Wikipedia:Reference desk/Archives/Mathematics/2011 November 24

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November 24

Integration of floor functions

Hey guys, I came across an interesting dillema and was wondering if anybody could tell me what I'm doing wrong.

There's no doubt that $\left\lfloor {\frac {2}{x}}\right\rfloor -2\left\lfloor {\frac {1}{x}}\right\rfloor >=0$

so therefore,$\int _{0}^{1}(\left\lfloor {\frac {2}{x}}\right\rfloor -2\left\lfloor {\frac {1}{x}}\right\rfloor )dx>0$

But when you separate $\int _{0}^{1}(\left\lfloor {\frac {2}{x}}\right\rfloor -2\left\lfloor {\frac {1}{x}}\right\rfloor )dx$ into $\int _{0}^{1}\left\lfloor {\frac {2}{x}}\right\rfloor dx-\int _{0}^{1}2\left\lfloor {\frac {1}{x}}\right\rfloor dx$

and substitute x=2t in the first integral, you get

$\int _{0}^{1/2}2\left\lfloor {\frac {1}{t}}\right\rfloor dt-\int _{0}^{1}2\left\lfloor {\frac {1}{x}}\right\rfloor dx$

which in turn, equals

$-\int _{1/2}^{1}2\left\lfloor {\frac {1}{x}}\right\rfloor dx$ which is definitely negative. This conclusion contradicts with my first statement where I said this integral is positive.

Can anybody point out where I've gone wrong? Because no matter how many times I go over it, the steps I've taken seem perfectly logical.

P.S. What is with wikipedia and its unbelievably difficult way of typing up mathematical formulas? Took me nearly half an hour typing this up! Grrr Johnnyboi7 (talk) 11:31, 24 November 2011 (UTC)

The problem is separating the one integral into a difference of two. The original expression is something between 0 and 1; each term in the difference is infinite. You can't separate an integral into pieces unless both pieces converge. --121.74.125.249 (talk) 12:14, 24 November 2011 (UTC)

Your inequality assumes x≠0. The integrals should go from epsilon to one rather than from zero to one, and then take the limit epsilon towards zero. Bo Jacoby (talk) 12:30, 24 November 2011 (UTC).

The integral of 1/x is the log function which tends to minus infinity at 0. So separating the factors means you are subtracting one infinity from another. Dmcq (talk) 12:52, 24 November 2011 (UTC)

identity crisis

(a-b)2 = (b-a)2,taking sqrt both sides we have a-b= b-a, which is false for general a nd b as a-b= -(b-a).sometimes i m a fool,but pls help. 59.165.108.89 (talk) 13:29, 24 November 2011 (UTC)

You can't just take the square root of both of them, as you observe. That does not work. Remember that the square of a negative number is positive. So (2-4)^2 = (4-2)^2. You can kinda take the root values like this ... ABS(4-2)=ABS(2-4) where ABS is the absolute value - i.e. the difference between the two integers. If that helps. --Tagishsimon (talk) 13:56, 24 November 2011 (UTC)

(ec) The square root of a positive number has two values. Thus both (a-b) and -(a-b) = (b-a) are square roots of the left side of your equation. Your confusion probably arises from the "sqrt" function (on calculators and in computer languages) which gives only the positive root of a positive number and ignores the negative root. Cuddlyable3 (talk) 14:00, 24 November 2011 (UTC)

The general approach to this kind of equations is subtracting one side from both sides, which converts them to 'difference of squares equals zero':
    (a − b)² − (b − a)² = 0
Then we apply a simple binomial factorization
    [(a − b) − (b − a)] × [(a − b) + (b − a)] = 0
which reduces the problem to an alternative:
    (a − b) − (b − a) = 0
        or
    (a − b) + (b − a) = 0
equivalent to
    a − b = 0 or 0 = 0
which is always true due to the identity. Thus the equation is satisfied by any (a − b), consequently by every a and b. --CiaPan (talk) 22:05, 24 November 2011 (UTC)

a - b = −(b - a)

(−(b - a))² = (b - a)²

Therefore, (a - b)² = (b - a)²

Unit ball and sphere

Let $\mathbb {S} ^{n}$ and $\mathbb {D} ^{n}$ be the n sphere and n dimensional unit ball respectively, and give both of them the basepoint (1,0,...,0). There's obviously a basepoint preserving homemorphism from $\mathbb {D} ^{n}$ to $\mathbb {S} ^{n}$, but can someone give me the explicit formula for it? I need it so that the map sends the boundary $\mathbb {S} ^{n-1}$ of $\mathbb {D} ^{n}$ to the basepoint in $\mathbb {S} ^{n}$ Money is tight (talk) 14:30, 24 November 2011 (UTC)

S² and D² are homeomorphic? Icthyos (talk) 14:36, 24 November 2011 (UTC)

I think he means Sⁿ with a point deleted. If that's the case, spherical coordinates seem to be what he's after. Sławomir Biały (talk) 14:40, 24 November 2011 (UTC)

(Actually, geodesic polar coordinates centered at $(1,0,...,0$ on the sphere. Sławomir Biały (talk) 14:42, 24 November 2011 (UTC))

Oops... Sorry I didn't mean homeomorphism. What I wanted was the map like follows: wrap D^2 around S^2 starting from the north pole and end with the boundary S^1 of D^2 all stacked at the south pole. But I need a map that takes the basepoint (1,0,...,0) of D^n to S^n. Spherical coordinates won't do it because the inverse image of S^2 is the square [0,2pi] x [0,pi], whereas I need D^2. Money is tight (talk) 14:58, 24 November 2011 (UTC)

I mean geodesic polar coordinates. This is the exponential map for the sphere. Sławomir Biały (talk) 17:24, 24 November 2011 (UTC)

Ok thanks. Money is tight (talk) 08:03, 25 November 2011 (UTC)