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The meromorphic continuation iii of the zeta function is initially surprising, but can be interpreted either as a manifestation of the extremely regular spacing of the natural numbers occurring in the sum 1 , or as a consequence of various integral representations of or slight modifications thereof. We will focus in this set of notes on a particular representation of as essentially the Mellin transform of the theta function that briefly appeared in previous notes, and the functional equation iv can then be viewed as a consequence of the modularity of that theta function.
This in turn was established using the Poisson summation formula, so one can view the functional equation as ultimately being a manifestation of Poisson summation. For a direct proof of the functional equation via Poisson summation, see these notes. Henceforth we work with the meromorphic continuation of. The functional equation iv , when combined with special values of such as 2 , gives some additional values of outside of its initial domain , most famously.
From Theorem 3 and the non-vanishing nature of , we see Byjus Class 12 Maths Notes Solution that has simple zeroes known as trivial zeroes at the negative even integers , and all other zeroes the non-trivial zeroes inside the critical strip. The non-trivial zeroes are conjectured to all be simple, but this is hopelessly far from being proven at present.
As we shall see shortly, these latter zeroes turn out to be closely related to the distribution of the primes. The functional equation tells us that if is a non-trivial zero then so is ; also, we have the identity. Conjecture 4 Riemann hypothesis All the non-trivial zeroes of lie on the critical line. This conjecture would have many implications in analytic number theory, particularly with regard to the distribution of the primes.
Of course, it is far from proven at present, but the partial results we have towards this conjecture are still sufficient to establish results such as the prime number theorem. Return now to the original region where.
To take more advantage of the Euler product formula 3 , we take complex logarithms to conclude that. The series and that show up in the above formulae are examples of Dirichlet series, which are a convenient device to transform various sequences of arithmetic interest into holomorphic or meromorphic functions.
Here are some more examples:. Exercise 5 Standard Dirichlet series Let be a complex number with. Given the appearance of the von Mangoldt function , it is natural to reformulate the prime number theorem in terms of this function:. Theorem 6 Prime number theorem, von Mangoldt form One has. Let us see how Theorem 6 implies Theorem 1. Firstly, for any , we can write.
Exercise 7 Show that Theorem 1 conversely implies Theorem 6. The alternate form 8 of the Euler product identity connects the primes represented here via proxy by the von Mangoldt function with the logarithmic derivative of the zeta function, and can be used as a starting point for describing further relationships between and the primes. Most famously, we shall see later in these notes that it leads to the remarkably precise Riemann-von Mangoldt explicit formula :.
Theorem 8 Riemann-von Mangoldt explicit formula For any non-integer , we have. Actually, it turns out that this formula is in some sense too precise; in applications it is often more convenient to work with smoothed variants of this formula in which the sum on the left-hand side is smoothed out, but the contribution of zeroes with large imaginary part is damped; see Exercise Nevertheless, this formula clearly illustrates how the non-trivial zeroes of the zeta function influence the primes.
Indeed, if one formally differentiates the above formula in , one is led to the quite nonrigorous approximation. Comparing Theorem 8 with Theorem 6 , it is natural to suspect that the key step in the proof of the latter is to establish the following slight but important extension of Theorem 3 ii , which can be viewed as a very small step towards the Riemann hypothesis:.
Theorem 9 Slight enlargement of zero-free region There are no zeroes of on the line. It is not quite immediate to see how Theorem 6 follows from Theorem 8 and Theorem 9 , but we will demonstrate it below the fold.
Although Theorem 9 only seems like a slight improvement of Theorem 3 ii , proving it is surprisingly non-trivial. The basic idea is the following: if there was a zero at , then there would also be a different zero at note cannot vanish due to the pole at , and then the approximation 9 becomes.
Exercise 10 For the purposes of this exercise, assume Theorem 6 , but do not assume Theorem 9. For any non-zero real , show that. This equivalence can help explain why the prime number theorem is remarkably non-trivial to prove, and why the Riemann zeta function has to be either explicitly or implicitly involved in the proof. This post is only intended as the briefest of introduction to complex-analytic methods in analytic number theory; also, we have not chosen the shortest route to the prime number theorem, electing instead to travel in directions that particularly showcase the complex-analytic results introduced in this course.
For some further discussion see this previous set of lecture notes , particularly Notes 2 and Supplement 3 with much of the material in this post drawn from the latter. Read the rest of this entry �. We were saddened and horrified to learn that Ilya Dumanski, a brilliant young mathematician who has been admitted to our graduate programs at Stanford and MIT, has been imprisoned in Russia, along with several other mathematicians, for participation in a peaceful demonstration.
Our thoughts are with them. We urge their rapid release, and failing that, that they be kept in humane conditions. A petition in their support has been started at. NT Tags: complex tori , elliptic curves , j-invariant , modular discriminant , modular forms , modular functions , Weierstrass elliptic function by Terence Tao 32 comments. Previous set of notes: Notes 2.
Next set of notes: Notes 4. On the real line, the quintessential examples of a periodic function are the normalised sine and cosine functions , , which are -periodic in the sense that. What about periodic functions on the complex plane?
We can rescale as before. For holomorphic functions, we have the following characterisations:. Proposition 1 Description of singly periodic holomorphic functions i Every -periodic entire function has an absolutely convergent expansion.
As the map is a covering map from to , we see that will be holomorphic if and only if is. Thus must have a Laurent series expansion with coefficients obeying 2 , which gives 1 , and the inversion formula 5 follows from the usual contour integration formula for Laurent series coefficients. The converse direction to i also follows by reversing the above arguments. For part ii , we observe that the map is also a covering map from to the punctured disk , so we can argue as before except that now is a bounded holomorphic function on the punctured disk.
By the Riemann singularity removal theorem Exercise 35 of A Notes 3 extends to be holomorphic on all of , and thus has a Taylor expansion for some coefficients obeying 4.
The argument now proceeds as with part i. The additive cylinder and the multiplicative cylinder can both be identified on the level of smooth manifolds, at least with the geometric cylinder , but we will not use this identification here. Now let us turn attention to doubly periodic functions of a complex variable , that is to say functions that obey two periodicity relations.
Within the world of holomorphic functions, the collection of doubly periodic functions is boring:. Proposition 2 Let be an entire doubly periodic function with periods linearly independent over. Then is constant. In the language of Riemann surfaces, this proposition asserts that the torus is a non-hyperbolic Riemann surface; it cannot be holomorphically mapped non-trivially into a bounded subset of the complex plane.
Proof: The fundamental domain up to boundary enclosed by is compact, hence is bounded on this domain, hence bounded on all of by double periodicity. One could alternatively have argued here using the compactness of the torus. To obtain more interesting examples of doubly periodic functions, one must therefore turn to the world of meromorphic functions � or equivalently, holomorphic functions into the Riemann sphere.
As it turns out, a particularly fundamental example of such a function is the Weierstrass elliptic function. CA , math. Previous set of notes: Notes 1. Next set of notes: Notes 3. In Exercise 5 and Lemma 1 of A Notes 4 we already observed some links between complex analysis on the disk or annulus and Fourier series on the unit circle:.
This connection lets us interpret various facts about Fourier series through the lens of complex analysis, at least for some special classes of Fourier series.
For instance, the Fourier inversion formula becomes the Cauchy-type formula for the Laurent or Taylor coefficients of , in the event that the coefficients are doubly infinite and obey 2 for some , or singly infinite and obey 1 for some.
It turns out that there are similar links between complex analysis on a half-plane or strip and Fourier integrals on the real line, which we will explore in these notes. We first fix a normalisation for the Fourier transform. If is an absolutely integrable function on the real line, we define its Fourier transform by the formula.
Exercise 1 Fourier transform of Gaussian If is a coplex number with and is the Gaussian function , show that the Fourier transform is given by the Gaussian , where we use the standard branch for. The Fourier transform has many remarkable properties. Exercise 2 Decay of implies regularity of Let be an absolutely integrable function. Furthermore, this function continues to be defined by 3. In particular, if is supported in then. For uniqueness, use analytic continuation, or for part iv the Cauchy integral formula.
Later in these notes we will give a partial converse to part ii of this exercise, known as the Paley-Wiener theorem ; there are also partial converses to the other parts of this exercise. From 3 we observe the following intertwining property between multiplication by an exponential and complex translation: if is a complex number and is an absolutely integrable function such that the modulated function is also absolutely integrable, then we have the identity.
AP , math. However, as was first discovered in by my two coauthors , in these situations one can replace the rough potential by a smoother effective potential , with the eigenfunctions typically localised to a single connected component of the effective wells. In fact, a good choice of effective potential comes from locating the landscape function , which is the solution to the equation with reasonable behavior at infinity, and which is non-negative from the maximum principle, and then the reciprocal of this landscape function serves as an effective potential.
There are now several explanations for why this particular choice is a good effective potential. Closely related to this, we have the integration by parts identity. Our chosen model is that of an -matrix : self-adjoint positive definite matrices whose off-diagonal entries are negative. The analogue of the landscape function would then be the vector , where denotes the vector with all entries. In particular, we establish the inequality. It turns out that this argument extends without much difficulty to the -matrix setting.
The analogue of the crucial double commutator identity 2 is. Numerically we have also found some aspects of the landscape theory to persist beyond the -matrix setting, even though the double commutators cease being negative definite, so this may not yet be the end of the story, but it does at least demonstrate that utility the landscape does not purely rely on identities such as 1. CV Tags: gamma function , Hadamard factorisation theorem , Jensen's formula , Weierstrass factorisation theorem by Terence Tao 65 comments.
Previous set of notes: A Notes 5. Next set of notes: Notes 2. Suppose is a non-zero rational function , then by the fundamental theorem of algebra one can write. Exercise 1 Let be a complex polynomial of degree. There are a number of useful ways to extend these formulae to more general meromorphic functions than rational functions.
Then, if is neither a zero nor a pole of , we have. One can view 3 as a truncated or localised variant of 1. Note also that the summands are always non-positive.
Proof: By perturbing slightly if necessary, we may assume that none of the zeroes or poles of which form a discrete set lie on the boundary circle. By translating and rescaling, we may then normalise and , thus our task is now Byjus Maths Class 8 Notes Youtube to show that. Note that the fundamental theorem was invoked previously in this section, but only for motivational purposes, so the proof here is non-circular.
Show that. Exercise 6 Bounded type i If is a holomorphic function on that is not identically zero, show that. Functions of this form are said to be of bounded type and lie in the Nevanlinna class for the unit disk.
Exercise 7 Smoothed out Jensen formula Let be a meromorphic function on an open set , and let be a smooth compactly supported function. Here is a typical result:. Proposition 8 Let be an entire function, not identically zero, that obeys a growth bound for some and all.
Then there exists a constant such that has at most zeroes counting multiplicity for any. Entire functions that obey a growth bound of the form for every and where depends on are said to be of order at most.
The above theorem shows that for such functions that are not identically zero, the number of zeroes in a disk of radius does not grow much faster than. This is often a useful preliminary upper bound on the zeroes of entire functions, as the order of an entire function tends to be relatively easy to compute in practice. Proof: First suppose that is non-zero. From 6 applied with and one has. Just as 3 and 7 give truncated variants of 1 , we can create truncated versions of 2.
The following crude truncation is adequate for many applications:. Theorem 9 Truncated formula for log-derivative Let be a holomorphic function on an open neighbourhood of a disk that is not identically zero on this disk.
Suppose that one has a bound of the form for some and all on the circle. Let be constants. Then one has the approximate formula. Proof: To abbreviate notation, we allow all implied constants in this proof to depend on.
Firstly, we may translate and rescale so that and , so we have when , and our main task is to show that. Suppose has a zero with. If we factor , where is the Blaschke product 5 , then. Similarly, given a zero with , we have , so using Blaschke products to remove all of these zeroes also only affects the left-hand side of 8 by since the number of zeroes here is , with also modified by at most. Thus we may assume in fact that has no zeroes whatsoever within the unit disk.
We may then also normalise , then for all. A variant of the above argument allows one to make precise the heuristic that holomorphic functions locally look like polynomials:.
Exercise 11 Local Weierstrass factorisation Let the notation and hypotheses be as in Theorem 9. Then show that. Exercise 12 Preliminary Beurling factorisation Let denote the space of bounded analytic functions on the unit disk; this is a normed vector space with norm. Remark 13 The factorisation iii can be refined further, with being the Poisson integral of some finite measure on the unit circle. There are also extensions to larger spaces than which are to as is to , known as Hardy spaces.
We will not discuss this topic further here, but see for instance this text of Garnett for a treatment. CV by Terence Tao 11 comments. Just a short announcement that next quarter I will be continuing the recently concluded A complex analysis class as B. Topics I plan to cover:.
Blog at WordPress. Ben Eastaugh and Chris Sternal-Johnson. Subscribe to feed. What's new Updates on my research and expository papers, discussion of open problems, and other maths-related topics. Boosting the van der Corput inequality using the tensor power trick 27 February, in expository , math. In this previous blog post I noted the following easy application of Cauchy-Schwarz: Lemma 1 Van der Corput inequality Let be unit vectors in a Hilbert space.
Then Proof: The left-hand side may be written as for some unit complex numbers. By Cauchy-Schwarz we have and the claim now follows from the triangle inequality.
As a corollary, correlation becomes transitive in a statistical sense even though it is not transitive in an absolute sense : Corollary 2 Statistical transitivity of correlation Let be unit vectors in a Hilbert space such that for all and some. Then we have for at least of the pairs. Proof: From the lemma, we have The contribution of those with is at most , and all the remaining summands are at most , giving the claim. While working on an ongoing research project, I recently found that there is a very simple way to get around the latter problem by exploiting the tensor power trick : Corollary 3 Simultaneous statistical transitivity of correlation Let be unit vectors in a Hilbert space for and such that for all , and some.
Then there are at least pairs such that. In particular by Cauchy-Schwarz we have for all. We start with the identity where is the orthogonal projection to the complement of. This implies a Gram matrix inequality for each where denotes the claim that is positive semi-definite.
By the Schur product theorem, we conclude that and hence for a suitable choice of signs , One now argues as in the proof of Corollary 2. The inclusion-exclusion principle for commuting projections 23 February, in expository , math. Iterating this we obtain ii. Conversely, from ii and separating out the term, and grouping all the other terms based on the value of , we obtain 1 , and hence i. It is not difficult to see that Proposition 2 includes Proposition 1 as a special case, after verifying the combinatorial fact that the quantity is equal to when divides , and vanishes otherwise.
The classical version of this principle can be phrased in terms of indicator functions: if are subsets of some set , then In particular, if there is a finite measure on for which are all measurable, we have One drawback of this formula is that there are exponentially many terms on the right-hand side: of them, in fact.
It turns out that it is possible to use Proposition 2 to do this, in which one only needs to sum over chains in the resulting poset of intersections: Proposition 3 Hall-type inclusion-exclusion principle Let be subsets of some set , and let be the finite poset formed by intersections of some of the with the convention that is the empty intersection , ordered by set inclusion.
Then for any , one has where are understood to range in. Applying Proposition 2 , this is equivalent to the assertion that for all. But this amounts to the assertion that for each , there is precisely one in with the property that and for any in , namely one can take to be the intersection of all in such that contains. Example 4 If with , and are all distinct, then we have for any finite measure on that makes measurable that due to the four chains , , , of length one, and the three chains , , of length two.
Note that this expansion just has six terms in it, as opposed to the given by the usual inclusion-exclusion formula, though of course one can reduce the number of terms by combining the factors.
This may not seem particularly impressive, especially if one views the term as really being three terms instead of one, but if we add a fourth set with for all , the formula now becomes and we begin to see more cancellation as we now have just seven terms or ten if we count as four terms instead of terms.
Example 5 Variant of Legendre sieve If are natural numbers, and is some sequence of complex numbers with only finitely many terms non-zero, then by applying the above proposition to the sets and with equal to counting measure weighted by the we obtain a variant of the Legendre sieve where range over the set formed by taking least common multiples of the with the understanding that the empty least common multiple is , and denotes the assertion that divides but is strictly less than.
I am curious to know of this version of the Legendre sieve already appears in the literature and similarly for the other applications of Proposition 2 given here. Actually in our application we need an abstraction of the above formula, in which the indicator functions are replaced by more abstract idempotents : Proposition 6 Hall-type inclusion-exclusion principle for idempotents Let be pairwise commuting elements of some ring with identity, which are all idempotent thus for.
Using Proposition 2 as before, it suffices to show that for all all sums and products are understood to range in. We can expand where ranges over all subsets of that contain. For such an , if we write , then is the greatest lower bound of , and we observe that vanishes whenever fails to contain some with.
Thus the only that give non-zero contributions to 8 are the intervals of the form for some which then forms the greatest lower bound for that interval , and the claim 7 follows after noting that for any. Then or in asymptotic notation, as.
Initially, it is only defined on the half-plane : Definition 2 Riemann zeta function, preliminary definition Let be such that. Then we define Note that the series is locally uniformly convergent in the half-plane , so in particular is holomorphic on this region.
In previous notes we have already evaluated some special values of this function: However, it turns out that the zeroes and pole of this function are of far greater importance to analytic number theory, particularly with regards to the study of the prime numbers. The Riemann zeta function has several remarkable properties, some of which we summarise here: Theorem 3 Basic properties of the Riemann zeta function i Euler product formula For any with , we have where the product is absolutely convergent and locally uniform in and is over the prime numbers.
Furthermore, the Riemann xi function is an entire function of order after removing all singularities. The function is an entire function of order one after removing the singularity at. Equivalently, we have for all. The equivalence between the 5 and 6 is a routine consequence of the Euler reflection formula and the Legendre duplication formula, see Exercises 26 and 31 of Notes 1.
Proof: We just prove i and ii for now, leaving iii and iv for later sections. Writing this representation as , we see that whenever , , and consists of all the natural numbers of the form for some. Sending and to infinity, we conclude from monotone convergence and the geometric series formula that whenever is real, and then from dominated convergence we see that the same formula holds for complex with as well. Local uniform convergence then follows from the product form of the Weierstrass -test Exercise 19 of Notes 1.
This can be viewed as a weak version of the prime number theorem, and already illustrates the potential applicability of the Riemann zeta function to control the distribution of the prime numbers. The functional equation iv , when combined with special values of such as 2 , gives some additional values of outside of its initial domain , most famously If one formally compares this formula with 1 , one arrives at the infamous identity although this identity has to be interpreted in a suitable non-classical sense in order for it to be rigorous see this previous blog post for further discussion.
The functional equation tells us that if is a non-trivial zero then so is ; also, we have the identity for all by 1 , hence for all except the pole at by meromorphic continuation. Thus if is a non-trivial zero then so is. We conclude that the set of non-trivial zeroes is symmetric by reflection by both the real axis and the critical line. We have the following infamous conjecture : Conjecture 4 Riemann hypothesis All the non-trivial zeroes of lie on the critical line. The contribution of the higher prime powers is negligible in practice, and as a first approximation one can think of the von Mangoldt function as the indicator function of the primes, weighted by the logarithm function.
Here are some more examples: Exercise 5 Standard Dirichlet series Let be a complex number with. Use this and i to provide an alternate proof of the identity 8. Thus we see that 8 is really just another encoding of the fundamental theorem of arithmetic. Given the appearance of the von Mangoldt function , it is natural to reformulate the prime number theorem in terms of this function: Theorem 6 Prime number theorem, von Mangoldt form One has or in asymptotic notation, as.
Firstly, for any , we can write The sum is non-zero for only values of , and is of size , thus Since , we conclude from Theorem 6 that as. Next, observe from the fundamental theorem of calculus that Multiplying by and summing over all primes , we conclude that From Byjus Class 10 Maths Notes Reading Theorem 6 we certainly have , thus By splitting the integral into the ranges and we see that the right-hand side is , and Theorem 1 follows.
Most famously, we shall see later in these notes that it leads to the remarkably precise Riemann-von Mangoldt explicit formula : Theorem 8 Riemann-von Mangoldt explicit formula For any non-integer , we have where ranges over the non-trivial zeroes of with imaginary part in.
Furthermore, the convergence of the limit is locally uniform in. Indeed, if one formally differentiates the above formula in , one is led to the quite nonrigorous approximation or writing Thus we see that each zero induces an oscillation in the von Mangoldt function, with controlling the frequency of the oscillation and the rate to which the oscillation dies out as.
Comparing Theorem 8 with Theorem 6 , it is natural to suspect that the key step in the proof of the latter is to establish the following slight but important extension of Theorem 3 ii , which can be viewed as a very small step towards the Riemann hypothesis: Theorem 9 Slight enlargement of zero-free region There are no zeroes of on the line.
The basic idea is the following: if there was a zero at , then there would also be a different zero at note cannot vanish due to the pole at , and then the approximation 9 becomes But the expression can be negative for large regions of the variable , whereas is always non-negative.
This conflict eventually leads to a contradiction, but it is not immediately obvious how to make this argument rigorous. All these factors point out that Jahangir was obsessed with portrait just like a modern teenager with selfies on Instagram.
Guessmastergiri TM : Not Applicable. PreSetA Q1. Jagirdars were holders of land assignments in lieu of judicial and police duties, whereas Zamindars were holders of revenue rights without obligation to perform any duty other than revenue collection. Land assignments to Jagirdars were hereditary and revenue rights of Zamindars were not hereditary.
Select the correct answer using the code given below. PreSetA Q4. It ended the trade monopoly of the East India Company in India except for trade in tea and trade with China. It asserted the sovereignty of the British Crown over the Indian territories held by the Company. The revenues of India were now controlled by the British Parliament. Which of the statements given above are correct?
Depending your memory power of all the charter acts and their features. PreSetA Q5. With reference to Swadeshi Movement, consider the following statements : 1. It contributed to the revival of the indigenous artisan crafts and industries. PreSetA Q6. Consider the following pairs : Movement: Organization Leader 1. Self Respect Movement: E. Big organizations, given in all routine books. ANS: Rajiv. With reference to the British colonial rule in India, consider the following statements : 1.
Consequent upon the breaking of Salt Law by Indian people, the Indian National Congress was declared illegal by the colonial rulers. Because these are less known factoids asked. War conference not specifically mentioned in routine books. Neogy : Member, The Constituent Assembly 3.
Consequent upon breaking of salt law, INC was banned but not declared illegal. Sir please clarify on this. Sir�TN books are now updated with new edition �which one shall I follow for ancient and medieval.. The que no. And one query sir preparing for do I relay on TN board for ancient and medieval will it sail through�. Your email address will not be published. Same pattern continued.
Last year 21 Qs so this year 15 Qs. Meaning, the Byjus App Class 6 Maths Notes difficulty level has increased. Asked 2 Qs from this genre: Vishti and Iqta. There would be at least one question related to Southern dynasties or poets. That Tokenism not continued this year. Tough Qs from Buddhism, Jainism, Whose answers are not found in the routine preparation books. Examiner had a change of heart! Relatively easy question on Mahayan Buddhism. Thankfully that lunacy is not repeated this time. Asked one question from this genre.
Match the following organisation with their founders and Communist Prem by asking about Trade disputes Act, Hind Mazdoor sabha.
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