FIRST MIDTERM MATH 104B, UCSD, WINTER 18 You have 80 minutes. There are 4 problems, and the total number of points is 70. Show all your work. Please make your work as clear and easy to follow as possible. Name: Signature: Student ID #: Problem Points Score 1 15 2 15 3 30 4 10 5 10 6 10 Total 70 1
1. (15pts) (i) Give the definition of π(). The number of primes up to. (ii) Give the definition of the Möbius function. The function µ: N N defined by the rule { ( 1) ν if n is the product of ν distinct primes µ(n) = 0 otherwise. (iii) Give the definition of the fractional part. If is real number and is the largest integer less than then the fractional part is {} = 1
2. (15pts) Let, 1 and 2 be real numbers and let n be an integer. Prove that (i) +n = +n. As it follows that + n + n. As + n + n it follows that + n n. As the LHS is an integer it follows that + n n. Adding n to both sides we get + n +n. As we have an inequality both ways we must have an equality +n = +n.. (ii) 1 + 2 1 + 2. Since i i for i = 1,2 we have 1 + 2 1 + 2. As the LHS is an integer it follows that 1 + 2 1 + 2. (iii) Assuming that n is a natural number, prove that = n n. As But then n n as the LHS is an integer. On the other hand, as so that n n it follows that so that n n. n n, it follows that n. n and so n n n n. As we have an inequality both ways, we have equality. 2
3. (30pts) Let f: N C be a function and define F: N C by the rule F(n) = f(d). d n (i) Show that if f is multiplicative then F is multiplicative. Suppose that m and n are coprime. Note that if d divides mn then d = d 1 d 2 where d 1 divides m and d 2 divides n. We have F(mn) = d mnf(d) = d 1 m,d 2 n f(d 1 d 2 ) = d 1 m d2 nf(d 1 )f(d 2 ) = d 1 m f(d 1 ) d 2 nf(d 2 ) Thus F is multiplicative. = F(m)F(n). 3
(ii) If the function M: N Z is defined by M(n) = d n µ(d) then show that M(n) = { 1 if n = 1 0 otherwise. Consider M(n) = d n µ(d). As µ(n) is multiplicative both sides are multiplicative. If n = p e is a power of a prime then M(p e ) = µ(1)+µ(p)+µ(p 2 )+ = 1 1+0+ +0 = 0. Thus M(n) = 0 unless n = 1 in which case M(1) = 1. (iii) Show that f(n) = d n µ(d)f( n d ). We have µ(d)f( n d ) = µ(d 1 )F(d 2 ) d n d 1 d 2 =n = µ(d 1 ) f(d) d d 2 d 1 d 2 =n = d 1 d nµ(d 1 )f(d) = d n = d n f(d) µ(d 1 ) d 1 n/d f(d)m(n/d) = f(n). 4
4. (10pts) Show that π(n) logn 2log2. Let r = π(n) and let p 1,p 2,...,p r be the first r primes, so that p 1,p 2,...,p r are the primes up to n. Note that we may form 2 r distinct square-freenaturalnumbersmwhichareonlydivisiblebyp 1,p 2,...,p r. For each prime p i we either choose m coprime to p i or divisible by p i. On other hand there are at most n perfect squares up n. Now any natural number l is the product of a perfect square and a square-free number m. If l n then m n and so m is divisible by p 1,p 2,...,p r. Thus there are at most 2 r n numbers up to n. On the other hand there are n natural numbers up to n. Thus 2 π(n) n n so that 2 π(n) n. Taking logs we see that π(n)log2 = log2 π(n) log n = 1 2 logn. Thus π(n) logn 2log2. 5
5. Show that 1 = +O(). p i p j p p i p j,p i <p i p j j i<j=1 Note that where i,j=1:p i <p j p i p j = E p p i p j,p i <p i p j j p i p j,p i <p j p i p j,p i <p j 1. p i p j E, The term on the RHS is the number of ways to pick two primes p i, p j such that p i < p j and p i p j. Let y = p i p j. Then y determines p i and p j by unique factorisation and y is a natural number between 1 and so that y. Thus E is at most and so E = O(1). 6
Bonus Challenge Problems 6. (10pts) Show that there is a constant γ such that n ( ) 1 1 = logn+γ +O. k n Let Note that k=1 α k = logk log(k 1) 1 k Note also that k k 1 1 γ n = and γ n = n α k. k=2 1 d = [log]k k 1 = logk log(k 1) n k=1 1 k logn. is the area under the curve y = 1/ over the interval k 1 k. On the other hand 1/k is the area over the interval k 1 k inside the largest rectangle inscribed between the -ais and the curve y = 1/. It follows that α k is the difference between these two areas, so that α k is positive. Note that if we drop these areas down to the region between = 0 and = 1 then all of these areas fit into the unit square bounded by y = 0 and y = 1. Thus 0 < 1 γ n < 1 is bounded and monotonic increasing. It follows that 1 γ n tends to a limit. Define γ by the formula: lim (1 γ n) = 1 γ. n Finally note that the difference γ n γ = (1 γ) (1 γ n ) = k=n+1 is represented by an area which fits inside a bo with one side 1 and the other side 1/n, so that it is less than 1/n. Thus ( ) 1 γ n γ = O. n α k 7
7. (10pts) Show that ( π() = O ). The number of integers up to not divisible by the first r primes p 1,p 2,...,p r is r A(,r) = + + +( 1) r. p i p i p j p 1 p 2...p r i=1 i j r by inclusion-eclusion. If we approimate this by ignoring the round downs the error is at most ( ) ( ) ( ) r r r 1+ + + + = 2 r, 1 2 r and so π() r+ r ) (1 1pi +2 r. i=1 Now 1 1 1 = (1+ 1p + 1p ) +.... p p 2 p If we epand the RHS we get the sum of the reciprocals of all numbers divisible by p 1,p 2,...,p r, which is at least 1 k > logn. k n Thus π() r+ log +2r. If we take r = log then π() 2 log+2 + take r = log = O ( 2 log) + ( ) o + ( ) = O. as log2 < 1 8