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Below are the 7 most recent journal entries recorded in
MathProofs' LiveJournal:
Thursday, December 6th, 2007  8:56 pm [lainlain29] 
S.O.S., emergency! ALGEBRA AND THEORY OF NUMBERS dear members of community, i need your help a lot! the problem is  i can't find any info on topic "theory of divisibility in the ring of polynomials with TWO variables and their usage" i've looked through all catalogues of our uni library but in vain, there is only one Russian book which is not enough for my research. if you can advise relevant links on free ematerials or send me the scanned books on topic (bereginya4891@ukr.net), i'll be veryvery grateful! THANKS IN ADVANCE!  Tuesday, January 24th, 2006  10:34 pm [pandora1017]

is this sort of thing allowed? some basic set theory
i'm not sure where to start with this. i get the ideas in my head, but don't know how to lay out the proof. is this an element wise proof? i've always been bad at those.... Let I be the set of all real numbers greater than 0. For each x in I, let A_{x} be the open interval (0, x). Prove that the the intersection over x in I of A_{x} is the empty set and the union of x in I of A_{x} is I. Similarly, for each x in I, let B_{x} be the closed interval [0, x]. Prove the intersection over x in I of B_{x} is {0} and the union of x in I of B_{x} is I union {0}. thanks in advance....  Monday, July 25th, 2005  4:38 pm [ihatelivejrnls]

Two Ideals in Every Field
This has been sitting in my lj for a while so I thought I'd share it here, just for fun :) Because any ideal of a field must be a subring, it must contain a 0 element. If the 0 element itself is the ideal then it is the 0 ideal. If there is some other element k in the ideal then, because k is an element of a field, its reciprocal must also be in the field. since the product of any element of the ideal and any other element in the field must also be in the ideal, k(1/k) must be in the ideal. But this product is 1 and any subring which contains 1 must in fact be the entire ring. Therefore, the only two possible ideals of a field are the 0 ideal and the unit ideal.  Friday, July 22nd, 2005  10:56 pm [amplimax]

 Saturday, July 23rd, 2005  8:40 pm [pinterface] 
On Hexagons I'm sketching out a construction project for my current obsession and ran into some difficulty because I want to be sure of something and suck at proofs (my general math is also out of practice, but I digress). My problem is probably stupidly simple for mathematical genii such as yourselves, so feel free to laugh at my incompetence where appropriate. ;)
Given the above diagram (hexagon 𝒽), and that all three rectangles are congruent, what is the smallest regular hexagon ℋ that will enclose the shape? In other words: given a hexagon 𝒽 with parallel but nonequal sides, what is the smallest enclosing regular hexagon ℋ?
I strongly suspect the incircle of ℋ will be the same as the incircle of the three shortest sides (min(𝓁, 𝒸)) of 𝒽, but have no idea how to go about proving it. Has this been shown before? If so, please point me in the right direction. If not, do you have any pointers as to how to prove it?
Update: Now with more math!
Assuming min(𝓁, 𝒸) = 𝓁, and given 𝓇 as the inradius of the triangle formed by the rectangles:
𝓇′ = 𝓇 + 𝓌
(𝓇′ is inradius of the triangle which would be formed if you extended the outer 𝓁s), can it be shown that 𝓇′ is the inradius of ℋ? Current Mood: calculating  Friday, July 22nd, 2005  1:42 pm [htrland] 
A Rigorous Proof of the Symmetric Difference Quotient
Here's my rigorous proof that the limit (as h approaches 0) of [f(x+h)f(xh)]/(2h) is equal to the limit (as h approaches 0) of [f(x+h)f(x)]/h (assuming that the limit exists): From the definition of the derivative, we know that the limit (as h approaches 0) of [f(x+h)f(x)]/h = f'(x). In addition, it is also true that the limit (as h approaches 0) of [f(x)f(xh)]/h = f'(x). The average of the two quantities would be: the limit (as h approaches 0) of [f(x+h)f(x)+f(x)f(xh)]/(2h) = the limit (as h approaches 0) of [f(x+h)f(xh)]/(2h). Since the average of two equal quantities must be equal to either of the original two quantities, therefore the limit (as h approaches 0) of [f(x+h)f(xh)]/(2h) is equal to the limit (as h approaches 0) of [f(x+h)f(x)]/h.  Wednesday, July 13th, 2005  7:17 pm [imuniterator]

f(x)=F'(x) define some unknown function F to be the area under the curve f(x) from the origin to x. then the area from some x _{k} to x _{k}+Δx _{k} is given by F(x _{k}+Δx _{k})F(x _{k}) and may be approximated by the area of trapezoid PQRS where P(x,0), Q(x+Δx,0), R(x+Δx,f(x+Δx)), S(x,f(x)) given by A _{t}=(1/2)(Δx)(f(x)+f(x+Δx)) \R »(1/2)(Δx)(f(x)+f(x+Δx)) notice that as Δx decreases, ½RA _{t}½ decreases. In fact, lim _{Δx®0}A _{t}=R, and lim _{Δx®0}(1/2)(Δx)(f(x)+f(x+Δx))=lim _{Δx®0}f(x)Δx=lim _{Δx®0}F(x+Δx)F(x) \f(x)=lim _{Δx®0}[F(x+Δx)F(x) ]/Δx \f(x)=F'(x) by the definition of Derivative. 
