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Thread: Try this, tough guys!

  1. #1 Try this, tough guys! 
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    As I took the day off sick, through boredom I will post the following challenge just for fun (proper mathematicians are barred for the moment).

    Let S be a circle with radius r and circumference C. Then we know that C = 2πr.

    We easily see that π= C/2r = C/d

    where d = 2r = diameter.

    Suppose the C is some integer. Suppose further that d is integer. Then π = C/d, a rational number. Therefore, π is rational!

    Whoops, what's wrong?


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  3. #2  
    Forum Bachelors Degree Demen Tolden's Avatar
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    This is what I see.

    First of all you set off by defining the value of C by using 2πr, and then claim later that C is an integer. The only way that this can be possible is if r = x/π (where x is an integer) so that the π cancels out, however 2r = d, which means that d = 2(x/π), which means d = 2x/π. If x is an integer than 2x/π is not a rational number, and therefore C may be an integer or d may be an integer, but both C and d may not be integers.

    I am not a proper mathematician am I? I am still learning calculus. I'll erase the answer if you think I should.


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  4. #3  
    Forum Bachelors Degree Demen Tolden's Avatar
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    Here are more math problems to keep the sick and bored entertained.

    http://mathproblems.info/group1.html
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  5. #4  
    Forum Bachelors Degree Demen Tolden's Avatar
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    This problem seems pretty easy. Why would you have to know any calculus for it?

    Integral calculus required. A dart is thrown at a circular dart board of radius one. The dart can land at any place on the dartboard with equal probability. What is the mean distance between where the dart hits and the center of the board?
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  6. #5  
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    What is the mean distance between where the dart hits and the center of the board?
    0
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  7. #6  
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    I'm not so sure you can do it without calculus. And the answer is not 0... the distance to the origin is positive away from the origin, and the integral of a positive function over a set of positive measure is positive.
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  8. #7  
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    The bulls eye is merely used as a point of reference and the whole board has to be taken into account. So I stick with the mean being 0, i.e. the centre of the board.

    edit: or make it ->0.
    Disclaimer: I do not declare myself to be an expert on ANY subject. If I state something as fact that is obviously wrong, please don't hesitate to correct me. I welcome such corrections in an attempt to be as truthful and accurate as possible.

    "Gullibility kills" - Carl Sagan
    "All people know the same truth. Our lives consist of how we chose to distort it." - Harry Block
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  9. #8  
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    Spoiler :

    The answer is 2/3. This is a simple case of calculating the cumulated distribution function for the problem (F(r) = r^2) and taking its derivative to find the probability density function (f(r) = 2r) and then calculating the mean by integrating r f(r) from 0 to 1

    End Spoiler
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  10. #9  
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    Ok. It's been 13 years since I did any integrals
    Disclaimer: I do not declare myself to be an expert on ANY subject. If I state something as fact that is obviously wrong, please don't hesitate to correct me. I welcome such corrections in an attempt to be as truthful and accurate as possible.

    "Gullibility kills" - Carl Sagan
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    "It is the mark of an educated mind to be able to entertain a thought without accepting it." - Aristotle
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  11. #10  
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    Excellent solution, river_rat. It's (one integration) easier and more intuitive than taking a double integral over the circle (and normalizing the measure to 1, of course).
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  12. #11  
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    You are drinking a beer. When the can is full, its center of mass in the middle. As you drink some of the beer you notice it becomes more stable when you set it down because its center of mass is lower. When the can is completely empty, the center of mass is in the middle again. Given the height and weight of the empty can and full can, at what level of beer is the center of mass a minimum? Don't use calculus.
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  13. #12  
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    What kind of beer? In what size can?
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  14. #13  
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    Quote Originally Posted by Demen Tolden
    What kind of beer? In what size can?
    It's Grain Belt in the 16 ounce can.
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  15. #14  
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    Heck, I must have been sicker than I thought, as I assumed the proof of pi's irrationality would merely require showing that radius and circumference cannot both be integer. This was Demen's approach.

    But no, that approach doesn't work. I seems that Niven's proof is the gold standard. It's a bit hard to follow, but there are some explanations out there (or here?) if anyone is interested.
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  16. #15  
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    I'd be happy to answer any questions about Niven's proof that anyone may have!
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  17. #16  
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    OK, here are mine - I'm working from memory here.

    Niven defines the function f(x) in his first line. This seems pulled out of his ear (or worse). Would it be fair to say that any function where f(0) = f(1) = 0, with f(x) peaking in [0, 1] would do just as well?

    Then he makes the opaque statement that "the coefficients of n!f(x) are integral". What can he mean? Integer coefficients? That doesn't seem quite right. Integrals of what then?

    He defines the polynomial F(x), and uses what he calls "elementary calculus" to show what I seem to recognize as the Fundamental Theorem. As I understood it (incidentally, I'm crap at analysis), under this theorem, F(x) is the anti-derivative of f(x). Is that what he means?
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  18. #17  
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    Let's deal with the integrality condition first. Integral does indeed mean integer-valued. The fact that n! f(x) has integral coefficients is no surprise--f is defined as:

    f(x) = (x^n (a-bx)^n)/n!

    where a and b are integers such that a/b = π, and so n! f(x) is just x^n (a-bx)^n. The fact following this is a bit more difficult: he claims that f(0) is integral and, moreover, f<sup>(i)</sup>(0) is integral for all i. To see this, first note that f(x) has no terms of degree less than n. (Its lowest degree term is (ax)^n /n!.) Thus, for all derivatives with i < n, the constant term of f<sup>(i)</sup> is 0, whence f<sup>(i)</sup>(0) = 0. For derivatives with i ≥ n, the constant term is i! times the coefficient on x^i in f. The denominator of this coefficient is a divisor of n!, as n! f(x) has integral coefficients, and n! divides i!, so i! clears the denominator of the coefficient. We thus have f<sup>(i)</sup>(0) is integral.
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  19. #18  
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    Now let's deal with the "elementary calculus" statement. The function F that Niven comes up with is actually an alternating sum of the even derivatives of f:

    F = f - f'' + f<sup>(4)</sup> - ... + (-1)<sup>n</sup>f<sup>(2n)</sup>

    He cooks it up like this so that it satisfies:

    F + F'' = f

    (Note that the 2n+2nd derivative of f is 0.) Then the elementary calculus he talks about is just the product rule:

    (d/dx)[F' sin - F cos] = F'' sin + F' cos - F' cos + F sin = (F'' + F)sin = f sin

    And the fundamental theorem tells us to integrate f sin from 0 to π, we just plug in the values into an antiderivative--for example, F' sin - F cos.
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  20. #19  
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    Nobody bit on the beer can problem, so I’ll just give the answer. Let’s say there is 12 oz of beer and the can weighs 1 oz.

    M1 = 1 oz, the mass of the can
    x = the liquid level in inches
    M2 = 12x/4 =3x, the mass of the liquid in ounces
    R1 = 2 inches - The center of mass of the can, half the can height
    R2 = 0.5 x, The center of mass of the liquid, half the liquid level
    R = (M1R1+M2R2)/(M1+M2) = the center of mass of the system, a weighted average of the two centers of mass.

    R = (2+3x*0.5x)/(1+3x) = (2+1.5x^2)/(1+3x)
    We could differentiate R with respect to x and set that to zero to find a minimum, but how can it be done without calculus?

    The center of mass will be increased by adding mass above the center of mass, or removing mass below. If the level is at the center of mass, any additional level will be added above, and any removed will be removed below the center of mass. Therefore the minimum is where R=x
    (2+1.5x^2)/(1+3x) = x
    1.5x^2 + x -2 = 0
    This can be worked out using the quadratic formula to find
    x = about 0.87 inches
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  21. #20  
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    I regret to say I cannot give a good reason why Niven would think up such a function. But the intuition behind what he accomplishes is this: he makes a function whose derivative is too small for its own good--i.e., he shows that a particular function has to take (distinct) integral values at two points, and then shows that its derivative is too small over the interval between for the fundamental theorem of calculus to work out. He manages to do this by abusing two varying properties of π: the first is an analytic property of π, namely that π is the smallest positive number so that sin(π) = 0; the second is the purported rationality of π. The former allows him to construct functions which satisfy various differential relations; the latter allows him to force certain integrals to be integers.

    I think a historical search should show that this is not the first proof of this type. There are similar looking proofs for the transcendence of e and π, and I believe these predate Niven's proof, in which case I would imagine his proof is somehow related to ideas that were already known. More advanced things were certainly known at the time--for example, the Gelfond-Schreier theorem which tells about the transcendence of numbers a^b, where a and b are algebraic, a neither 0 nor 1, and b irrational. You can prove neat things from this--for example, e^π is transcendental, because if it were algebraic, than (e^π)^i = e^iπ = -1 would be transcendental. A proof of a generalized version of this theorem works similar to the ideas above--if you have a bunch of functions which "don't look like each other" (e.g., polynomials and trig functions) but which satisfy some sort of controlled differential equations with respect to each other (e.g., the relation between F, f, sin, and cos) and take on particular special values at specified points (e.g., the assumption that F(0) and F(π) are integers and the knowledge that sin and cos at these points are 0 or ±1), then you have severe constraints on what the functions can look like (e.g., f has to be small)--so much so that you can often derive a contradiction if you set things up right. Or something like that... I gave a talk on this a couple of months ago and forget the details.
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  22. #21  
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    The answer is 42.
    "If you wish to make an apple pie from scratch, you must first invent the universe". - Carl Sagan
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  23. #22  
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    I'm going to take that as a compliment.
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  24. #23  
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    As we're talking about 1947 here, I think it extremely unlikely it's the first proof of it's type.

    His function f(x) = 1/n![x<sup>n</sup>(a - bx)<sup>n</sup>] of course becomes f(x) = b<sup>n</sup>/n![x<sup>n</sup>(π - x)<sup>n</sup>] when π = a/b, therefore f(x) = f(π - x) and f'(x) = -f'(π - x) and f''(x) = +f''(π - x) etc

    It also has some very nice other features that he cannot have been unaware of;

    it is a polynomial of degree not exceeding 2n in x for some n, therefore derivatives higher than the 2n-th vanish;

    derivatives up to the n-th for some n have no constant term;

    the n th derivative of x<sup>n</sup> is n!, which, as you pointed out, is cancelled by 1/n! so the remaining constant terms for the derivatives higher than the n-th are integer;

    f(0) = f(π) = 0, and f(x) is positive when 0 < x < π

    for all 0 < x < π, as n increases, n! grows faster than x<sup>n</sup>.

    Seeing all this, and one or two other consequences of his function, the impending contradiction almost jumps out and grabs you by the throat!

    I saw a proof based on Niven's that makes use of all these properties, and moreover, did not make explicit use of his definition of the poly, F(x); it was, for me at least, easier to follow.
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  25. #24 Re: Try this, tough guys! 
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    Quote Originally Posted by Guitarist
    As I took the day off sick, through boredom I will post the following challenge just for fun (proper mathematicians are barred for the moment).

    Let S be a circle with radius r and circumference C. Then we know that C = 2πr.

    We easily see that π= C/2r = C/d

    where d = 2r = diameter.

    Suppose the C is some integer. Suppose further that d is integer. Then π = C/d, a rational number. Therefore, π is rational!

    Whoops, what's wrong?
    This is not difficult. is not equal to . It is equal to , that is the circumference of a given circle divided by the same circle's diameter. Both of these values will never be integers.
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  26. #25 Re: Try this, tough guys! 
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    Quote Originally Posted by Ellatha
    Quote Originally Posted by Guitarist
    As I took the day off sick, through boredom I will post the following challenge just for fun (proper mathematicians are barred for the moment).

    Let S be a circle with radius r and circumference C. Then we know that C = 2πr.

    We easily see that π= C/2r = C/d

    where d = 2r = diameter.

    Suppose the C is some integer. Suppose further that d is integer. Then π = C/d, a rational number. Therefore, π is rational!

    Whoops, what's wrong?
    This is not difficult. is not equal to . It is equal to , that is the circumference of a given circle divided by the same circle's diameter. Both of these values will never be integers.
    Now that's what I'd call a thread revitalisation...

    Why would both values not be integers? One could be. Suppose then .

    Since is irrational (ir), I wondered whether the following table is true (ra = rational):

    C = ir then always d = ra
    C = ra then always d = ir

    Clearly not, since an obvious exception is already: (and is proven to be irrational).
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  27. #26 Re: Try this, tough guys! 
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    Quote Originally Posted by accountabled
    Quote Originally Posted by Ellatha
    Quote Originally Posted by Guitarist
    As I took the day off sick, through boredom I will post the following challenge just for fun (proper mathematicians are barred for the moment).

    Let S be a circle with radius r and circumference C. Then we know that C = 2πr.

    We easily see that π= C/2r = C/d

    where d = 2r = diameter.

    Suppose the C is some integer. Suppose further that d is integer. Then π = C/d, a rational number. Therefore, π is rational!

    Whoops, what's wrong?
    This is not difficult. is not equal to . It is equal to , that is the circumference of a given circle divided by the same circle's diameter. Both of these values will never be integers.
    Now that's what I'd call a thread revitalisation...

    Why would both values not be integers? One could be. Suppose then .

    Since is irrational (ir), I wondered whether the following table is true (ra = rational):

    C = ir then always d = ra
    C = ra then always d = ir

    Clearly not, since an obvious exception is already: (and is proven to be irrational).
    You are correct in your first statement. One of the two could possibly be an integer, but never both, and since both are required to be integers to form a rational number, is irrational.

    You are correct in your latter statement as well. The circumference of a circle can be found to be rational as well in the same way you found that the diameter could be rational. Suppose that , than the diameter would be .
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  28. #27  
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    To entertain more ur boredom
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