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August 25[edit]

Logarithmic Spiral[edit]

What's the reason why the logarithmic spiral is so common in nature? The article mentions self-similarity, but could someone be more specific? 74.15.137.168 (talk) 02:49, 25 August 2011 (UTC)[reply]

Only God knows the true reason why, but man can guess. Symmetries may reuse code in the DNA. These symmetries are often broken in later stages of development or evolution. There is a high level of symmetry between arm and leg on a human, and even higher symmetry between right arm and left arm. There is also a kind of symmetry between a small boy and a big boy. The organs grow. The change of teeth is an exception. The logarithmic spiral has a continuous symmetry group, and so it may grow using a short piece of DNA code. Bo Jacoby (talk) 08:00, 25 August 2011 (UTC).[reply]

I think the fundamental underlying reason is that exponential growth and exponential decay are common in nature, and when something changes exponentially while circling around a fixed center, you naturally get a logarithmic spiral. Looie496 (talk) 16:03, 25 August 2011 (UTC)[reply]

Oh okay, so because

{\frac {dr}{d\theta }}={\frac {dr}{dt}}{\frac {dt}{d\theta }}={\frac {v}{\omega }}\propto r

, logarithmic spirals are common? 74.15.137.168 (talk) 17:18, 25 August 2011 (UTC)[reply]

That's a reasonable way of looking at it, but it also ignores all the processes (developmental, evolutionary, geo-chemical, etc) that go into forming such spirals in nature (which others allude to above). You may be interested in reading up on phyllotaxis and links therein (this is the start of the whole 'pineapple bracts follow the fibonacci sequence' business). You might also get better answers on the science desk, this isn't really a math question :) SemanticMantis (talk) 20:29, 25 August 2011 (UTC)[reply]

Proof too easy?[edit]

Good morning. I was to show rigourously that f(x)=x³-4x+6 is not one to one (injective, I think it's also called?). This is my proof (contradiction), but something doesn't sit right with me about it; it seems too easy. Have a look:
Suppose that f(m)=f(n) if and only if m=n
$m^{3}-4m+6=n^{3}-4n+6$ (m=n)
$(m^{3}-n^{3})-(4m-4n)=0$ (6s cancel, bring both to one side)
$(m-n)(m^{2}+mn+n^{2})-(m-n)4=0$ (factoring)
$(m-n)(m^{2}+mn+n^{2}-4)=0$ (grouping)
$m^{2}+mn+n^{2}-4=0$ (The expression is zero when either the first or the second factor is zero; I'm not interested in the first because it is the same as the hypothesis)
By the quadratic formula:
$(*)\qquad m={\frac {-n\pm {\sqrt {n^{2}-4(n^{2}-4)}}}{2}}$
$n={\frac {-n\pm {\sqrt {n^{2}-4(n^{2}-4)}}}{2}}$ (m=n)
$\rightarrow n={\frac {2{\sqrt {3}}}{3}}$ However, putting n back into $(*)$ , I get $m={\frac {2{\sqrt {3}}}{3}},{\frac {-4{\sqrt {3}}}{3}}$ , contradiction Therefore, f(m)=f(n) for at least one pair (m,n): m≠n 203.117.33.23 (talk) 23:24, 25 August 2011 (UTC) Is this proof legit? Thanks so much, and sorry for wasting so much of your time. PS: What happened to all the other pairs, because in the graph there is obviously more than one. THanks again.[reply]

Didn't look too closely at your proof above, but you're probably meant to just notice that f(2) = f(0). Staecker (talk) 00:00, 26 August 2011 (UTC)[reply]

OK now I looked at your proof. It's definitely suspicious. Your assumption, in order to derive a contradiction, is: "f(n)=f(m) iff n=m". By your argument (which is fine) this is equivalent to "

(m-n)(m^{2}+mn+n^{2}-4)=0

iff m=n". Your use of the quadratic formula is fine, but why then substitute n for m? You're not assuming that m=n, you're assuming the more complicated "iff" statement. Staecker (talk) 00:06, 26 August 2011 (UTC)[reply]

The first derivative has two distinct real roots, so there are portions with positive and negative slope, which can't be true for a one to one function. I don't know what your standard of rigour has to be, but something could be worked up round this fact.←86.155.185.195 (talk) 12:03, 26 August 2011 (UTC)[reply]