Nicholas Roberts

pretty much the same as what i wrote

oh yeah, much easier lol

@Koro Suppose $f(y) = 0$ and take any $x < y$. Then $0 < \frac{f(x)-f(y)}{x-y} = \frac{f(x)}{x-y}$. Since the denominator is negative, and the quotient is positive, the numerator must also be negative, that is, $f(x) < 0$ which is a contradiction to the first part of the proof. Does this work?

yup, just making sure

just to make sure, you took epsilon to be $-f(y)$ in the assumption of the limit being 0 at negative infinity, right?

@Koro thanks, i filled in the details of what you wrote. will work on $f(y)=0$ now

@ko

statement*

Do you mean to fill in the missing details? Because that seems like it proves the statemewnt

@Koro

i have a basic question, can someone give me a hint or a step in the right direction? Suppose I have a strictly increasing function with $\lim_{x \to -\infty}f(x) = 0$. Can we deduce that $f(x) > 0$? Seems like it should be true

Yeah, that'll do it. thanks

so how come the statement isnt usually presented with this relaxed condition?

Hi all, in the statement of the dominated convergence we need a dominating function $g$ such that $|f_n| \leq g$ for all $n \in \mathbb{N}$. Can this condition be relaxed to there exists an $N \in \mathbb{N}$ such that $|f_n| \leq g$ for all $n \geq N$? It seems like that should work.

Yeah, maybe by messing with the sum/difference identites. Anyways, thanks!

I was wondering if there was a way to explicitly solve for these values. Or find like a closed form of them as opposed to proving existence of such

Ahh, intermediate value theorem.

Hm, well $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$

I'm trying. Can't seem to find any

Would it be possible to find some values $\theta$ and $\phi$ such that $\cos(\theta + \phi) = \cos(\theta)+\cos(\phi)$? In other words, do there exist values in which any of the trig functions are additive?

very cool. thanks guys

@Thorgott

Thanks, have you read it? is it any good?

@Thorgott springer.com/gp/book/9780817646769 is this the book you speak of?

linear functionals on compactly supported differential m-forms

does anyone know a good resource for learning about currents?

Hi all, can anyone care to point me in the right direction regarding proving this trig identity? $\dfrac{1 + \cos(x/2) - \sin(x/2)}{1 - \cos(x/2) - \sin(x/2)} = -\cot(x/4)$

hi can someone verify/critique my proof

@Semiclassical hmmmm ok

@LeakyNun regarding a PID

@LeakyNun Can you scroll up and look at my message?

Suppose I have a subring R of $\mathbb{C}[x]$ with the property that $f(1) = f(-1)$ for all $f \in R$. Can anyone show me or give me a hint to show that R is not principal ideal domain?

reffering to*?

When you say Big Rudin, what book are we referring?

ok thanks!

@TedShifrin random question, but do you know any good Functional Analysis books? My only background in the subject is Chapter 5 in Folland (which is very brief) and I'm interested in going deeper in it.

very nice @RScrlli ! Depending on your background in analysis, Stein may be good for you to look at.

@RScrlli The second book I mentioned, Stein and Shakarchi, is more forgiving than Folland. I see it as a nice bridge between Undergrad and Graduate analysis. Folland is definitely a graduate text.

Oh!

@TedShifrin True. I'm not aware of @RScrlli 's background

@RScrlli May I suggest Folland's Real Analysis for a nice exposition on Measure Theory as well as Stein and Shakarchi's Book 3 Real Analysis. Both have served me very well

But 4m+1 is not equal to 4k + 1, 4m + 1 is an element of your set R with m = 2k

To gain entrance into the set R, it needs to be of the form 4m + 1 for a positive integer m. Surely, 2k is a positive integer. So x = 8k + 1 = 4(2k) + 1 is an element of R

@krauser126 exactly! Good job

@krauser126 If $n = 8k + 1$ for some positive integer $k$, do you see a way re-write the expression $8k + 1$ as $4m + 1$ where $m$ is another positive integer?

I was thinking this: $|f| = |\lim f_n| = \lim |f_n| < \lim g = g$ but im not sure

Hi all, I have somewhat of a basic question. Consider of a seqence $\{f_n\}$ such that there is a $g$ with $|f_n| < g$. Suppose $f_n \rightarrow f$ pointwise. Will it follow that $|f| < g$? And how to prove this?

Oh right, ok. Thanks guys

What is MVP?