@Brody: It's surprisingly difficult to get graphics software to draw good pictures of global phenomena. Of course, there are some functions where it does just fine without effort.
Yeah, that's the right idea, @Brody, but there's some twisting going on at the origin. (This curve is in fact a prototype for what generic space curves look like very locally. You'll learn that in differential geometry.)
@TedShifrin Since $dim(C(\mathbb{K})) \geq 2$ so we have $c_1 \neq c_2 \in \mathbb{K}$ consider $x \mapsto c_1$ and $x \mapsto c_2$ then the parralleogram identity isn't satisfied.
@Null I think the example I gave earlier is in the spirit of that. You give two objects and show that one of them is a counterexample, but you can't say which one of them is the counterexample..
I think I'm going to get named the room meanie. I keep telling people "that makes no sense." I just need a meme that says it. Where is @Hippa when I need him?
@TedShifrin Ted, remember when I asked you how to know to which number a series converges, and you told me there only was a formula for geometrical ones ? Isn't there something like partial sums ?
I didn't quite say that, @Maks. I know ways of doing other ones, but only special ones. And people smarter than I am at this stuff can do way more with all sorts of trickery.
There are series you can figure out by recognizing that they're related to Taylor series. You can either plug in and evaluate or differentiate/integrate or do something tricky and then plug in.
That's not as mean as you dummy; that makes no sense! :D
@Maks Ryan Reynolds is more appropriate for when someone does something that, although potentially valid, seems pointless and/or unnecessarily complicated
@TedShifrin We saw geometrical and something about partial sums, which isnt really explained, they just gave us an example, but I cant figure out how does it work
Suppose $\lim\limits_{n\to\infty}n\left(f(1/n)-f(0)\right) = 1$ and $f$ is continuous at $0$. Can you conclude that the right-hand derivative $f'_+(0)=1$?
@TedShifrin I have to use that for studying $f(x)=\frac{ax+b}{cx+d}$, but right now I feel like the determinant method works because Mathematicians said so.
Suppose $f$ is continuous near $0$ and let $a_n=f(1/n)$. Prove that (a) $\sum a_n$ converges $\implies f(0)=0$. (b) If $f'(0)$ exists and $\sum a_n$ converges, then $f'(0)=0$. (c) If $f''(0)$ exists and $f(0)=f'(0)=0$, then $\sum a_n$ converges. (d) Suppose $\sum a_n$ converges. Must $f'(0)$ exist? (e) Suppose $f(0)=f'(0)=0$; must $\sum a_n$ converge?$
@Semiclassical I think I we shouldn't be exposed to something in Math until we can rigorously prove and explain why it works and why is it defined the way it is.
@TedShifrin does the following then even make sense? Lines can in many ways be linked with vectorspaces. We analyse the line $G$ which is given by $y=1+\frac{x}{2}$. a) State a subspace $U\subset \mathbb{R}^2$, such that $G$ is an element of $R^2/ U$.
@TedShifrin $f'(0)=\lim(f(h)-f(0))/h=\lim f(h)/h$, which, if it exists, is $\lim f(1/n)n=\lim na_n$. If that's nonzero, we can bound $\sum a_n$ below by a multiple of the harmonic series, so it must be zero.
That doesn't make sense. I'm sure we can have all derivatives at 0 and no derivatives anywhere else. Multiply any nowhere-continuous but bounded function by $e^{-x^2}$. @TedShifrin
No, @Null. It's analogous to modding out $\Bbb Z$ by $3\Bbb Z$. Then the cosets are $3\Bbb Z$, $3\Bbb Z + 1$, $3\Bbb Z + 2$. "Parallel sets" of integers.
