Recent content by paniurelis

Take a look here http://www.phys.uu.nl/~thooft/theorist.html The page above is written by famous theoretical physicist about answers to excatly your questions.

Mathematics competition usually require a preparation in the form of solving A LOT of competition type problems. So, if you are not doing well in competition, it only means you didn't do your training. For example, if you would work on competition problems two hours per day for a year, yours...

Let me introduce myself: I am 29, studying as a part timer computer science in eastern europe, Lithuania. Math is a hobby for me. I have been solving elementary math contest type problems for a year before enrolling into bachelour program in local university, basically for review of elementary...

Thank you, gel, I have corrected x_n with ! in my first post. Also, I have found error in my original post: \lim_{n\rightarrow\infty}z_n=e \Longrightarrow \lim_{n\rightarrow\infty}x_n=e Should be: \lim_{n\rightarrow\infty}z_n=\frac{1}{e} \Longrightarrow e\geq...

And another related problem with one above: Prove sequence (y_n), y_n=\frac{n}{\left(\sqrt[n]{n!}\right)^2} is decreasing and \lim y_n = 0 . I see y_n=\frac{x_n^2}{n} and \lim y_n = \frac{\lim x_n^2}{\lim_{n\rightarrow\infty}n}=\frac{e^2}{\lim_{\rightarrow\infty}n}=0. But how to show...

Prove that sequence x_n=\frac{n}{\sqrt[n]{n!}} is increasing and \lim_{n\rightarrow\infty}{x_n} = e My attempt: First, I try to prove \forall n\in N:\frac{x_{n+1}}{x_n}>1. \forall n\in :\frac{x_{n+1}}{x_n}=\frac{n+1}{n}\frac{\sqrt[n]{n!}}{\sqrt[n+1]{(n+1)!}}...

Computer science research and math Such topics as machine learning, pattern recognition, neural networks are based heavily on mathematics. If I understand correctly, the researchers of these fields should be proficient in mathematics ? They must be mathematicians, not the computer scientists...

I have corrected error in my post above, was \frac{1}{k^2}<\frac{1}{k(k+1)}, corrected \frac{1}{k^2}<\frac{1}{(k-1)k}

\exists N=16 \forall n > N : n^4 < 2^n \forall n > N : \sum_{k = 1} ^ {n} \frac{1}{2 ^ {\sqrt{k}}} = \sum_{k = 1} ^ {N} \frac{1}{2 ^ {\sqrt{k}}} + \sum_{l = N+1} ^ {n} \frac{1}{2 ^ {\sqrt{l}}} < \sum_{k = 1} ^ {N} \frac{1}{2 ^ {\sqrt{k}}} + \sum_{l = N+1} ^ {n} \frac{1}{l ^ 2}<\sum_{k = 1} ^...

Thanks morphism, it is a good idea. The problem is, I have an impression, the solution should need only the knowledge of limit theory and some elementary algebra, because I found this problem in the problem book for calculus I, before the chapters with problems about integrals... IMHO, there is...

Let's have a sequence x_n=\sum_{k=1}^{k=n}{\frac{1}{2^{\sqrt{k}}}. We must prove it is convergent. First thought, let's try to prove it is monotonic and bounded, which means convergence of sequence. Monotonicity is easy, \forall n \in N: x_{n+1}-x_n = \frac{1}{2^{\sqrt{n+1}}} > 0 So...

Hello, I am not a native english speaker, so be warned :-). This problem is simple really. For any real number we have (1-x)^2>=0, so 1+x^2>=2x; (1-y)^2>=0, so 1+y^2>=2y; (1-z)^2>=0, so 1+z^2>=2z. Lets plug these inequalities into the left part of inequality we want to prove...