analytic continuation of zeta function
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This is defined as: So, if we want to find the sum of 1 + 2 + 3 … then we need to substitute z = -1 into the above summation.
There is also a fully typed up mark scheme. We also prove that the multiple Lucas zeta values at negative integer arguments are rational. <> is -1/12.
IB Maths Resources from British International School Phuket, IB HL Paper 3 Practice Questions Exam Pack, Complex Numbers as Matrices: Euler’s Identity, Sierpinski Triangle: A picture of infinity, The Tusi couple – A circle rolling inside a circle, Classical Geometry Puzzle: Finding the Radius, Further investigation of the Mordell Equation. Exponential and trigonometric regression. = -1/12 ? Includes: Fourteen full investigation questions – each one designed to last around 1 hour, and totaling around 35 pages and 500 marks worth of content. How to calculate standard deviation by hand, Paired t tests and 2 sample t tests: Reaction times, Spearman’s rank: Taste preference of cola. Fill in your details below or click an icon to log in: You are commenting using your WordPress.com account. But we can notice that the sum of a geometric sequence formula allows us to calculate f(z) in a different way: Here we have used the formula for summing a geometric, with the first term 1 and common ratio z. f(z) = g(z) when -1 < z< 1 , but g(z) is complex differentiable for all values except for z = 1 (when the denominator is 0).
In particular, we prove the meromorphic continuation of the multiple Lucas zeta functions of depth d:∑0
Therefore we can regard g(z) as the analytic continuation of f(z), and we have extended the domain of f(z) from all values except -1, to all values of z. ( Log Out / We prove that a series derived using Euler's transformation provides the analytic continuation of ((s) for all complex s ^ 1 . Unbelievable: 1+2+3+4….
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Change ), You are commenting using your Facebook account. Follow IB Maths Resources from British International School Phuket on WordPress.com. Blog at WordPress.com.Ben Eastaugh and Chris Sternal-Johnson. Remember complex numbers are of the form a + bi and can be thought of as coordinate points in an x,y axis. ~����ǧ��2��JO*!��k<9���ӓ�QV_��^�Drp����_*. However we can write a new function: Now, g(z) = f(z) for all z when z is not -1, but g(z) also exists when z = -1. �H{�4���Gw�^�����f���g� �p�+q_�]���}���:N�Gj��#g��Kr�_�����3J+p�I�}��@)4W���I��Q�n~�f�U��E�ix3� 絏=z����ы��G/e� ����DF��ΪT%�ֹu*�I������]��+���w�y��;�"3-��f��H��H����wy��f�B��wWMHa�sÃ�!�G��։2Y���o��q��$��&I�r��ӁkRoY��[Dԁ��6q�QG�IS?�G�&]��*����^��(�')s��s�����`r��+��j�M�m�s��K���[�=�F��h>�).z��f����)�f|ZǘR%���������iz�gz9cm�y��j�g��b�?a2endstream �o��Ў�x���HsI��qn��/�cG�X�r��jʍDBM0�3�C�-������[�tu�~��e�=�J�����\���7��-v�����X�EK�Hw\s��_khWУ��AI���W��e�D�-d-�2� e"y�Q�3���c-�9��]��]�U��JSi5��yWt��-+�CL0$:ٯaV;R��)���t���##��[8�#�`n�O_��QM)՟Zd��7�Z�����q�������/�2_3u�qj�O�y���eKZ�q�$:J��h��E_��Ҷ��'����Q$dw��4��5����`�=:�N���ȍ�ꕓ_�|./������K]R]M-����x�H��s4���"e7�\�S����}%��
endobj �{v�8��(�|�$B�8���������5E��W��"��H�S0Q^Ȧ������i{"������O�Qn�%������ A�-8V���|�ӯ��$�r�*��YQsR^=FL���E����M�v:16�uR~Q�٢�cY����K�n��X��CIc\�R�m��L�j���k�X�|�kpta��%Uech���8v�4E��9&i�5��g\�\&�[�@f�Ʊ �z�$�X����ذՃ"����@�?�R�2�Ī����5l,t�=��˼!�F�vZ��$��R���A6,�ȵ!螧k~���(Ğ��8*�Gq�b�E��I-&�$�y��@������BLC0�'}� a�H:�D8�v �`?+ ��[̌JS(&̋�;�N��1���20�d��Y�GP�Ȫ������)�Z6�$b�/ʏ�����d��j��s9R���>��.ǽ�Oi�Ϧ��D�H�b�(۸ȓ�H;]�v���IT��N�L[s��eJ�ֲ���� �x�5���1�g_�����yJ���@��?����@�)q��$;(��A�Oc��O;����/��oE`�����|tq^yD�����E2�MM���u� ���)�K\�q:A.&ϓ\��j���ĵ��迉�����Pn�ߴ�މ�(�gIPry9&�*'��i����)x�0 �n�־e����ډ���䃔��6_l^#�B���wY���!��u�:$��ps����7�=����������|�O����v�6v_L�+��|�} ��Ȋ9ݧ��|2��r�c�T*L �8��i�8�W��fB,��m6�⛋O5���hD2r0��z��z;�0�jɾ"����:5"RR�Fv'��������R>�>�?d���MS�%�P����� �_E$�$5}� o��3K99'�-\��@'S����v$�E;D�Vx_b�4�p�[�@.�$4����Ĺ"t�B�о^d ����yd0e��O�J�f\��/~_"�E��q3U����g���d�$El���b�7����'�U:��Z2ͽ3&����a���Y�tXc� Z�E��b��eN6;(ˉv��;�*{��N�;i���#�~1+TБW�����H�G( ASTOUNDING: 1 + 2 + 3 + 4 + 5 +... = -1/12 Analytic Continuation and the Riemann Zeta Function Analytic Continuation is a very important mathematical technique which allows us to extend the domain of functions. �Z��B�x6�(�k7�g This extends by analytic continuation the complex plane into the complex plane plus infinity. Many thanks! (Don’t worry about how this is calculated – though it is related to the domain of convergence).
by trying and error or some mathematic ways, Please direct me, thank you. The proof revolves around the Riemann Zeta function, (Riemann is pictured above). Analytic Continuation is a very important mathematical technique which allows us to extend the domain of functions. From the above analytic continuation of Euler numbers, we consider ( Log Out / If you have found these resources useful please consider making a voluntary donation to help support the process of creating more resources for IB students. Modelling the spread of Coronavirus (COVID-19), Rational Approximations to Irrational Numbers – A 78 Year old Conjecture Proved, Hollow Cubes and Hypercubes investigation, Ramanujan’s Taxi Cab and the Sum of 2 Cubes, Finding the volume of a rugby ball (or American football), The Shoelace Algorithm to find areas of polygons, IB Applications and Interpretations SL and HL Resources, IB Analysis and Approaches SL and HL Resources, Stacking cannonballs – solving maths with code, Normal Numbers – and random number generators, The Gini Coefficient – measuring inequality, Zeno’s Paradox – Achilles and the Tortoise. Together this is around 100 pages of content.
It’s a bit difficult, but try and understand the general method! ( Log Out / ANALYTIC CONTINUATION OF RIEMANN'S ZETA FUNCTION AND VALUES AT NEGATIVE INTEGERS VIA EULER'S TRANSFORMATION OF SERIES JONATHAN SONDOW (Communicated by William W. Adams) Abstract. Proof of the R.H. We compute a complete list of poles and their residues. How? Fourier Transforms – the most important tool in mathematics? © 2018 Elsevier Inc. All rights reserved. Quadratic regression and cubic regression. Riemann’s second proof of the analytic continuation of the Riemann Zeta function Andreas Steiger Seminar on Modular Forms, Winter term 2006 1 Abstract The Riemann zeta-function ζ(s) is defined by ζ… We also prove that the multiple Lucas zeta values at negative integer arguments are rational. endobj Advice on using Geogebra, Desmos and Tracker. Analytic continuation concerns functions of the form: f(z) where z is a complex number and f(z) is (complex) differentiable. ScienceDirect ® is a registered trademark of Elsevier B.V. ScienceDirect ® is a registered trademark of Elsevier B.V. Analytic continuation of the multiple Lucas zeta functions. Change ). Another example is used in showing that the sum of natural numbers is -1/12. �Xxóy�K�O���8�5�M_(־� � �;@B�˚��f!�m0X�"����ѐ��%���C�(�/ߖ��a�r���w.Fw�0��)-'h$N�|/I�B����/��Z��x�-ŀuh�. <>
On analytic continuation of various multiple zeta-functions Kohji Matsumoto Abstract In this article we describe the development of the problem of analytic continuation of multiple zeta-functions. b�+a�4wN{�a�ODnMζ,51���A`w��n2.�HEm������R�h߬.�,$ %�쏢 The idea of analytic continuation is to take an original function with a restricted domain, then to find another function which is the same within that restricted domain, but also is valid outside that domain.
300 IB Maths Exploration ideas, video tutorials and Exploration Guides, November 9, 2014 in puzzles | Tags: analytic continuation, Riemann Zeta, Analytic Continuation and the Riemann Zeta Function. It is essential in higher level mathematics and physics and leads to some remarkable results. We begin with the work of E. W. Barnes and H. Mellin, and then discuss the Euler sum and its multi-variable generalization. I’m going to try and talk through another proof of this result. (9.1) Our purpose in this chapter is to extend this definition to the entire complex s-plane, and show that the Riemann zeta function is analytic everywhere except at s = 1, where it has a simple pole of residue 1. 25 0 obj
We define the Hurwitz zeta function for with and with by see [3, 7–10, 12]. In particular, we prove the meromorphic continuation of the multiple Lucas zeta functions of depth d: ∑ 0 < n 1 < ⋯ < n d 1 U n 1 s 1 ⋯ U n d s d, where U n is the n-th Lucas number of first kind and ∑ i = j d Re (s i) > 0 for 1 ≤ j ≤ d. We compute a complete list of poles and their residues. Includes: IB Exploration Modelling and Statistics Guide. ��N���N�������7�p��G��Z;�$(��eX��j�Z�U=]�T�0��vT��ʼ��-��j��F���jU$�]D�^*N�4_�G��m���w5�^���@�}�x��kNv���#E~LP��?TwssA^��LB }���d[�]���|�J��_ӎ�ޥ �_�ZIV�/������ For the purposes of this post we will only look at real values of z (real numbers are still a subset of complex numbers). ( Log Out / %PDF-1.2 Journal of Mathematical Analysis and Applications, https://doi.org/10.1016/j.jmaa.2018.08.063. stream Useful websites for use in the exploration, A selection of detailed exploration ideas.
Change ), You are commenting using your Google account. We denote it by ζ (s).
A result that at first glance looks ridiculous – and yet can be shown to be correct. One example of analytic continuation that I’ve written about before is the Riemann Sphere. In this article, we consider more general situation. This sounds very complicated – but let’s look at a couple of examples: This is a function which is defined for all values except for z = -1. Change ), You are commenting using your Twitter account. For example, by using analytic continuation we can prove that the sum of the natural numbers (1 + 2 + 3 + ….) We have taken as the definition of the Riemann zeta function ζ(s) = X∞ n=1 1 ns, Res > 1. What is the Riemann Hypothesis – and how solving it can win you $1 million. It is essential in higher level mathematics and physics and leads to some remarkable results. If you enjoyed this post you might also like: The Riemann Hypothesis Explained.
When z = -1 we have zero on the denominator so the function doesn’t exist. 5 0 obj As you know well, the Riemann zeta function ∑ n = 1 ∞ 1 n s for R e (s) > 1 has a meromorphic continuation to the whole complex plane and has a simple pole at s = 1.
Through analytic continuation (where we extend the domain from z > 1 to all complex numbers apart from -1) we can rewrite the zeta function as: and substituting z = 2 into this formula, so that we end up with zeta(-1) we get: We have proved that 1 + 2 + 3 … = -1/12 !
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