Comments for Felix' Math Place http://math.fontein.de Focussed on, but not limited to Computational Number Theory Fri, 03 May 2013 13:23:07 +0000 hourly 1 http://wordpress.org/?v=3.3.2 Comment on A Cute Identity. by Felix Fontein http://math.fontein.de/2011/07/30/a-cute-identity/comment-page-1/#comment-3221 Felix Fontein Fri, 03 May 2013 13:23:07 +0000 https://math.fontein.de/?p=869#comment-3221 Hi Gerard, yes, that indeed looks <b>much</b> better :-) I wonder why I overlooked that... Thanks a lot! Best, Felix Hi Gerard,
yes, that indeed looks much better :-) I wonder why I overlooked that…
Thanks a lot!
Best, Felix

]]> Comment on A Cute Identity. by Gerard http://math.fontein.de/2011/07/30/a-cute-identity/comment-page-1/#comment-3220 Gerard Fri, 03 May 2013 13:07:01 +0000 https://math.fontein.de/?p=869#comment-3220 Hey Felix, If you write $Y_0=1$ and $Y_i=1+\sum_{j=1}^i x_j$, then the RHS becomes [align]1-\sum_{i=1}^n \frac{Y_i-Y_{i-1}}{Y_iY_{i-1}} ={} & 1-\sum_{i=1}^n \biggl(\frac{1}{Y_{i-1}}-\frac{1}{Y_i}\biggr) \\ {}={} & 1-\frac{1}{Y_0}+\frac{1}{Y_n} = \frac{1}{Y_n}[/align] and you have it. Something slightly more insightful? You tell me :-) G. Hey Felix,

If you write Y_0=1 and Y_i=1+\sum_{j=1}^i x_j, then the RHS becomes
1-\sum_{i=1}^n \frac{Y_i-Y_{i-1}}{Y_iY_{i-1}} ={} & 1-\sum_{i=1}^n \biggl(\frac{1}{Y_{i-1}}-\frac{1}{Y_i}\biggr) \\ {}={} & 1-\frac{1}{Y_0}+\frac{1}{Y_n} = \frac{1}{Y_n}

and you have it. Something slightly more insightful? You tell me :-)

G.

]]> Comment on The Probability That Two Numbers Are Coprime. by Felix Fontein http://math.fontein.de/2012/07/10/the-probability-that-two-numbers-are-coprime/comment-page-1/#comment-1900 Felix Fontein Wed, 28 Nov 2012 14:09:41 +0000 https://math.fontein.de/?p=884#comment-1900 An alterantive approach can be found <a href="http://arxiv.org/abs/1211.6246" rel="nofollow">here</a> (Proposition 3.2). There, we analyzed the probability that integers $0 \le x, y \le n$ are coprime (as opposed to $1 \le x, y \le n$ as here). An alterantive approach can be found here (Proposition 3.2). There, we analyzed the probability that integers 0 \le x, y \le n are coprime (as opposed to 1 \le x, y \le n as here).

]]> Comment on The Hasse derivative. by Felix Fontein http://math.fontein.de/2009/08/12/the-hasse-derivative/comment-page-1/#comment-371 Felix Fontein Wed, 07 Nov 2012 12:48:37 +0000 http://math.fontein.de/?p=277#comment-371 Yes, you are right. Thank you very much! Yes, you are right. Thank you very much!

]]> Comment on The Hasse derivative. by Michael Forbes http://math.fontein.de/2009/08/12/the-hasse-derivative/comment-page-1/#comment-370 Michael Forbes Tue, 06 Nov 2012 23:41:26 +0000 http://math.fontein.de/?p=277#comment-370 I believe your Faa di Bruno formula has a small error. In particular, I believe it should be: $\displaystyle D^{(k)} (f \circ g) = \sum \binom{c_1+\cdots+c_k}{ c_1, \dots, c_k} (D^{(c_1+\cdots+c_k)} f) \circ g \cdot \prod_{j=1}^k (D^{(j)} g)^{c_j}$ where the sum ranges over all $c_i$ where $\sum_{i=1}^k i c_i = k$. In particular, in the derivation above for $x^n$ you end with a $g^{c_0}$ term, but this term is not equal to $D^{c_0}(f)(g)$, but is rather equal to $D^{n-c_0}(f)(g)/\binom{n}{c_0}$. When you do this substitution, you get what I list above. Incidentally, if you look on Wikipedia for this formula (for normal partial derivatives) it agrees with what I put above, when you ignore the factorials. I believe your Faa di Bruno formula has a small error. In particular, I believe it should be:

\displaystyle D^{(k)} (f \circ g) = \sum \binom{c_1+\cdots+c_k}{ c_1, \dots, c_k} (D^{(c_1+\cdots+c_k)} f) \circ g \cdot \prod_{j=1}^k (D^{(j)} g)^{c_j}

where the sum ranges over all c_i where

\sum_{i=1}^k i c_i = k.

In particular, in the derivation above for x^n you end with a g^{c_0} term, but this term is not equal to D^{c_0}(f)(g), but is rather equal to D^{n-c_0}(f)(g)/\binom{n}{c_0}. When you do this substitution, you get what I list above.

Incidentally, if you look on Wikipedia for this formula (for normal partial derivatives) it agrees with what I put above, when you ignore the factorials.

]]> Comment on The Power of the Distributive Law. by Felix Fontein http://math.fontein.de/2012/03/11/the-power-of-the-distributive-law/comment-page-1/#comment-331 Felix Fontein Wed, 11 Jul 2012 15:59:43 +0000 https://math.fontein.de/?p=878#comment-331 Thanks for the info, Jens. I didn't knew about "near-rings" yet. Also the example - endomorphisms of a group - is very helpful. I knew that they don't form a ring (unless the group is abelian, or maybe in some other random cases), but didn't knew the structure they form had a name. Thanks for the info, Jens. I didn’t knew about “near-rings” yet. Also the example – endomorphisms of a group – is very helpful. I knew that they don’t form a ring (unless the group is abelian, or maybe in some other random cases), but didn’t knew the structure they form had a name.

]]> Comment on The Power of the Distributive Law. by Jens http://math.fontein.de/2012/03/11/the-power-of-the-distributive-law/comment-page-1/#comment-326 Jens Thu, 29 Mar 2012 00:21:42 +0000 https://math.fontein.de/?p=878#comment-326 Hi Felix, I just came across your math blog incidentally and found (again) something interesting. :) There is actually the notion of "near-ring", http://en.wikipedia.org/wiki/Near-ring, which is - besides semirings - another generalization of rings. Here, only a one-sided distributive law is postulated, and hence the addition might actually be noncommutative. The endomorphisms of a group form such a near-ring, which is probably the most important example. Hi Felix, I just came across your math blog incidentally and found (again) something interesting. :)
There is actually the notion of “near-ring”, http://en.wikipedia.org/wiki/Near-ring, which is – besides semirings – another generalization of rings. Here, only a one-sided distributive law is postulated, and hence the addition might actually be noncommutative.
The endomorphisms of a group form such a near-ring, which is probably the most important example.

]]> Comment on Multiplicity of the Determinant. by Felix Fontein http://math.fontein.de/2010/11/10/multiplicity-of-the-determinant/comment-page-1/#comment-314 Felix Fontein Mon, 19 Sep 2011 18:10:41 +0000 http://math.fontein.de/?p=790#comment-314 And again I regret that I didn't start studying in Oldenburg two years earlier... :) And again I regret that I didn’t start studying in Oldenburg two years earlier… :)

]]> Comment on Multiplicity of the Determinant. by Jens http://math.fontein.de/2010/11/10/multiplicity-of-the-determinant/comment-page-1/#comment-313 Jens Mon, 19 Sep 2011 17:39:47 +0000 http://math.fontein.de/?p=790#comment-313 Hi! This proof looked too familiar to me... and indeed: Quebbemann (WS 99/00) used the same proof in his Lineare Algebra. :) Hi! This proof looked too familiar to me… and indeed: Quebbemann (WS 99/00) used the same proof in his Lineare Algebra. :)

]]> Comment on On a Certain Determinant. by Felix Fontein http://math.fontein.de/2011/03/25/on-a-certain-determinant/comment-page-1/#comment-286 Felix Fontein Mon, 20 Jun 2011 07:49:46 +0000 https://math.fontein.de/?p=814#comment-286 I just noticed that this determinant fits into a more general scheme: the matrix can be written as the sum of $D = diag(x_1, \dots, x_n)$ and $u v^T$, where $u = v$ is the vector with all coefficients being 1. According to <a href="http://planetmath.org/encyclopedia/DeterminantsOfSomeMatricesOfSpecialForm.html" rel="nofollow">planetmath.org</a>, we have $\det(D + u v^T) = \det D + v^T D^\# u$ for arbitrary $D \in K^{n \times n}$, $u, v \in K^n$, where $D^\#$ is the <a href="http://en.wikipedia.org/wiki/Adjugate_matrix" rel="nofollow">adjugate</a> of $D$. For a diagonal $D$, $D^\#$ is a diagonal matrix with the $i$-th diagonal entry being $\prod_{j \neq i} x_j$, whence one obtains the result I've shown above. I just noticed that this determinant fits into a more general scheme: the matrix can be written as the sum of D = diag(x_1, \dots, x_n) and u v^T, where u = v is the vector with all coefficients being 1. According to planetmath.org, we have \det(D + u v^T) = \det D + v^T D^\# u for arbitrary D \in K^{n \times n}, u, v \in K^n, where D^\# is the adjugate of D.
For a diagonal D, D^\# is a diagonal matrix with the i-th diagonal entry being \prod_{j \neq i} x_j, whence one obtains the result I’ve shown above.

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