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October 19

Smallest Polymino whose Convex Hull is not tilable...

What is the Smallest Polymino whose convex Hull is not tilable? is it the quadmino "T", that one has a convex hull with sides 3-1-root2-1-root2? Naraht (talk) 03:00, 19 October 2014 (UTC)[reply]

It looks like all the tretrominos have tilable convex hulls. But the convex hull of the Y-pentomino does not tile. It has one side of length √5 and this can only be placed next to another tile which is rotated 180 degrees. When you then try to match the side with length √2, there are two ways but both leave a bay that can't be filled. There are probably other pentominoes as well, I haven't checked them all, but that's enough to answer the question as to the smallest. --RDBury (talk) 07:28, 19 October 2014 (UTC)[reply]

PS & correction. Actually the convex hull of the T-tetromino does not tile. When you place a tile next to the side of length √2, there are two ways to do it but both leave a bay that can't be filled. After a bit of doodling I found that the convex hulls of 5 of the 12 pentominoes, Y, F, T, X and W do not tile, while the convex hulls of the remaining 7, I, L, N, V, P, U, Z, do. Perhaps someone could check this as it would probably take more effort than it's worth to turn my doodles into something rigorous. --RDBury (talk) 07:57, 19 October 2014 (UTC)[reply]

Real Analysis: Bounded Sequences

The question I am having trouble with is Prove that sn is bounded where sn=sin(n)/n.

I know I have to use the definition of bounded sequence. I know sin(n) converges and therefore is bounded and 1/n is bounded, but I am not really sure where to go with this. A suggestion about where to begin would be very helpful. — Preceding unsigned comment added by Pinterc (talk • contribs) 13:44, 19 October 2014 (UTC)[reply]

sin(n), although bounded, does not converge. The sequence 1/n converges to zero and so is bounded. sin(n)/n is the product of two bounded sequences, so it's bounded. (In fact, you can prove that it converges to zero.) Sławomir Biały (talk) 14:31, 19 October 2014 (UTC)[reply]

Is the following elementary proof of Fermat's Last Theorem correct?

The theorem states there exists no integral (all positive integers) solutions for the equation, 1. x^n+y^n=z^n for n > 2. I begin the proof by assuming there exists an integral (positive integer) solution to equation one for some n > 2. Equation one becomes with some algebraic manipulation, 2. x^n=z^n-y^n = (z^(n/2)+y^(n/2))*(z^(n/2)-y^(n/2)).

Now that I have factored the right side of equation two, Fermat, the great French mathematician and respectable jurist, made I believe the next logical and crucial step. He factored the left side as well, x^n, with the help of an extra real variable, Ɛ, such that 0 < Ɛ < n . I have the following equation, x^n = x^(n/2+Ɛ/2)* x^(n/2-Ɛ/2) = (z^(n/2)+y^(n/2))*(z^(n/2)-y^(n/2) ). This equation implies x^(n/2+Ɛ/2)= z^(n/2)+y^(n/2) and x^(n/2-Ɛ/2) = z^(n/2)-y^(n/2). (Note: The latter two equations are of the form, ab = c + d and a/b = c - d, respectively. And they are quite reasonable given the constraints or parameters of this problem and given the reader has a sufficient understanding of high school algebra.)

And next I have, 3. x^(n/2+Ɛ/2)/ x^(n/2-Ɛ/2) = x^Ɛ = (z^(n/2)+y^(n/2) )/(z^(n/2)-y^(n/2) ). Using equation 3 and applying some algebraic manipulation and simplification to it, I generate the equation, 4. z^n=y^n *((x^Ɛ+1)/(x^Ɛ – 1))^2.

And finally, by combining equations one and four, I generate the following equation after some more algebraic manipulation and simplification, 5. y = (1/4)^(1/n)*(x^(n-Ɛ))^(1/n)*(x^Ɛ – 1)^(2/n).

However, (1/4)^(1/n) is not a rational number, a ratio of two whole numbers, for n > 2. This implies the right side of equation five is not a positive integer. This contradicts my assumption that y is a positive integer. Thus, Fermat’s Last Theorem is true, and Fermat was right! Thank God! Praise God!--David Cole, primesdor@gmail.com Primesdegold (talk) 19:59, 19 October 2014 (UTC)[reply]

There are several very elementary errors in this proof. Are you a troll, or just an idiot? Sławomir Biały (talk) 20:06, 19 October 2014 (UTC)[reply]

WP:AGF. There are many people who think they've proved some open problem in math, and presenting their work here is not necessarily a sign of trolling or idiocy. These questions could just as easily be based in naivety and zeal ;) If anyone wants to help OP find their mistakes, I don't think that's a problem. SemanticMantis (talk) 14:45, 20 October 2014 (UTC)[reply]

We should not tolerate trolls like this. I think we should just tell the idiot to go away and bother some other corner of the internet. Sławomir Biały (talk) 15:06, 20 October 2014 (UTC)[reply]

I agree, this person is wasting everyone's time with their "proofs". WP:ABAN? Neuroxic (talk) 11:02, 26 October 2014 (UTC)[reply]

If any elementary proof were correct, then it would have been found in the eighteenth century (or Fermat could have written it in the margin in the seventeenth century). Robert McClenon (talk) 15:21, 20 October 2014 (UTC)[reply]

I absolutely agree! I once worked in Graduate admissions in the math dept, and we had people claiming to trisect angles in their applications, which is even worse, because that is not merely open, but provably impossible... The unsatisfying thing about this reasoning though, is it ends up being basically an appeal to authority, which is of course not how math really works, and not something we really should rely on when answering questions about math. Though I don't have the time or interest to find errors on these so-called 'proofs', I still think others should be allowed to do so if they wish. Sławomir seems certain this is a troll, but I think a crank need not be a troll. It's a distinction of intent. I try not to ascribe malice to online users where simple ignorance and naivety may provide sufficient explanation. Maybe I'm in the minority, but I see it as our goal here to help educate the naive. SemanticMantis (talk) 15:29, 20 October 2014 (UTC)[reply]

First, SemanticMantis is obviously referring to trisection with Euclidean tools, where he or she correctly notes that the construction has been proved impossible. Trisection of an angle is easy if one is not limited to Euclidean tools. Archimedes trisected the angle using an expanded toolset. Doubling a cube is similarly impossible with Euclidean tools and easy with better tools. Squaring the circle is a different, more difficult problem. Robert McClenon (talk) 15:42, 20 October 2014 (UTC)[reply]

Second, I agree with SemanticMantis that the OP is not a troll, but an editor whose enthusiasm exceeds his knowledge and his knowledge of the limits of his knowledge. Robert McClenon (talk) 15:42, 20 October 2014 (UTC)[reply]

I find it funny that the OP has the skill to carry out some elaborate algebraic manipulations correctly, and yet seems to think that

ab=cd

implies

a=b,\ c=d

.

This could be a result of a phenomenon that is all too common among students, of treating math too formally - they learn by rote manipulations and problem solving techniques, but they have no idea what the concepts actually mean. -- Meni Rosenfeld (talk) 05:10, 21 October 2014 (UTC)[reply]

We need a template {{FLT-Proof-Postcard}} for the equivalent of what Edmund Landau did. Note, if we can find a good source, that should be added to Landau's page.

Request permission to add a category to the Fractal subheading 'Common techniques for generating fractals'

If you don't mind, I'd like to add 'Sirsty-Firsty' to the Fractal subheading 'Common techniques for generating fractals'.

Sirsty-Firsty, as in www.sirsty-firsty.org This is a method for generating designs based on moving points around in spirals.

Any thoughts?

Thanks, --InverseSubstance (talk) 22:40, 19 October 2014 (UTC)[reply]

That subsection of the article is not about specific software implementations, but general methods. Is the method used in the software that you linked to discussed in the peer reviewed literature? If not, then it should not be added to the article. Sławomir Biały (talk) 12:53, 20 October 2014 (UTC)[reply]

You could perhaps add a link in the section Fractal#Fractal-generating_programs. You could also post this question at Talk:Fractal. SemanticMantis (talk) 14:49, 20 October 2014 (UTC)[reply]

Is it obvious that these things are fractals? I can't see that it is self evident that these things exhibit an obvious self similarity across all scales. — Preceding unsigned comment added by 86.156.222.88 (talk) 23:25, 21 October 2014 (UTC)[reply]

Well that's the question. Sirsty-Firsty is definitely some kind of design generator. It's based on a simple function repeated over and over again. There is a kind of repeating design, only it happens in terms of the color persistence chosen. If an image sums up a persistence of 1,2,4,8,16... then the design would appear repeatedly at the scale of 1,2,4,8,16... Anyway, if this isn't a fractal, maybe someone could suggest what category of computer design it does fit in? Thanks, --InverseSubstance (talk) 06:35, 22 October 2014 (UTC)[reply]

Fractals don't have to be self-similar at all scales, that's a popular misconception. They just have to have fractional dimension. Lots of things are fractal but not self-similar. Heck even the very famous Mandelbrot set "in general is not strictly self-similar but it is quasi-self-similar, as small slightly different versions of itself can be found at arbitrarily small scales." I won't go through the details of this linked algorithm, but I don't think it's unreasonable to think that the red colored regions in the example pics constitute a set whose boundary has fractal dimension. SemanticMantis (talk) 15:28, 22 October 2014 (UTC)[reply]