Calvin's Chat Room - 2020-10-01

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Calvin Khor

10:56

@Stupidquestioninc because I'm back to work proper :) Sorry! I'll try to keep hanging around

@DeadGuy whats that article about?

(more of a leading question; its nice to add a sentence or two when sharing a link :) )

Stupid question inc

11:10

@CalvinKhor OK so I have some questions about complex number by bombelli

A: Bombelli's Problem

When a polynomial equation with integer coefficients arises in your research, e.g., $$x^3-15x-4=0\ ,$$ then you "naturally" check whether it has integer solutions. In the case at hand the constant $4$ indicates that such solutions would be from the set $$\{-4,-2, -1,1,2,4\}\ .$$ It is easily seen...

I think I got like 3 stupid answers

I hope you can answer well

Calvin Khor

lol

Sonofgreek

https://math.stackexchange.com/questions/1728376/proof-limit-of-frac1x2-1-1-using-epsilon-delta-as-x-0-t Trying to solve exactly the same problem as this one. I defined $\delta = \frac{1}{2}$, which means that $|x| < 1/2$. And $|x^2 - 1| > \frac{3}{4}$. But how does one proceed from this to create the relationship between $\delta$ and $\epsilon$?

Calvin Khor

@Stupidquestioninc I do not know a better answer than those 3

@Sonofgreek you should not choose $\delta=1/2$, but rather $\delta\le 1/2$

Sonofgreek

Ah, that makes sense. To limit $\delta$

Stupid question inc

Now I can go back to corner and start crying like I did when I was in kindergarten

Calvin Khor

11:20

yup, basically you need to avoid the bad points $x=\pm1$

@Stupidquestioninc sry m8. Sometimes theres a cute algebraic trick you can guess and all u can do is stare at it

Sonofgreek

However, I am struggling to proceed from this. Do you have any advice?

Calvin Khor

@Sonofgreek have you tried Andre's suggestion?

when you're doing a limit $x\to x_0$ problem you should keep an eye out for positive powers of $|x-x_0|$, these help you prove the limit exist

in this case its powers of $|x|$ and the nicest thing to do is probably waht Andre suggested

Sonofgreek

I tried making sense of it so, $|x^2-1| > 3/4$, which means that $\frac{x^2}{x^2-1} \lt \frac{4x^2}{3}$ which in turn means that $|f(x) - L| \le \frac{4x^2}{3}$. But how do I go forward from here?

Sorry if I am asking very obvious questions, I am having a hard time understanding Andre's suggestion

Calvin Khor

@Sonofgreek You need to impose $|x|<C$ so that $|f(x)-L| \le 4x^2/3$ becomes $|f(x) -L| \le \epsilon$. Any guesses for $C$?

Set $4x^2/3 \le \epsilon$ and solve

Sonofgreek

11:52

Hmm

If I I set $4x^2/3 \le \epsilon$ I get $x \le \sqrt{3 \epsilon} / 2$

Any further advice? Sorry for the stupidity

1 hour later…

Sonofgreek

13:04