + + a 1 z + a 0 = a z + + a 1 z + a 0 = a z + + a 1 z + a 0 sice a m real = a z + + a 1 z + a 0 = f(z. Thus if f(z = 0 the f( z = 0. Thus the oreal roots of f come i complex cojugate pairs. Example 1.6 (AIME1 009/: There is a complex umber z with imagiary part 164 ad a positive iteger such that Fid. z z + = 4i. 3
4 Solutio. We ca write z = a + 164i where a is real. The Clearig the deomiator gives a + 164i (a i = 4i. a + 164i = (a + i. Hece matchig the real ad imagiary parts, a = 656 ad 164 = 4(a+ = 4( We get 41 = or = Problem Set 1 1. [1] Simplify a+bi, where a, b, c, d are real ad c + di 0. c+di. [1] Factor x + y ad x 3 + y 3 over the complex umbers. 3. [1] Fid i [1.5] (AMC 1B, 004/16 A fuctio f is defied by f(z = i z. How may values of z satisfy z = 5 ad f(z = z? 5. [1.5] a ad b are complex umbers such that x a + b = x a + x b is a idetity, true for all values of x. Fid all possible values of a b. 6. [1.5] (AMC 1A, 007/18 The polyomial f(x = x 4 + ax 3 + bx + cx + d has real coefficiets, ad f(i = f( + i = 0. What is a + b + c + d? 7. [1.5] (AMC 1B, 008/19 A fuctio f is defied by f(z = (4 + iz + αz + γ for all complex umbers z, where α ad γ are complex umbers ad i = 1. Suppose that f(1 ad f(i are both real. What is the smallest possible value of α + γ? 8. [1.5] (AIME1 007/3 The complex umber z is equal to 9+bi, where b is a positive real umber. Give that the imagiary parts of z ad z 3 are equal, fid b. 9. [] (AIME1 00/1 Let F (z = z+i for all complex umbers z i, ad let z i z = F (z 1 for all positive itegers. Give that z 0 = 1 + i ad z = a + bi, where a ad b are real, fid a + b. 10. [.5] Let Λ = {a + bi a, b Z, (a, b 0}. Evaluate λ Λ 1 λ 6. (Note: You may assume the sum is absolutely coverget, i.e. well-defied. 11. [4] (USAMO 1989/3 Let P (z = z + c 1 z c be a polyomial i z, with real coefficiets c k. Suppose that P (i < 1. Prove that there exist real umbers a ad b such that P (a + bi = 0 ad (a + b + a < 4b
5 Polar form.1 Basic Facts ad Examples Cosider the complex plae. Whe we write a complex umber as a + bi, we graph the poit by goig horizotally a uits ad vertically b uits. This is called the rectagular form of the complex umber. Additio of complex umbers correspods icely to additio of vectors. Cojugatio correspods to reflectio over the x-axis (real axis. But what does multiplicatio ad divisio? To aswer this questio, we eed to write our complex umbers i a differet form. We could istead specify a locatio z o the complex plae by the distace r from 0, ad the agle θ made with the positive real axis. Drawig a triagle, we see that z = r(cos θ + i si θ which we abbreviate as z = r cis θ. This is called the polar form, z is called the modulus, ad θ is called the argumet. Let s try to multiply complex umbers i this form. Let z 1 = r 1 cis α ad z = r cis β. The usig the additio idetities for sie ad cosie, z 1 z = r 1 r (cos α + i si α(cos β + i si β We state this as a theorem. Theorem.1 (De Moivre: I particular, = r 1 r [(cos α cos β si α si β + i(cos α si β + cos β si α] = r 1 r [cos(α + β + i si(α + β]. (r 1 cis α(r cis β = r 1 r cis(α + β. (r cis θ = r cis(θ. Usig this theorem we ca fid th roots of a complex umber z, i.e. fid the solutios to x = z. Writig z = r cis θ ad x = s cis φ, this is equivalet to (s cis φ = r cis θ. Usig De Moivre s Theorem, this is equivalet to s cis φ = r cis θ. Hece we eed s = r ad φ equal to θ as agles. This meas they are allowed to differ by a multiple of πi. Hece r = s ad φ = θ + πik φ = θ + πik for some iteger k. Takig k = 0, 1,..., 1 the possibilities for k modulo give distict possible values of φ (other values differ from oe of these by a multiple of πi. Thus each ozero complex umber has th roots. For example, the third roots of i = 8 cis π are cis ( π + πk 6 3 for k = 0, 1,, i.e. cis π 6 = i cis 5π 6 = i cis 3π = i. 5
6 Example.: Let z 0 ad > 1. Show that the sum of the th roots of z equals 0. Solutio 1. Let x be a th root of z, ad ω = cis π. The 1, ω,..., ω 1 are the th roots of 1 so z, zω,..., zω 1 are the th roots of z. So the sum is z(1 + ω + + ω 1 = z ω 1 ω 1 = 0 sice ω = 1. Solutio. Let x 1,..., x be the th roots of z ad ω = cis π above. The x 1 ω,..., x ω are also the th roots of uity, so are equal to x 1,..., x i some order. Thus the sum s equals both x x ad ω(x x. Sice s = sω, s = 0. Example.3 (AIME 008/9: A particle is located o the coordiate plae at (5, 0. Defie a move for the particle as a couterclockwise rotatio of π radias about the origi 4 followed by a traslatio of 10 uits i the positive x-directio. Give that the particle s positio after 150 moves is (p, q, fid the greatest iteger less tha or equal to p + q. Solutio. Thik of the coordiate plae as the complex plae, ad let z be the particle s positio after moves, as a complex umber. A couterclockwise rotatio of π 4 correspods to multiplicatio by t = cis π, ad traslatio by 10 uits i the positive 4 x-directio correspods to addig 10. Thus Calculatig the first few terms, z 0 = 5 z 1 = 5t + 10 z = tz z = (5t + 10t + 10 = 5t + 10t + 10 so we see that z = 5t + 10t , easily proved by iductio. Sice t is a eighth root of 1, by Example (. the sum of the eighth roots 1 + t + t + + t 7 equals 0. Sice t 8 = 1, Hece 0 = 1 + t + + t 7 = t t 15 = = t t 151. z 150 = 5t t = 10[(1 + t + + t (t t 15 ] 10t 151 5t 150 = 10t 7 5t 6 = 10 cis 7π 4 5 cis 3π ( = 10 i + 5i. Thus p + q = = The aswer is 19. Example.4: Let P (x = (x + (x + 1 (x + 1 (x 1(x (x. Show that all the zeros of P (x are purely imagiary, i.e. have real part 0. 6
7 Proof. Sice a polyomial of degree at most 1 has at most 1 zeros, it suffices to show that there are 1 pure imagiary roots. Let x = yi ad θ k (y = ta 1 y k. Makig this substitutio ad rewritig i polar form, P (x = 0 is equivalet to (iy + (iy + 1 (iy + 1 = (iy 1(iy (iy cis θ (y cis θ 1 (y = cis(π θ 1 (y cis(π θ (y cis(θ 1 (y + θ (y = cis(π (θ 1 (y + + θ (y (θ 1 (y + + θ (y π (mod π. (1 Note that each θ k is a cotiuous fuctio of y, ad that (θ 1 (y + + θ (y π = π as y (θ 1 (y + + θ (y π = π as y Thus (θ 1 (y + + θ (y attais the value π kπ for k = 1,,..., 1. Hece there are 1 values of y such that (1 holds ad hece there are 1 purely imagiary roots of P, as eeded.. Trigoometry The polar form of complex umbers ca be used to simplify trig expressios ad prove trig idetities. Example.5: Show that cos 3θ = 4 cos 3 θ 3 cos θ ad si 3θ = 3 si θ 4 si 3 θ. Solutio. By De Moivre s Theorem, cos 3θ + i si 3θ = (cos θ + i si θ 3 Matchig the real ad imagiary parts, ad = cos 3 θ + 3 cos θ(i si θ + 3 cos θ(i si θ + (i si θ 3 = (cos 3 θ 3 cos θ si (θ + (3 cos θ si θ si 3 θi cos 3θ = cos 3 θ 3 cos θ si (θ = cos 3 θ 3 cos θ(1 cos (θ = 4 cos 3 θ 3 cos θ si 3θ = 3 cos θ si θ si 3 θ = 3(1 si θ si θ si 3 θ = 3 si θ 4 si 3 θ. Example.6: Show that cos 0 + cos 1 + cos + + cos 89 = 1+cot.5. Solutio. Let ω = cos 1 + i si 1. The ω = cos + i si. Hece the desired sum equals the real part of 1 + ω + ω + + ω 89. Usig the geometric series formula, 1 + ω + ω + + ω 89 = ω90 1 ω 1 = i 1 (cos i si 1 = (i 1((cos 1 1 i si 1 (cos si 1 = (i 1((cos 1 1 i si 1 cos 1 = 1 cos 1 + si 1 + (cos si 1 i cos 1. 7
8 This has real part 1 cos 1 + si 1 = 1 si 1 (1 + = cos 1 1 cos 1 where we used the trig idetity cot θ =.3 Calculus viewpoit si θ. 1 cos θ 1 + cot.5 Usig calculus we ca write cos θ+i si θ i a more suggestive form. The Taylor expasios of cos ad si are Thus cos x = 1 x! + x4 4! x6 6! + si x = x x3 3! + x5 5! x7 7! + cos x + i si x = 1 + ix x! ix3 3! + x4 4! + ix5 5! + = 1 + (ix + (ix! = e ix. + (ix3 3! + (ix4 4! + (ix5 5! Therefore we ca write cis θ as e iθ. Thus De Moivre s Theorem simply correspods to the fact that e iθ 1 e iθ = e i(θ 1+θ the familiar rule for expoets..4 Problem set 1. [1] Show that cos θ = eiθ + e iθ si θ = eiθ e iθ. i. [] Prove that cos θ = cos θ si θ = / ( = ( 1 k k k=0 cos 1 θ = ( 1 ( 1/ k=0 ( cos θ si θ + cos k θ si k θ. ( ( 1 k k + 1 ( cos 3 θ si 3 θ + 3 cos k 1 θ si k+1 θ. 8
9 3. [] Evaluate si π + si π ( 1π + + si. 4. [] (AIME 000/9 Give that z is a complex umber such that z + 1 z = cos 3, fid the least iteger that is greater tha z z [] (AMC 1B, 005/ sequece of complex umbers z 0, z 1,... is defied by the rule z +1 = iz z. Suppose that z 0 = 1 ad z 005 = 1. How may possible values are there for z 0? 6. [] (AMC 1A 008/5 A sequece (a 1, b 1, (a, b, (a 3, b 3, of poits i the coordiate plae satisfies (a +1, b +1 = ( 3a b, 3b + a. Suppose that (a 100, b 100 = (, 4. What is a 1 + b 1? 7. [] (AIME 005/9 For how may positive itegers less tha or equal to 1000 is true for all real t? (si t + i cos t = si t + i cos t 8. [.5] (AIME 1994/11 The equatio x 10 + (13x 1 10 = 0 has 10 complex roots r 1, r 1, r, r,..., r 5, r 5. Fid r 1 r 1 r 5 r 5 9. [.5] (AIME 001/14 There are complex umbers that satisfy both z 8 z 8 1 = 0 ad z = 1. These umbers have the form z m = cos θ m + i si θ m where 0 θ 1 < < θ < 360 ad agles are measured i degrees. Fid the value of θ + θ θ. 10. [3] Fid a closed form for the sum ( + 0 ( + 3 ( + 6 Hit: Oe way to evaluate ( ( is the followig. By the biomial formula, we kow ( ( ( ( ( = (1 + 1 = = (1 1 = 0 ( ( 1 ( 3 + ( 3 Addig these two gives (( ( ( = ( ( ( 1 = How ca you geeralize this? ( 4
10 11. [4] (MOSP 007 Let a, b 1,..., b, c 1,..., c be real umbers such that x + ax ax + 1 = (x + b 1 x + c 1 (x + b x + c (x + b x + c. Prove that c 1 = c = = c = [5] (APMO 009/5 Larry ad Rob are two robots travelig i oe car from Argovia to Zillis. Both robots have cotrol over the steerig ad steer accordig to the followig algorith: Larry makes a 90 left tur after every l kilometers drivig from the start; Rob makes a 90 right tur after every r kilometers drivig from the start, where l ad r are relatively prime positive itegers. I the evet of both turs occurig simultaeously, the car will keep goig straight without chagig directio. Assume that the groud is flat ad the car ca move i ay directio. Let the car start from Argovia facig towards Zillis. For which choices of the pair (l, r is the car guarateed to reach Zillis, regardless of how far it is from Argovia? 3 Roots of Uity The th roots of uity, i.e. roots of 1, satisfy the equatio x 1 = 0. The th roots of uity, ot coutig 1, satisfy x 1 x 1 = x x + 1 = 0. If we plug i x = cis π, the terms of this sum are all the th roots of uity, ad we get Example.. This simple idea the sum of the th roots of uity satisfy the equatios above, ad that their sum is zero ca be very useful. Example 3.1: poits Q 1,..., Q are equally spaced o a circle of radius 1 cetered at O. Poit P is o ray OQ 1 so that OP =. Fid the product P Q k i closed form, i terms of. k=1 Solutio. Lettig ray OQ 1 be the positive real axis, Q i represet the th roots of uity ω i i the complex plae. Hece P Q i equals ω i. The roots of x 1 = 0 are just the th roots of uity, so x 1 = 1 i=0 (x ωi. Pluggig i x = gives k=1 P Q i = 1. Pluggig i roots of uity ca help i factorig polyomials ad establishig divisibility, as the followig example shows: Example 3.: Fid all itegers 1 such that x +1 + x + 1 is divisible by x + x + 1. Solutio. I order for x +1 + x + 1 to be divisible by x ± x + 1, we must have that all zeros of x + x + 1 are zeros of x +1 + x + 1. The zeros are exactly the two third roots of uity ot equal to 1. Let ω be such a root of uity. The we eed ω +1 + ω + 1 = 0. However, we kow ω 3 = 1, so the above equatio is equivalet to ω (+1 mod 3 + ω mod = 0. For 0 (mod 3 this equals ω +, for 1 (mod 3 this equals ω + ω + 1 = 0, ad for (mod 3 this equals ω +. Hece the aswer is all with (mod 3. 10
11 Example 3.3: [MOSP 007] Let be a positive iteger which is ot a prime power. Prove that there exists a equiagular polygo whose side legths are 1,..., i some order. Proof. Sice is ot a prime power, we ca write = pq, where p ad q are relatively prime. Let ω = cis π. The existece of such a equiagular polygo is equivalet to the existece of a permutatio a 1,..., a of the umbers 1,..., such that 1 a k ω k = 0 ( k=0 Ideed, give such a equiagular polygo, place it o the complex plae so that the real axis be parallel to oe of its sides. Thik of the sides of the polygo as vectors formig a loop, ad traslate them to the origi. The we get vectors equally spacced apart; they are i the directio of the th roots of uity ad have legths 1,..., i some order. The sum of the vectors must be 0 because they formed a loop. Hece we get the equatio above. Coversely, we ca easily reverse the above costructio. We use the two factors of to cause double cacelatio. Note that (pr + 1ω rp + (pr + ω rp+q + + (pr + pω rp+(p 1q (3 = prω rp (1 + ω q + + ω (p 1q + ω rp (1 + ω + + pω p 1 = ω rp (1 + ω q + + pω (p 1q. We used the fact 1 + ω q + + ω (p 1q = 0 as the terms are the pth roots of uity. Now addig the above expressio for r = 1,..., q gives q ω rp (1 + ω q + + pω (p 1q = (1 + ω p + + ω (q 1p (1 + ω q + + pω (p 1q = 0 r=1 becuase 1 + ω p + + ω (q 1p = 0 as the terms are the qth roots of uity. However, addig up (3 without simplifyig for r = 1,..., q, we get the pq expoets of ω rage over the umbers rp + kq, where 1 r q ad 0 k q. Sice p, q are relatively prime, these umbers are all distict modulo = pq. (If r 1 p + k 1 q = r p + k q the (r 1 r p = (k k 1 q with r 1 r < q, k k 1 < p so they must equal 0. Thus takig the expoets modulo pq, we get a sum of the form (, as eeded. 3.1 Problems 1. [1] Factor x 5 + x [] (AIME 1996/11 Let P be the product of the roots of z 6 + z 4 + z 3 + z + 1 = 0 that have positive imagiary part, ad suppose that P = r(cos θ + i si θ where 0 < r ad 0 θ < 360. Fid θ. 3. [.5] (AIME 1997/14 Let v ad w be distict, radomly chose roots of the equatio z = 0. Let m be the probability that + 3 v + w, where m ad are relatively prime positive itegers. Fid m +. 11
12 4. [.5] (AIME1 004/13 The polyomial P (x = (1 + x + x + + x 17 x 17 has 34 complex zeros of the form z k = r k [cos(πα k +i si(πα k ], k = 1,, 3,..., 34, with 0 < α 1 α α 34 < 1 ad r k > 0. Give that α 1 + α + α 3 + α 4 + α 5 = m/ where m ad are relatively prime positive itegers, fid m [.5] (AIME 003/15 Let 3 P (x = 4x 4 + (4 j(x 4 j + x 4+j. j=1 Let z 1,..., z r be the distict zeros of P (x ad let z k = a k + b k i for k = 1,,..., r, where a k ad b k are real umbers. Let b k = m + p k=1 where m,, ad p are itegers ad p is ot divisible by the square of ay prime. Fid m + + p. 6. [.5] (AIME 009/13 Let A ad B be the edpoits of a semicircular arc of radius. The arc is divided ito seve cogruet arcs by six equally spaced poits C 1, C,..., C 6. All chords of the form AC i or BC i are draw. Let be the product of the legths of these twelve chords. Fid the remaider whe is divided by [3] Fid all ordered pairs (m, such that x +1 + x + 1 divides x m+1 + x m [5] (IMO 1990/6 Prove that there exists a covex 1990-go such that all its agles are equal ad the legths of the sides are the umbers 1,,..., 1990 i some order. Ca you geeralize the statemets of this problem ad Example 3.3? 9. [5] (USAMO 1999/3 Let p > be prime ad let a, b, c, d be itegers ot divisible by p, such that {ra/p} + {rb/p} + {rc/p} + {rd/p} = for ay iteger r ot divisible by p. Prove that at least two of the umbers a + b, a + c, a + d, b + c, b + d, c + d are divisible by p. (Note: {x} = x x deotes the fractioal part of x. 4 Complex umbers i combiatorics We give two applicatios of complex umbers to combiatorics problems. beautiful problems of a similar ature, the reader is referred to [1, 8]. For other Example 4.1: Show that a a b rectagle ca be tiled by 1 blocks if ad oly if either a or b. 1
13 Solutio. If a or b the the rectagle ca obviously be tiled. Now suppose that a a b rectagle ca be tiled by 1 blocks. Label the squares of the rectagle (x, y with 0 x a 1, 0 y b 1. Let ω be a primitive th root of uity. Label (x, y with ω x+y. Each 1 tile covers umbers of the form ω t, ω t+1,..., ω t+ 1 for some t; these umbers sum to 0. However the sum of all umbers i the board is (1 + ω + + ω a 1 (1 + ω + + ω b 1 = ωa 1 ω 1 ωb 1 ω 1. This is 0 oly if ω a = 1 or ω b = 1, i.e. a or b. Example 4. (TST 004/: Assume is a positive iteger. Cosider sequeces a 0, a 1,..., a for which a i {1,,..., } for all i ad a = a 0. (a Call a sequece good if for all i = 1,,...,, a i a i 1 i (mod. Suppose that is odd. Fid the umber of good sequeces. (b Call a sequece great if for all i = 1,,...,, a i a i 1 i, i (mod. Suppose that is a odd prime. Fid the umber of great sequeces. Solutio. Let f be a fuctio from {1,,..., } to the set of subsets of {0, 1,..., }. Defie a o-f sequece to be a sequece a 0, a 1,..., a such that 1. a i {1,,..., } for i = 0,...,,. a = a 0, ad 3. a i a i 1 is ot cogruet to a umber i f(i modulo. Claim 4.3: The umber of o-f sequeces is times the umber of sequeces d 1, d,..., d such that 1. d i {0, 1,,..., 1} for each i,. d 1 + d 0 (mod, ad 3. d i f(i. Proof. Give a o-f sequece a i, associate with it a sequece as above, with d i = (a i a i 1 mod. Each sequece d i satisfyig the above is associated with o-f sequeces: a 0 ca be chose arbitrarily (there are choices, ad oce a i 1 has bee defied, a i must be the uique iteger i {0, 1,..., 1} so that a i a i 1 + d i (mod. Claim 4.4: Let P (x = 1 + x + + x 1, ad defie b k such that P (x x j = b k x k. i=1 k 0 j f(i The the umber of valid sequeces {d i } i=1 is j 0 b j. 13
14 Proof. Note that P (x j f(i xj cotais oly powers of x whose expoets are allowable values for d i. Take a term i the expasio of the left-had-side, suppose it takes the term x d i from P (x j f(i xj. The the term is x d 1+d + +d. Now {d i } i=1 is a valid sequece iff d 1 + d + + d 0 (mod. Hece summig the coefficiets of x k for k gives the umber of valid sequeces. Combiig this with the previous claim, the umber of o-f sequeces i j 0 b j. (a A good sequece is exactly a o-f sequece for f(i = {i mod }. Let Q(x = P (x 1 [ = ] P (x x i. Now i=1 j f(i x j i=0 1 1 Q(1 = (P (1 1 = ( 1 = ( 1 i=0 ad for ay ω 1 a th root of uity, i=0 1 Q(ω = (P (ω ω i i=0 1 = ( ω i i=0 = ω ( 1 = 1 sice the fact that is odd gives ( 1. Now Q(x + 1 is zero for all th roots of uity ω 1, so is divisible by P (x. Writig Q(x + 1 = k 0 c kx k, the fact that P (x Q(x gives that the sums S r = k r (mod c k are all equal (fill i the details basically ay term times P (x is spread amog terms whose expoets make up a complete residue class modulo. Hece, sice the sum of coefficiets of Q(x + 1 is Q(1 + 1, k 0 c k = Q(1 + 1 = ( The the desired sum for Q(x is ( 1 +1 are ( good sequeces. 1 ad multiplyig by we get there (b A great sequece is a o-f sequece for f(i = {i mod, i mod }. The calculatios are left to the reader. The followig lemma will be useful below. Lemma 4.5: If ω 1 is pth root of uity, ad a 0,..., a p 1 are ratioal umbers such that a 0 + a 1 ω + + a p 1 ω p 1 = 0 the a 0 = = a p 1. This lemma follows from the fact that 1 + x x p 1 is the irreducible (miimal polyomial of ω, which we will prove i a later lecture. 14
15 4.1 Problem set 4 1. [3] Cosider a rectagle which ca be tiled with a fiite combiatio of 1 m or 1 rectagles, where m ad are atural umbers. (The tiles caot be rotated. Prove that it is possible to tile this rectagle with oly 1 m or oly with 1 rectagles.. [4] TST 004/b above. 3. [4, with the above preparatio] (IMO 1995/6 Let p > be a prime umber ad A = {1,,..., p}. Fid the umber of subsets of A, havig p elemets ad with sum of elemets divisible by p. Refereces [1] Adreescu, T.; Dospiescu, G.: Problems from The Book. XYZ Press, Alle TX,
Complex Numbers Solutions
Complex Numbers Solutios Joseph Zoller February 7, 06 Solutios. (009 AIME I Problem ) There is a complex umber with imagiary part 64 ad a positive iteger such that Fid. [Solutio: 697] 4i + + 4i. 4i 4i
More informationAPPENDIX F Complex Numbers
APPENDIX F Complex Numbers Operatios with Complex Numbers Complex Solutios of Quadratic Equatios Polar Form of a Complex Number Powers ad Roots of Complex Numbers Operatios with Complex Numbers Some equatios
More informationAppendix F: Complex Numbers
Appedix F Complex Numbers F1 Appedix F: Complex Numbers Use the imagiary uit i to write complex umbers, ad to add, subtract, ad multiply complex umbers. Fid complex solutios of quadratic equatios. Write
More informationComplex Numbers. Brief Notes. z = a + bi
Defiitios Complex Numbers Brief Notes A complex umber z is a expressio of the form: z = a + bi where a ad b are real umbers ad i is thought of as 1. We call a the real part of z, writte Re(z), ad b the
More informationStanford Math Circle January 21, Complex Numbers
Staford Math Circle Jauary, 007 Some History Tatiaa Shubi (shubi@mathsjsuedu) Complex Numbers Let us try to solve the equatio x = 5x + x = is a obvious solutio Also, x 5x = ( x )( x + x + ) = 0 yields
More informationCALCULUS BASIC SUMMER REVIEW
CALCULUS BASIC SUMMER REVIEW NAME rise y y y Slope of a o vertical lie: m ru Poit Slope Equatio: y y m( ) The slope is m ad a poit o your lie is, ). ( y Slope-Itercept Equatio: y m b slope= m y-itercept=
More informationPresentation of complex number in Cartesian and polar coordinate system
a + bi, aεr, bεr i = z = a + bi a = Re(z), b = Im(z) give z = a + bi & w = c + di, a + bi = c + di a = c & b = d The complex cojugate of z = a + bi is z = a bi The sum of complex cojugates is real: z +
More information2 Geometric interpretation of complex numbers
2 Geometric iterpretatio of complex umbers 2.1 Defiitio I will start fially with a precise defiitio, assumig that such mathematical object as vector space R 2 is well familiar to the studets. Recall that
More information3.2 Properties of Division 3.3 Zeros of Polynomials 3.4 Complex and Rational Zeros of Polynomials
Math 60 www.timetodare.com 3. Properties of Divisio 3.3 Zeros of Polyomials 3.4 Complex ad Ratioal Zeros of Polyomials I these sectios we will study polyomials algebraically. Most of our work will be cocered
More informationMath 155 (Lecture 3)
Math 55 (Lecture 3) September 8, I this lecture, we ll cosider the aswer to oe of the most basic coutig problems i combiatorics Questio How may ways are there to choose a -elemet subset of the set {,,,
More informationANSWERS SOLUTIONS iiii i. and 1. Thus, we have. i i i. i, A.
013 ΜΑΘ Natioal Covetio ANSWERS (1) C A A A B (6) B D D A B (11) C D D A A (16) D B A A C (1) D B C B C (6) D C B C C 1. We have SOLUTIONS 1 3 11 61 iiii 131161 i 013 013, C.. The powers of i cycle betwee
More informationC. Complex Numbers. x 6x + 2 = 0. This equation was known to have three real roots, given by simple combinations of the expressions
C. Complex Numbers. Complex arithmetic. Most people thik that complex umbers arose from attempts to solve quadratic equatios, but actually it was i coectio with cubic equatios they first appeared. Everyoe
More informationPUTNAM TRAINING, 2008 COMPLEX NUMBERS
PUTNAM TRAINING, 008 COMPLEX NUMBERS (Last updated: December 11, 017) Remark. This is a list of exercises o Complex Numbers Miguel A. Lerma Exercises 1. Let m ad two itegers such that each ca be expressed
More informationTEACHER CERTIFICATION STUDY GUIDE
COMPETENCY 1. ALGEBRA SKILL 1.1 1.1a. ALGEBRAIC STRUCTURES Kow why the real ad complex umbers are each a field, ad that particular rigs are ot fields (e.g., itegers, polyomial rigs, matrix rigs) Algebra
More informationMATH 304: MIDTERM EXAM SOLUTIONS
MATH 304: MIDTERM EXAM SOLUTIONS [The problems are each worth five poits, except for problem 8, which is worth 8 poits. Thus there are 43 possible poits.] 1. Use the Euclidea algorithm to fid the greatest
More informationWe are mainly going to be concerned with power series in x, such as. (x)} converges - that is, lims N n
Review of Power Series, Power Series Solutios A power series i x - a is a ifiite series of the form c (x a) =c +c (x a)+(x a) +... We also call this a power series cetered at a. Ex. (x+) is cetered at
More informationMath 451: Euclidean and Non-Euclidean Geometry MWF 3pm, Gasson 204 Homework 3 Solutions
Math 451: Euclidea ad No-Euclidea Geometry MWF 3pm, Gasso 204 Homework 3 Solutios Exercises from 1.4 ad 1.5 of the otes: 4.3, 4.10, 4.12, 4.14, 4.15, 5.3, 5.4, 5.5 Exercise 4.3. Explai why Hp, q) = {x
More informationMAT 271 Project: Partial Fractions for certain rational functions
MAT 7 Project: Partial Fractios for certai ratioal fuctios Prerequisite kowledge: partial fractios from MAT 7, a very good commad of factorig ad complex umbers from Precalculus. To complete this project,
More informationSeunghee Ye Ma 8: Week 5 Oct 28
Week 5 Summary I Sectio, we go over the Mea Value Theorem ad its applicatios. I Sectio 2, we will recap what we have covered so far this term. Topics Page Mea Value Theorem. Applicatios of the Mea Value
More information[ 11 ] z of degree 2 as both degree 2 each. The degree of a polynomial in n variables is the maximum of the degrees of its terms.
[ 11 ] 1 1.1 Polyomial Fuctios 1 Algebra Ay fuctio f ( x) ax a1x... a1x a0 is a polyomial fuctio if ai ( i 0,1,,,..., ) is a costat which belogs to the set of real umbers ad the idices,, 1,...,1 are atural
More information(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3
MATH 337 Sequeces Dr. Neal, WKU Let X be a metric space with distace fuctio d. We shall defie the geeral cocept of sequece ad limit i a metric space, the apply the results i particular to some special
More informationChapter 13: Complex Numbers
Sectios 13.1 & 13.2 Comple umbers ad comple plae Comple cojugate Modulus of a comple umber 1. Comple umbers Comple umbers are of the form z = + iy,, y R, i 2 = 1. I the above defiitio, is the real part
More informationThe z-transform. 7.1 Introduction. 7.2 The z-transform Derivation of the z-transform: x[n] = z n LTI system, h[n] z = re j
The -Trasform 7. Itroductio Geeralie the complex siusoidal represetatio offered by DTFT to a represetatio of complex expoetial sigals. Obtai more geeral characteristics for discrete-time LTI systems. 7.
More informationPatterns in Complex Numbers An analytical paper on the roots of a complex numbers and its geometry
IB MATHS HL POTFOLIO TYPE Patters i Complex Numbers A aalytical paper o the roots of a complex umbers ad its geometry i Syed Tousif Ahmed Cadidate Sessio Number: 0066-009 School Code: 0066 Sessio: May
More informationThe natural exponential function
The atural expoetial fuctio Attila Máté Brookly College of the City Uiversity of New York December, 205 Cotets The atural expoetial fuctio for real x. Beroulli s iequality.....................................2
More information6 Integers Modulo n. integer k can be written as k = qn + r, with q,r, 0 r b. So any integer.
6 Itegers Modulo I Example 2.3(e), we have defied the cogruece of two itegers a,b with respect to a modulus. Let us recall that a b (mod ) meas a b. We have proved that cogruece is a equivalece relatio
More informationMATH 31B: MIDTERM 2 REVIEW
MATH 3B: MIDTERM REVIEW JOE HUGHES. Evaluate x (x ) (x 3).. Partial Fractios Solutio: The umerator has degree less tha the deomiator, so we ca use partial fractios. Write x (x ) (x 3) = A x + A (x ) +
More informationMath 220A Fall 2007 Homework #2. Will Garner A
Math 0A Fall 007 Homewor # Will Garer Pg 3 #: Show that {cis : a o-egative iteger} is dese i T = {z œ : z = }. For which values of q is {cis(q): a o-egative iteger} dese i T? To show that {cis : a o-egative
More informationHomework 3. = k 1. Let S be a set of n elements, and let a, b, c be distinct elements of S. The number of k-subsets of S is
Homewor 3 Chapter 5 pp53: 3 40 45 Chapter 6 p85: 4 6 4 30 Use combiatorial reasoig to prove the idetity 3 3 Proof Let S be a set of elemets ad let a b c be distict elemets of S The umber of -subsets of
More informationCSE 1400 Applied Discrete Mathematics Number Theory and Proofs
CSE 1400 Applied Discrete Mathematics Number Theory ad Proofs Departmet of Computer Scieces College of Egieerig Florida Tech Sprig 01 Problems for Number Theory Backgroud Number theory is the brach of
More informationInfinite Sequences and Series
Chapter 6 Ifiite Sequeces ad Series 6.1 Ifiite Sequeces 6.1.1 Elemetary Cocepts Simply speakig, a sequece is a ordered list of umbers writte: {a 1, a 2, a 3,...a, a +1,...} where the elemets a i represet
More informationRADICAL EXPRESSION. If a and x are real numbers and n is a positive integer, then x is an. n th root theorems: Example 1 Simplify
Example 1 Simplify 1.2A Radical Operatios a) 4 2 b) 16 1 2 c) 16 d) 2 e) 8 1 f) 8 What is the relatioship betwee a, b, c? What is the relatioship betwee d, e, f? If x = a, the x = = th root theorems: RADICAL
More informationUnit 4: Polynomial and Rational Functions
48 Uit 4: Polyomial ad Ratioal Fuctios Polyomial Fuctios A polyomial fuctio y px ( ) is a fuctio of the form p( x) ax + a x + a x +... + ax + ax+ a 1 1 1 0 where a, a 1,..., a, a1, a0are real costats ad
More informationZeros of Polynomials
Math 160 www.timetodare.com 4.5 4.6 Zeros of Polyomials I these sectios we will study polyomials algebraically. Most of our work will be cocered with fidig the solutios of polyomial equatios of ay degree
More informationChapter 1. Complex Numbers. Dr. Pulak Sahoo
Chapter 1 Complex Numbers BY Dr. Pulak Sahoo Assistat Professor Departmet of Mathematics Uiversity Of Kalyai West Begal, Idia E-mail : sahoopulak1@gmail.com 1 Module-2: Stereographic Projectio 1 Euler
More informationSolutions to Homework 1
Solutios to Homework MATH 36. Describe geometrically the sets of poits z i the complex plae defied by the followig relatios /z = z () Re(az + b) >, where a, b (2) Im(z) = c, with c (3) () = = z z = z 2.
More informationINEQUALITIES BJORN POONEN
INEQUALITIES BJORN POONEN 1 The AM-GM iequality The most basic arithmetic mea-geometric mea (AM-GM) iequality states simply that if x ad y are oegative real umbers, the (x + y)/2 xy, with equality if ad
More informationWHEN IS THE (CO)SINE OF A RATIONAL ANGLE EQUAL TO A RATIONAL NUMBER? 1. My Motivation Some Sort of an Introduction
WHEN IS THE (CO)SINE OF A RATIONAL ANGLE EQUAL TO A RATIONAL NUMBER? JÖRG JAHNEL 1. My Motivatio Some Sort of a Itroductio Last term I taught Topological Groups at the Göttige Georg August Uiversity. This
More informationPUTNAM TRAINING INEQUALITIES
PUTNAM TRAINING INEQUALITIES (Last updated: December, 207) Remark This is a list of exercises o iequalities Miguel A Lerma Exercises If a, b, c > 0, prove that (a 2 b + b 2 c + c 2 a)(ab 2 + bc 2 + ca
More informationComplex Analysis Spring 2001 Homework I Solution
Complex Aalysis Sprig 2001 Homework I Solutio 1. Coway, Chapter 1, sectio 3, problem 3. Describe the set of poits satisfyig the equatio z a z + a = 2c, where c > 0 ad a R. To begi, we see from the triagle
More informationSeptember 2012 C1 Note. C1 Notes (Edexcel) Copyright - For AS, A2 notes and IGCSE / GCSE worksheets 1
September 0 s (Edecel) Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets September 0 Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets September 0 Copyright
More informationSolutions to Math 347 Practice Problems for the final
Solutios to Math 347 Practice Problems for the fial 1) True or False: a) There exist itegers x,y such that 50x + 76y = 6. True: the gcd of 50 ad 76 is, ad 6 is a multiple of. b) The ifiimum of a set is
More informationLemma Let f(x) K[x] be a separable polynomial of degree n. Then the Galois group is a subgroup of S n, the permutations of the roots.
15 Cubics, Quartics ad Polygos It is iterestig to chase through the argumets of 14 ad see how this affects solvig polyomial equatios i specific examples We make a global assumptio that the characteristic
More informationREVISION SHEET FP1 (MEI) ALGEBRA. Identities In mathematics, an identity is a statement which is true for all values of the variables it contains.
The mai ideas are: Idetities REVISION SHEET FP (MEI) ALGEBRA Before the exam you should kow: If a expressio is a idetity the it is true for all values of the variable it cotais The relatioships betwee
More informationElementary Algebra and Geometry
1 Elemetary Algera ad Geometry 1.1 Fudametal Properties (Real Numers) a + = + a Commutative Law for Additio (a + ) + c = a + ( + c) Associative Law for Additio a + 0 = 0 + a Idetity Law for Additio a +
More informationProperties and Tests of Zeros of Polynomial Functions
Properties ad Tests of Zeros of Polyomial Fuctios The Remaider ad Factor Theorems: Sythetic divisio ca be used to fid the values of polyomials i a sometimes easier way tha substitutio. This is show by
More informationIn number theory we will generally be working with integers, though occasionally fractions and irrationals will come into play.
Number Theory Math 5840 otes. Sectio 1: Axioms. I umber theory we will geerally be workig with itegers, though occasioally fractios ad irratioals will come ito play. Notatio: Z deotes the set of all itegers
More informationSolutions for May. 3 x + 7 = 4 x x +
Solutios for May 493. Prove that there is a atural umber with the followig characteristics: a) it is a multiple of 007; b) the first four digits i its decimal represetatio are 009; c) the last four digits
More informationProblem Cosider the curve give parametrically as x = si t ad y = + cos t for» t» ß: (a) Describe the path this traverses: Where does it start (whe t =
Mathematics Summer Wilso Fial Exam August 8, ANSWERS Problem 1 (a) Fid the solutio to y +x y = e x x that satisfies y() = 5 : This is already i the form we used for a first order liear differetial equatio,
More informationChapter 2. Periodic points of toral. automorphisms. 2.1 General introduction
Chapter 2 Periodic poits of toral automorphisms 2.1 Geeral itroductio The automorphisms of the two-dimesioal torus are rich mathematical objects possessig iterestig geometric, algebraic, topological ad
More informationMathematics Extension 1
016 Bored of Studies Trial Eamiatios Mathematics Etesio 1 3 rd ctober 016 Geeral Istructios Total Marks 70 Readig time 5 miutes Workig time hours Write usig black or blue pe Black pe is preferred Board-approved
More informationThe Advantage Testing Foundation Solutions
The Advatage Testig Foudatio 202 Problem I the morig, Esther biked from home to school at a average speed of x miles per hour. I the afteroo, havig let her bike to a fried, Esther walked back home alog
More informationP.3 Polynomials and Special products
Precalc Fall 2016 Sectios P.3, 1.2, 1.3, P.4, 1.4, P.2 (radicals/ratioal expoets), 1.5, 1.6, 1.7, 1.8, 1.1, 2.1, 2.2 I Polyomial defiitio (p. 28) a x + a x +... + a x + a x 1 1 0 1 1 0 a x + a x +... +
More information1. By using truth tables prove that, for all statements P and Q, the statement
Author: Satiago Salazar Problems I: Mathematical Statemets ad Proofs. By usig truth tables prove that, for all statemets P ad Q, the statemet P Q ad its cotrapositive ot Q (ot P) are equivalet. I example.2.3
More informationMAT1026 Calculus II Basic Convergence Tests for Series
MAT026 Calculus II Basic Covergece Tests for Series Egi MERMUT 202.03.08 Dokuz Eylül Uiversity Faculty of Sciece Departmet of Mathematics İzmir/TURKEY Cotets Mootoe Covergece Theorem 2 2 Series of Real
More information( ) 2 + k The vertex is ( h, k) ( )( x q) The x-intercepts are x = p and x = q.
A Referece Sheet Number Sets Quadratic Fuctios Forms Form Equatio Stadard Form Vertex Form Itercept Form y ax + bx + c The x-coordiate of the vertex is x b a y a x h The axis of symmetry is x b a + k The
More information11. FINITE FIELDS. Example 1: The following tables define addition and multiplication for a field of order 4.
11. FINITE FIELDS 11.1. A Field With 4 Elemets Probably the oly fiite fields which you ll kow about at this stage are the fields of itegers modulo a prime p, deoted by Z p. But there are others. Now although
More information6.003 Homework #3 Solutions
6.00 Homework # Solutios Problems. Complex umbers a. Evaluate the real ad imagiary parts of j j. π/ Real part = Imagiary part = 0 e Euler s formula says that j = e jπ/, so jπ/ j π/ j j = e = e. Thus the
More informationSolutions to Problem Set 8
8.78 Solutios to Problem Set 8. We ow that ( ) ( + x) x. Now we plug i x, ω, ω ad add the three equatios. If 3 the we ll get a cotributio of + ω + ω + ω + ω 0, whereas if 3 we ll get a cotributio of +
More informationLesson 10: Limits and Continuity
www.scimsacademy.com Lesso 10: Limits ad Cotiuity SCIMS Academy 1 Limit of a fuctio The cocept of limit of a fuctio is cetral to all other cocepts i calculus (like cotiuity, derivative, defiite itegrals
More informationSection 10.3 The Complex Plane; De Moivre's Theorem. abi
Sectio 03 The Complex Plae; De Moivre's Theorem REVIEW OF COMPLEX NUMBERS FROM COLLEGE ALGEBRA You leared about complex umbers of the form a + bi i your college algebra class You should remember that "i"
More informationDe Moivre s Theorem - ALL
De Moivre s Theorem - ALL. Let x ad y be real umbers, ad be oe of the complex solutios of the equatio =. Evaluate: (a) + + ; (b) ( x + y)( x + y). [6]. (a) Sice is a complex umber which satisfies = 0,.
More informationMath 299 Supplement: Real Analysis Nov 2013
Math 299 Supplemet: Real Aalysis Nov 203 Algebra Axioms. I Real Aalysis, we work withi the axiomatic system of real umbers: the set R alog with the additio ad multiplicatio operatios +,, ad the iequality
More informationLESSON 2: SIMPLIFYING RADICALS
High School: Workig with Epressios LESSON : SIMPLIFYING RADICALS N.RN.. C N.RN.. B 5 5 C t t t t t E a b a a b N.RN.. 4 6 N.RN. 4. N.RN. 5. N.RN. 6. 7 8 N.RN. 7. A 7 N.RN. 8. 6 80 448 4 5 6 48 00 6 6 6
More informationMath 113 Exam 3 Practice
Math Exam Practice Exam 4 will cover.-., 0. ad 0.. Note that eve though. was tested i exam, questios from that sectios may also be o this exam. For practice problems o., refer to the last review. This
More informationREVISION SHEET FP1 (MEI) ALGEBRA. Identities In mathematics, an identity is a statement which is true for all values of the variables it contains.
the Further Mathematics etwork wwwfmetworkorguk V 07 The mai ideas are: Idetities REVISION SHEET FP (MEI) ALGEBRA Before the exam you should kow: If a expressio is a idetity the it is true for all values
More informationBeurling Integers: Part 2
Beurlig Itegers: Part 2 Isomorphisms Devi Platt July 11, 2015 1 Prime Factorizatio Sequeces I the last article we itroduced the Beurlig geeralized itegers, which ca be represeted as a sequece of real umbers
More informationSEQUENCE AND SERIES NCERT
9. Overview By a sequece, we mea a arragemet of umbers i a defiite order accordig to some rule. We deote the terms of a sequece by a, a,..., etc., the subscript deotes the positio of the term. I view of
More informationThe Binomial Theorem
The Biomial Theorem Robert Marti Itroductio The Biomial Theorem is used to expad biomials, that is, brackets cosistig of two distict terms The formula for the Biomial Theorem is as follows: (a + b ( k
More information4 The Sperner property.
4 The Sperer property. I this sectio we cosider a surprisig applicatio of certai adjacecy matrices to some problems i extremal set theory. A importat role will also be played by fiite groups. I geeral,
More informationR is a scalar defined as follows:
Math 8. Notes o Dot Product, Cross Product, Plaes, Area, ad Volumes This lecture focuses primarily o the dot product ad its may applicatios, especially i the measuremet of agles ad scalar projectio ad
More informationFall 2013 MTH431/531 Real analysis Section Notes
Fall 013 MTH431/531 Real aalysis Sectio 8.1-8. Notes Yi Su 013.11.1 1. Defiitio of uiform covergece. We look at a sequece of fuctios f (x) ad study the coverget property. Notice we have two parameters
More informationNAME: ALGEBRA 350 BLOCK 7. Simplifying Radicals Packet PART 1: ROOTS
NAME: ALGEBRA 50 BLOCK 7 DATE: Simplifyig Radicals Packet PART 1: ROOTS READ: A square root of a umber b is a solutio of the equatio x = b. Every positive umber b has two square roots, deoted b ad b or
More informationAlgebra II Notes Unit Seven: Powers, Roots, and Radicals
Syllabus Objectives: 7. The studets will use properties of ratioal epoets to simplify ad evaluate epressios. 7.8 The studet will solve equatios cotaiig radicals or ratioal epoets. b a, the b is the radical.
More information4.1 Sigma Notation and Riemann Sums
0 the itegral. Sigma Notatio ad Riema Sums Oe strategy for calculatig the area of a regio is to cut the regio ito simple shapes, calculate the area of each simple shape, ad the add these smaller areas
More informationThe Structure of Z p when p is Prime
LECTURE 13 The Structure of Z p whe p is Prime Theorem 131 If p > 1 is a iteger, the the followig properties are equivalet (1) p is prime (2) For ay [0] p i Z p, the equatio X = [1] p has a solutio i Z
More informationComplex Number Theory without Imaginary Number (i)
Ope Access Library Joural Complex Number Theory without Imagiary Number (i Deepak Bhalchadra Gode Directorate of Cesus Operatios, Mumbai, Idia Email: deepakm_4@rediffmail.com Received 6 July 04; revised
More informationBertrand s Postulate
Bertrad s Postulate Lola Thompso Ross Program July 3, 2009 Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 1 / 33 Bertrad s Postulate I ve said it oce ad I ll say it agai: There s always a
More information62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 +
62. Power series Defiitio 16. (Power series) Give a sequece {c }, the series c x = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + is called a power series i the variable x. The umbers c are called the coefficiets of
More informationFLC Ch 8 & 9. Evaluate. Check work. a) b) c) d) e) f) g) h) i) j) k) l) m) n) o) 3. p) q) r) s) t) 3.
Math 100 Elemetary Algebra Sec 8.1: Radical Expressios List perfect squares ad evaluate their square root. Kow these perfect squares for test. Def The positive (pricipal) square root of x, writte x, is
More informationIn algebra one spends much time finding common denominators and thus simplifying rational expressions. For example:
74 The Method of Partial Fractios I algebra oe speds much time fidig commo deomiators ad thus simplifyig ratioal epressios For eample: + + + 6 5 + = + = = + + + + + ( )( ) 5 It may the seem odd to be watig
More informationMath 61CM - Solutions to homework 3
Math 6CM - Solutios to homework 3 Cédric De Groote October 2 th, 208 Problem : Let F be a field, m 0 a fixed oegative iteger ad let V = {a 0 + a x + + a m x m a 0,, a m F} be the vector space cosistig
More informationHomework 2 January 19, 2006 Math 522. Direction: This homework is due on January 26, In order to receive full credit, answer
Homework 2 Jauary 9, 26 Math 522 Directio: This homework is due o Jauary 26, 26. I order to receive full credit, aswer each problem completely ad must show all work.. What is the set of the uits (that
More information18th Bay Area Mathematical Olympiad. Problems and Solutions. February 23, 2016
18th Bay Area Mathematical Olympiad February 3, 016 Problems ad Solutios BAMO-8 ad BAMO-1 are each 5-questio essay-proof exams, for middle- ad high-school studets, respectively. The problems i each exam
More informationMath 4107: Abstract Algebra I Fall Webwork Assignment1-Groups (5 parts/problems) Solutions are on Webwork.
Math 4107: Abstract Algebra I Fall 2017 Assigmet 1 Solutios 1. Webwork Assigmet1-Groups 5 parts/problems) Solutios are o Webwork. 2. Webwork Assigmet1-Subgroups 5 parts/problems) Solutios are o Webwork.
More informationMATH 205 HOMEWORK #2 OFFICIAL SOLUTION. (f + g)(x) = f(x) + g(x) = f( x) g( x) = (f + g)( x)
MATH 205 HOMEWORK #2 OFFICIAL SOLUTION Problem 2: Do problems 7-9 o page 40 of Hoffma & Kuze. (7) We will prove this by cotradictio. Suppose that W 1 is ot cotaied i W 2 ad W 2 is ot cotaied i W 1. The
More informationlim za n n = z lim a n n.
Lecture 6 Sequeces ad Series Defiitio 1 By a sequece i a set A, we mea a mappig f : N A. It is customary to deote a sequece f by {s } where, s := f(). A sequece {z } of (complex) umbers is said to be coverget
More informationChapter 4. Fourier Series
Chapter 4. Fourier Series At this poit we are ready to ow cosider the caoical equatios. Cosider, for eample the heat equatio u t = u, < (4.) subject to u(, ) = si, u(, t) = u(, t) =. (4.) Here,
More informationG r a d e 1 1 P r e - C a l c u l u s M a t h e m a t i c s ( 3 0 S )
G r a d e 1 1 P r e - C a l c u l u s M a t h e m a t i c s ( 3 0 S ) Grade 11 Pre-Calculus Mathematics (30S) is desiged for studets who ited to study calculus ad related mathematics as part of post-secodary
More informationAddition: Property Name Property Description Examples. a+b = b+a. a+(b+c) = (a+b)+c
Notes for March 31 Fields: A field is a set of umbers with two (biary) operatios (usually called additio [+] ad multiplicatio [ ]) such that the followig properties hold: Additio: Name Descriptio Commutativity
More informationEnd-of-Year Contest. ERHS Math Club. May 5, 2009
Ed-of-Year Cotest ERHS Math Club May 5, 009 Problem 1: There are 9 cois. Oe is fake ad weighs a little less tha the others. Fid the fake coi by weighigs. Solutio: Separate the 9 cois ito 3 groups (A, B,
More informationBasic Sets. Functions. MTH299 - Examples. Example 1. Let S = {1, {2, 3}, 4}. Indicate whether each statement is true or false. (a) S = 4. (e) 2 S.
Basic Sets Example 1. Let S = {1, {2, 3}, 4}. Idicate whether each statemet is true or false. (a) S = 4 (b) {1} S (c) {2, 3} S (d) {1, 4} S (e) 2 S. (f) S = {1, 4, {2, 3}} (g) S Example 2. Compute the
More informationLecture 23 Rearrangement Inequality
Lecture 23 Rearragemet Iequality Holde Lee 6/4/ The Iequalities We start with a example Suppose there are four boxes cotaiig $0, $20, $50 ad $00 bills, respectively You may take 2 bills from oe box, 3
More informationPHYSICS 116A Homework 2 Solutions
PHYSICS 6A Homework 2 Solutios I. [optioal] Boas, Ch., 6, Qu. 30 (proof of the ratio test). Just follow the hits. If ρ, the ratio of succcessive terms for is less tha, the hits show that the terms of the
More information3. Z Transform. Recall that the Fourier transform (FT) of a DT signal xn [ ] is ( ) [ ] = In order for the FT to exist in the finite magnitude sense,
3. Z Trasform Referece: Etire Chapter 3 of text. Recall that the Fourier trasform (FT) of a DT sigal x [ ] is ω ( ) [ ] X e = j jω k = xe I order for the FT to exist i the fiite magitude sese, S = x [
More informationj=1 dz Res(f, z j ) = 1 d k 1 dz k 1 (z c)k f(z) Res(f, c) = lim z c (k 1)! Res g, c = f(c) g (c)
Problem. Compute the itegrals C r d for Z, where C r = ad r >. Recall that C r has the couter-clockwise orietatio. Solutio: We will use the idue Theorem to solve this oe. We could istead use other (perhaps
More information1. Complex numbers. Chapter 13: Complex Numbers. Modulus of a complex number. Complex conjugate. Complex numbers are of the form
Comple umbers ad comple plae Comple cojugate Modulus of a comple umber Comple umbers Comple umbers are of the form Sectios 3 & 32 z = + i,, R, i 2 = I the above defiitio, is the real part of z ad is the
More information6.3 Testing Series With Positive Terms
6.3. TESTING SERIES WITH POSITIVE TERMS 307 6.3 Testig Series With Positive Terms 6.3. Review of what is kow up to ow I theory, testig a series a i for covergece amouts to fidig the i= sequece of partial
More informationMATH 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 informationsubcaptionfont+=small,labelformat=parens,labelsep=space,skip=6pt,list=0,hypcap=0 subcaption ALGEBRAIC COMBINATORICS LECTURE 8 TUESDAY, 2/16/2016
subcaptiofot+=small,labelformat=pares,labelsep=space,skip=6pt,list=0,hypcap=0 subcaptio ALGEBRAIC COMBINATORICS LECTURE 8 TUESDAY, /6/06. Self-cojugate Partitios Recall that, give a partitio λ, we may
More information