Recent content by nuuskur

I missed a memo or something. E=mgh , ## E = mgh ##, E = mgh In some of my past posts the displaymode tex environment doesn't even include the tex brackets at all, they're replaced by some br /br thingamajig. As a result, the tex doesn't render. Is the ## still functional ? edit: and again...

The column of zeros would make the set of column vectors linearly dependent. But that does not imply the set of row vectors would be linearly dependent. There is no designation like vector x is linearly dependent on vectors y,z,w. A system of vectors is either linearly dependent or linearly...

This cannot be a definition, there will be only the trivial solution only if the rank of the system is equal to the number of variables. Undetermined? Otherwise correct, any subset that contains the zero is linearly dependent. Curious, which textbook that is. What does undetermined mean? Ah...

Specifically this part. What I understand is we consider the set of integer points bounded by the ellipse ## \frac{x^2}{R^2} + \frac{y^2}{\left(\frac{R}{\sqrt{2}} \right)^2} = 1 ## Are you saying that ##N(R)## is equivalent to ##cR^2 + \mbox{Error}(R)## as ##R\to\infty##? Why is that true and...

it always be floridians o0)

A point to consider is why the function is well defined, i.e, why does the minimum exist for ##x>0##. (If it doesn't then change minimum to infimum) Assuming the function is well defined and the limit exists, then the limit must be zero, because we'll just pick a sequence ##x_n =...

A textrobot compiles a list of data points and generates a (convex) combination of them based on your query. It is incapable of generating new data points. But humans (and other sentients) have this ability.

Turn the volume knob up gradually when holding the C key.

Constipation

Why isn't Lithium red or purple? :nb)

Reformulate in topological terms, then: ##\mathbb R = f((a-\delta,a+\delta)) \not\subseteq (f(a)-1, f(a)+1)##. There is no difference between this and epsilon delta language :oops:

It's not continuous at any point. As I mentioned, the image of any open interval is ##\mathbb R##. As a consequence, we take ##\varepsilon =1##, for example, and for any ##a\in\mathbb R## and ##\delta >0##, we find an ##x\in (a-\delta,a+\delta)## such that ##|f(x)-f(a)| \geqslant 1##.