STUDENT S COMPANIONS IN BASIC MATH: THE SECOND. Basic Identities in Algebra. Let us start with a basic identity in algebra:


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1 STUDENT S COMPANIONS IN BASIC MATH: THE SECOND Bsic Idetities i Algebr Let us strt with bsic idetity i lgebr: 2 b 2 ( b( + b. (1 Ideed, multiplyig out the right hd side, we get 2 +b b b 2. Removig the mixed terms, which ccel ech other, we get the left hd side. QUESTION 1. Wht is the result of multiplyig out ( (b 1 + b 2? QUESTION 2. Sme questio for (x (x b. QUESTION 3. Sme questio for ( (b 1 + b 2. QUESTION 4. Sme questio for ( (b 1 + b 2 (c 1 + c 2. Similr to (1, we hve, 3 b 3 ( b( 2 + b + b 2. EXERCISE 5. Check the lst idetity. I geerl, for y positive iteger, we hve b ( b( b + + b 2 + b 1 (2 QUESTION 6. Wht does (2 give for 4, 5, 6? EXERCISE 7. Prove (2 by multiplyig out its right hd side. Switchig to + 1, (2 becomes +1 b +1 ( b( + 1 b + + b 1 + b. Usig the summtio ottio, we write + 1 b + + b 1 + b So we get other form of (2: s k0 k b k. k0 k b k +1 b +1 b 1. (3
2 Lettig 1 d b x, (3 becomes k0 xk 1 + x + x x 1 x+1 1 x, (4 This idetity tells us how to fid the sum of geometric progressio. It should be memorized. Notice tht, whe x < 1, x +1 coverges to zero s d hece the right hd side of (4 coverges to 1/(1 x. Thus we hve for x < 1. 0 x 1 + x + x x, EXERCISE 8. Check the followig idetities 3 + b 3 ( + b( 2 b + b b 5 ( + b( 4 3 b + 2 b 2 b 3 + b 4. Mke guess of similr idetity for + b where is odd. EXERCISE 9. Show tht, whe is odd, + b ( + b( 1 2 b b 3 b 2 + b 1 (5 (Hit: Replce b by b i (2. Idetities (2 d (5 re ofte used i fctoriztio. EXAMPLE 10. We re sked to fctorize 12 b 12. We should mke wise choice i our first step i order ot to jeoprdize the rest of our work. Here is out first step: 12 b 12 ( 6 2 (b 6 2 ( 6 b 6 ( 6 + b 6. Now we tret 6 b 6 d 6 + b 6 seprtely: 6 b 6 ( 3 2 (b 3 2 ( 3 b 3 ( 3 + b 3 ( b( 2 + b + b 2 ( + b( 2 b + b 2. d 6 + b 6 ( (b 2 3 ( 2 + b 2 ( 4 2 b 2 + b 4. 2
3 (If we ccept the pperce of irrtiol umbers, we c go further by completig squre : 4 2 b 2 + b b 2 + b b 2 ( 2 + b 2 2 ( 3 b 2 ( 2 + b 2 3 b( 2 + b b. But let s ot tke this step. So, filly, 12 b 12 ( b( + b( 2 + b 2 ( 2 + b + b 2 ( 2 b + b 2 ( 4 2 b 2 + b 4. EXERCISE 11. Fctorize b 2 + b 4. (Hit: Rewrite it s b 2 + b 4 2 b 2. QUESTION 12. From (5 we see tht, whe is odd, + b is fctor of + b. Whe is eve, why is + b ot fctor of + b? EXERCISE 13. Fctorize ( + b 5 5 b 5 by extrctig fctor of + b first. Our ext topic is the biomil formul. Let s begi with ( + b b + b 2 ( + b b + 3b 2 + b 3. ( + b b b 2 + 4b 3 + b 4. EXERCISE 14. Verify the bove idetities. EXERCISE 15. Verify ( + b + c b 2 + c 2 + 2b + 2c + 2bc. Estblish similr idetity for ( + b + c + d 2. For y positive iteger, we hve ( + b ( k k0 + k b k ( 1 1 b + ( ( 2 b b k 1 + b, 2 1 (6 where ( k! k!(! with k! (k 1 k (7 3
4 d, by covetio 0! 1. This is clled the biomil formul. We give very brief resoig isted of proof. (For complete proof, see the EIGHTH COMPANION. The expressio ( is red s choose k. It stds for the umber of wys of choosig k objects from give objects. Now, multiplyig out ( + b, we get 2 terms. Ech of them is product of the form c 1 c 2 c, where ech fctor c j (1 j is either or b. If there re exctly k fctors equl to b, the ll of the rest re equl to d hece c 1 c 2 c becomes k b k. Let us fix some k. Amog these 2 terms, how my re equl to k b k? Well, it is the umber of wys of ssigig k fctors i c 1 c 2 c to b d ssigig the rest to. So there re exctly ( of 2 terms equl to k b k. For ech k, collect ll terms of the form k b k i the sum of 2 terms obtied by multiplyig out ( + b. The you get (6. About choose k (, the followig two idetities re bsic: ( k (, k ( + k 1 ( k ( + 1 k. (8 (The secod idetity tells us how to costruct the Pscl trigle. To see the vlidity of the first idetity, let us cosider the umber of wys to divide objects betwee Liz d Lrry, k of them for Liz d k of them for Lrry. We c let Liz to choose her k objects first. There re ( ( wys to do so. This tells us tht there re wys to divide the objects. If we let Lrry choose first, we hve ( ( wys. Sice both d ( c represet the umber of wys of dividig objects betwee Liz d Lrry, these two umbers must be equl. QUESTION 16. Cosider the umber of wys of choosig k objects from + 1 objects, which is equl to ( +1 k. Tke y oe from these + 1 objects d put tg o it. ( Cosider two groups of +1 k choices: the first group cosists of choices icludig the tgged oe d the secod group cosists of choices excludig it. How my choices re i ech group? Now, do you see why the secod idetity i (8 is vlid? QUESTION 17. Why is the idetity (k + 1! k! (k + 1 vlid? EXERCISE 18. Verify tht ( give i (7 ideed stisfies the idetities i (8. (You eed the idetity metioed i the QUESTION 17. 4
5 We metio the followig multiomil formul: ( p k 1 +k 2 + +k p where ( k 1 k 2 k p ( k 1 k 2 k p! k 1!, k 2! k p!. k 1 1 k 2 2 k p p, (You will be sked to prove this by your SEVENTH COMPANION s problem. Occsiolly we ecouter some simple but truly beutiful idetities; for exmple ( 2 + b 2 (c 2 + d 2 (c + bd 2 + (d bc 2. (9 This is very esy to verify. (A expressio of the form 2 + b 2 is clled sum of two squres for obvious reso. Idetity (9 sys the product of sums of two squres is lso sum of two squres. There re similr but much more itriguig idetities bout sums of four squres d sums of eight squres. The questio is, how c oe cojure up such et idetity? Or, wht is behid this ispirig someoe to write this dow? (See the SEVENTH COMPANION o complex umbers to fid the swer. EXERCISE 19. Check (9. Next item: lgebr of frctios. The sum of two frctios is give by: b + c d d + bc. (10 bd O the right hd side, the terms of the umertor re obtied by cross multiplictio. To see this, otice tht bd is multiple for both deomitors o the left hd side. This fct suggests bd for the commo deomitor. I detil, b + c d b d d + c d b b d bd + bc bd d + bc. bd For the differece of two frctios, we hve somethig similr to (10: b c d d bc. (11 bd 5
6 I prctice, we ofte use some commo multiple of b d d to simplify our steps. We should try to get commo multiple s smll s possible, or get the lest commo multiple if possible. Assume tht m is commo multiple of b d d. Tht m is multiple of b mes tht we c write m bp for some p. Similrly we hve m dq for some q. So, fidig the sum of two frctios /b d c/d is crried out like this: b + c d b p p + c d q q p bp + cq dq p m + bq m The lst expressio SHOULD NOT be memorized! p + bq m. EXAMPLE 20. We re sked to simplify t x 2 csc 2x. First, we covert this expressio to oe which oly ivolves si x d cos x, which re esier to del with. Write t x s si x/ cos x, d csc 2x s 1/ si 2x d Use the trig idetity si 2x 2 si x cos x: t x + 2 csc 2x si x cos x 2 2 si x cos x si x cos x 1 si2 x 1 si x cos x cos2 x si x cos x cos x si x si x cos x cot x. EXERCISE 21. Simplify the followig expressios. (NO CALCULATOR! ( ; (b x x (x 1 2. Multiplictio for frctios is esy: Divisio of frctios ivolves flip i the deomitor: b c d c bd. (12 b c d b d c d bc. (13 We metio the followig two specil cses b c bc d c d d c. We must be very creful i mipultig complicted frctios. Pitflls re everywhere. 6
7 EXERCISE 22. Give sequece { } 1 { 1, 2, 3,... }, we defie sequece of cotiued frctios {P ( 1, 2,..., } 1 Check tht P ( 1 1 1, P ( 1, 2,..., iductively s follows: P ( 2, 3,...,. P ( 1, d P ( 1, 2, A similr expressio for P ( 1, 2, 3, 4 is quite complicted. Fid it. QUESTION 23. I the previous exercise, let 1 for ll. Wht do you thik the limit L lim P ( 1, 2,..., should be? Compute P ( 1,..., 4 d compre it with the umericl vlue of L. By rtiol umber we me umber which c be writte i the form /b, where d b re itegers with b 0. Idetities (10, (11, (12 d (13 tell us tht, if r d s re rtiol umbers, the so re sum r + s, differece r s, product rs, s well s their quotiet r/s, provided tht s is ozero. Becuse of this property, we sy tht rtiol umbers form field. The stdrd ottio for the field of rtiol umbers is Q. All rel umbers lso form field, deoted by R. PROBLEM 24. Show tht ll umbers which c be writte s r + s 2, where r d s re rtiol umbers, form field. (Remember: i mth, show mes prove. Fuctios such s 2x 2 + x 1, x 3 5x 2 + 6x + 9 re clled polyomils. I geerl, polyomil is fuctio of the form p(x 0 x + 1 x x x +. where 0, 1, 2,..., re some costts. Whe 1 d 0 0, we sy tht the degree of the polyomil p is. (If p is ozero costt, we sy tht the degree of p is zero. The zero fuctio s polyomil hs degree. If f d g re polyomils where g is ocostt, we c use the usul method of log divisio to divide f by g for obtiig the quotiet q d the remider r such tht f(x q(xg(x + r(x. 7
8 Here, if r is ot zero, the the degree of r is less th the degree of g. EXERCISE 25. Divide x 4 2x 3 + 3x 2 + x + 2 by x 2 + x + 2 usig log divisio. A fuctio f is clled rtiol fuctio if there exist polyomils p d q such tht f(x p(x/q(x for ll x. Certily idetities (10 to (13 bove lso holds whe, b, c, d re iterpreted s polyomils. Hece ll rtiol fuctios form field. It is clled the field of rtiol fuctios. (Wht else c you cll? QUESTION 26. How do you simplify x3 x x 2 + x? EXERCISE 27. Simplify 1 x + 1 x x 2 + x. Next, we del with roots (squre roots, cube roots, etc. d frctiol powers. A commo mistke is writig 2. The correct oe is 2 (the bsolute vlue of. Normlly, whe we write, we tcitly ssume 0. It is correct to put 2, keepig i mid tht this is oly for 0. It is lso correct to write b b. But writig b b rises some cocer bout the sigs of d b. QUESTION 28. Why is tht so? QUESTION 29. Why is ot legitimte expressio? EXERCISE 30. Simplify Whe is positive iteger, the th power u of u is defied recursively by u 1 u; u +1 u u ( 1. Next, we exted u for y iteger. For 0, we defie u 0 1, provided tht u is ozero. The expressio 0 0 is udefied; (it is clled idetermite form 8
9 i clculus, which is to be used to idicte type of limits. For egtive k where k > 0 d defie, we write u u k 1 u k (for u 0. For exmple, u 1 1/u, u 2 1/u 2 etc. For y positive iteger, the expressio u 1/ stds for the th root of u. For exmple, u 1/2 u. Whe is eve umber, we hve to ssume tcitly tht u 0. Whe is rtiol umber, sy m/, where m, re itegers with > 0, u is defied for u > 0 vi u (u 1/ m (u m 1/. For the geerl u where u > 0 d is y rel umber, we eed the help of the expoetil fuctio e x d the logrithm fuctio l x to ssig proper meig: u ( (e l u e l u. It follows from the bsic dditive formul e s e t e s+t (e s t e st for the expoetil fuctio tht d the multiplictive formul u u b u +b, (u b u b (14 for ll rel umbers, b, u with u > 0. Let me emphsize gi the importce of the restrictio u > 0 here. Urestricted usge of (12 is wepo of mth destructio. It is TERRIFYING. QUESTION 31. Wht is the mistke mde i the followig proof of 1 1? /2 (1 2 1/2 (( 1 2 1/2 ( 1 2 1/2 ( EXERCISE 32. Derive (14 from e s e t e s+t d (e s t e st. Sometimes the swer to lgebric expressio depeds o cses. For exmple, if is iteger, the ( 1 is 1 or 1, depedig o whether is eve or odd. So we write { ( 1 1 if is eve; 1 if is odd. 9
10 This formt for mthemticl expressio is ot rre. EXAMPLE 33. Fid ( 1 (p 1/2, where p is odd iteger. To swer this questio, we first otice tht, whe iteger is divided by 4, its possible remiders re 0, 1, 2 d 3. So ech iteger c be writte i oe of the followig forms: 4k, 4k + 1, 4k + 2, 4k + 3. Clerly 4k d 4k + 2 re eve umbers. So we re left with two cses to cosider: p 4k + 1 d p 4k + 3. Whe p 4k + 1, we hve (p 1/2 (4k + 1 1/2 2k, which is eve umber d hece ( 1 (p 1/2 ( 1 2k 1. Whe p 4k + 3, we hve (p 1/2 (4k + 3 1/2 2k + 1 which is clerly odd umber d hece ( 1 (p 1/2 ( 1 2k+1 1. We coclude: ( 1 (p 1/2 { 1 if p 4k + 1 for some iteger k; 1 if p 4k + 3 for some iteger k. EXERCISE 34. Fid ( 1 m++m, where m d re itegers. The soclled Kroecker s delt δ jk, commoly used i the sciece literture, is expressio with double subscripts defied by δ jk { 1 if j k; 0 if j k. As you c tell, the (j, etry of the idetity mtrix I is δ jk (1 j, k. 10
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