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I believe the Domesday Book used words for numbers ("one hundred and forty") instead of either Arabic or Roman numerals. StuRat09:43, 18 September 2006 (UTC)[reply]
[Edit conflict] More generally, if x is any number, then for all a other than 0, . If we want it to hold for a = 0 as well, we will get . Hence you will sometimes see 0/0 mentioned as being equal to anything. This is usually little more than a memory trick - as a stand-alone expression, as StuRat explained, it is almost never defined. More details can be found in the article division by zero. -- Meni Rosenfeld (talk) 12:33, 18 September 2006 (UTC)[reply]
However, when the top and bottom of a fraction both approach zero, there may very well be an answer:
X/X, as X approaches zero from either side, is equal to 1.
X/2X, as X approaches zero from either side, is equal to 1/2.
X/X^2, as X appoaches zero from the positive side, equals positive infinity.
X/X^2, as X appoaches zero from the negative side, equals negative infinity.
X^2/X, as X appoaches zero from either side, equals zero.
There is such information on the main page of Mathematics portal ("Did You know..."), but there is no further information in the article "Klein bottle". Anybody has any further information? At first glance it seems impossible for a space with nontrivial fundamental group to cover itself...
Double cover indication I've sketched up an easy construction of a double cover and it should be visible on the right. Note that the square on the right is one patch, just drawn twice. The square on the left can be put in bijection with the square on the right by translation. This situation is simple enough to demonstrate by diagram. Note that this can be continued to any n-covering by stretching even further (or by replacing sub-diagrams on the right by copies of the diagram on the right). This is equivalent to the n-covering of the torus by the torus. Unlike the torus, this construction only demonstrates odd coverings in the other direction (vertically in the drawing) but not even. -- Fuzzyeric15:51, 18 September 2006 (UTC)[reply]
I guess this could go either here or at the science desk, but I imagine you guys will be of more help. Basically I'd like to know some more abstract mathematical concepts that have been discovered and understood mathematically before we knew there was a practical application. Sort of like non-euclidean geometry was worked out before we discovered that there actually existed the configuration in the universe. Thanks--152.23.204.7615:00, 18 September 2006 (UTC)[reply]
How about things from quantum mechanics? Much of the theory can be quite abstract, but still useful. Stuff from operator theory say. Group theory is also quite useful in chemistry - you can classify the types of spectra molecules will have with group theory.
Yes, Fourier series seem like a perfect example (even though the order seems a bit off). I'd like specific examples from Quantum mechanics. I understand there is some complex math involved, but I'd like to read some more about examples of this. And in reply to Maelin, yep primes are also a good example along with anything involved with cryptography.--152.23.204.7617:32, 18 September 2006 (UTC)[reply]
Wave packet theory, describing the location of electrons, is a good example. Mathematically, it says the position of electrons is indeterminate. Figuring out how this can be physically the case is a bit more challenging. StuRat18:17, 18 September 2006 (UTC)[reply]
For me, in addition to group theory, one of the classic examples is good old Hilbert space. Quantum mechanics is basically a theory about linear operators in Hilbert space, and many fields of applied mathematics (e.g. control, optimization and filtering problems in time series) are now formulated using Hilbert space techniques. –Joke20:15, 18 September 2006 (UTC)[reply]
How do you do problems like these without using the guess and check method?
I know the answer. I just want to know if there is a better way to get is other than substituting random numbers for x and checking it. If, in other problems, x was irrational, guess and check would be useless in getting an excact answer. --Yanwen20:54, 18 September 2006 (UTC)[reply]
This kind of problems cannot be solved with elementary functions. You'll have to use functions like Lambert W function (substitute ). However, you can use Newton's method to quickly find numerical solutions. Also, if the numerical solution turns out to appear to be a nice number, you can check that this number actually is a solution. -- Meni Rosenfeld (talk) 21:24, 18 September 2006 (UTC)[reply]
We can solve equations algebraically by applying the inverse of the operation in the equation. If the x is being added onto, we add the opposite number to both sides of the equation to solve for x. If the x is being multiplied, we multiply both sides of the equation by the reciprocal number. If x is an exponent in an exponential equation, we take the logarithm of the appropriate base of both sides of the equation. If x is the base of a power equation, we take the appropriate root of both sides of the equation. If x is the angle in a trigonometric equation, we take the inverse trig function of both sides of the equation.
However if x occurs in more than one type of function in an equation, most of the time we have no way to solve it algebraically. However, sometimes there are methods to change a multifunction equation into a single-function equation. For example, in an equation with two different trig functions, we may be able to use a trig identity to rewrite it with only one trig function. In solving polynomial equations, we may be able to factor the polynomial in to linear factors and solve each one. The theory of polynomial equations has developed formulas and methods for solving any equation of degree 2, 3 or 4; and has shown that there are no algebraic methods to solve all polynomial equations of higher degree. MathMan6423:27, 18 September 2006 (UTC)[reply]
About those high degree polynomial equations, what is really meant by "algebraic methods"? Or, from the other perspective, if it's impossible to solve it with "algebraic methods and taking roots" (as I believe it is, right?), what methods are left? —Bromskloss11:22, 19 September 2006 (UTC)[reply]
For integer solutions, Category:modular arithmetic can (sometimes) work. In the given instance, the equation has to be tru to any modulus, so we might be able to lift a solution all the way to the integers...
First, if x<0 and an integer, then the left-hand side of the equation is negative, so x>=0. Also, x=0 isn't a solution, so x is positive. Both terms on the left are >37 is x>37, so 0<x<=37.
implies
, or "x is odd". Replace x with 2xx+1.
xor (which isn't a solution)
is equivalent to
, taking mods again...
implies
, or xx is even. Replace xx with 2xxx.
implies xxx is odd.
implies xxxx is a multiple of 4.
Now x = 2xx+1, xx = 2xxx, xxx = 2xxxx+1, and xxxx=4k for some k. So, xxx = 8k+1, xx = 16k+2, and x = 32k+5. The only values of k compatible with 0<x<=37 are k=0 or k=1. Equivalently, x=5, or x=37. Five works. Thirty-seven doesn't.
Therefore, x = 5 is the solution in integers. I also know that there are an infinite number of solutions in the complex plane, but this method won't find them without considerable extension. -- Fuzzyeric16:17, 19 September 2006 (UTC)[reply]