The Riemann Hypothesis for High School Studentsby Luca Tags: dirichlet eta, riemann hypothesis, riemann zeta 

#1
Jun1309, 06:15 PM

P: 11

Hi All,
I would like to present what I believe to be a simple way to convey the essence of the Riemann Hypothesis to High School students. I hope you like it, and reply with suggestions for further improvements. Note for teachers: the rationale behind the graphs lays with the geometric meaning of complex numbers, and with the equivalence of the zeros of the Riemann Zeta function with the zeros of the Dirichlet Eta function (more details at the bottom). The required level of math literacy is the following:  you are familiar with natural logarithms [tex]\ln[/tex]  you are familiar with angles measured in radians ([tex]\pi \Leftrightarrow 180[/tex]°)  you are familiar with the meaning of fractional powers, such as [tex]\sqrt{n}=n^{\frac{1}{2}} \;\;\; \sqrt[3]{n}=n^{\frac{1}{3}} \;\;\; \sqrt[5]{n^3}=n^{\frac{3}{5}} \;\;\; ,[/tex] etc. The explanation goes as follows (refer to Figure_1.pdf):
What are the zeros of the Riemann Zeta Function ? said zeros are those particular values of [tex]t[/tex] that will bring you back where you started from, that is: the point X=0, Y=0 (see examples in Fig. 2 and 3). What does the Riemann Hypothesis state ? that you may have chances for finding values of [tex]t[/tex] bringing you back where you started from, if and only if the operation you carry out at the denominator for calculating the length of segment n is exactly the square root, no other root will ever work (examples: [tex]\sqrt[3]{n}[/tex] or [tex]\sqrt[4]{n}[/tex] or [tex]\sqrt[9]{n}[/tex] or etc. etc. will not work, and will never, ever allow you to go back where you started from). In other words: if we write the length of segment n as [tex] \frac{1}{n^{\sigma}} \;\;\; with \;\;\; 0 < \sigma < 1 [/tex] the only hope we will ever have to find values of t eventually bringing us back where we started from is that [tex]\sigma = \frac{1}{2}[/tex] Note for teachers: each of the segments making up the paths depicted in the attached figures actually corresponds to one of the terms of the following alternating sign infinite sum (the Dirichlet Eta function) [tex] \eta(s) = \sum_{n=1}^\infty\frac{(1)^{n1}}{n^s} = 1\frac{1}{2^s}+\frac{1}{3^s}\frac{1}{4^s}+\ldots [/tex] where [tex] s = \sigma + i t[/tex] each term is therefore a complex number, which can be represented by a vector, whose polar representation is [tex](1)^{n1}\frac{1}{n^{\sigma}} \;\; e^{it \ln n}[/tex] If we wish to be strictly rigorous, the equivalent definition given above for the zeros of the Riemann Zeta function is in reality referring to zeros of the Dirichlet Eta function. But of course, in the interior of the critical strip the nontrivial zeros of the Riemann Zeta function coincide with the zeros of the Dirichlet Eta function, so that said equivalent definition is indeed a rigorous and correct definition. 



#2
Jun1409, 06:20 PM

P: 150

[tex]\zeta (a+bi) = \sum_{n=1}^{\infty}{1 \over n^{(a+bi)}} = \sum_{n=1}^{\infty}{\cos (b \ln n)  i \sin(b \ln n) \over n^a}[/tex]
and like luca said, this sum only seems to converge to zero when a = 1/2, and the Riemann Hypothesis (RH) says ALL the zeros have real part 1/2 or a = 1/2. 



#3
Jun1509, 07:58 AM

P: 891

Luca, thanks for the nice description. Now maybe a high school student (and my math background is about on par with that of a high school student) might ask what does this have to do with the natural numbers, in particular the primes?




#4
Jun1509, 02:09 PM

P: 104

The Riemann Hypothesis for High School StudentsThat sum is not the continuation of the zeta function in 0<a<1, that sum actually diverges. As the OP states one must use the equivalence relation with the Dirichlet eta function here. @ramsey2879: The "first" indication that the zeta function had 'something to do with primes' was noted by Euler and was that the zeta function could be 'factored' into a product of primes (in Re s>1) [tex]\zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s} = \prod_p \frac{1}{1p^{s}}[/tex] Where p runs over the primes. This I believe was Riemanns inspiration and starting point for his work with the zeta function. He later used it to show a really nice explicit formula which said something about the growth of the prime counting function "[tex]\pi(x)=\sum_{p\leq x} 1[/tex]". In a nutshell, RH (if true) shows us that [tex]\pi(x)[/tex] is 'quite nicely' approximated by an integral we call [tex]\text{li}(x)=\int_2^x \frac{dx}{\log(x)}[/tex] 



#5
Jun1709, 12:19 AM

P: 150

Luca, in pdf figure 1, you wrote this equation, but I cant to seem the get answer or am converting wrong..
you wrote, for t=38, that [tex]\theta_2 = 38 \ln 2 + \pi = 1.935 rad = 110.9^o[/tex] but I seem to get [tex]\theta_2 = 38 \ln 2 + \pi = 23.198[/tex] ??? can you tell me what Im doing wrong or how you're converting the 23 to 110.9 degrees? 



#7
Jun1709, 03:09 AM

P: 11

 take for example 740°, that is 2 x 360° + 20°, which is the same angular position as 20° (as any added 360° turn brings you back to the same angular position)  of course, the same is true for radians, with [tex] 2 \pi [/tex] corresponding to a complete turn therefore [tex] 38 \ln 2 + \pi = 26.3396 + \pi = (4\ * \ 2\pi + 1.2069) + \pi = 1.2069 + \pi = 1.935 rad = 180 \ * \ 1.935/\pi = 110.9deg[/tex] Hope this helps Luca 


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