Recent content by VKnopp

The eccentric mathematician Paul Erdos believed in a deity known as the SF (supreme fascist). He believed the SF teased him by hiding his glasses, hiding his Hungarian passport and keeping mathematical truths from him. He also believed that the SF has a book that consists of all the most...

A good bound that has been found for the logarithm (in a form that might be useful to you) is the following, ##\ln x \leq x^{\frac{1}{e}}## That of course implies, ##|x \ln x| \leq |x^{\frac{e+1}{e}}|## Now review the mathematical definition of ##f(x)=O(g(x))##. You'll get the answer for...

What do you mean equivalent? I beg to differ. ##\mathbb{Z}## is much more dense then your set. I think what you mean is the cardinal is the same. You can prove this via 1 to 1 correspondence.

Okay, thank you for the advice. I believe I was told not to do this somewhere else, it apparently makes me look pretentious and arrogant.

It put it because it's an interesting thing about myself and I really don't have much to share about myself, so I decided to add it in.

I agree with you. I was just introducing myself.

Is it gravity? I should assume that gravity pulls objects down at a rate of ##9.8 ms^{-2}##.

Of course. I get it know, thank you very much.

I have decided to self-teach myself A-Level Physics. I am confused about one particular step of a solution to a certain problem that the textbook provided. Problem: A stone is thrown vertically upwards at ##15 ms^{-1}##. How long is the stone in the air until hitting the ground? There are a...

I forgot to formally introduce myself on this forum. I am VKnopp. I am 14 year old maths enthusiast with Asperger's Syndrome. I self-educated myself all the way up to Calculus III with a little bit of number theory, linear algebra, complex analysis and real analysis supplements. I am in the top...

This is similar to what SteamKing wrote. The ##\cos z## and ##\sin z## functions respectively are just real and imaginary parts of ##e^{iz}##. We can calculate the sine and cosine functions (in radians) by calculating the real and imaginary parts of the series, \sum_{n=0}^{\infty}...

Using the zeta function: $$\sum_{n=0}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}$$Multiply by six and take the positive square root. You can approximate by choosing the number of terms in the partial sum.