Comments on: Where you been? http://blog.drscottfranklin.net/2006/03/30/where-you-been/ Ramblings of a Christian Mathematician and Bioinformaticist Tue, 06 Jan 2009 02:52:20 +0000 http://wordpress.org/?v=2.7 hourly 1 By: jhqpc4h@hotmail.com http://blog.drscottfranklin.net/2006/03/30/where-you-been/comment-page-1/#comment-256 jhqpc4h@hotmail.com Mon, 09 Oct 2006 04:59:24 +0000 http://blog.drscottfranklin.net/2006/03/30/where-you-been/#comment-256 ringtones free ringtones free

]]> By: Vito Prosciutto http://blog.drscottfranklin.net/2006/03/30/where-you-been/comment-page-1/#comment-255 Vito Prosciutto Thu, 30 Mar 2006 18:33:29 +0000 http://blog.drscottfranklin.net/2006/03/30/where-you-been/#comment-255 It's worth noting that this is in fact an if and only if statement. Another way to think about this is to consider the roots of x^2+bx+c=0 (we can take ax^2+bx+c=0 and make it monic by dividing all terms by a) Let these roots by m and n. Then we have c=mn and b=-(m+n). If the roots are real, then we have familiar results, so we can focus on the case where m is the complex conjugate of n, that is m=p+qi and n=p-qi. Then we have mn=p^2+q^2, which is nonnegative (and only zero if m=n=0) and m+n=2p=2Re(m)=2Re(n) so if we have complex roots, c must be positive and the sign of b is the opposite of the sign of the real part of the roots. For real roots, following rules from algebra I, we have c positive gives us two roots of the same sign, with the sign determined by the opposite of the sign of b, and c negative gives us two roots of differing signs. Yes, I am a bit of an Algebra freak It’s worth noting that this is in fact an if and only if statement. Another way to think about this is to consider the roots of x^2+bx+c=0 (we can take ax^2+bx+c=0 and make it monic by dividing all terms by a) Let these roots by m and n. Then we have c=mn and b=-(m+n). If the roots are real, then we have familiar results, so we can focus on the case where m is the complex conjugate of n, that is m=p+qi and n=p-qi. Then we have mn=p^2+q^2, which is nonnegative (and only zero if m=n=0) and m+n=2p=2Re(m)=2Re(n) so if we have complex roots, c must be positive and the sign of b is the opposite of the sign of the real part of the roots. For real roots, following rules from algebra I, we have c positive gives us two roots of the same sign, with the sign determined by the opposite of the sign of b, and c negative gives us two roots of differing signs.

Yes, I am a bit of an Algebra freak

]]> By: SplineGuy http://blog.drscottfranklin.net/2006/03/30/where-you-been/comment-page-1/#comment-254 SplineGuy Thu, 30 Mar 2006 16:53:50 +0000 http://blog.drscottfranklin.net/2006/03/30/where-you-been/#comment-254 Thanks for posting. I just love it when the simple stuff comes through as truly relevant in more advanced classes. This came up in helping a student with a homework assignment where we were basically verifying the stability of the mass-spring oscillator. While looking at the three cases for roots of the characteristic polynomial, we discovered this property. Then, we proved the theorem in class. Nice corollary by the way. I'll point that out in the next class Thanks for posting. I just love it when the simple stuff comes through as truly relevant in more advanced classes. This came up in helping a student with a homework assignment where we were basically verifying the stability of the mass-spring oscillator. While looking at the three cases for roots of the characteristic polynomial, we discovered this property. Then, we proved the theorem in class.

Nice corollary by the way. I’ll point that out in the next class

]]> By: Vito Prosciutto http://blog.drscottfranklin.net/2006/03/30/where-you-been/comment-page-1/#comment-253 Vito Prosciutto Thu, 30 Mar 2006 16:35:47 +0000 http://blog.drscottfranklin.net/2006/03/30/where-you-been/#comment-253 [reposted to fix inequalities turning into HTML] This actually doesn't get a lot more involved than high school algebra (although you do need to know and understand a bit about complex numbers, but not much). The statement here is in fact equivalent to \sqrt{b^2-4ac}<b Since b is positive, we can square both sides and get b^2-4ac<b^2 which then reduces to -4ac<0 which is true as long since a and c are positive. But, the sharp-eyed student in the front row may object, what about if a and c are both negative? Well, we kind of glossed over the fact that there's the denominator of 2a in the denominator, which changes all of our signs which gives us the corrolary that ax^2-bx+c=0 with a,b,c>0 implies Re(x) >0 (and oh my do I hope that I didn't make a mistake in coming up with that corollary). [reposted to fix inequalities turning into HTML]

This actually doesn’t get a lot more involved than high school algebra (although you do need to know and understand a bit about complex numbers, but not much).

The statement here is in fact equivalent to
\sqrt{b^2-4ac}<b

Since b is positive, we can square both sides and get
b^2-4ac<b^2
which then reduces to
-4ac<0
which is true as long since a and c are positive.

But, the sharp-eyed student in the front row may object, what about if a and c are both negative?

Well, we kind of glossed over the fact that there’s the denominator of 2a in the denominator, which changes all of our signs which gives us the corrolary that
ax^2-bx+c=0 with a,b,c>0 implies Re(x) >0

(and oh my do I hope that I didn’t make a mistake in coming up with that corollary).

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