Let's discuss Mathematics

Discussion in 'Science & Technology' started by ParadigmShifter, Mar 16, 2009.

1. OlleusDeity

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The closure of a set is always closed, no?

2. onejayhawkAfflicted with reason

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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

3. KaitzillaLord Croissant

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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

Pistols at dawn sirrah
Math should always be precise.

Last edited: Jan 12, 2019
4. onejayhawkAfflicted with reason

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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

5. KaitzillaLord Croissant

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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

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

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!

Last edited: Jan 12, 2019
6. onejayhawkAfflicted with reason

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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

7. Lohrenswald老仁森林

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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

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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.

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9. Lohrenswald老仁森林

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Thanks a lot, it was very helpful!

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

10. Lohrenswald老仁森林

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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)

Last edited: Jan 21, 2019
11. CKSDeity

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That is what you are supposed to do.

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

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13. Lohrenswald老仁森林

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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

14. SamsonDeity

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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.

15. Lohrenswald老仁森林

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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

16. Lohrenswald老仁森林

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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

17. CKSDeity

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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.

18. Lohrenswald老仁森林

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yea so haha funny story I was wrong

the limits are 0 and (an arbitrary) a

19. Lohrenswald老仁森林

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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

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I assume the set is
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.