The Toeplitz Conjecture (better known as the inscribed square problem) states that all Jordan curves have an inscribed square. It has been stated in the early 1900's and remains an open problem.
I drew a square and then making a ton of curves that touch its four vertices:
This shows that the...
There's a ton of "Stans" that aren't there own nations. Here are a few more: Dagestan, Tatarstan, Khusestan, Nuristan, Baluchistan, and Sistan.
The region of Afghanistan, Pakistan, and Kashmir are very ethnically diverse themselves. I have a theory that this happens because of the numerous...
He probably did know in his heart since he didn't find any spices or valuables like in China. But if he said that he didn't go to China and went to somewhere else, then Spain won't fund his voyages. I have also heard that he made his crew swear that one of the landmasses he visited (which we now...
I don't know how to write the smallest value of ##\pi^{-1}(x)## in set notation. But I did go ahead and graph a few values of ##R(x)##: https://www.desmos.com/calculator/vacrq5jxg1
OH. Well, the inverse prime function actually doesn't exist since ##\pi(x)## is a step function. So now what? Maybe if we think of ##\pi^{-1}## as a set of numbers and take the smallest one then we are fine?
I asked Wolfram to find the inverse function of ##\frac{x}{\ln (x)}## (which is an approximation for ##\pi (x)##) and it gave me ##-xW(-\frac{1}{x})##. So an approximation for ##R(x)## is ##-(\pi(x)+1)W(-\frac{1}{\pi(x)+1}) - x##
Hi PF!
I created a function ##R(x)## that gives the gap between the largest two primes less than or equal to ##x##. To define it, I used this property: $$\pi(x+R(x))=\pi(x)+1$$ Which is true since the ##x## distance between ##\pi(x)## and ##\pi(x)+1## is ##R(x)##. If we solve for ##R(x)## we...
Hello PF!
If ##\Re (s)## is the real part of ##s## and ##\Im (s)## is the imaginary part, then t is very easy to prove that $$\zeta (s) = \zeta ( \Re (s) ) \zeta ( \Im (s)i) - \displaystyle\sum_{n=1}^\infty \frac{1}{n^{\Re (s)}} [\displaystyle\sum_{k \in S, \mathbb{Z} \S = n}...
Yes the second question is true if you are considering just the circumference. Now dimensions aren't the same thing as degrees of freedom, but they kind of have the same analogy as shapes and dimensions.
Hi PF!
Everyone knows that: $${\varphi }^2 - \varphi - 1 = 0$$ But guess what? $${\varphi}^3-2{\varphi}^2+1=0$$ Generalizing this for all n-bonacci numbers: $$x^{n+1}+1 = 2x^n$$ where ##x## is the n-bonacci number and ##n## is the degree of the polynomial that the n-bonacci number is a root of...
Okay I watched it again and he said that the classical wave equation only works for waves and physicists wanted to find an equation that works for both waves and particles. They wanted this new equation to have the same solution as the classical wave equation that works for particle properties...
https://www.udemy.com/course/quantumphysics/?src=sac&kw=quantum+physics But it's for money. Maybe I misunderstood, I will watch the lecture (motivating the Schrodinger equation) again to clarify.
Oh wait, finding the particle function of a wave doesn't make sense. I was thinking of the opposite of the wave function, but when I searched up what it exactly measures (since I only learned the mathematical definition), I realized we couldn't make a counterpart.
I think I got scammed, Dr. Borge Gobel on his Quantum Physcis Udemy course said that the wave equation couldn't be used because of Heisenberg's uncertainty and that quantum physicists had to come up with a new equation. I was kind of confused with this, but I decided to just go with it.
Some...
That's why I put derived in quotation marks. The Schrodinger equation is motivated by the wave equation.
I didn't limit particle functions to electromagnetic fields. I am just talking in general.
The following is the wave equation from Electrodynamics: $$\frac{\partial^2 \Psi}{\partial t^2} = c^2\frac{\partial^2 \Psi}{\partial x^2}$$ Where ##\Psi## is the wave function. But because of Heisenberg's Uncertainty, physicists had to come up with another equation (the Schrodinger equation)...
It was part internet part this book: https://www.amazon.com/dp/0395977258/?tag=pfamazon01-20 I borrowed this book from my seventh grade math teacher and she let me keep it until the end of the year. But I learned about the Basel problem series from 3blue1brown.
##\pi=\sqrt{6\displaystyle\sum^\infty_{n=1} \frac{1}{n^2}}##. So if we calculate the first terms in the sum, we can get an approximation of pi. But it will take a ton of terms just to get the second decimal digit.
If I did my calculations correctly, there are 5,000 digits in each page (except the one with a three on it). There is an indian man who memorized over seventy thousand digits of pi, which is like memorizing fourteen pages of the book.
I made the following estimations of ##\pi## using elliptic curves and brute force:
##\sqrt{10}## (probably known already)
third root of 31
fourth root of 97
fifth root of 306
sixth root of 961
...
Here's how I came up with the approximations: https://www.desmos.com/calculator/yokgiknjuj
I...
OOPS! $$\displaystyle\sum_{n=1}^\infty \frac{1}{n^x}*(\displaystyle\sum_{k \in S, \mathbb{Z} \S =n} \frac{1}{k^{yi}})$$ There you go. Note that ##\mathbb{Z} \S =n## is actually ##\mathbb{Z}##\S = n, I couldn't fix the bug.
How do I re-write $$\displaystyle\sum_{n=1}^\infty \frac{1}{n^x} *(\displaystyle\sum_{k \in S, \mathbb{Z}\S =n})$$ in terms of ##\zeta (x)## ? I want to solve for ##\zeta (x)## and simplifying the above expression in terms of ##\zeta (x)## would avoid repetition.
Does this already exist:
##K=\frac{\tan C (a^2+b^2-c^2)}{4}## I don't know tbh
After thinking about it, I won't use this formula for Davidson Institute since it requires four variables
Thanks for the advice!