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February 23[edit]

Linking Geometry to Combinatorics[edit]

Is there a branch of mathematics that (systematically) connects these two fields with one another ?
Are there any books, papers, articles (both online or offline) that treat this connection systematically ?

For instance, we know that geometric shapes of the form $x^{n}+y^{m}=r^{n}$ are connected with binomial coefficients, inasmuch as the former's area is nothing else than the latter's reciprocal, or multiplicative inverse. (Wallis' integrals also come here to mind). Likewise, Vandermonde's identity $\sum _{0}^{n}{n \choose k}^{2}={2n \choose n}$ gives, for $n={\tfrac {1}{2}},$ a formula for π : $\sum _{0}^{\infty }\left[{\frac {(2k-3)!!}{(2k)!!}}\right]^{2}={\frac {4}{\pi }}.$

Are there any other such similar results, connecting these two fields together ? — 79.113.198.225 (talk) 16:58, 23 February 2014 (UTC)[reply]

Your examples are more like connections between calculus and combinatorics, so it's not clear exactly what you're trying to get. There's a whole field of discrete geometry which explores combinatorial aspects of geometry/geometrical aspects of combinatorics. --RDBury (talk) 11:59, 24 February 2014 (UTC)[reply]

Do you also have any suggestions concerning possible connections with calculus ? I was thinking, maybe generating functions ? — 79.114.140.233 (talk) 00:13, 25 February 2014 (UTC)[reply]

In addition to discrete geometry, there is also geometric combinatorics. In analysis, people have looked at analytic properties of generating functions in combinatorics, .e.g, Analytic Combinatorics—A Calculus of Discrete Structures. Analytic number theory also has a number of connections to combinatorics. --Mark viking (talk) 01:22, 25 February 2014 (UTC)[reply]

Equations[edit]

Through what process do these two equations...

{\begin{aligned}N\cdot \!\,sin\alpha \ &=m{\frac {v^{2}}{r}}\ \\N\cdot \!\,cos\alpha \ &=m\cdot \!\,g\\\end{aligned}}

...become this:

{\begin{aligned}{\frac {N\cdot \!\,sin\alpha }{N\cdot \!\,cos\alpha }}\ &={\frac {m{\frac {v^{2}}{r}}}{m\cdot \!\,g}}\ \\\end{aligned}}

Th4n3r (talk) 20:49, 23 February 2014 (UTC)[reply]

Er, division? So long as

N\cos \alpha \neq 0

… --Tardis (talk) 21:55, 23 February 2014 (UTC)[reply]

To elaborate slightly, either (A) The two sides of eq. 1 are equal, so if each side is divided by the same quantity, then they remain equal - but the two sides of eq. 2 are equal (i.e. are the same quantity), so it's a valid move to divide side 1 of eq. 1 by side 1 of eq. 2 and side 2 of eq. 1 by side 2 of eq. 2 - or alternatively (B) Divide both sides of eq. 1 by N cos(α), then substitute for N cos(α) on the right hand side only using eq. 2. --catslash (talk) 23:01, 23 February 2014 (UTC)[reply]

Thank you. Th4n3r (talk) 23:23, 23 February 2014 (UTC)[reply]

I find it helps to rephrase the equations as A=B, C=D, and then A/C=B/D. HTH, 164.11.203.58 (talk) 08:31, 24 February 2014 (UTC)[reply]

You can do the same with all sorts of operations, for instance squaring and adding you get:

\left(N\cdot \!\,sin\alpha \ \right)^{2}+\left(N\cdot \!\,cos\alpha \ \right)^{2}=\left(m{\frac {v^{2}}{r}}\right)^{2}+\left(m\cdot \!\,g\right)^{2}

which simplifying and taking the square root becomes:

N=m\cdot {\sqrt {\left({\frac {v^{2}}{r}}\right)^{2}+g^{2}}}

which eliminates the dependence on

\alpha

. Dmcq (talk) 08:54, 24 February 2014 (UTC)[reply]