Math thread! Come in and have some pi...

[ctownballer04]

So does anyone want to fight over if a zero ring is a real ring or not

I have you know, I took 4 weeks of Ring Theory and I think the trivial ring is indeed a ring.

[ctownballer04]

Miscmath explained metric spaces very well. To expound a bit on why Cauchy is useful for you:

In the specific case of the space R with metric $d(x,y)=|x-y|$, the biggest difference between Cauchy and convergent sequence is the following:

In practice, to prove a sequence is convergent, you need to know (or at least guess) what its limit may be. You can prove a sequence is Cauchy using only the definition of its terms without having to know its limit.

For instance, let $a_n=\sum_1^n\,1/i^2$. You may or may not know what $\lim_{n\rightarrow\infty}a_n$ is... and even if you do, proving convergence using the limit is actually quite hard. However, it's not too hard to prove that that sequence is Cauchy (but it needs integration*, which would be "cheating" for you since you haven't formally introduced integration yet).

* : indeed, $|a_n-a_m|=\sum_{n+1}^m\,1/i^2$ and you can use a geometric "area under the curve" argument to show that $\sum_{n+1}^m\,1/i^2\leq\int_n^m\,1/x^2\,dx$ which clearly converges to zero...

You know D is the (estimated) difference between the autist brain and the normal brain. By the definitions seen in your class, you know that D' is the (estimated) rate of change of D.

Since D' can be negative - as you've shown in b) - then its minimum will be its smallest negative value. In other words, it represents the moment(s) where D decreases the fastest, and what its rate of decrease will be.

Therefore, find the minimum of D'...

will revisit after classes tomorrow, too sleepy to follow atm.

[ckuyook]

I have you know, I took 4 weeks of Ring Theory and I think the trivial ring is indeed a ring.

Agreed. As you know, ring theory is an equational theory so its axioms are negation-free and the consequences of negation-free statements are equivalent to negation-free statements. Equational theories can be interpreted in any category with finite products so that’s how, say, topological rings work - they are ring objects in top

[succinctly]

Yes cool, I love maths!

Has anyone seen the youtube video about turning spheres inside out?

[NephilimRising]

So my muli-variable calc teacher briefly went into "differential forms"......sounds interesting but it's also confusing as hell. "Wedge Product" "0-form, 1-form, 2-form" ...........u wut m8? I looked at it some more on wikipedia and i saw some familiar words such as "pull-back" "push-forward" etc from reading the conversations between MiscMathematician and YokedBrah.............

dude I don't know how you go from doing multi-variable calc/diff eq/linear algebra and then jump straight to fukking differential geometry and topology. Chit is insane.

[NephilimRising]

Agreed. As you know, ring theory is an equational theory so its axioms are negation-free and the consequences of negation-free statements are equivalent to negation-free statements. Equational theories can be interpreted in any category with finite products so that’s how, say, topological rings work - they are ring objects in top

Staahp Ron Staahp

[ctownballer04]

Man, Number Theory could not end soon enough. Our assigned homework problems are sooooo tedious, it's driving me nuts.

[ctownballer04]

Just landed a decent FT job in a math department. Not as a lecturer but Ill still be able to teach as an adjunct there.



can't be that bad!

ayyy! congrats man!

It's not too bad of a class, I have 100% between the first midterm and 5 homework assignments lolol.... I don't find it too interesting though. Analysis is a lot harder, and more interesting but honestly I never thought I'd say that I'd miss Group Theory. Group Theory/Ring Theory much more entertaining than Analysis/Number Theory imo.

[charity4thepoor]

Just landed a decent FT job in a math department. Not as a lecturer but Ill still be able to teach as an adjunct there.



can't be that bad!

Did you put Misc Math Helper on your Resume?

[SCAR-H]

Preparing my angus for this summer quarter statistics class.


Any comments on how this schedule looks? [Final is 80% of grade]

[SCAR-H]

^ introductory stats is extremely basic on the mathematics side. there may be formulas you have to know but it's mostly plug & chug... the only thing to worry about is the pace of the class, it seems to go quickly, not leaving much time to digest the material. spend a few hours on it after classes to reflect on material/ do homework. make sure you understand how to use the formulas, practice a lot, and you're golden. do not cram.

10 hours of class per week is what has me preparing the angus. Thanks for the heads up!

[chrismophat]

oh ****, feeling now more than ever this is not within my field of expertise lol good luck

[Slacker23]

That feel when you go on a hiking trip and the weather decides to be a bish...

We'll probably stay at the motel and get drunk today, fml.

Just landed a decent FT job in a math department. Not as a lecturer but Ill still be able to teach as an adjunct there.

Das it mane.

I've put off sending out resumes until October/November since I'm hoping to hear back about a few of my articles by then, but job-search time is looming for me too. Peppering angus.

[JimFromRaleigh1]

[(x-3)/(x+3)]^4

Find the first derivative

I got it down to [24/(x+3)^2] * [(x-3)/(x+3)]^3

I left my answer like that, and the professor docked points and told me to simplify more. How could I simplify my answer anymore?

[ctownballer04]

[(x-3)/(x+3)]^4

Find the first derivative

I got it down to [24/(x+3)^2] * [(x-3)/(x+3)]^3

I left my answer like that, and the professor docked points and told me to simplify more. How could I simplify my answer anymore?

Seems weird you'd lose points for that as you showed the calculus and the last step if just an algebraic simplification.

Just multiply them together. The RHS quantity cubed is equivalent to the cube of the numerator and the cube of the denominator.
So (24*(x-3)^3)/(x+3)^5

[JimFromRaleigh1]

Seems weird you'd lose points for that as you showed the calculus and the last step if just an algebraic simplification.

Just multiply them together. The RHS quantity cubed is equivalent to the cube of the numerator and the cube of the denominator.
So (24*(x-3)^3)/(x+3)^5

5 points docked on a 15pt problem. Thank you.

[ctownballer04]

So I'm looking at a solution to a problem and they knew that a^4=1 (mod4) and a^2=1 (mod2) and from knowing these two things they jumped from a^6=1 (mod8)

Anyone know what property they used here? I can't seem to figure out why this is legal.

[DesEsseintes]

So I'm looking at a solution to a problem and they knew that a^4=1 (mod4) and a^2=1 (mod2) and from knowing these two things they jumped from a^6=1 (mod8)

Anyone know what property they used here? I can't seem to figure out why this is legal.

Note the squares (mod 2) are 0 and 1, so that you know 'a' must be odd. In particular, we thus have 'a^3' is odd. It is well known that the square of an odd integer leaves remainder 1 modulo 8.

[DesEsseintes]

I think it would be difficult to use 'properties' (though am willing to stand corrected).

Note a^6 - 1 = (a^2)^3 - 1 = (a^2 -1)(a^4 + a^2 +1)

Similarly, a^6 - 1 = (a^3)^2 - 1 = (a^3 -1)(a^3 +1)

It's not clear to me how to show any of these factorisations is divisible by 8 without considering the parity of 'a'.

[ctownballer04]

I think it would be difficult to use 'properties' (though am willing to stand corrected).

Note a^6 - 1 = (a^2)^3 - 1 = (a^2 -1)(a^4 + a^2 +1)

Similarly, a^6 - 1 = (a^3)^2 - 1 = (a^3 -1)(a^3 +1)

It's not clear to me how to show any of these factorisations is divisible by 8 without considering the parity of 'a'.

And just like that, we lose a wizard. RIP in peace.

[ctownballer04]

Any advice for proving continuity in greater generality than individual points? I've been proving functions are continuous at specific points. Now on my homework I have two functions that are piecewise defined from R-->R and I'm asked to determine at which points the function is continuous and then prove it's continuous at these points and prove it's not continuous at all other points.

For example, one of the functions is x+1 for all x>=2 and 3-x^2 for all x<2. Intuitively I know this is certainly continuous everywhere but at x=2. So this means I need to prove it's continuous everywhere else which I'm not sure how to do since we've only been proving individual points and now I have an infinite number.



edit: I found a good article online that has better examples than my text so let me try a little longer. I think I can figure it out now.

[NephilimRising]

Need some advice guys..........

There's one student that is really smart but really annoying; He interrupts teachers and sometimes argues with them. Even the most patient teachers groan when he begins to open his mouth. He once interrupted a student in the middle of her presentation (in a govt class) and this student told him to die. He still hasn't gotten the clue. I'm glad that he's so perceptive that he can instantly think outward about new material but when the class turns into an hour of his stream of consciousness.....it hinders my ability and everyone else's ability to learn anything. I've only had one class with him and I had to drop it. I'm gonna have 2 classes with him next semester and I'm already dreading it.

How should I deal with this? Who can I talk to? Should I talk to the professors in private and express my concerns about this student? The dean? etc...?

I feel like teachers should exercise their right to ask students to leave their classes. It's not nearly done enough.

[ctownballer04]

Any advice for proving continuity in greater generality than individual points? I've been proving functions are continuous at specific points. Now on my homework I have two functions that are piecewise defined from R-->R and I'm asked to determine at which points the function is continuous and then prove it's continuous at these points and prove it's not continuous at all other points.

For example, one of the functions is x+1 for all x>=2 and 3-x^2 for all x<2. Intuitively I know this is certainly continuous everywhere but at x=2. So this means I need to prove it's continuous everywhere else which I'm not sure how to do since we've only been proving individual points and now I have an infinite number.



edit: I found a good article online that has better examples than my text so let me try a little longer. I think I can figure it out now.

Okay I'm stuck and need to go eat something.. I think I have shown that f(x) is continuous on R\{2}. Now I just need to show that f(x) is discontinuous at x=2, however, for whatever reason I haven't been able to do this rigorously. I think I want to do some sort of contradiction based on the Sequential Characterization of Limits or the theorem that says the limit as x approaches a limit point equals L iff the limit as x approaches that limit point from the right and left equals L... However, I've failed miserably at actually producing a proof for this.

Is my proof for f(x) being continuous on R\{2} okay? How do I finish? I don't have an actual definition for discontinuity yet so I think I just need to use one of the two theorems above and do a contradiction but this hasn't worked for me :/

[iDontExistHere]

[ctownballer04]

Some comments:

1) Why do you assume delta is less than 1? Make sure your grader knows what you're saying.

2) x is your variable, right? I think you wanted to separate the cases depending on whether or not x_0 > 2 or x_0 < 2.

3) Writing "for all x>2 is an element of R" is not grammatically what you want. You already know 2 is an element of R. It is always better to use words and full sentences than symbols anyway.

4) How do you conclude |x+x_0| < 2|x_0|+1 when x_0 < 2? You dont want your professor guessing at your deduction


For the discontinuity at 2, contradiction should be fine. Assume there is a limit. Choose epsilon to be something less than the distance of the jump discontinuity and it will work out.

Thanks for the comments. You're absolutely right, that is a huge blunder on my part, the x>2 and x<2 should be x_0>2 and x_0<2.

My first step is always looking for an appropriate delta but I omitted that work for some reason. When I do a more formal write up for the hw I should probably leave it on there at the top to see where I'm pulling numbers from.

So I had two cases,
x_0>=2:
When I search for the delta 1 I get
|x+1-(x_0+1)|=|x-x_0|<epsilon
so choose delta=epsilon

x_0<2:
When I search for the delta 2 I get
|3-x^2-(3-x_0^2)|=|x+x_0|*|x-x_0|
If |x-x_0|<1, then |x+x_0|<1+2|x_0|

So, |x+x_0|*|x-x_0|<|x-x_0|*(1+2|x_0|)<epsilon if |x-x_0|<epsilon/(1+2|x_0|)

Then choose the actual delta to equal the min(delta1,delta2)=delta2

This is all the work I omitted. This should clarify all my "leaps" in the proof I uploaded.

EDIT:
I'm just going to write it up again how I would write it up for my Hw assignment. That way it'll be the best work I can do on my own, and then if somethings still off it won't be because I'm lazy, it'll be because I didn't understand it well enough.

[ctownballer04]

However, I'm still very confused on the contradiction.

I see what you're saying about choosing epsilon less than the distance of the jump discontinuity, that makes sense from a geometric perspective. My confusion revolves around the structure I guess.

I just don't know what I'm even setting up I guess. Maybe through example I can explain my confusion better than in words.
So choose epsilon=1
|f(x)-L|<1

I don't know what I'm setting up since i have two portions of a piece wise function. So am I not guessing a limit? Am I keeping it arbitrary?

So something like:
|x+1-L|<1
and
|3-x^2-L|<1

and then try to show that there doesn't exist a delta where |x-2|<delta and this holds

I don't think this post makes any sense lol.

[iDontExistHere]

Any errors?

[ctownballer04]

tfw disappearing posts

[ctownballer04]

What is your definition of continuity? Normally you use f(a) instead of L, although showing the limit does not exist would also be sufficient. Here we have f(2)=3.

Suppose there is a d>0 such that |x-2|<d implies |f(x)-3|<1. Replacing d with min(1,d), this will still hold. Then
2<f(x)<4 for x in (2-d, 2+d)

Choose a=(4-d)/2=2-d/2 (the average of 2 and 2-d, so it's less than 2 and belongs to this interval).
Then, as d<1, f(a) = 3-(2-d/2)^2 < 3-(2-1/2)^2=3/4 < 1.

So there is a real number a in (2-d,2+d) such that 2<f(a)<4 is false.



lol I had the function as x^2-3 instead of 3-x^2

ughh..
I've been reading this post for nearly an hour now and I'm still confused.

I have like three questions:
1) I don't really follow this let d=min(1,d) maneuver you did.
2) @Bold, what was the point of saying 3/4<1, I don't know where this one came from.
3) This is sort of in addition to question 1, but when you say d=min(1,d) and then use the assumption that d<1, isn't this fixing d, which would invalidate our proof since we need to show this can't work for any d?


I really appreciate the help man. I think I'm about to call it a night from math, maybe tomorrow will go better.. This class is so frustrating to me, I started doing math nearly 8 hours ago and I finished one problem on my homework set -_____-. I'm not cut out for this chit.

[ctownballer04]

I didn't let d=min(1,d). Here is a more full explanation. The assumption that |f(x)-3|<1 leads to 2<f(x)<4. I wanted to contradict this. I know that the function's values when x<2 is nowhere near the range of 2 to 4.

If there exists a d'>0 such that |x-2|<d' implies |f(x)-3|<1, then for any d'' < d' it is also true that |x-2|<d'' implies |f(x)-3|<1 (since this is a subset of x-values from a larger interval). Now I'm letting d=min(d',1).



I was just showing that f((4-d)/2) < 2 by a string of inequalities. The last part was not necessary, or I probably meant the 1 to be a 2.





it takes time. those eureka moments do come if you're patient enough.

Thank you so much! I'm pretty sure I follow everything now. Going to try and write it up on my own w/o looking again loll.


edit: Actually let me clarify again lol.

So you chose d=min{d',1} because if d=1, we'd have 1<x<3 => 2<f(x)<4 which is obviously not true because when x<2, more specifically, when x is in the interval (1,2), 2<f(x)<4 is a false inequality. So this shows that no delta greater than 1 could work, and then we chose delta to equal the minimum of 1 and some arbitrary d' which will take care of all the very small cases where d is less than 1. Then you showed if delta is<1, we are at 3/4 which is certainly less than 2..

So we are done.

Is my logic all okay here or am I misstating something?