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Let's discuss Mathematics

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

  1. Olleus

    Olleus Warlord

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    The closure of a set is always closed, no?
     
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  2. onejayhawk

    onejayhawk Afflicted 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. Kaitzilla

    Kaitzilla Lord 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. onejayhawk

    onejayhawk Afflicted 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. Kaitzilla

    Kaitzilla Lord 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? :hmm:


    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? :dunno:
    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? :hmm:

    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. onejayhawk

    onejayhawk Afflicted 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

    Lohrenswald stupid idiot

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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
     
  8. Petek

    Petek Alpha Centaurian Administrator Supporter

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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

    Lohrenswald stupid idiot

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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

    Lohrenswald stupid idiot

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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. CKS

    CKS Chieftain

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    That is what you are supposed to do.
     
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  12. Petek

    Petek Alpha Centaurian Administrator Supporter

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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

    Lohrenswald stupid idiot

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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. Samson

    Samson Chieftain

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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

    Lohrenswald stupid idiot

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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

    Lohrenswald stupid idiot

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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. CKS

    CKS Chieftain

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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

    Lohrenswald stupid idiot

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    yea so haha funny story I was wrong :blush:

    the limits are 0 and (an arbitrary) a
     

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