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About renerpho

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  1. In that case, you should doubt your data first; if that does not help, you can still doubt the literature. Maybe the formula is not the one you actually need? Maybe your data is flawed? Or there was a problem with your setup? Or there's a problem in the calculations that I missed. Don't just suppose that the literature is wrong. Or, maybe you're misinterpreting the literature and the result you observed is actually what should happen? Check again if the conclusions you made are right. Can you repeat the experiment? Can you ask someone else who actually did the experiment?
  2. If one could calculate the Galois group of the septic equation [math]x^{7}+x^{5}+a{_0}=0[/math], there is a nice test: The septic is solveable by radicals if and only if its Galois group is either the cyclic group of order 7, the dihedral group of order 14 or the metacyclic group of order 21 or 42. Septics that have the Galois group [math]L(3,2)[/math] of order 168 can be solved using elliptic functions. All other septics (with Galois groups of higher order, 2520 or 5040) can not be solved with radicals or elliptic functions alone. Unfortunately, to calculate the Galois group of a septic eq
  3. I'm not entirely sure about the correct sign (the question is not clear about that). So it is [math]\vec{q}(n)=\vec{q}(0) \pm \frac{2n}{3} (\vec{a}+\vec{b}+\vec{c})[/math], depending on the direction in which the force is acting. I will leave vector arrows aside from now on; the following refers to vectors in [math]R^{2}[/math]. The continuous system can be modelled analytically, too. Note that [math]q=q(t)[/math]. You have [math]F(t,q)=q \pm \frac{2}{3} (a+b+c)[/math]. From Newton's second law, we have [math]F(t,q)=m \cdot \ddot{q}[/math], where [math]m[/math] is the mass of the test pa
  4. [latex]\begin{pmatrix} -1 \\ 2 \end{pmatrix}[/latex] [math]e^{\frac {t}{\sqrt{m}}}[/math] [latex]\begin{pmatrix} m \ddot{x}-x \\ m \ddot{y}-y \end{pmatrix}=\pm \frac{2}{3} \begin{pmatrix} a{_x}+b{_x}+c{_x} \\ \ a{_y}+b{_y}+c{_y} \end{pmatrix}[/latex]
  5. I understand it as follows: You start with a vector [math]\vec{q}(0)[/math], and (simultaneously) apply forces [math]\frac{2}{3} (\vec{a}-\vec{q}(0))[/math], [math]\frac{2}{3} (\vec{b}-\vec{q}(0))[/math] and [math]\frac{2}{3} (\vec{c}-\vec{q}(0))[/math]. You repeat [math]n[/math] times. That means, in the n-th step you apply forces [math]\frac{2}{3} (\vec{a}-\vec{q}(n-1))[/math], [math]\frac{2}{3} (\vec{b}-\vec{q}(n-1))[/math] and [math]\frac{2}{3} (\vec{c}-\vec{q}(n-1))[/math] on [math]\vec{q}(n-1)[/math] to get [math]\vec{q}(n)[/math]. You can give an explicit formula for the result: [math
  6. Actually we think that the primes behave essentially like a pseudo-random number sequence (with a few known differences that are already well-understood). The Riemann hypothesis would confirm some that (at least in parts). It would allow to make a lot of predictions about the behaviour of primes (because many methods used to study random number sequences could be used to tackle prime numbers). It's a common misconception that the Riemann hypothesis would result in hidden patterns in the prime numbers. The opposite is true: The reason why there are so many unproven conjectures about primes is
  7. Hey steq, The calculations look fine, including the integration part. Your Excel sheet seems to calculate what is given in formula (5) and (6), although I have no clue if the result makes sense, physically. I suggest to test it by using an alternative integration method (try the rectangle rule)! Just to see if the integration method has a significant effect.
  8. Hello Pawel. The mimimum [math]Q_{min}=\max(KB-1,0)[/math] is calculated as follows: [math]\sum_{i=1}^{K}|B-A{_i}|\geq 0[/math] is trivial, and reached if all [math]A{_i}[/math] are equal to [math]B[/math]. If [math]KB-1>0[/math] then the minimum is not reached at 0, but at [math]\sum_{i=1}^{K}|B-A{_i}|\stackrel{|x|\geq x}{\geq}\sum_{i=1}^{K}(B-A{_i})\[/math] [math]=KB-\sum_{i=1}^{K}A{_i}\ \stackrel{\sum_{i=1}^{K}A{_i}=1}{=}KB-1[/math]. This minimum value is reached if and only if [math]A{_i}<B \forall i[/math]. In the case [math]KB-1<0[/math], you have [math]B<\frac{1}{K
  9. Perhaps - but notice that, even if your formula is related to Fermat's Last Theorem for the case [math]n=3[/math], it can not attack the general case. And there already are easy and elementary proofs for the case [math]n=3[/math].
  10. Primes of that form are are quite rare, see https://oeis.org/A004023 to find numbers of the form [math]111 \dots 111=\frac{10^n-1}{9}[/math] that are prime. It turns out that this number is prime for n=2, 19, 23, 317, 1031, 49081, 86453, 109297 or 270343.
  11. For real numbers [math]x,y>1[/math], the solutions to the equation [math]x^y=y^x[/math] are given by the trivial [math]x=y[/math] and the more interesting [math]y=\frac{-x}{\ln(x)}W\left ( \frac{-x}{\ln(x)} \right )[/math], where [math]W[/math] is the product log function, see https://en.wikipedia.org/wiki/Lambert_W_function Examples: [math]x=3\textup{, }y\approx 2.47805 \dots[/math] [math]x=4\textup{, }y=2[/math] [math]x=5\textup{, }y\approx 1.76492 \dots[/math] For [math]0<x \leqslant 1[/math], there is only the trivial solution. For negative [math]x[/math] the term [math]x^y[/mat
  12. (1) An idea to reduce the amount of guesswork in my previous ansatz: Because the infinite sum converges (this is easy to show), [math]Q{_1}(m)[/math] has to be a constant, and it will be equal to the value of the infinite sum. That's because [math]\lim_{m\to\infty} \frac{{Q{_1}(m)}2^m+Q{_2}(m)}{2^m} = \lim_{m\to\infty} {Q{_1}(m)}[/math], and as [math]Q[/math] is a polynomial this limit only exists if [math]Q[/math] is constant and is equal to the value of the infinite sum. If you already suspect the infinite sum to be equal to 26 then you can save some work by setting [math]a=26[/math], reduc
  13. Remark: Bailey, Borwein et al. (2006) give a nice heuristic argument why this formula might be true, which is related to double Euler sums and 4-dimensional geometry, as well as Quantum Physics. See p.11 of http://crd-legacy.lbl.gov/~dhbailey/dhbpapers/tenproblems.pdf.
  14. If you want to keep the proof elementar, you will have to put some ideas into it. Here is one possible approach: Notice that your sum is of the form [math]\sum_{n=1}^{m}\frac{P(n)}{2^n}[/math] where [math]P[/math] is a polynomial. You start your proof with an educated guess, that the sum will be of similar form, namely [math]\sum_{n=1}^{m}\frac{n^3}{2^n}\stackrel{?!}{=}\frac{{Q{_1}(m)}2^m+Q{_2}(m)}{2^m}[/math] where [math]Q{_1}[/math] and [math]Q{_2}[/math] are themselves polynomials (you include a polynomial term multiplied by [math]2^m[/math] to increase your chances of success). There is n
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