Let's discuss Mathematics

[Olleus]

The closure of a set is always closed, no?

 

[onejayhawk]

The closure of a set is always closed, no?

The closure of a set is defined as the set joined with its limit points. Since you simply added the point 1 to the set [0,1) you assumed the conclusion. However, I did rephrase to make my objection clearer.

Once again, the fact that lim (0.9+0.09+0.009+...) = 1 is insufficient to show that 0.999... = 1.

J

 

[Kaitzilla]

I've seen this one before.

1 divided by 3 is 0.333....
Multiply by 3 and the answer is 0.999...

1 / 3 * 3 = 0.999... = 1

So as long as you believe that multiplication and division are inverse operations, 0.999... = 1.



The solution is obviously to abolish the use of decimals and switch entirely to fractions.
When asked what each decimal place means, the teacher always uses fractions like 1/10th, 1/100th, 1/1000th etc. to explain anyway.
Fractions are far superior to decimals!

https://abcnews.go.com/Technology/story?id=4194473&page=1

A few years ago, Dennis DeTurck, an award-winning professor of mathematics at the University of Pennsylvania, stood at an outdoor podium on campus and proclaimed, "Down with fractions!"

"Fractions have had their day, being useful for by-hand calculation," DeTurck said as part of a 60-second lecture series. "But in this digital age, they're as obsolete as Roman numerals are."

Pistols at dawn sirrah
Math should always be precise.

 

[onejayhawk]

That's one approach. The problem is that 1/3 is not equal to 0.333... for the same reason that 1 is not equal to 0.999.... It works as a practical matter because Rational numbers are dense on the Real numbers.

J

 

[Kaitzilla]

The problem is that 1/3 is not equal to 0.333... for the same reason that 1 is not equal to 0.999..

1 is between [0,1) and (1,2] right?

If 0.999... is not equal to 1, and is in fact less than 1, then it must be part of [0,1)
By that logic, an infinitely small value must exist, and 1 - 0.999... is equal to the infinitely small value.
1 plus the infinitely small value must exist too, and would be written as 1.000...
This 1.000... with an infinite amount of zeros is part of (1,2]

I find it hard to imagine the 1.000... with infinite zeros is somehow greater than 1, but it must be if the infinitely small value exists since it was added to 1 to create 1.000...
In addition, the infinitely small value must exist, because how do we count from 0 to 1 without it?


I think I see your objection to 0.999... = 1
https://en.wikipedia.org/wiki/0.999...

It is something like this maybe?
https://en.wikipedia.org/wiki/Infinitesimal

Pioneering works based on Abraham Robinson's infinitesimals include texts by Stroyan (dating from 1972) and Howard Jerome Keisler (Elementary Calculus: An Infinitesimal Approach). Students easily relate to the intuitive notion of an infinitesimal difference 1-"0.999...", where "0.999..." differs from its standard meaning as the real number 1, and is reinterpreted as an infinite terminating extended decimal that is strictly less than 1.[14][15]

They tried to make math work where 0.999... did not equal 1 with Non-Standard Analysis I think?
https://en.wikipedia.org/wiki/Non-standard_analysis

Here is some talk about it I found.
http://math.coe.uga.edu/tme/Issues/v21n2/5-21.2_Norton & Baldwin.pdf

The Hyperreals
The argument that 0.999... only approximates 1 has grounding in formal mathematics.
In the 1960’s, a mathematician, Abraham Robinson, developed nonstandard analysis (Keisler, 1976).
In contrast to standard analysis, which is what we normally teach in K–16 classrooms, nonstandard analysis posits the existence of infinitely small numbers (infinitesimals) and has no need for limits.
In fact, until Balzano formalized the concept of limits, computing derivatives relied on the use of infinitesimals and related objects that Newton called “fluxions” (Burton, 2007).
These initially shaky foundations for Calculus prompted the following whimsical remark from fellow Englishman, Bishop George Berkeley: “And what are these fluxions? ... May we not call them ghosts of departed quantities?” (p. 525).
Robinson’s work provided a solid foundation for infinitesimals that Newton lacked, by extending the field of real numbers to include an uncountably infinite collection of infinitesimals (Keisler,1976).
This foundation (nonstandard analysis) requires that we treat infinite numbers like real numbers that can be added and multiplied.
Nonstandard analysis provides a sound basis for treating infinitesimals like real numbers and for rejecting the equality of 0.999...and 1(Katz & Katz, 2010).
However, we will see that it also contradicts accepted concepts, such as the Archimedean property.

Shouldn't we use the current math system with limits where 0.999... = 1 if the math is easier to understand and gives more useful results?
The other math system with infinitesimals where 0.999...≠ 1 requires learning the current math system first I think?

I know there should only be one truth for 0.999... = 1 or 0.999... ≠ 1, but non-math people should be spared too much pain!

 

[onejayhawk]

1 is between [0,1) and (1,2] right?

If 0.999... is not equal to 1, and is in fact less than 1, then it must be part of [0,1)
By that logic, an infinitely small value must exist, and 1 - 0.999... is equal to the infinitely small value.

You lost it here. For this to work the distance between [0,1) and 1 must be nonzero. It is not even defined.

You are correct that 0.9999... is in [0,1) and that 1 is not. This provides the contradiction of the assumption that 0.999... = 1.

J

 

[Lohrenswald]

Alright so I've got some like repetition style problems I evidentally absolutely need



so obviously the answer is f(x)=exp(bx)
but I don't remember the like formalism to reach that answer (haven't really solved any differential equations in at least a year, probably more)

so like, what is the formal way to do it, again?

sorry for annoying question

 

[Petek]

For simplicity, write y for f(x) and assume that y <> 0. Then your equation is

dy/dx = by

Rewrite as

dy/y = bdx

Integrate both sides to get

log (y) = b(x + C)

where C is a constant. Exponentiate both sides:

y = exp(b(x + C)) = k exp(bx)

where k = exp(C) is a constant. Therefore, the most general solution is

y = k exp(bx)

For some constant k.

 

[Lohrenswald]

Thanks a lot, it was very helpful!

(also stupid of me to forget the C constant so thanks a lot for that as well)

 

[Lohrenswald]

another banal question



I'm supposed to transform to spherical coordinates. Is it simply to change x to r*cos(fi)*sin(theta) etc, change the limits to 0 to pi for fi and to 2pi for theta etc and multiply with the jacobi determinant r^2*sin(theta) or is there something I've gotten wrong or am forgetting?

also, I don't expect anyone to do this for me but just in case that someone wants to try it, the problem gives this identity as a hint

though I expect that to be fairly easy to do once I manage to change to spherical coordinates

(which when I examine it very closely now using cos^2+sin^2=1 makes it very easy if my assumptions are right)

 

[CKS]

That is what you are supposed to do.

 

[Petek]

Thanks a lot, it was very helpful!

(also stupid of me to forget the C constant so thanks a lot for that as well)

You're welcome. I forgot to add a constant after integrating dy/y, but it doesn't change the result.

 

[Lohrenswald]

does anyone have any neat tricks on how to integrate something like a(x^2)sin^2(bx) (from minus to plus infinity)?
like it does seem symetrical so I can multilpy the integral from 0 to infinity by 2 and get the same result, but that doesn't help terribly much

 

[Samson]

I do not think it counts as a neat trick, but these days when I have to do something like this I use wolfram alpha. Their answer, use a series expansion, does not look very elegant.

 

[Lohrenswald]

yea
this is from a university class with weekly assignments and just based on that I doubt that's the way I'm meant to go

also for the record if it helps any this is trying to find the expectation value of x^2 in an infinite square well potential in quantum physics

I was able to say the the expectation value for x was zero because axsin^2(bx) is an antisymetrical function, but with x^2 this gets worse

 

[Lohrenswald]

managed to solve it with sin^2(x) = (1 - cos(2x)) / 2

actually not quite because I got an answer that's obviously wrong lol but this is pretty clearly the technique you need to use

 

[CKS]

If the limits are + and - infinity, it seems that your answer must be infinity - it is going to blow up as x gets big. However, your limits shouldn't be infinity if you are looking at a square-well potential, but instead 0 and L or +/- L/2.

As a student, I'd have looked it up in a table. Wolfram Alpha suggests converting sin^2 bx to 0.5 (1-cos(2bx)) to do the integral if you want to do it yourself, but it does give the answer. (Note: for me, it chokes on what Samson typed in, but it works fine when I type it in as "integrate ax^2sin^2(bx) dx" instead. I don't know why it doesn't like the extra parentheses.)

Edit: well, my help was too slow. Sorry.

 

[Lohrenswald]

If the limits are + and - infinity

yea so haha funny story I was wrong

the limits are 0 and (an arbitrary) a

 

[Lohrenswald]

alright check this set of equations

a^2+bc=1

d^2+bc=1

b(a+d)=0

c(a+d)=0

b=c* (the problem starts assuming all of a, b, c and d are complex, but through verious unrelated stuff it turns out a and d have to be real, and like I'm certain here b and c also are real so b=c really but I technically don't know that yet)

so consider a+d=0 (because b=c=0 I've figured out)

obviously a=-d and a^2=d^2

now twice I've felt for certain I've seen something that means a or d have to be 1 in this case, only to have it slip seconds later

but like how do I solve this, I guess is my question

 

[AdrienIer]

I assume the set is

a^2+bc=1

d^2+bc=1

b(a+d)=0

c(a+d)=0

If you're looking at a+d=0 then b and c can be whatever you want. There are many quadruplets of solutions, like a=sqrt(2), d=-sqrt(2), b=i and c=i or (if you prefer real solutions) a=sqrt(2), d=-sqrt(2), b=1 and c=-1. It's probably possible to make a 3D visualization of which a, b, c triplets work (d and a being linked d is not needed).

If b=c then bc=b² and the first two equations look much better.