Estimating the Growth Rate of the Zeta Function Using Exponent Pairs - PowerPoint PPT Presentation

Estimating the Growth Rate of the Zeta Function Using Exponent Pairs Shreejit Bandyopadhyay July 06, 2015

The Riemann Zeta Function • The Riemann-zeta function is of prime interest in Number Theory and also in many other branches of mathematics.

The Riemann Zeta Function • The Riemann-zeta function is of prime interest in Number Theory and also in many other branches of mathematics. 1 � • It can be defined either by the infinite summation ζ ( s ) = n s n ∈ N p (1 − 1 p s ) − 1 . or as the product �

The Riemann Zeta Function • The Riemann-zeta function is of prime interest in Number Theory and also in many other branches of mathematics. 1 � • It can be defined either by the infinite summation ζ ( s ) = n s n ∈ N p (1 − 1 p s ) − 1 . or as the product � • The equivalence of these two definitions follows by factoring n as a product of primes and the fundamental theorem of arithmetic.

The Riemann Zeta Function • The Riemann-zeta function is of prime interest in Number Theory and also in many other branches of mathematics. 1 � • It can be defined either by the infinite summation ζ ( s ) = n s n ∈ N p (1 − 1 p s ) − 1 . or as the product � • The equivalence of these two definitions follows by factoring n as a product of primes and the fundamental theorem of arithmetic. • Both the sum and the product converge for σ = Re ( s ) > 1. We note that | n − s | = | n − σ | .

The Zeta Function cont.

The Zeta Function cont. Other representations of the zeta function include � ∞ x s − 1 1 • (Integral form): ζ ( s ) = e x − 1 dx , Γ( s ) 0

The Zeta Function cont. Other representations of the zeta function include � ∞ x s − 1 1 • (Integral form): ζ ( s ) = e x − 1 dx , Γ( s ) 0 • (Functional equation): ζ ( s ) = 2 s π s − 1 sin ( π s 2 )Γ(1 − s ) ζ (1 − s ). We often write χ ( s ) = 2 s π s − 1 sin ( π s 2 )Γ)(1 − s ) to write the functional equation in a more compact form as ζ ( s ) = χ ( s ) ζ (1 − s ).

The Zeta Function cont. Other representations of the zeta function include � ∞ x s − 1 1 • (Integral form): ζ ( s ) = e x − 1 dx , Γ( s ) 0 • (Functional equation): ζ ( s ) = 2 s π s − 1 sin ( π s 2 )Γ(1 − s ) ζ (1 − s ). We often write χ ( s ) = 2 s π s − 1 sin ( π s 2 )Γ)(1 − s ) to write the functional equation in a more compact form as ζ ( s ) = χ ( s ) ζ (1 − s ). • (Approximate functional equation):With s = σ + it and t = 2 π xy , this is 1 1 1 � � n 1 − s + O ( x − σ log | t | ) + O ( | t | 2 − σ y σ − 1 ). ζ ( s ) = n s + χ ( s ) n ≤ x n ≤ y

Zeta Function in the Critical Strip

Zeta Function in the Critical Strip • The critical strip is the infinite strip 0 < σ < 1. We wish to study the rate of growth of the zeta function in this strip.

Zeta Function in the Critical Strip • The critical strip is the infinite strip 0 < σ < 1. We wish to study the rate of growth of the zeta function in this strip. • For a given real part σ , we define µ ( σ ) = inf { α > 0 | ζ ( σ + it ) << O ( t α ) } .

Zeta Function in the Critical Strip • The critical strip is the infinite strip 0 < σ < 1. We wish to study the rate of growth of the zeta function in this strip. • For a given real part σ , we define µ ( σ ) = inf { α > 0 | ζ ( σ + it ) << O ( t α ) } . • An important hypothesis in this direction is the Lindelof hypothesis. It claims that ζ ( 1 2 + it ) << O ( t ǫ ) for any ǫ > 0.

Zeta Function in the Critical Strip cont.

Zeta Function in the Critical Strip cont. • It has been shown that the Lindelof hypothesis is equivalent to showing that the number of zeroes of the zeta function in the critical strip with real part greater than 1 2 and imaginary part between t and t + 1 is o ( log t ).

Zeta Function in the Critical Strip cont. • It has been shown that the Lindelof hypothesis is equivalent to showing that the number of zeroes of the zeta function in the critical strip with real part greater than 1 2 and imaginary part between t and t + 1 is o ( log t ). • The Riemann hypothesis, of course, claims that there are no such zeroes at all. So the Lindelof hypothesis is much weaker than the Riemann hypothesis.

Zeta Function in the Critical Strip cont. • It has been shown that the Lindelof hypothesis is equivalent to showing that the number of zeroes of the zeta function in the critical strip with real part greater than 1 2 and imaginary part between t and t + 1 is o ( log t ). • The Riemann hypothesis, of course, claims that there are no such zeroes at all. So the Lindelof hypothesis is much weaker than the Riemann hypothesis. • But, despite the efforts of many people, even the Lindelof hypothesis is far from being proved.

Zeta Function as Exponential Sum

Zeta Function as Exponential Sum • Our goal will be to reduce the sum involved in the zeta function to an exponential sum.

Zeta Function as Exponential Sum • Our goal will be to reduce the sum involved in the zeta function to an exponential sum. • We see that n − s = n − σ − it << N − σ n − it | . � � � N < u ≤ M | max N < n ≤ M N < n ≤ M N < n ≤ u

Zeta Function as Exponential Sum • Our goal will be to reduce the sum involved in the zeta function to an exponential sum. • We see that n − s = n − σ − it << N − σ n − it | . � � � N < u ≤ M | max N < n ≤ M N < n ≤ M N < n ≤ u n − it then � • This holds because if S ( u ) = N < n ≤ u � M n − σ − it = S ( M ) M − σ + σ � S ( u ) u − σ − 1 du by partial N N < n ≤ M summation. Putting the value of S ( u ), we get our estimate.

Zeta Function as Exponential Sum • Our goal will be to reduce the sum involved in the zeta function to an exponential sum. • We see that n − s = n − σ − it << N − σ n − it | . � � � N < u ≤ M | max N < n ≤ M N < n ≤ M N < n ≤ u n − it then � • This holds because if S ( u ) = N < n ≤ u � M n − σ − it = S ( M ) M − σ + σ � S ( u ) u − σ − 1 du by partial N N < n ≤ M summation. Putting the value of S ( u ), we get our estimate. e ( − logn t n − it = e 2 π i ( − logn t 2 π ) = � � � • Writing 2 π ), we get an exponential sum!

Zeta Function as Exponential Sum cont.

Zeta Function as Exponential Sum cont. • Using partial summation, we can also show that � n − σ − it | + t 1 − 2 σ log t . | ζ ( σ + it ) | << | n ≤ t

Zeta Function as Exponential Sum cont. • Using partial summation, we can also show that � n − σ − it | + t 1 − 2 σ log t . | ζ ( σ + it ) | << | n ≤ t • This allows us to look at the zeta function in terms of exponential sums.

Exponential Sums

Exponential Sums � • An exponential sum is any sum of the form e ( f ( n )) where f n ∈ I is a function, I a real interval and e ( x )) stands for e 2 π ix . Often, we shall have to assume some nice additional properties of f like continuity, differentiability, etc.

Exponential Sums � • An exponential sum is any sum of the form e ( f ( n )) where f n ∈ I is a function, I a real interval and e ( x )) stands for e 2 π ix . Often, we shall have to assume some nice additional properties of f like continuity, differentiability, etc. • By the triangle inequality, � � | S | = | e ( f ( n )) | ≤ | e ( f ( n )) | ≤ | I | if the interval I has n ∈ I n ∈ I integer endpoints. We see that equality holds if f ( n ) = xn + y and x is an integer and the interval I is closed and has integer endpoints. So we need additional conditions on f to improve our bound.

Exponential Sums cont.

Exponential Sums cont. • Weyl introduced a trick of dealing with such sums. He wrote | S | 2 = � � � e ( f ( m ) − f ( n )) = e ( f ( n + h ) − f ( n )) a < m , n ≤ b | h | < b − a n ∈ I h where I r = { n | a < n , n + r ≤ b } . This holds because | S | 2 = � � e ( f ( m )) e ( f ( n )) = a < m ≤ b a < n ≤ b � � � e ( f ( m )) e ( − f ( n )) = e ( f ( m ) − f ( n )). a < m ≤ b a < n ≤ b a < m , n ≤ b

Exponential Sums cont. • Weyl introduced a trick of dealing with such sums. He wrote | S | 2 = � � � e ( f ( m ) − f ( n )) = e ( f ( n + h ) − f ( n )) a < m , n ≤ b | h | < b − a n ∈ I h where I r = { n | a < n , n + r ≤ b } . This holds because | S | 2 = � � e ( f ( m )) e ( f ( n )) = a < m ≤ b a < n ≤ b � � � e ( f ( m )) e ( − f ( n )) = e ( f ( m ) − f ( n )). a < m ≤ b a < n ≤ b a < m , n ≤ b • This helps us because the difference function f ( n + h ) − f ( n ) is usually easier to handle than f ( n ) itself. For example, if f is a polynomial, it gives a polynomial of one degree less in the exponent. Using this repeatedly, we may want to get a linear polynomial in the exponent.