Farey Fractions. Rickard Fernström. U.U.D.M. Project Report 2017:24. Department of Mathematics Uppsala University

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "Farey Fractions. Rickard Fernström. U.U.D.M. Project Report 2017:24. Department of Mathematics Uppsala University"

Transcription

1 U.U.D.M. Project Report 07:4 Frey Frctions Rickrd Fernström Exmensrete i mtemtik, 5 hp Hledre: Andres Strömergsson Exmintor: Jörgen Östensson Juni 07 Deprtment of Mthemtics Uppsl University

2

3 Frey Frctions Uppsl University Rickrd Fernström June, 07

4 Introduction The Frey sequence of order n is the sequence of ll reduced frctions etween 0 with denomintor less thn or equl to n, rrnged in order of incresing size. The properties of this sequence hve een thoroughly investigted over the yers, out of intrinsic interest. The Frey sequences lso ply n importnt role in vrious more dvnced prts of numer theory. In the present tretise we give detiled development of the theory of Frey frctions, following the presenttion in Chpter 6.- of the ook MNZ = I. Niven, H. S. Zuckermn, H. L. Montgomery, An Introduction to the Theory of Numers, fifth edition, John Wiley & Sons, Inc., 99, ut filling in mny more detils of the proofs. Note tht the definition of Frey sequence Frey frction which we give elow is priori different from the one given ove; however in Corollry 7 we will see tht the two definitions re in fct equivlent. Frey Frctions Frey Sequences We will ssume tht frction is the quotient of two integers, where the denomintor is positive (every rtionl numer cn e written in this wy. A reduced frction is frction where the gretest common divisor of the 3.5 numertor denomintor is. E.g. 4 is not frction, ut 7 is oth 8 frction reduced frction (even though we would normlly sy tht = 7. Also 8 0 re not frctions, since their denomintors re negtive. We will construct tle in the following wy, where the frctions in ech row of the tle re in the specified order. The st row only contins the frctions 0. If the nth row hs een constructed, then the (n + st row is constructed y copying the nth row then for ech pir of consecutive frctions tht exist in the nth row tht stisfy + n +,

5 the frction + is inserted in the (n + st row etween. For exmple, the nd row is constructed y copying the st row then inserting = etween 0, so tht the nd row is 0,,. When constructing the 3rd row, = 3 is inserted etween = 3 is inserted etween, so tht the 3rd row ecomes 0, 3,, 3,. 0 + When constructing the 4th row, + 3 = 4 is inserted etween = 3 4 is inserted etween 3, ut = is not inserted 5 etween 3 + ecuse = 3 5 is not inserted etween 3 ecuse 5 4. So the 4th row is 0, 4, 3,, 3, 3 4,. The first six rows in the tle re: Definition (Frey sequence. The sequence of frctions in the nth row in the ove tle is clled the Frey sequence of order n. Definition (Frey frction of order n. A Frey frction of order n is frction in the Frey sequence of order n. 3

6 Definition 3 (Frey frction. A Frey frction is Frey frction of some order, i.e. frction in the Frey tle. Theorem (Theorem 6. Corollry 6.3 in MNZ. If re consecutive frctions in the nth row with to the left of, then =. The frctions in the nth row re lso listed in order of their size (in strictly scending order. Proof. Bse cse (n = : 0 = the frctions in the first row re clerly listed in order of their size (in strictly scending order. re frctions Induction hypothesis: Assume for some n N >0 tht if in the nth row with to the left of, then =. Also ssume the frctions in the nth row re listed in order of their size (in strictly scending order. Induction step: We wnt to show tht the frctions in the (n + st row re listed in order of their size (in strictly scending order. We know tht the (n + st row is constructed y copying the nth row then for ech pir of consecutive frctions in the nth row, insert or 0 frctions etween those frctions we lso know y the induction hypothesis tht the frctions in the nth row re listed in order of their size (in strictly scending order. So it s sufficient to show tht for ech pir of consecutive frctions p q, r in the s nth row with p q to the left, p q < p + r q + s < r s. So let p q r e consecutive s frctions in the nth row with p to the left. Then q p q < p + r q + s p(q + s q(q + s < q(p + r q(q + s p(q + s < q(p + r rq ps > 0. But the lst inequlity holds ecuse it follows from the induction hypothesis tht rq ps =. Hence p q < p + r. Similrly it cn e shown tht q + s 4

7 p + r q + s < r. So the frctions in the (n + st row re listed in order of their s size (in strictly scending order from tht it follows tht no frction cn pper twice in the (n + st row. Let e consecutive frctions in the (n + st row with to the left of. We wnt to show tht =. If they re lso consecutive frctions in the nth row with to the left, then it follows from the induction hypothesis tht =. So ssume they re not consecutive frctions in the nth row with to the left. We wnt to show it cn t e the cse tht oth exist in the nth row. Assume for contrdiction tht oth exist in the nth row. If they re not consecutive frctions in the nth row, then there is some frction in the nth row somewhere etween them, sy p. But ecuse of how q p rows re constructed, q must e etween in the (n + st row, which contrdicts eing consecutive frctions in the (n + st row. Assume insted for contrdiction tht re consecutive frctions in the nth row, ut with to the left. Either no frction ws dded etween them when constructing the (n + st row, in which cse will e consecutive frctions in the (n + st row, ut with to the left. This leds to contrdiction. If insted frction ws dded etween when constructing the (n + st row, then re not even consecutive frctions in the (n + st row, which is contrdiction. So it cn t e the cse tht oth exist in the nth row when they re consecutive frctions in the (n + st row with to the left they re not consecutive frctions 5

8 in the nth row with to the left. We wnt to show tht it cn t e the cse tht neither of the frctions exist in the nth row. Assume or contrdiction tht neither of the frctions exist in the nth row. Since they exist in the (n + st row, it must e the cse tht they were oth inserted etween frctions in the nth row. But ecuse t most frction is inserted etween ech pir of consecutive frctions, they must hve een inserted etween distinct pirs of frctions in the nth row. But then it s cler tht some frction must exist etween in the (n + st row, which contrdicts them eing consecutive frctions in the (n + st row. So it s impossile tht neither of the frctions exist in the nth row when they re consecutive frctions in the (n + st row with to the left. The remining cse is when one of the frctions exist in the nth row ut the other doesn t. Assume it s the left frction tht exists in the nth row (if insted it s the right frction tht exists in the nth row, then the proof cn e done nlogously. The right frction must hve een inserted in the (n + st row etween the frction in the nth row directly to the right of, sy p. Then the nth (n + st row will look something like q p Row n: q this: + p Row n + :... + q = p.... q Then = ( + p ( + q = p q =. {y induction hypothesis} 6

9 Corollry (Corollry 6. in MNZ. Every frction reduced form, i.e. gcd(, =. in the tle is in Proof. Let n N >0. We wnt to show every frction in the nth row hs gcd(, =. Let e frction in the nth row. Since there is more thn frction in the nth row, there is frction, sy p q next to in the nth row. Assume p q is to the right of (if it is to the left of, then the proof cn e done nlogously. By Theorem, p q =. Consider the diophntine eqution x + y =. It hs solution iff gcd(, =. But x = p, y = q is solution, so gcd(, =. Theorem 3. n N >0 : if row, then + n +. re consecutive frctions in the nth Proof. For n =, + = +, so it s true for n =. Induction hypothesis: Assume it s true for some n N >0. Induction step: We wnt to show it s true for n +. Let e consecutive frctions in the (n+st row with to the left. If they re lso consecutive frctions in the nth row then y the induction hypothesis, + n +. It cn t e the cse tht + = n +, ecuse if tht were the cse then + n +, so in the (n + st row frction should hve een inserted etween, ut then cn t e consecutive frctions in the (n + st row, contrdiction. So + > n +, i.e. + n +. Otherwise if they re not consecutive frctions in the nth row, exctly of the frctions pper in the nth row like in Theorem. Sy ppers in the nth row p q is the frction in the nth row directly to the right of. Then is the frction etween p q, so = + q. By the induction hypothesis, +q n+. And hence + = ++q ++q +(n+ = n+. 7

10 Theorem 4. Let n N >0 let p e frction in the nth row. Then q n q with equlity iff p q does not exist in ny previous row. Proof. We will egin y showing the first prt, i.e. frction in the nth row, then q n. n N >0 : if p q is For n = the only frctions re 0,. Induction hypothesis: Assume for some n N >0 tht for every frction p q in the nth row, q n. Induction step: Let p q tht q n +. p Cse : exists in the nth row s well. e frction in the (n + st row. We wnt to show q q n < n +. Cse : p q does not exist in the nth row. Then p q Then y induction hypothesis, must lie etween frctions tht exist in the nth row, sy. Then q = + ecuse of the rules of how rows re constructed, + n +, so q n +. Now we wnt to prove the second prt, i.e. n N >0 : if p q is frction in the nth row, then q = n iff p q does not exist in ny previous row. For n =, we see tht q = p does not exist in ny previous row, so q the theorem holds for n =. Sy n > let p q e frction in the nth row. If p exists in previous q row, then since rows re constructed y copying the previous row, p q must exist in the (n st row. This mens we re in cse of the proof of the previous prt of this theorem, so q < n, i.e. q n. If insted p doesn t exist in previous row, then we re in cse of the proof q p of the previous prt of this theorem, i.e. q lies etween the frctions tht exist in the (n st row. Then q = + n. But ecuse of 8

11 Theorem 3, + n. Since q n q n we must hve tht q = n. Lemm 5. Let n N >0 let x N e such tht 0 x n + gcd(x, n + =. Let e the gretest frction in the nth row tht is less x thn thn let e the smllest frction in the nth row tht is greter n + x n +. Then + = x + = n +. Proof. It cn t e the cse tht x n + would contrdict Theorem 4 (n + n. So then frctions in the nth row with to the left < is frction in the nth row, since tht x n + <. re consecutive We see tht < x x (n + > 0 x (n +. n + Consider the diophntine eqution u v = k, where k := x (n + u v re the unknowns. It hs prticulr solution u, v = k, k since k k = k( = k. { y Theorem } We hve tht gcd(, = ecuse of Corollry, so it hs the generl solution u, v = k + e, k + e, e Z. But we lso know it hs prticulr solution u, v = x, n +, so m Z : x = k + m n + = k + m. Let m Z e such tht k + m = n + ( k + m = x. ( 9

12 We wnt to show tht k = m =. Since 0 x n + gcd(x, n + =, it must e the cse tht x > 0. We hve tht x n + < n + > x (n + x > 0 (n + x (k + m (k + m {y ( (} ( m m. { y Theorem } This shows tht m. We lso know tht k. If we ssume for contrdiction tht k > or m >, then + < m + k = n +, {y (} ut tht s impossile, since y Theorem 3, + n +. This shows tht k m. Since lso m k, it follows tht k = m =. Eqution ( now ecomes + = n + eqution ( ecomes + = x. Theorem 6 (Theorem 6.5 in MNZ. If 0 x y, gcd(x, y =, then the frction x ppers in the yth ll lter rows. y Proof. It is cler tht if x y ppers in the yth row then it lso ppers in ll lter rows, so it is enough to show tht x y ppers in the yth row. If y = then either x = 0 or x =. Both 0 pper in the st row, so the theorem holds for y =. We wnt to show the theorem holds when y, i.e. when y = n + for some n N >0. Let n N >0, 0 x n + gcd(x, n + =. If x = 0 or x = n + then gcd(x, n + = n + >, so it must e the cse tht 0 < x < n +. 0

13 x We wnt to show tht n + ppers in the (n + st row. Let e the x gretest frction in the nth row tht is less thn let e the n + x smllest frction in the nth row tht is greter thn. By Lemm 5 we n + know tht + = x + = n +. From the first few lines of the proof x of Lemm 5 we know tht n + does not exist in the nth row, so must e consecutive frctions in the nth row. Becuse + = x + n + x how rows re constructed, we know tht must e inserted in the n + (n + st row etween. Corollry 7 (Corollry 6.6 in MNZ. The nth row consists exctly of ll reduced frctions such tht 0 0 < n. The frctions re listed in order of their size. Proof. Let n N >0 let e reduced frction such tht 0 0 < n. We wnt to show tht ppers in the nth row. Clerly 0 n. By Theorem 6 we hve tht ppers in the th ll lter rows, so ppers in the nth row. So every reduced frction such tht 0 0 < n exist in the nth row, they re listed in order of their size y Theorem. We wnt to show tht the nth row doesn t contin ny other frctions thn these. Let p e frction in the q nth row. The frction p must e reduced y Corollry. By Theorem 4 q we hve tht q n q > 0 y the definition of frctions. So 0 < q n. Becuse rows re constructed y copying the previous row then inserting frctions etween the frctions the fct tht rows re ordered y size, we cn never get row with frction tht is strictly greter thn less thn 0. So 0 p q. or strictly

14 Theorem 8. If re Frey frctions with = ( thus <, then re consecutive Frey frctions of some order. Proof. Let e Frey frctions with = (3 ( thus <. Let c d e Frey frction such tht < c d <. We wnt to show tht d > mx(,. From < c d c d < { c d > 0 d c > 0. it follows tht Let k := c d l := d c. Then k l re positive integers { c d = k (4 d c = l. Consider the diophntine eqution We know tht (k (k = k( x y = k. (5 = k. {y (3} Hence x, y = k, k is prticulr solution to (5. Becuse gcd(, = (this follows from Corollry, the generl solution to (5 is x, y = k + m, k + m = k, + m,, m Z. Consider the diophntine eqution Since (l + (l = l( x + y = l. (6 = l {y (3}

15 it follows tht x, y = l, l is prticulr solution to (6. We hve tht gcd(, = ecuse of Corollry, so the generl solution to (6 is x, {( y = ( l } + n, l + n = l, + n,, n Z. We know tht, is sis for the vector spce R, since ( det = = ( = {y (3} 0. ( ( u u But then every vector R cn e written in unique wy s = ( ( v v c +c, where c, c R. From (4 it follows tht x, y = c, d is prticulr solution( to oth (5 ( (6. ( By looking t the generl solution c of (5 we see tht = k d + m for some m Z. By looking ( ( ( c t the generl solution of (6 we see tht = l + n d for some ( c n Z. By compring these wys of writing using the fct tht {( ( } d, is sis for R we conclude tht l = m k = n. Thus ( ( ( c = k d + l. But then d = k + l + > mx(, =: m. In the ove equtions we used the fct tht k l re positive integers. By Corollry 7 it follows tht c does not exist in the mth row, since d > m. d But c d ws n ritrry Frey frction lying strictly etween. So no Frey frction lying strictly etween exist in the mth row, ut oth exist in the mth row. So must e consecutive Frey frctions in the mth row. 3

16 Corollry 9. Let e Frey frctions such tht <. Then = iff re consecutive Frey frctions of some order. Proof. = : Assume =. Then it follows from Theorem 8 tht re consecutive Frey frctions of some order. = : Assume re consecutive Frey frctions of some order. Then it follows from Theorem tht =. Theorem 0 (Theorem 6.4 in MNZ. If re consecutive frctions in the nth row (with to the left, then mong ll frctions p such tht q < p q <, p q = + is the unique frction with smllest denomintor. + Proof. Let n N >0 let e consecutive frctions in the nth row, with to the left. There re possiilities. Cse : + n +. Then + n + y Theorem 3, so + = n +. We hve tht + will e inserted in the (n + st row etween +. Cse : + > n +. Then for ll integers k such tht n k < + : will e consecutive frctions in the kth row + will e + inserted etween in the ( + th row. This is ecuse the kth row is constructed y copying the (k st row inserting frctions etween the frctions. If re consecutive frctions in the (k st row + ws not inserted etween them in the kth row (i.e. + k, then + will e consecutive frctions in the kth row s well. If re consecutive frctions in the (k st row then frction is only inserted etween them in row k if + k. 4

17 In oth cses we get tht, + + re 3 consecutive frctions in the ( + th row tht re consecutive frctions in the ( + st row. Let x y e reduced frction such tht < x y <. Since re frctions in the nth row, it follows from Corollry 7 tht 0 0. Since < x y <, we hve tht 0 x. If we ssume for y contrdiction tht y < +, then y Theorem 6, x y ppers in the (+ st row. But tht s impossile since we know from Theorem tht frctions re listed in order of their size there is no frction strictly etween in the (+ st row. So y +, i.e. reduced frction tht lies strictly etween cn t hve denomintor smller thn +. Suppose x y hs miniml denomintor, i.e. y = +. By Theorem 6, x exists in the y ( + th row. But since, + re 3 consecutive frctions in the + (+ th row the (+ th row is ordered y size y Theorem, the only frction in the ( + th row tht lies strictly etween x y must e the frction + +. is + +. So If you hve frction p q tht lies strictly etween tht is not reduced, then it cn t e the cse tht q +, since then you could simplify p q into reduced frction where the denomintor is strictly less thn +, ut tht s impossile. Sy you hve frction r s tht lies strictly etween. Then s +, since if r s is not reduced, then s > + if r s is reduced, then s +. Also since + exists in the ( + th row it must e reduced frction + ecuse of Corollry. So the smllest denomintor r s cn hve is +, 5

18 in tht cse r must e reduced frction. So if you hve frction tht s lies strictly etween with smllest denomintor, then it must e reduced frction with denomintor + (since + is the smllest denomintor. But the only reduced frction with denomintor + tht lies strictly etween is +. So mong ll frctions tht lie strictly etween +, + is the unique frction with smllest denomintor. + Proposition (Prolem 6.. in MNZ. Let n e positive integer such tht n > let e the Frey frctions immeditely to the left the right of respectively in the Frey sequence of order n. Then = n = +, i.e. is the gretest odd integer n. It is lso true tht + =. Proof. By Corollry 7 we know tht exists in the Frey sequence of order n since n. We will prove the Proposition y induction. Bse cse (n = : Then we see from the Frey tle on pge 3 tht is 0 is. We hve tht =, is the gretest odd integer 0 + =. Induction hypothesis: Assume for some n tht if re the Frey frctions immeditely to the left the right of the frction respectively in the Frey sequence of order n, then = is the gretest odd integer n + =. Induction step: Let e the Frey frctions immeditely to the left respectively in the Frey sequence of order n. the right of the frction Let c d c d e the Frey frctions immeditely to the left the right of the frction respectively in the Frey sequence of order n +. We wnt to 6

19 show tht d = d tht d is the gretest odd integer n+ c+c = d. Cse : + = + = n + ( = y induction hypothesis. Then when constructing the (n + st row the frction + + is inserted etween lso the frction + + is inserted etween. Then it must e the cse tht c + c is the frction is the frction +, so then d + d + d = + d = +. Also from the induction hypothesis we see tht re odd, since + = n + we hve tht n = +, i.e. n is even. From the induction hypothesis we know tht is the gretest odd integer n. Since d = + we need to show tht + is the gretest odd integer n +, ut this is ovious, since we know tht n is even. Since d = + d = + = y induction hypothesis, it follows tht d = d, so d = d d is the gretest odd integer n +. We lso see tht c + c = {c = + c = + } = + {y induction hypothesis} = d. Cse : + = + > n + ( = y induction hypothesis. Then no frction will e inserted etween or etween constructing the (n + st row. So c d is the frction. We know tht n + < + n +, { n y Theorem 4} when c is the frction d so it follows tht + = n +, i.e. = n. But is odd y the induction hypothesis, so n must e odd. We know from the induction hypothesis tht is the gretest odd integer n d =. We need to show tht is the gretest odd integer n + But tht is ovious. 7

20 Since = y induction hypothesis, = d = d it follows tht d = d. So d = d d is the gretest odd integer n +. We lso see tht c + c = + { = c = c } = {y induction hypothesis} = d. There is no cse where + = + < n + ecuse of Theorem 3. Theorem (Prolem 6..9 in MNZ. For ech Frey frction ( ( let C denote the circle in the plne of rdius ( center, (. These circles re clled the Ford circles. The interior of Ford circle contins ( ( no point of ny other Ford circle two Ford circles C C re tngent if only if order. re consecutive Frey frctions of some Proof. First ssume tht re distinct Frey frctions such tht ( < (. We wnt to show tht neither of the Ford circles C C ( contin point tht is n interior point of the other Ford circle. C hs rdius center r := (, ( =, r. 8

21 ( C hs rdius center s := ( (, ( ( =, s. ( ( Let d e the distnce etween the center of C the center of C. Then ( d = + (s r = ( + s + r rs. (7 ( ( ( In order to show tht the Ford circles C C don t contin ny interior points of the other Ford circle, it is sufficient to show tht d r + s, i.e. d r + s + rs. Using (7 we see tht we must show tht ( + s + r rs r + s + rs ( ( ( 4rs = ( (. (8 It cn t e the cse tht = 0, since then = 0 = =, {divide oth sides y } which contrdicts the Frey frctions eing distinct. But then is non-zero integer, hence its squre must e greter thn or equl to, i.e. (8 ( ( is true. So the Ford circles C C don t contin ny interior 9

22 points of the other Ford circle d r + s. ( ( We wnt to show tht C C re tngent iff consecutive Frey frctions of some order. If re the sme Frey ( ( frction, then the Ford circles C C re the sme re therefore not tngent re not consecutive Frey frctions of some order, so the Theorem holds in tht cse. So ssume re distinct. Without loss of generlity ssume tht < like efore. We know ( ( tht C C re tngent if only if d = r + s or d = r s. Since r s re positve, it follows tht r s mx(r, s < r + s. But efore we showed tht d r + s, so it cn t e the cse tht d = r s. ( ( Therefore C C re tngent if only if d = r + s. But d = r + s iff d = r + s + rs. Using (7 we see tht d = r + s + rs ( ( ( ( = 4rs = ( ( = = ±. + s + r rs = r + s + rs But it cn esily e shown tht < iff > 0, since we ssumed tht <, it must e the cse tht > 0. Therefore = ± =. re From Corollry 9 it follows tht = re consecutive 0

23 ( Frey frctions of some order. So C C if re consecutive Frey frctions of some order. ( re tngent if only Remrk. We will now use n lterntive method to show tht if ( ( re consecutive Frey frctions of some order, then C C re tngent. When using this method we will find explicitly the coordintes for the intersection of the Ford circles. Proof. Let e consecutive Frey frctions of order n, with <. ( ( We wnt to show tht C C re tngent. The eqution for ( C is ( x ( + y = (9 4 4 ( the eqution for C is ( (x + y = ( 4(. (0 4 ( Then the x-coordinte of center of C is less thn (i.e. to the left of the ( x-coordinte of the center of C. Let L e the line tht goes through

24 ( the center of C the center of C ( ( ( / ( ( ( = / ( ( ( ( = / ( = (. ( The eqution for L is y ( = ( (x ( L intersects C. It s slope is {y Theorem }. We wnt to see where, so we plug the eqution for L into (0 get ( (x ( + (x = 4( ( 4 ( ( ( + x = 4( 4 ( 4 ( 4 ( + ( ( + (( (x = 4 ( 4( 4 ( + ( (x = 4( 4 (x = ( ( + ( x = ± ( ( + ( x = ± ( + (. We only cre out the left intersection, i.e. where x = ( + (.

25 If we plug this x-vlue into the eqution for L solve for y we get So x = ( C y = ( + ( ( + ( = ( ( ( ( + ( + ( = ( ( + ( ( ( ( + ( = + (. ( + ( y = + ( is the left intersection of L (. Plugging. We wnt to show tht this point lso elongs to C 3

26 the x-vlue y-vlue into the left h side of (9 gives ( ( + ( ( + + ( ( = ( + ( ( + ( + ( + ( ( + ( ( ( = ( + {y Theorem } ( + ( ( + ( ( + ( = ( + ( ( + ( ( ( + ( + ( ( ( 3 ( ( ( 4 = + ( ( + ( ( ( + ( = 4 ( ( 4 ( 6 + ( ( 4 ( + ( = ( 4 ( + ( 4 4 ( 4 ( + ( = 4, 4 ( which mens tht ( + (, elongs to C + ( ( ( L, so it must e the cse tht C C (, C re tngent. ( Theorem 3 (Prolem 6.. in MNZ. The numer of Frey frctions of order n stisfying the inequlities 0 n is + φ(j their sum is exctly hlf this vlue. Proof. We wnt to prove the first prt of the theorem. Let n e positive integer. Every Frey frction of order n must stisfy the inequlities 4 j=

27 0 y Corollry 7, so the Frey frctions of order n stisfying the inequlities 0 re precisely the Frey frctions of order n. By Corollry 7 the Frey frctions of order n re precisely the reduced frctions such tht 0 0 < n. For integers j such tht j n, define A(j := the numer of reduced frctions j such tht 0 j. Also let F (n := the numer of Frey frctions of order n. n Then clerly F (n = A(j. It is cler tht A(j is the numer of Frey j= frctions of order n with denomintor j. We hve tht A( =, ecuse 0 re the only reduced frctions with denomintor tht lie etween 0. The definition of A(j cn e rewritten y multiplying the lst inequlities y j using the definition of reduced frction to otin A(j = the numer of frctions j such tht gcd(, j = 0 j. Different vlues for give different frctions, so we cn rewrite A(j s j A(j = the numer of integers such tht gcd(, j = 0 j. For j we hve tht gcd(0, j = j, so for j we cn rewrite A(j s So A(j = the numer of integers such tht gcd(, j = j F (n = n A(j = A( + j= n A(j = + j= n φ(j j= = + φ( + = + n φ(j j= n φ(j. j= = φ(j. 5

28 We wnt to prove the second prt of the theorem. Let S(n e ( the sum of the Frey frctions of order n. We wnt to show tht S(n = n + φ(j. S(n = is Frey frction of order n If is Frey frction of order n, then = is lso Frey frction of order n. This is ecuse if is Frey frction of order n then y Corollry 7 we hve tht 0, so 0 if gcd(, =, then gcd(, = hence y Corollry 7, is lso Frey frction of order n. If 3, then it cn t e the cse tht = if is Frey frction of order n, since tht would imply =, i.e., ut lso = >, so tht is common divisor of tht is strictly greter thn, which mens tht gcd(,. So S(n = + is Frey frction of order n. j= is Frey frction of order n > ( The terms in the right sum cn e pired up s,. The terms in ech pir dd up to. If we ssume n then the only Frey frctions of order n with denomintor re 0, n, so there re + φ(j 3 j= ( n terms eing dded in the right sum, so there re φ(j pirs eing j= ( n summed in the second sum. Thus the right sum equls φ(j. The left sum equls 3, so ( n ( S(n = 3 + φ(j = + j= 6 j= n φ(j. j=.

29 ( If n =, then S(n = 0 + = = + for ll n. φ(j, so the theorem holds j= 3 Rtionl Approximtions Theorem 4 (Theorem 6.7 in MNZ. Let c e Frey frctions of d order n such tht no other Frey frction of order n lies etween them. Then + c + d = ( + d (n + c d + c + d = d( + d d(n +. Proof. For the first formul we hve + c + d = ( + d ( + c ( + d d c = ( + d { } = y Theorem ( + d. {since + d n + y Theorem 3} (n + The second formul is otined in similr wy. Theorem 5 (Theorem 6.8 in MNZ. Let n N >0 x R. Then there is rtionl numer such tht 0 < n x (n +. 7

30 Proof. Let n N >0 x R. Let k e the unique integer such tht 0 x + k <. There re cses. Cse : x+k hs the sme vlue s some Frey frction of order n. Let p q e the Frey frction of order n tht x+k simplifies to. Let the rtionl numer e p k q. Since p is Frey frction of order n, q n y Theorem 4. q q So is rtionl numer such tht 0 < n x = x p k q q = x + k p k q k q q q = p q p q = 0 (n +. Cse : x + k does not hve the sme vlue s some Frey frction of order n. It cn t e the cse tht x + k = 0, since 0 is Frey frction of order n. So 0 < x + k <. Let p e the gretest Frey frction of order n tht is q smller thn x + k let r s e the smllest Frey frction of order n tht is greter thn x + k. There is no Frey frction of order n tht lies etween p q r s. Since p q r re Frey frctions of order n, it follows from Theorem s 4 tht q n s n. There re cses now. p + r Cse. : q + s x + k. Let the rtionl numer e r k s. We hve s 8

31 tht is rtionl numer such tht 0 < n x = x r k s x s = r + k s = r s (x + k r s p + r { q + s since p + r q + s x + k < r } s { } y Theorem 4 s(n + = (n +. p + r Cse. : q + s > x + k. Let the rtionl numer e p k q. We hve q tht is rtionl numer such tht 0 < n x = x p k q q = x + k p q p + r q + s p { q since p q < x + k < p + r } q + s = p q p + r q + s { } y Theorem 4 q(n + = (n +. Theorem 6 (Theorem 6.9 in MNZ. If ξ R \ Q, then there re infinitely mny distinct rtionl numers such tht ξ <. 9

32 Proof. Let ξ R \ Q. For ech n N >0, let { ξ A n := Q 0 < n }. (n + Let X := {A n n N >0 } By Theorem 5, ech A n is non-empty, so tht X is set of non-empty sets. By the xiom of choice, there exists choice function f : X X such tht A X : f(a A. So let f : X X e function such tht A X : f(a A. For ech n N >0, let n e the rtionl numer f(a n. Then ξ n n (n + n n < n {since n + > n } for ll n N >0. We wnt to show there re infinitely mny distinct n. Assume for contrdiction there re only finitely mny distinct n n. Then there re only finitely n mny distinct vlues for ξ n. Let d := min ξ n. Clerly d 0 d 0, since d = 0 would imply tht ξ = k n n N >0 k n for some k N >0, ut tht contrdicts ξ eing irrtionl. So d > 0. Then for ll n N >0 we hve tht d ξ n n n (n + n +. {since n } But if we let n e sufficiently lrge (n d will work, then n + < n d. But then d < d, contrdiction. Lemm 7 (Lemm 6.0 in MNZ. If x y re positive integers then not oth of the inequlities xy ( 5 x + y x(x + y ( 5 x + (x + y cn hold. 30

33 Proof. We will rewrite the inequlities. We see tht xy ( 5 x + y xy y + x 5 (xy { 5xy y + x multiply oth sides y } 5(xy x(x + y ( 5 x + (x + y x(x + y (x + y + x 5 (x(x + y 5x(x + y (x + y + x. {multiply oth sides y 5(x(x + y } Assume for contrdiction tht oth 5xy y + x ( 5x(x + y (x + y + x ( re true. By dding the inequlities ( ( we get 5(x + xy 3x + xy + y 3x + xy + y 5(x + xy (3 5x ( 5 xy + y 0 (5 5 + x 4( 5 xy + 4y 0 {multiply y } (y ( 5 x 0 y ( 5 x = 0 5 = y + x, x ut tht contrdicts 5 eing irrtionl. 3 {since squres re non-negtive} { solve for } 5

34 Lemm 8. If c d re consecutive Frey frctions of order n with to the left, then c d n. Proof. The frctions 0 n, n, n,, n n, n n (3 re Frey frctions of order n if they re simplified y Corollry 7 (the simplified frctions lie etween 0 the denomintor of ech simplified frction lies etween n. It cn t e the cse tht < j n c d > j n for some j n in the sequence (3, since then j would e Frey frction of order n n (if simplified tht lies strictly etween c, which would contrdict d c d eing consecutive Frey frctions of order n. So c must oth d lie etween consecutive elements in (3, which mens c d n, since the distnce etween consecutive elements of (3 is lwys n. Theorem 9 (Theorem 6. in MNZ. Given ny ξ R \ Q, there exist infinitely mny different rtionl numers h such tht k ξ h k <. 5k Proof. Let µ e the unique integer such tht 0 ξ + µ <. It cn t e the cse tht ξ + µ = 0, since tht would imply tht ξ is rtionl, which is not the cse. So 0 < ξ + µ <. Let λ := ξ + µ. Then 0 < λ <. Let n e positive integer. Let n e the gretest Frey frction of order n tht is less thn λ let c n d n thn λ. Then n n n e the smllest Frey frction of order n tht is greter c n d n re consecutive frctions in the nth row. We wnt 3

35 to show tht t lest one of the numers n, c n n + c n will work s h n d n n + d n k in the theorem (when ξ = λ. Assume for contrdiction this is not the cse. Then λ n, (4 n 5 n Cse : n + c n n + d n < λ. Then (6 ecomes c n λ (5 d n 5d n λ n + c n n + d n 5(n + d n. (6 λ n + c n n + d n 5(n + d n. By dding the inequlities (4 (5 we get c n n ( + d n n 5 n d n nc n n d n ( + n d n 5 n d n ( + n d n 5 n d n { y Theorem } (7 y dding the inequlities (5 (6 we get c n n + c n ( + d n n + d n 5 d n ( n + d n c n( n + d n d n ( n + c n d n ( n + d n nc n n d n d n ( n + d n ( + 5 d n d n ( n + d n 5 ( d n + ( + 5 d n ( n + d n ( n + d n. ( n + d n { y Theorem } (8 33

36 But not oth the inequlities (7 (8 cn e true y Lemm 7 (if we let x = d n y = n in the lemm, so we get contrdiction. Cse : n + c n n + d n > λ. Then (6 ecomes n + c n n + d n λ 5(n + d n. Just like in cse we cn dd the inequlities (4 (5 to get (7. By dding the inequlities (4 (6 we get n + c n n ( + n + d n n 5 n ( n + d n n( n + c n n ( n + d n ( + n ( n + d n 5 n nc n n d n n ( n + d n ( + 5 n ( n + d n n ( n + d n ( +. 5 n ( n + d n ( n + d n { y Theorem } (9 But not oth inequlities (7 (9 cn e true y Lemm 7 (if we let x = n y = d n in the lemm, so we get contrdiction. Thus one of the numers n n, c n d n in the the- n + c n n + d n orem (when ξ = λ. So for ech n N >0, let h n c n d n n + c n n + d n such tht k n λ h n <, 5k n k n will work s h k e one of the frctions n n, where n is the gretest Frey frction of order n tht is less thn λ c n n d n is the smllest Frey frction of order n tht is greter thn λ. We wnt to show tht for every ɛ > 0 there exists Frey frction h n k n (s 34

37 k n defined ove such tht λ h n < ɛ. Let ɛ > 0 let n e positive integer such tht n < ɛ (i.e. n > ɛ. We know tht n n c n d n re consecutive frctions of order n, so we know y Lemm 8 tht c n n < ɛ. Clerly d n n n λ n n < ɛ λ c n d n < ɛ. Also we know tht n < n + c n < c n, so n n + d n d n λ h n < ɛ, since h n is one of the frctions n, c n n + c n. k n n d n n + d n k n We wnt to show tht there re infinitely mny h n with distinct vlues. k n Assume for contrdiction tht there re only finitely mny h n. Then let { } k n hn A := k n n N >0 ɛ := min hn λ h n kn A k n. But s we ve shown there is Frey frction h n such tht k n λ h n < ɛ, which contrdicts ɛ eing miniml. So there re infinitely mny h n k n with distinct vlues. For every n N >0 : k n λ h n k n < 5k n λ µ h n µ k n k n < 5k n ξ h n µ k n k n < 5k n there re infinitely mny h n µ k n k n infinitely mny h n with distinct vlues. This is ecuse h n µ k n k n k n the function g : R R, x x µ is ijective. with distinct vlues since there re = h n k n µ 35

38 Theorem 0 (Theorem 6. in MNZ. The constnt 5 in Theorem 9 is the est possile, i.e. Theorem 9 does not hold if 5 is replced y ny lrger vlue. Proof. It s enough to find one ξ such tht 5 cnnot e replced y ny lrger vlue in Theorem 9. Let ξ := + 5. We see tht ( (x ξ x ( 5 = x + ( 5 x 5 = x x. So if we let h k e integers with k > 0, then h k ξ h k ξ + 5 (0 ( ( h = k ξ h k 5 = h k h k = k h hk k. ( We know tht (0 cn t e zero since tht would imply ξ = h k or h k = ξ 5, ut neither ξ nor ξ 5 re rtionl. We know tht h hk k is nonnegtive integer. It cn t e the cse tht h hk k = 0, since then eqution ( is zero hence eqution (0 is zero. Since h hk k it must e the cse tht h hk k ecuse eqution ( k k is equl to eqution (0 it follows tht h k ξ h k ξ + 5 k. ( Let m e positive rel numer ssume we hve n infinite sequence of rtionl numers h j, k j > 0 such tht k j h j ξ k j < mk j. (3 36

39 If we multiply y k j on oth sides of (3 we get From this we cn see tht h j k j ξ < mk j. k j ξ mk j < h j < k j ξ + mk j, so tht for ech rtionl numer h j k j in the sequence there cn only e finitely mny other frctions in the sequence with the sme denomintor, ut different numertor. It follows tht lim j k j =. We see tht k j h j ξ h j k j ξ + 5 k j {y (} < h j mkj ξ + 5 k j {y (3} ( h j ξ mkj k j + 5 {y -inequlity} < ( + 5. {y (3} mkj mkj Multiplying y mk j we get Therefore ( m lim j mkj m < mk j + 5 = { } lim k j = j So if m > 5 then there is no infinite sequence of rtionl numers h j, k j k j > 0 such tht h j ξ k j < mk j 37

40 for every h j k j in the sequence. 38

MTH 505: Number Theory Spring 2017

MTH 505: Number Theory Spring 2017 MTH 505: Numer Theory Spring 207 Homework 2 Drew Armstrong The Froenius Coin Prolem. Consider the eqution x ` y c where,, c, x, y re nturl numers. We cn think of $ nd $ s two denomintions of coins nd $c

More information

set is not closed under matrix [ multiplication, ] and does not form a group.

set is not closed under matrix [ multiplication, ] and does not form a group. Prolem 2.3: Which of the following collections of 2 2 mtrices with rel entries form groups under [ mtrix ] multipliction? i) Those of the form for which c d 2 Answer: The set of such mtrices is not closed

More information

UniversitaireWiskundeCompetitie. Problem 2005/4-A We have k=1. Show that for every q Q satisfying 0 < q < 1, there exists a finite subset K N so that

UniversitaireWiskundeCompetitie. Problem 2005/4-A We have k=1. Show that for every q Q satisfying 0 < q < 1, there exists a finite subset K N so that Problemen/UWC NAW 5/7 nr juni 006 47 Problemen/UWC UniversitireWiskundeCompetitie Edition 005/4 For Session 005/4 we received submissions from Peter Vndendriessche, Vldislv Frnk, Arne Smeets, Jn vn de

More information

QUADRATIC EQUATIONS OBJECTIVE PROBLEMS

QUADRATIC EQUATIONS OBJECTIVE PROBLEMS QUADRATIC EQUATIONS OBJECTIVE PROBLEMS +. The solution of the eqution will e (), () 0,, 5, 5. The roots of the given eqution ( p q) ( q r) ( r p) 0 + + re p q r p (), r p p q, q r p q (), (d), q r p q.

More information

8. Complex Numbers. We can combine the real numbers with this new imaginary number to form the complex numbers.

8. Complex Numbers. We can combine the real numbers with this new imaginary number to form the complex numbers. 8. Complex Numers The rel numer system is dequte for solving mny mthemticl prolems. But it is necessry to extend the rel numer system to solve numer of importnt prolems. Complex numers do not chnge the

More information

MATH 573 FINAL EXAM. May 30, 2007

MATH 573 FINAL EXAM. May 30, 2007 MATH 573 FINAL EXAM My 30, 007 NAME: Solutions 1. This exm is due Wednesdy, June 6 efore the 1:30 pm. After 1:30 pm I will NOT ccept the exm.. This exm hs 1 pges including this cover. There re 10 prolems.

More information

MORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.)

MORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.) MORE FUNCTION GRAPHING; OPTIMIZATION FRI, OCT 25, 203 (Lst edited October 28, 203 t :09pm.) Exercise. Let n be n rbitrry positive integer. Give n exmple of function with exctly n verticl symptotes. Give

More information

Improper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows:

Improper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows: Improper Integrls The First Fundmentl Theorem of Clculus, s we ve discussed in clss, goes s follows: If f is continuous on the intervl [, ] nd F is function for which F t = ft, then ftdt = F F. An integrl

More information

Section 6.1 Definite Integral

Section 6.1 Definite Integral Section 6.1 Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot e determined

More information

Convert the NFA into DFA

Convert the NFA into DFA Convert the NF into F For ech NF we cn find F ccepting the sme lnguge. The numer of sttes of the F could e exponentil in the numer of sttes of the NF, ut in prctice this worst cse occurs rrely. lgorithm:

More information

Regular Language. Nonregular Languages The Pumping Lemma. The pumping lemma. Regular Language. The pumping lemma. Infinitely long words 3/17/15

Regular Language. Nonregular Languages The Pumping Lemma. The pumping lemma. Regular Language. The pumping lemma. Infinitely long words 3/17/15 Regulr Lnguge Nonregulr Lnguges The Pumping Lemm Models of Comput=on Chpter 10 Recll, tht ny lnguge tht cn e descried y regulr expression is clled regulr lnguge In this lecture we will prove tht not ll

More information

USA Mathematical Talent Search Round 1 Solutions Year 21 Academic Year

USA Mathematical Talent Search Round 1 Solutions Year 21 Academic Year 1/1/21. Fill in the circles in the picture t right with the digits 1-8, one digit in ech circle with no digit repeted, so tht no two circles tht re connected by line segment contin consecutive digits.

More information

QUADRATIC EQUATION EXERCISE - 01 CHECK YOUR GRASP

QUADRATIC EQUATION EXERCISE - 01 CHECK YOUR GRASP QUADRATIC EQUATION EXERCISE - 0 CHECK YOUR GRASP. Sine sum of oeffiients 0. Hint : It's one root is nd other root is 8 nd 5 5. tn other root 9. q 4p 0 q p q p, q 4 p,,, 4 Hene 7 vlues of (p, q) 7 equtions

More information

along the vector 5 a) Find the plane s coordinate after 1 hour. b) Find the plane s coordinate after 2 hours. c) Find the plane s coordinate

along the vector 5 a) Find the plane s coordinate after 1 hour. b) Find the plane s coordinate after 2 hours. c) Find the plane s coordinate L8 VECTOR EQUATIONS OF LINES HL Mth - Sntowski Vector eqution of line 1 A plne strts journey t the point (4,1) moves ech hour long the vector. ) Find the plne s coordinte fter 1 hour. b) Find the plne

More information

MATH 101A: ALGEBRA I PART B: RINGS AND MODULES 35

MATH 101A: ALGEBRA I PART B: RINGS AND MODULES 35 MATH 101A: ALGEBRA I PART B: RINGS AND MODULES 35 9. Modules over PID This week we re proving the fundmentl theorem for finitely generted modules over PID, nmely tht they re ll direct sums of cyclic modules.

More information

Lecture 1. Functional series. Pointwise and uniform convergence.

Lecture 1. Functional series. Pointwise and uniform convergence. 1 Introduction. Lecture 1. Functionl series. Pointwise nd uniform convergence. In this course we study mongst other things Fourier series. The Fourier series for periodic function f(x) with period 2π is

More information

1B40 Practical Skills

1B40 Practical Skills B40 Prcticl Skills Comining uncertinties from severl quntities error propgtion We usully encounter situtions where the result of n experiment is given in terms of two (or more) quntities. We then need

More information

Review of Gaussian Quadrature method

Review of Gaussian Quadrature method Review of Gussin Qudrture method Nsser M. Asi Spring 006 compiled on Sundy Decemer 1, 017 t 09:1 PM 1 The prolem To find numericl vlue for the integrl of rel vlued function of rel vrile over specific rnge

More information

Homework 4. 0 ε 0. (00) ε 0 ε 0 (00) (11) CS 341: Foundations of Computer Science II Prof. Marvin Nakayama

Homework 4. 0 ε 0. (00) ε 0 ε 0 (00) (11) CS 341: Foundations of Computer Science II Prof. Marvin Nakayama CS 341: Foundtions of Computer Science II Prof. Mrvin Nkym Homework 4 1. UsetheproceduredescriedinLemm1.55toconverttheregulrexpression(((00) (11)) 01) into n NFA. Answer: 0 0 1 1 00 0 0 11 1 1 01 0 1 (00)

More information

Chapter 9 Definite Integrals

Chapter 9 Definite Integrals Chpter 9 Definite Integrls In the previous chpter we found how to tke n ntiderivtive nd investigted the indefinite integrl. In this chpter the connection etween ntiderivtives nd definite integrls is estlished

More information

2 b. , a. area is S= 2π xds. Again, understand where these formulas came from (pages ).

2 b. , a. area is S= 2π xds. Again, understand where these formulas came from (pages ). AP Clculus BC Review Chpter 8 Prt nd Chpter 9 Things to Know nd Be Ale to Do Know everything from the first prt of Chpter 8 Given n integrnd figure out how to ntidifferentite it using ny of the following

More information

Convex Sets and Functions

Convex Sets and Functions B Convex Sets nd Functions Definition B1 Let L, +, ) be rel liner spce nd let C be subset of L The set C is convex if, for ll x,y C nd ll [, 1], we hve 1 )x+y C In other words, every point on the line

More information

13.3 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS

13.3 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS 33 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS As simple ppliction of the results we hve obtined on lgebric extensions, nd in prticulr on the multiplictivity of extension degrees, we cn nswer (in

More information

Improper Integrals. Introduction. Type 1: Improper Integrals on Infinite Intervals. When we defined the definite integral.

Improper Integrals. Introduction. Type 1: Improper Integrals on Infinite Intervals. When we defined the definite integral. Improper Integrls Introduction When we defined the definite integrl f d we ssumed tht f ws continuous on [, ] where [, ] ws finite, closed intervl There re t lest two wys this definition cn fil to e stisfied:

More information

APPENDIX. Precalculus Review D.1. Real Numbers and the Real Number Line

APPENDIX. Precalculus Review D.1. Real Numbers and the Real Number Line APPENDIX D Preclculus Review APPENDIX D.1 Rel Numers n the Rel Numer Line Rel Numers n the Rel Numer Line Orer n Inequlities Asolute Vlue n Distnce Rel Numers n the Rel Numer Line Rel numers cn e represente

More information

2. VECTORS AND MATRICES IN 3 DIMENSIONS

2. VECTORS AND MATRICES IN 3 DIMENSIONS 2 VECTORS AND MATRICES IN 3 DIMENSIONS 21 Extending the Theory of 2-dimensionl Vectors x A point in 3-dimensionl spce cn e represented y column vector of the form y z z-xis y-xis z x y x-xis Most of the

More information

How do we solve these things, especially when they get complicated? How do we know when a system has a solution, and when is it unique?

How do we solve these things, especially when they get complicated? How do we know when a system has a solution, and when is it unique? XII. LINEAR ALGEBRA: SOLVING SYSTEMS OF EQUATIONS Tody we re going to tlk out solving systems of liner equtions. These re prolems tht give couple of equtions with couple of unknowns, like: 6= x + x 7=

More information

Math 554 Integration

Math 554 Integration Mth 554 Integrtion Hndout #9 4/12/96 Defn. A collection of n + 1 distinct points of the intervl [, b] P := {x 0 = < x 1 < < x i 1 < x i < < b =: x n } is clled prtition of the intervl. In this cse, we

More information

Quadratic reciprocity

Quadratic reciprocity Qudrtic recirocity Frncisc Bozgn Los Angeles Mth Circle Octoer 8, 01 1 Qudrtic Recirocity nd Legendre Symol In the eginning of this lecture, we recll some sic knowledge out modulr rithmetic: Definition

More information

Polynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230

Polynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230 Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given

More information

Harvard University Computer Science 121 Midterm October 23, 2012

Harvard University Computer Science 121 Midterm October 23, 2012 Hrvrd University Computer Science 121 Midterm Octoer 23, 2012 This is closed-ook exmintion. You my use ny result from lecture, Sipser, prolem sets, or section, s long s you quote it clerly. The lphet is

More information

Section 4: Integration ECO4112F 2011

Section 4: Integration ECO4112F 2011 Reding: Ching Chpter Section : Integrtion ECOF Note: These notes do not fully cover the mteril in Ching, ut re ment to supplement your reding in Ching. Thus fr the optimistion you hve covered hs een sttic

More information

approaches as n becomes larger and larger. Since e > 1, the graph of the natural exponential function is as below

approaches as n becomes larger and larger. Since e > 1, the graph of the natural exponential function is as below . Eponentil nd rithmic functions.1 Eponentil Functions A function of the form f() =, > 0, 1 is clled n eponentil function. Its domin is the set of ll rel f ( 1) numbers. For n eponentil function f we hve.

More information

Homework Solution - Set 5 Due: Friday 10/03/08

Homework Solution - Set 5 Due: Friday 10/03/08 CE 96 Introduction to the Theory of Computtion ll 2008 Homework olution - et 5 Due: ridy 10/0/08 1. Textook, Pge 86, Exercise 1.21. () 1 2 Add new strt stte nd finl stte. Mke originl finl stte non-finl.

More information

Math 360: A primitive integral and elementary functions

Math 360: A primitive integral and elementary functions Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. DeTurck Mth 360 001 2017C: Integrl/functions 1 / 32 Setup for the integrl prtitions Definition:

More information

SOME INTEGRAL INEQUALITIES OF GRÜSS TYPE

SOME INTEGRAL INEQUALITIES OF GRÜSS TYPE RGMIA Reserch Report Collection, Vol., No., 998 http://sci.vut.edu.u/ rgmi SOME INTEGRAL INEQUALITIES OF GRÜSS TYPE S.S. DRAGOMIR Astrct. Some clssicl nd new integrl inequlities of Grüss type re presented.

More information

PARTIAL FRACTION DECOMPOSITION

PARTIAL FRACTION DECOMPOSITION PARTIAL FRACTION DECOMPOSITION LARRY SUSANKA 1. Fcts bout Polynomils nd Nottion We must ssemble some tools nd nottion to prove the existence of the stndrd prtil frction decomposition, used s n integrtion

More information

Homework 3 Solutions

Homework 3 Solutions CS 341: Foundtions of Computer Science II Prof. Mrvin Nkym Homework 3 Solutions 1. Give NFAs with the specified numer of sttes recognizing ech of the following lnguges. In ll cses, the lphet is Σ = {,1}.

More information

REVIEW Chapter 1 The Real Number System

REVIEW Chapter 1 The Real Number System Mth 7 REVIEW Chpter The Rel Number System In clss work: Solve ll exercises. (Sections. &. Definition A set is collection of objects (elements. The Set of Nturl Numbers N N = {,,,, 5, } The Set of Whole

More information

CS 301. Lecture 04 Regular Expressions. Stephen Checkoway. January 29, 2018

CS 301. Lecture 04 Regular Expressions. Stephen Checkoway. January 29, 2018 CS 301 Lecture 04 Regulr Expressions Stephen Checkowy Jnury 29, 2018 1 / 35 Review from lst time NFA N = (Q, Σ, δ, q 0, F ) where δ Q Σ P (Q) mps stte nd n lphet symol (or ) to set of sttes We run n NFA

More information

Continuous Random Variables Class 5, Jeremy Orloff and Jonathan Bloom

Continuous Random Variables Class 5, Jeremy Orloff and Jonathan Bloom Lerning Gols Continuous Rndom Vriles Clss 5, 8.05 Jeremy Orloff nd Jonthn Bloom. Know the definition of continuous rndom vrile. 2. Know the definition of the proility density function (pdf) nd cumultive

More information

MA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.

MA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp. MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27-233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.

More information

Lecture 3. Limits of Functions and Continuity

Lecture 3. Limits of Functions and Continuity Lecture 3 Limits of Functions nd Continuity Audrey Terrs April 26, 21 1 Limits of Functions Notes I m skipping the lst section of Chpter 6 of Lng; the section bout open nd closed sets We cn probbly live

More information

Linear Systems with Constant Coefficients

Linear Systems with Constant Coefficients Liner Systems with Constnt Coefficients 4-3-05 Here is system of n differentil equtions in n unknowns: x x + + n x n, x x + + n x n, x n n x + + nn x n This is constnt coefficient liner homogeneous system

More information

State Minimization for DFAs

State Minimization for DFAs Stte Minimiztion for DFAs Red K & S 2.7 Do Homework 10. Consider: Stte Minimiztion 4 5 Is this miniml mchine? Step (1): Get rid of unrechle sttes. Stte Minimiztion 6, Stte is unrechle. Step (2): Get rid

More information

10. AREAS BETWEEN CURVES

10. AREAS BETWEEN CURVES . AREAS BETWEEN CURVES.. Ares etween curves So res ove the x-xis re positive nd res elow re negtive, right? Wrong! We lied! Well, when you first lern out integrtion it s convenient fiction tht s true in

More information

Calculus Module C21. Areas by Integration. Copyright This publication The Northern Alberta Institute of Technology All Rights Reserved.

Calculus Module C21. Areas by Integration. Copyright This publication The Northern Alberta Institute of Technology All Rights Reserved. Clculus Module C Ares Integrtion Copright This puliction The Northern Alert Institute of Technolog 7. All Rights Reserved. LAST REVISED Mrch, 9 Introduction to Ares Integrtion Sttement of Prerequisite

More information

Equations and Inequalities

Equations and Inequalities Equtions nd Inequlities Equtions nd Inequlities Curriculum Redy ACMNA: 4, 5, 6, 7, 40 www.mthletics.com Equtions EQUATIONS & Inequlities & INEQUALITIES Sometimes just writing vribles or pronumerls in

More information

20 MATHEMATICS POLYNOMIALS

20 MATHEMATICS POLYNOMIALS 0 MATHEMATICS POLYNOMIALS.1 Introduction In Clss IX, you hve studied polynomils in one vrible nd their degrees. Recll tht if p(x) is polynomil in x, the highest power of x in p(x) is clled the degree of

More information

1 From NFA to regular expression

1 From NFA to regular expression Note 1: How to convert DFA/NFA to regulr expression Version: 1.0 S/EE 374, Fll 2017 Septemer 11, 2017 In this note, we show tht ny DFA cn e converted into regulr expression. Our construction would work

More information

Problem Set 9. Figure 1: Diagram. This picture is a rough sketch of the 4 parabolas that give us the area that we need to find. The equations are:

Problem Set 9. Figure 1: Diagram. This picture is a rough sketch of the 4 parabolas that give us the area that we need to find. The equations are: (x + y ) = y + (x + y ) = x + Problem Set 9 Discussion: Nov., Nov. 8, Nov. (on probbility nd binomil coefficients) The nme fter the problem is the designted writer of the solution of tht problem. (No one

More information

SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 2014

SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 2014 SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 014 Mrk Scheme: Ech prt of Question 1 is worth four mrks which re wrded solely for the correct nswer.

More information

Diophantine Steiner Triples and Pythagorean-Type Triangles

Diophantine Steiner Triples and Pythagorean-Type Triangles Forum Geometricorum Volume 10 (2010) 93 97. FORUM GEOM ISSN 1534-1178 Diophntine Steiner Triples nd Pythgoren-Type Tringles ojn Hvl bstrct. We present connection between Diophntine Steiner triples (integer

More information

AP Calculus BC Chapter 8: Integration Techniques, L Hopital s Rule and Improper Integrals

AP Calculus BC Chapter 8: Integration Techniques, L Hopital s Rule and Improper Integrals AP Clulus BC Chpter 8: Integrtion Tehniques, L Hopitl s Rule nd Improper Integrls 8. Bsi Integrtion Rules In this setion we will review vrious integrtion strtegies. Strtegies: I. Seprte the integrnd into

More information

Problem Set 3

Problem Set 3 14.102 Problem Set 3 Due Tuesdy, October 18, in clss 1. Lecture Notes Exercise 208: Find R b log(t)dt,where0

More information

15 - TRIGONOMETRY Page 1 ( Answers at the end of all questions )

15 - TRIGONOMETRY Page 1 ( Answers at the end of all questions ) - TRIGONOMETRY Pge P ( ) In tringle PQR, R =. If tn b c = 0, 0, then Q nd tn re the roots of the eqution = b c c = b b = c b = c [ AIEEE 00 ] ( ) In tringle ABC, let C =. If r is the inrdius nd R is the

More information

Lecture 2 : Propositions DRAFT

Lecture 2 : Propositions DRAFT CS/Mth 240: Introduction to Discrete Mthemtics 1/20/2010 Lecture 2 : Propositions Instructor: Dieter vn Melkeeek Scrie: Dlior Zelený DRAFT Lst time we nlyzed vrious mze solving lgorithms in order to illustrte

More information

P 1 (x 1, y 1 ) is given by,.

P 1 (x 1, y 1 ) is given by,. MA00 Clculus nd Bsic Liner Alger I Chpter Coordinte Geometr nd Conic Sections Review In the rectngulr/crtesin coordintes sstem, we descrie the loction of points using coordintes. P (, ) P(, ) O The distnce

More information

Section 3.1: Exponent Properties

Section 3.1: Exponent Properties Section.1: Exponent Properties Ojective: Simplify expressions using the properties of exponents. Prolems with exponents cn often e simplied using few sic exponent properties. Exponents represent repeted

More information

dx dt dy = G(t, x, y), dt where the functions are defined on I Ω, and are locally Lipschitz w.r.t. variable (x, y) Ω.

dx dt dy = G(t, x, y), dt where the functions are defined on I Ω, and are locally Lipschitz w.r.t. variable (x, y) Ω. Chpter 8 Stility theory We discuss properties of solutions of first order two dimensionl system, nd stility theory for specil clss of liner systems. We denote the independent vrile y t in plce of x, nd

More information

Math 259 Winter Solutions to Homework #9

Math 259 Winter Solutions to Homework #9 Mth 59 Winter 9 Solutions to Homework #9 Prolems from Pges 658-659 (Section.8). Given f(, y, z) = + y + z nd the constrint g(, y, z) = + y + z =, the three equtions tht we get y setting up the Lgrnge multiplier

More information

8 Laplace s Method and Local Limit Theorems

8 Laplace s Method and Local Limit Theorems 8 Lplce s Method nd Locl Limit Theorems 8. Fourier Anlysis in Higher DImensions Most of the theorems of Fourier nlysis tht we hve proved hve nturl generliztions to higher dimensions, nd these cn be proved

More information

5: The Definite Integral

5: The Definite Integral 5: The Definite Integrl 5.: Estimting with Finite Sums Consider moving oject its velocity (meters per second) t ny time (seconds) is given y v t = t+. Cn we use this informtion to determine the distnce

More information

1. For each of the following theorems, give a two or three sentence sketch of how the proof goes or why it is not true.

1. For each of the following theorems, give a two or three sentence sketch of how the proof goes or why it is not true. York University CSE 2 Unit 3. DFA Clsses Converting etween DFA, NFA, Regulr Expressions, nd Extended Regulr Expressions Instructor: Jeff Edmonds Don t chet y looking t these nswers premturely.. For ech

More information

Introduction to Olympiad Inequalities

Introduction to Olympiad Inequalities Introdution to Olympid Inequlities Edutionl Studies Progrm HSSP Msshusetts Institute of Tehnology Snj Simonovikj Spring 207 Contents Wrm up nd Am-Gm inequlity 2. Elementry inequlities......................

More information

The Modified Heinz s Inequality

The Modified Heinz s Inequality Journl of Applied Mthemtics nd Physics, 03,, 65-70 Pulished Online Novemer 03 (http://wwwscirporg/journl/jmp) http://dxdoiorg/0436/jmp03500 The Modified Heinz s Inequlity Tkshi Yoshino Mthemticl Institute,

More information

#A29 INTEGERS 17 (2017) EQUALITY OF DEDEKIND SUMS MODULO 24Z

#A29 INTEGERS 17 (2017) EQUALITY OF DEDEKIND SUMS MODULO 24Z #A29 INTEGERS 17 (2017) EQUALITY OF DEDEKIND SUMS MODULO 24Z Kurt Girstmir Institut für Mthemtik, Universität Innsruck, Innsruck, Austri kurt.girstmir@uik.c.t Received: 10/4/16, Accepted: 7/3/17, Pulished:

More information

4 VECTORS. 4.0 Introduction. Objectives. Activity 1

4 VECTORS. 4.0 Introduction. Objectives. Activity 1 4 VECTRS Chpter 4 Vectors jectives fter studying this chpter you should understnd the difference etween vectors nd sclrs; e le to find the mgnitude nd direction of vector; e le to dd vectors, nd multiply

More information

The Leaning Tower of Pingala

The Leaning Tower of Pingala The Lening Tower of Pingl Richrd K. Guy Deprtment of Mthemtics & Sttistics, The University of Clgry. July, 06 As Leibniz hs told us, from 0 nd we cn get everything: Multiply the previous line by nd dd

More information

NWI: Mathematics. Various books in the library with the title Linear Algebra I, or Analysis I. (And also Linear Algebra II, or Analysis II.

NWI: Mathematics. Various books in the library with the title Linear Algebra I, or Analysis I. (And also Linear Algebra II, or Analysis II. NWI: Mthemtics Literture These lecture notes! Vrious books in the librry with the title Liner Algebr I, or Anlysis I (And lso Liner Algebr II, or Anlysis II) The lecture notes of some of the people who

More information

, MATHS H.O.D.: SUHAG R.KARIYA, BHOPAL, CONIC SECTION PART 8 OF

, MATHS H.O.D.: SUHAG R.KARIYA, BHOPAL, CONIC SECTION PART 8 OF DOWNLOAD FREE FROM www.tekoclsses.com, PH.: 0 903 903 7779, 98930 5888 Some questions (Assertion Reson tpe) re given elow. Ech question contins Sttement (Assertion) nd Sttement (Reson). Ech question hs

More information

UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE

UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE 1. Pointwise Convergence of Sequence Let E be set nd Y be metric spce. Consider functions f n : E Y for n = 1, 2,.... We sy tht the sequence

More information

Name Ima Sample ASU ID

Name Ima Sample ASU ID Nme Im Smple ASU ID 2468024680 CSE 355 Test 1, Fll 2016 30 Septemer 2016, 8:35-9:25.m., LSA 191 Regrding of Midterms If you elieve tht your grde hs not een dded up correctly, return the entire pper to

More information

440-2 Geometry/Topology: Differentiable Manifolds Northwestern University Solutions of Practice Problems for Final Exam

440-2 Geometry/Topology: Differentiable Manifolds Northwestern University Solutions of Practice Problems for Final Exam 440-2 Geometry/Topology: Differentible Mnifolds Northwestern University Solutions of Prctice Problems for Finl Exm 1) Using the cnonicl covering of RP n by {U α } 0 α n, where U α = {[x 0 : : x n ] RP

More information

Math 113 Exam 2 Practice

Math 113 Exam 2 Practice Mth Em Prctice Februry, 8 Em will cover sections 6.5, 7.-7.5 nd 7.8. This sheet hs three sections. The first section will remind you bout techniques nd formuls tht you should know. The second gives number

More information

The final exam will take place on Friday May 11th from 8am 11am in Evans room 60.

The final exam will take place on Friday May 11th from 8am 11am in Evans room 60. Mth 104: finl informtion The finl exm will tke plce on Fridy My 11th from 8m 11m in Evns room 60. The exm will cover ll prts of the course with equl weighting. It will cover Chpters 1 5, 7 15, 17 21, 23

More information

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES PAIR OF LINEAR EQUATIONS IN TWO VARIABLES. Two liner equtions in the sme two vriles re lled pir of liner equtions in two vriles. The most generl form of pir of liner equtions is x + y + 0 x + y + 0 where,,,,,,

More information

Math 8 Winter 2015 Applications of Integration

Math 8 Winter 2015 Applications of Integration Mth 8 Winter 205 Applictions of Integrtion Here re few importnt pplictions of integrtion. The pplictions you my see on n exm in this course include only the Net Chnge Theorem (which is relly just the Fundmentl

More information

Triangles The following examples explore aspects of triangles:

Triangles The following examples explore aspects of triangles: Tringles The following exmples explore spects of tringles: xmple 1: ltitude of right ngled tringle + xmple : tringle ltitude of the symmetricl ltitude of n isosceles x x - 4 +x xmple 3: ltitude of the

More information

63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1

63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1 3 9. SEQUENCES AND SERIES 63. Representtion of functions s power series Consider power series x 2 + x 4 x 6 + x 8 + = ( ) n x 2n It is geometric series with q = x 2 nd therefore it converges for ll q =

More information

ENGI 3424 Engineering Mathematics Five Tutorial Examples of Partial Fractions

ENGI 3424 Engineering Mathematics Five Tutorial Examples of Partial Fractions ENGI 44 Engineering Mthemtics Five Tutoril Exmples o Prtil Frctions 1. Express x in prtil rctions: x 4 x 4 x 4 b x x x x Both denomintors re liner non-repeted ctors. The cover-up rule my be used: 4 4 4

More information

CS 330 Formal Methods and Models

CS 330 Formal Methods and Models CS 0 Forml Methods nd Models Dn Richrds, George Mson University, Fll 2016 Quiz Solutions Quiz 1, Propositionl Logic Dte: Septemer 8 1. Prove q (q p) p q p () (4pts) with truth tle. p q p q p (q p) p q

More information

9.4. The Vector Product. Introduction. Prerequisites. Learning Outcomes

9.4. The Vector Product. Introduction. Prerequisites. Learning Outcomes The Vector Product 9.4 Introduction In this section we descrie how to find the vector product of two vectors. Like the sclr product its definition my seem strnge when first met ut the definition is chosen

More information

LINEAR ALGEBRA APPLIED

LINEAR ALGEBRA APPLIED 5.5 Applictions of Inner Product Spces 5.5 Applictions of Inner Product Spces 7 Find the cross product of two vectors in R. Find the liner or qudrtic lest squres pproimtion of function. Find the nth-order

More information

Algebra II Notes Unit Ten: Conic Sections

Algebra II Notes Unit Ten: Conic Sections Syllus Ojective: 10.1 The student will sketch the grph of conic section with centers either t or not t the origin. (PARABOLAS) Review: The Midpoint Formul The midpoint M of the line segment connecting

More information

Analytically, vectors will be represented by lowercase bold-face Latin letters, e.g. a, r, q.

Analytically, vectors will be represented by lowercase bold-face Latin letters, e.g. a, r, q. 1.1 Vector Alger 1.1.1 Sclrs A physicl quntity which is completely descried y single rel numer is clled sclr. Physiclly, it is something which hs mgnitude, nd is completely descried y this mgnitude. Exmples

More information

38 Riemann sums and existence of the definite integral.

38 Riemann sums and existence of the definite integral. 38 Riemnn sums nd existence of the definite integrl. In the clcultion of the re of the region X bounded by the grph of g(x) = x 2, the x-xis nd 0 x b, two sums ppered: ( n (k 1) 2) b 3 n 3 re(x) ( n These

More information

Project 6: Minigoals Towards Simplifying and Rewriting Expressions

Project 6: Minigoals Towards Simplifying and Rewriting Expressions MAT 51 Wldis Projet 6: Minigols Towrds Simplifying nd Rewriting Expressions The distriutive property nd like terms You hve proly lerned in previous lsses out dding like terms ut one prolem with the wy

More information

1 Nondeterministic Finite Automata

1 Nondeterministic Finite Automata 1 Nondeterministic Finite Automt Suppose in life, whenever you hd choice, you could try oth possiilities nd live your life. At the end, you would go ck nd choose the one tht worked out the est. Then you

More information

z TRANSFORMS z Transform Basics z Transform Basics Transfer Functions Back to the Time Domain Transfer Function and Stability

z TRANSFORMS z Transform Basics z Transform Basics Transfer Functions Back to the Time Domain Transfer Function and Stability TRASFORS Trnsform Bsics Trnsfer Functions Bck to the Time Domin Trnsfer Function nd Stility DSP-G 6. Trnsform Bsics The definition of the trnsform for digitl signl is: -n X x[ n is complex vrile The trnsform

More information

STRAND B: NUMBER THEORY

STRAND B: NUMBER THEORY Mthemtics SKE, Strnd B UNIT B Indices nd Fctors: Tet STRAND B: NUMBER THEORY B Indices nd Fctors Tet Contents Section B. Squres, Cubes, Squre Roots nd Cube Roots B. Inde Nottion B. Fctors B. Prime Fctors,

More information

Intuitionistic Fuzzy Lattices and Intuitionistic Fuzzy Boolean Algebras

Intuitionistic Fuzzy Lattices and Intuitionistic Fuzzy Boolean Algebras Intuitionistic Fuzzy Lttices nd Intuitionistic Fuzzy oolen Algebrs.K. Tripthy #1, M.K. Stpthy *2 nd P.K.Choudhury ##3 # School of Computing Science nd Engineering VIT University Vellore-632014, TN, Indi

More information

Similarity and Congruence

Similarity and Congruence Similrity nd ongruence urriculum Redy MMG: 201, 220, 221, 243, 244 www.mthletics.com SIMILRITY N ONGRUN If two shpes re congruent, it mens thy re equl in every wy ll their corresponding sides nd ngles

More information

CS 330 Formal Methods and Models

CS 330 Formal Methods and Models CS 330 Forml Methods nd Models Dn Richrds, section 003, George Mson University, Fll 2017 Quiz Solutions Quiz 1, Propositionl Logic Dte: Septemer 7 1. Prove (p q) (p q), () (5pts) using truth tles. p q

More information

Homework Assignment 3 Solution Set

Homework Assignment 3 Solution Set Homework Assignment 3 Solution Set PHYCS 44 6 Ferury, 4 Prolem 1 (Griffiths.5(c The potentil due to ny continuous chrge distriution is the sum of the contriutions from ech infinitesiml chrge in the distriution.

More information

MATH34032: Green s Functions, Integral Equations and the Calculus of Variations 1

MATH34032: Green s Functions, Integral Equations and the Calculus of Variations 1 MATH34032: Green s Functions, Integrl Equtions nd the Clculus of Vritions 1 Section 1 Function spces nd opertors Here we gives some brief detils nd definitions, prticulrly relting to opertors. For further

More information

Handout: Natural deduction for first order logic

Handout: Natural deduction for first order logic MATH 457 Introduction to Mthemticl Logic Spring 2016 Dr Json Rute Hndout: Nturl deduction for first order logic We will extend our nturl deduction rules for sententil logic to first order logic These notes

More information

Chapter 4 Contravariance, Covariance, and Spacetime Diagrams

Chapter 4 Contravariance, Covariance, and Spacetime Diagrams Chpter 4 Contrvrince, Covrince, nd Spcetime Digrms 4. The Components of Vector in Skewed Coordintes We hve seen in Chpter 3; figure 3.9, tht in order to show inertil motion tht is consistent with the Lorentz

More information

ECON 331 Lecture Notes: Ch 4 and Ch 5

ECON 331 Lecture Notes: Ch 4 and Ch 5 Mtrix Algebr ECON 33 Lecture Notes: Ch 4 nd Ch 5. Gives us shorthnd wy of writing lrge system of equtions.. Allows us to test for the existnce of solutions to simultneous systems. 3. Allows us to solve

More information

Example 1: Express as a sum of logarithms by using the Product Rule. (By the definition of log)

Example 1: Express as a sum of logarithms by using the Product Rule. (By the definition of log) Section 5. Properties of Logrithmic Functions Section 5. Properties of Logrithmic Functions This section covers some properties of rithmic function tht re very similr to the rules for exponents. Properties

More information