How can I show that the function ${\displaystyle f(x)=x+{\tfrac {\sin x}{2}}}$ is injective, but without using it's derivative? I've tried using the inequality ${\displaystyle |\sin x|\leq |x|}$ and showing that ${\displaystyle 0<|x-y|\,\Longrightarrow \,0<|f(x)-f(y)|}$, but with no success. (Using the derivative is simple, because f'>0 ⇒ f is strictly increasing ⇒ f is injective). Thanks. 109.67.31.31 (talk) 04:42, 23 March 2011 (UTC)[reply]

Hint: Use the trigonometric identity for ${\displaystyle \sin x-\sin y}$. -- Meni Rosenfeld (talk) 09:40, 23 March 2011 (UTC)[reply]

Could you please be more specific? :) I've already tried using the method ${\displaystyle f(x)=f(y)\,\Longrightarrow \,x=y}$ with the identity of ${\displaystyle \sin x-\sin y}$, but I got stuck..

${\displaystyle f(x)=f(y)\,\Longrightarrow \,x+{\tfrac {\sin x}{2}}=y+{\tfrac {\sin y}{2}}\,\Longrightarrow \,x-y={\tfrac {\sin y-\sin x}{2}}\,\Longrightarrow \,x-y=\cos \left({\tfrac {y+x}{2}}\right)\sin \left({\tfrac {y-x}{2}}\right)\,\Longrightarrow \,??}$.

Thanks again. 79.180.13.32 (talk) 10:47, 23 March 2011 (UTC)[reply]

Great. Now let ${\displaystyle t=|x-y|}$ and use the last equation to show that ${\displaystyle t\leq 0}$. -- Meni Rosenfeld (talk) 11:04, 23 March 2011 (UTC)[reply]

I think I got it, using the inequality I've mentioned above:

${\displaystyle x-y=\cos \left({\tfrac {y+x}{2}}\right)\sin \left({\tfrac {y-x}{2}}\right)\,\Longrightarrow \,|x-y|=|\cos \left({\tfrac {y+x}{2}}\right)\sin \left({\tfrac {y-x}{2}}\right)|=|\cos \left({\tfrac {y+x}{2}}\right)|\cdot |\sin \left({\tfrac {y-x}{2}}\right)|\leq |\cos \left({\tfrac {y+x}{2}}\right)|\cdot {\tfrac {|y-x|}{2}}\,\Longrightarrow \,|y-x|\left(2-|\cos \left({\tfrac {y+x}{2}}\right)|\right)\leq 0}$

but ${\displaystyle |\cos \theta |\leq 1}$ for all θ, so ${\displaystyle 2-|\cos \left({\tfrac {y+x}{2}}\right)|\geq 1>0\,\Longrightarrow \,|y-x|\leq 0}$0\,\Longrightarrow \,|y-x|\leq 0}"/>, and finally ${\displaystyle 0\leq |y-x|\leq 0\,\Longrightarrow \,|x-y|=0\,\Longrightarrow \,x=y}$. Q.E.D. Is there a simpler way? :) 79.179.35.2 (talk) 12:07, 23 March 2011 (UTC)[reply]

It becomes slightly simpler if you use ${\displaystyle |\cos \theta |\leq 1}$ earlier:

${\displaystyle |x-y|=|\cos \left({\tfrac {y+x}{2}}\right)|\cdot |\sin \left({\tfrac {y-x}{2}}\right)|\leq |\sin \left({\tfrac {y-x}{2}}\right)|\leq {\tfrac {|x-y|}{2}}}$

And so ${\displaystyle |x-y|\leq 0}$. -- Meni Rosenfeld (talk) 12:28, 23 March 2011 (UTC)[reply]

Many, many thanks! May your functions always be integrable! :D 109.65.45.25 (talk) 12:51, 23 March 2011 (UTC)[reply]

In my culture, it is extremely vulgar to suggest I will only deal with the kind of mindless high-school problems that only require elementary antiderivatives! ;) :P

Thanks, take care. -- Meni Rosenfeld (talk) 14:22, 23 March 2011 (UTC)[reply]

Given a bezier curve in two dimensions defined by start point p0, two control points p1 and p2, and end point p3 is there a simple expression for the curve length?

78.245.228.100 (talk) 10:28, 23 March 2011 (UTC)[reply]

Don't I wish! It can be done in terms of Bessel functions(must have just written that because it also started with B :) ) elliptic integrals or their various equivalents - they enable you to integrate the square roots of fourth degree polynomials and your C maths library probably contains them. Not that I'd call those functions 'simple' but at least you wouldn't have to code them. I see our article on Bezier curves doesn't say anything, I'll have a look around for a source. Lots of people just break the curve into bits and approximate which is also what one has to do if doing things like finding the time if one has a different speed moving along the curve at different times. Dmcq (talk) 11:54, 23 March 2011 (UTC)[reply]

Elliptic integrals is probably our best article on those type integrals. Dmcq (talk) 12:01, 23 March 2011 (UTC)[reply]

Metapost has a function called arclength to compute this, so perhaps look at the source to see how it does it. That's no guarantee that there's a simple formula though. – b_jonas 21:59, 23 March 2011 (UTC)[reply]

How precisely do you need the arclength calculated? Perhaps the length of an inscribed open 7-gon is a sufficient approximation. Bo Jacoby (talk) 10:58, 24 March 2011 (UTC).[reply]

I believe that programs that use Bezier splines almost invariably do this by walking along the curve and computing the length of each segment, which is easy to do because Bezier curves are given parametrically. Even dividing the 0-1 range into 100 segments usually imposes a minimal computational burden. Looie496 (talk) 16:55, 24 March 2011 (UTC)[reply]

To put my question bluntly: why do we care about automorphism groups? I can think of lots of reasons (they're interesting, they have certain applications (for instance, twisting a group action by an automorphism of the group to obtain a new action...)), but what do they tell us about the actual group of which they are the automorphisms? I suppose they give some sense of flexibility of the group, by demonstrating the ways that we can map the group into itself, but how can we make that rigorous? For instance, what can we say about two groups if one has a finite (outer) automorphism group, but the other has an infinite one? Thanks, Icthyos (talk) 11:52, 23 March 2011 (UTC)[reply]

A group with a finite automorphism group is "close to" being an abelian group, since it can't have many inner automorphisms. Sławomir Biały (talk) 12:19, 23 March 2011 (UTC)[reply]

In many cases, the dimension of the automorphism group of an object is important. For instance, tori are the only connected compact Lie groups whose automorphism groups are zero dimensional (i.e., discrete), since any other Lie group will have many inner automorphisms. This is an example of what is known as a rigidity theorem: that in an extremal case of some invariant, we are able to conclude everything about the object. If we're allowed to generalize the question slightly, then oftentimes a geometrical space is known completely if only the dimension of its isometry group (a type of automorphism group) is known. For instance, the automorphism group of an n-dimensional Riemannian manifold has dimension at most n(n+1)/2, and if the dimension is exactly equal to this, then the manifold is the n dimensional Euclidean space. This is another example of a rigidity theorem. It's probably possible to rephrase this as a statement about reductive groups because of the connection with symmetric spaces. Sławomir Biały (talk) 14:13, 23 March 2011 (UTC)[reply]

Thanks for your reply. I was suspecting the answer might involve the intersection of group theory with some other area of maths. My question arose because I'm currently looking at the outer automorphism groups of a certain class of groups, and I was wondering what I could say, in any concrete terms, when one group has a finite number of outer automorphisms, while another has an infinite number of them. Icthyos (talk) 10:39, 24 March 2011 (UTC)[reply]

Hello, our Wikipedia page on Frank Nelson Cole says that it took him 'three years of Sundays' to find the factors of ${\displaystyle 2^{67}-1}$, attributing it to a BBC page, as opposed to a quote given in the Introduction of chapter 1 in Mathematica Navigator book by Heikki Ruskeepää, which says it took him twenty years of Sundays. Since there is a huge difference between the two claims, can anybody else verify it from some other historical sources? Thanks - DSachan (talk) 12:50, 23 March 2011 (UTC)[reply]

I think that BBC's h2g2 is not really considered to be reliable, since it's effectively an open wiki. But according to a Google books search, the three years figure is correct. Sławomir Biały (talk) 13:14, 23 March 2011 (UTC)[reply]

Brilliant, so I am going to send an email to Heikki about it so he corrects it in his next edition. I don't know how he got this 'twenty years' figure anyway. No other book on your Google search seems to suggest that. Thanks - DSachan (talk) 13:22, 23 March 2011 (UTC)[reply]

Unless of course he only worked on the problem on a Sunday. In which case every Sunday for 20 years is approximately 3 years where every day is a Sunday... -- SGBailey (talk) 14:16, 23 March 2011 (UTC)[reply]

I think this is a language problem. If taken literally three years of Sundays means twenty-one years (i.e. 20 years, give or take) between starting and completion (about 1000 Sundays). Of course, it would also mean he only worked Sundays for three years, about 150 Sundays. I don't know which one Cole intended/which one is true. Grandiose (me, talk, contribs) 19:07, 23 March 2011 (UTC)[reply]

Though, it's hard to imagine that he worked everyday for three years to look out for its factors. Spending Sunday only to find its factor sounds more reasonable. So, about 150 Sundays (Sundays in three years) seem more plausible. - DSachan (talk) 20:43, 23 March 2011 (UTC)[reply]