r/askmath • u/Hunter_rajan03 • 9d ago
Discrete Math I'm confused. Help me .
My maths profesor has given me this problem and said it is the Galactic hard question. I even afraid to attempt it . Like how does the logarithm gets even harder and confusion in all the steps which will be required.
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u/Shevek99 Physicist 9d ago
You have to find the points where the sum is equal to 1, being the first term of the sum equal to 1. What does that tells you about the rest?
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u/Fit_Fortune_7692 6d ago edited 6d ago
Das ist die Zetafunktion. Der Logarithmus kann nur 0 werden, wenn die Funktion 1 ergibt. Und das ist erst im Unendlichen der Fall. Re(s)=∞. Re heißt Realteil der komplexen Zahl. Der Limes der Zetafunktion von s, wenn der Realteil von s gegen Unendlich strebt, ist 1. Aber es ist nur eine Annäherung, unendlich wird nie erreicht. Wenn der Lehrer das Unendlichzeichen sehen wollte: bitte schön, laß die 8 umfallen
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u/Few_Fun_544 4d ago
Step 1: Rewrite the Equation
The expression inside the logarithm is the definition of the Riemann Zeta function, denoted as (\zeta(s)):(\zeta (s)=\sum _{n=1}{\infty }\frac{1}{n{s}}) This infinite series converges absolutely for any complex number (s) whose real part is greater than 1 ((\text{Re}(s) > 1)). Substituting (\zeta(s)) into your equation gives:(\log \big(\zeta (s)\big)=0)
Step 2: Eliminate the LogarithmTaking the exponential ((e{x})) of both sides eliminates the natural logarithm:(\zeta (s)=e{0})(\zeta (s)=1) Step 3: Find the Solutions for (s)Now, we look for all complex numbers (s) where the series equals (1):(1+\frac{1}{2{s}}+\frac{1}{3{s}}+\frac{1}{4{s}}+\dots =1)Subtracting (1) from both sides yields:(\frac{1}{2{s}}+\frac{1}{3{s}}+\frac{1}{4{s}}+\dots =0) Real Solutions: There are no real solutions. If (s) is a real number greater than (1), every term in the sum is strictly positive, meaning the sum can never equal zero.
Complex Solutions: There are infinitely many complex solutions. When (s = \sigma + it) is a complex number, the terms (n{-s} = n{-\sigma}e{-it\log(n)}) oscillate in direction on the complex plane, allowing them to destructively interfere and sum up to exactly (0).According to number theory research regarding the zeros of (\zeta(s) - 1), all of these infinitely many solutions lie within a vertical strip in the convergent region where (1 < {Re}(s) < 2 2).
Conclusion The equation has infinitely many complex solutions, all satisfying (1 < \Re}(s) < 2). Because these values are transcendental, they cannot be written out in a simple algebraic formula and must be computed using numerical root-finding algorithms (such as Newton's method).
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u/Trimutius 9d ago
It is a logarithm of Riemann Zeta Function if it helps...
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u/BrotherItsInTheDrum 9d ago
The zeta function is the analytic continuation of the sum, but the question doesn't tell you to take the analytic continuation. So you don't have to deal with the part of the complex plane where it diverges.
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u/Trimutius 9d ago edited 9d ago
Yes... but using zeta function will help with having analytic formula... (even if it is expressed as an integral)
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u/Euphoric_Key_1929 9d ago
My best guess is that your professor has grossly misunderstood the Riemann Hypothesis, or something got lost in translation. If the problem was instead "find all zeros of the analytic continuation of the infinite sum inside the logarithm", that would be a "galactic hard" question.