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| thyristor |
 Posted: Fri May 23, 2008 6:29 am Post subject: Evidence |
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Do you have any favourite evidence, which is extra neat and short but still deep. 
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| serpicojr |
| Posted: Sat May 24, 2008 3:57 pm Post subject: |
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Joined: 17 Jul 2007
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I like this proof that sqrt(2) is irrational. Suppose otherwise. Then you can write sqrt(2) = x/y for some integers x, y. We may assume that x is the smallest positive integer such that you can write sqrt(2) = x/y for some y. Note that x/y > 1, and so x > y. Now draw a square with side lengths x, and draw two squares with side lengths y in opposing corners of the first square:
x2 is clearly the area of the large square. But we may also calculate the area of the large square as:
2y2-D+2U
This is twice the area of either of the two smaller squares (2y2) minus the area that they both cover (D) plus the area that neither covers (2U). By definition of x and y, x2 = 2y2. So we must have D = 2U. But U = (x-y)2, and D = (2y-x)2. Thus (2y-x)2 = 2(x-y)2. Since x > y > 0, x-y is a positive integer smaller. Note that 2 > x/y, so 2y > x, and thus 2y-x is a positive integer, and we have ((2y-x)/(x-y))2 = 2, i.e. sqrt(2) = (2y-x)/(x-y). Since 2y < 2x, we have 2y-x < x. Thus we have found a representation of sqrt(2) with positive numerator smaller than x. This contradicts the choice of x. So sqrt(2) must be irrational.
Last edited by serpicojr on Sat May 24, 2008 5:33 pm; edited 1 time in total |
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| JaneBennet |
| Posted: Sat May 24, 2008 4:58 pm Post subject: |
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Forum Ph.D.
Joined: 06 Apr 2008
Posts: 874
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I like the application of the AM–GM inequality to the problem of finding the minimum surface area or the maximum volume of a cuboid. (That’s a solid with 6 rectangular faces such that adjacent faces are perpendicular to each other.) Suppose a cuboid has length, width and height l, w and h respectively. Its surface area is S = 2(lw+wh+hl) and its volume V = lwh.
By the AM–GM inequality,
i.e.
i.e.
Hence we can find the minimum surface area for a given volume, or the maximum volume for a given surface area. Moreover, equality occurs if and only if lw = wh = hl, i.e. if and only if l = w = h.
This is so much less messy than using calculus. 
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| thyristor |
| Posted: Sun May 25, 2008 1:51 am Post subject: |
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Forum Sophomore
Joined: 11 Feb 2008
Posts: 184
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Neat ones. I fancy the proof of the Fundamental theorem of calculus.
Unfortunately I can't write it here since I don't know how to write
the symbol of integral in the message.
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Last edited by thyristor on Mon May 26, 2008 10:47 am; edited 1 time in total |
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| river_rat |
| Posted: Sun May 25, 2008 10:32 pm Post subject: |
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Forum Ph.D.
Joined: 01 Jun 2006
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Finite sums theorem, its quite deep but the proof is really neat.
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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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| algebraic topology |
| Posted: Mon May 26, 2008 10:45 am Post subject: |
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Forum Freshman
Joined: 06 May 2008
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| The fact that you can solve problems in topology using methods from abstract algebra is IMHO one of the best things in pure mathematics. |
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| thyristor |
| Posted: Mon May 26, 2008 10:48 am Post subject: |
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Posts: 184
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How do I write an integration in my post?
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| serpicojr |
| Posted: Mon May 26, 2008 10:51 am Post subject: |
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Joined: 17 Jul 2007
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| algebraic topology wrote: |
| The fact that you can solve problems in topology using methods from abstract algebra is IMHO one of the best things in pure mathematics. |
And vice versa! |
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| serpicojr |
| Posted: Mon May 26, 2008 10:54 am Post subject: |
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Forum Professor
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| thyristor wrote: |
| How do I write an integration in my post? |
Copy and paste ∫. To get upper and lower limits, use HTML for sub- and superscript formatting (< sub > and < sup >). In general, appeal to this post for math symbols:
http://www.thescienceforum.com/More-non-Latex-codes-8625t.php
If you know latex, this site has free latex hosting:
http://math.b3co.com/
Ignore the fact that it says the service is down, as I've posted a few images from the site recently. |
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| thyristor |
| Posted: Tue May 27, 2008 9:06 am Post subject: |
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Forum Sophomore
Joined: 11 Feb 2008
Posts: 184
Location: Sweden
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Ok, so here comes the evidence ( if anyone doesn't know it)
F and f are functions.
F(x)=∫xx0 f(t)dt
Thus:
(F(x+deltax)-F(x))*1/deltax= (∫x+deltaxx0 f(t)dt -∫x+deltaxx f(t)dt)*1/deltax
Thus:
F'(x)=(∫x+deltaxx f(t)dt)*1/deltax
According to the mean value theorem there is a dot p such that f(p)*deltax=(∫x+deltaxx f(t)dt)
Thus:
F'(x)=f(p)*deltax/deltax F'(x)=f(p)
lim deltax -> 0 p=x
Thus:
F'(x)=f(x) (Which completes the proof)
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| bit4bit |
| Posted: Tue May 27, 2008 1:15 pm Post subject: |
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Forum Ph.D.
Joined: 14 Jul 2007
Posts: 625
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I like the derivation of the vector cross product.
v = <a,b,c> = ai+bj+ck
u = <d,e,f> = di+ej+fk
v x u = (ai+bj+ck)(di+ej+fk)
=adixi+aeixj+afixk+bdjxi+dejxj+bfjxk+cdkxi+cekxj+cfkxk
|ixi|, |jxj|, |kxk| = 0
so,
v x u = aeixj+afixk+bdjxi+bfjxk+cdkxi+cekxj
|ixj|=k, |jxi|=-k
|jxk|=i, |kxj|=-i
|kxi|=j, |ixk|=-j
so,
v x u = aek-afj-bdk+bfi+cdj-cei
= (bf-ce)i + (cd-af)j + (ae-bd)k
= (bf-ce)i - (af-cd)j + (ae-bd)k
= i | b c | - j | a c | + k | a b|
.....| e f |......| d f |.......| d e |
=
| i j k |
| a b c |
| d e f |
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| talanum1 |
| Posted: Fri Jun 20, 2008 2:15 am Post subject: |
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My >3D Vector product definition leads to a definition of non square determinants. A definition isn't always cheap if it must fit into already existing theorems and definitions.
Here is the secrets of non square determinants. Let A be nxm matrix with n<m then you develop it's determinant just like square determinants untill
reaching 2xp, p > 2 stage.
At this stage you write this as a sum of 2x2 matrices, one term for every combination of deleted coulumns (so that only 2 columns remain), marking the deleted columns in their correct place(s) by a symbol . Then you interchange columns untill all empty columns is (are) at rightmost and multiply the term by -1 if an odd amount of columns were interchanged.
That's it.
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Found cross product of two vectors in >3D. |
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| serpicojr |
| Posted: Fri Jun 20, 2008 7:31 am Post subject: |
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So the determinant satisfies these properties:
1. It's alternating in the columns of a matrix--i.e., if you swap two columns, the determinant changes by -1.
2. It's linear in each column--i.e., fixing all of the other columns, the determinant gives a linear map in the one unfixed column.
3. det(I) = 1
These properties uniquely specify the determinant. If you try to come up with something like this for mxn matrices with m ≠ n, you'll find that... well, first, the 3rd condition doesn't make sense. But if you get rid of the 3rd condition, then the determinant is uniquely specified up to a scalar multiple on nxn matrices, and so you can ask whether the same holds for mxn, m ≠ n.
The answer is no. When m > n, you'll always have at least two possible "determinants" which are not scalar multiples of each other. When m < n, there is necessarily a linear dependency amongst the columns of your matrix, and so anything satisfying 1 and 2 must be identically 0--in other words, there is not even a single candidate for determinant. |
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| serpicojr |
| Posted: Fri Jun 20, 2008 7:38 am Post subject: |
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Joined: 17 Jul 2007
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Let me illustrate what I'm talking about.
If v = [x,y]T is a column vector, then d(v) = x and d'(v) = y both give potential "determinants"--the alternating property is trivial, and the second property boils down to d being a linear map, since there's only one column. And clearly there does not exist a scalar so that d(v) = kd'(v) for all v.
Now let u = [z,w] be a row vector, then suppose d satisfies 1 and 2. Clearly d([0,0]) = 0, so we may assume that u ≠ [0,0]. Suppose z ≠ 0, and consider d on the vector [z,z]. Swapping columns preserves this vector, so:
d([z,z]) = -d([z,z])
Thus d([z,z]) = 0. But d is linear in any one column if you fix the other, so that:
d([z,w]) = d([z,(w/z)z]) = (w/z)d([z,z]) = 0
So d is trivial. A similar argument work if z ≠ 0. |
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| talanum1 |
| Posted: Sat Jun 21, 2008 5:09 am Post subject: |
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Forum Sophomore
Joined: 28 Jul 2007
Posts: 103
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Yes you get two determinants, one developed by column: D_C, the other one developed by row D_R.
However the basic properties (except D (A) = D (A transposed)) holds for some combination(s) of:
D_R, D_C, n<m, m<n
and never for neither D_R nor D_C.
Of course you have: D_R (A) = D_C (A transposed) and vice versa.
Your conditoin 3 just generalise as an nxn identity matrix with zero last column vector(s). Condition 1 holds for D_R A, A a row longest matrix and intercahnge of rows and D_C A, A a column longest and interchange of columns.
I just need to check condition 2 and the following post.
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Found cross product of two vectors in >3D. |
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