Hello! I'm trying to prove that ##d_1\left({x,y}\right)=\max\{\left|{x_j-y_j}\right|:j=1,2,...,k\}## is a metric. I know that since ##d_1\left({x,y}\right) = \left|{x_j-y_j}\right|## for some ##j## that ##d_1\left({x,y}\right) \geq 0## and since ##\left|{x_j-y_j}\right| =...
As far as I can figure, the way to prove it is this: ## \left|{a+b}\right| \leq \left|{a}\right| + \left|{b}\right| \implies \left|{a}\right| \geq \left|{a+b}\right| - \left|{b}\right| \implies \left|{\left({a+b}\right)-b}\right| \geq \left|{a+b}\right| - \left|{b}\right|
\implies...
I found that in general, a < b + 1 \nRightarrow \left| a \right| < \left| b \right| + 1. Setting c = 0 would give me \left|{s_n-s}\right| < \left|{s_n}\right| + \left|{s}\right|. I feel like what I really need to show is \left|{s}\right| - \left|{s_n}\right| \leq \left|{s_n-s}\right|.
Homework Statement
Use the triangle inequality to prove that \left| s_n - s \right| < 1 \implies \left| s_n \right| < \left| s \right| +1
Homework Equations
The triangle inequality states that \left| a-b \right| \leq \left| a-c \right| + \left| c-b \right|
The Attempt at a Solution...
Ok I think I'm with you. I was getting mixed up over the reversibility thing, but what your saying makes perfect sense to me now. Thank you! I will try to knock out the actual proof now.
Okay, I think what you're saying is this: \left| \frac{n - \sqrt{n^2+n}}{\sqrt{n^2+n}+n} \right| = \left| \frac{1- \sqrt{1+\frac{1}{n}}}{\sqrt{1+\frac{1}{n}}+1} \right| . The problem for me here is that I don't see how to get an inequality of the form n > f(\epsilon). Keep in mind I am not...
I think I have \left| \sqrt{n^2+n} - n - \frac{1}{2} \right| < 2\epsilon \Rightarrow \left( \sqrt{n+1} - \sqrt{n} \right)^2 < 2\epsilon but now I'm not sure where to go from here, or if this is even a useful place to be at. I get a nagging feeling that I am missing something very simple and...
Nope, still haven't solved this one. Came very close, filled up my hand-held whiteboard with stuff, but it fell apart at the last minute because of one little error at the beginning. I'm still working on it now.
I am trying to prove that \lim\left[\sqrt{n^2+n}-n\right]=\frac{1}{2}
Where n \in \mathbb{N} and \lim is the limit of a sequence as n\to\infty.
From the definition of a limit, I know that I need to show that \exists{N}:n>N\Rightarrow\left|\sqrt{n^2+n}-n-\frac{1}{2}\right| < \epsilon...
Thank you! I get:
\frac{dy}{dx} = \frac{dy}{dt} \frac{dt}{dx}
\frac{dy}{dx} = 2k^2 \sin t \cos t \frac{dt}{dx}
\cot t = 2k^2 \sin t \cos t \frac{dt}{dx}
1 = 2k^2 \sin^2 t \frac{dt}{dx}
dx = 2k^2 \sin^2 t dt
Your genius rivals that of Gauss himself!