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Suppose p(x,r) is a polynomial, where r is a parameter. For each r, you could evaluate the roots of p. As r changes, the roots of p changes continuous. What is the theorem that proves this? Example:
Started by on , 20 posts by 3 people.  
Therefore, all roots of must cross the imaginary ....
Suppose is a root of , and is a root, the relation is impossible to hold, therefore it cannot be a root.
The roots of lies on a complex plane.
Differentiated.
Imaginary bloodlust IC A searing flash of pain bounces through you head as you slowly awaken. Around you the ground seems burned. You clothes are in threads and the rest of your belongings are missing. And then you remembered.... You were just living...
Started by on , 30 posts by 5 people.  
He leaves his vengeance." Re: Imaginary bloodlust ....
Re: Imaginary bloodlust IC Spoiler Resource Food Magic Basic Amount 9.4 Amount Out .4 Re: Imaginary bloodlust IC Tycho looks around him after he handles the barbarian situation.
I was listening to the roots "Do you want more album" the other day, probably their best album or at least, the most representative I really feel the sound of this album is pretty unique, a perfect mix between jazz and soul with a perfect rhodes sound...
Started by on , 15 posts by 7 people.  
I love Hami - The Funky Descendant Its a jazzy ass album, not quite the same sense as The Roots, which is a fundamental element of the roots of jazz, which got a little lost after the bebop influence an expanded audience with the use of ....
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On Tue, 15 Nov 2011 23:40:04 +0000, Matthieu Schaller <matthieu.schaller@gmail.com Dear all, I have recently been playing a lot on numerical analysis problems containing complex numbers and functions. I have been using the SL std::complex<weren...
Started by on , 12 posts by 6 people.  
Answer Snippets (Read the full thread at omgili):
And a kind of imaginary://boost.2283326.n4.nabble....
Comparing imaginary and reals seems confusing to me ///Returns true if x equals y template<typename expect an implicit conversion to complex as an imaginary is also a complex.
Hi all, Is it possible to separate real and imaginary part of complex number only with basic operations (add/subtract/multiply/divide)? It seams that is not possible. Kind regards, Vladimir
Started by on , 6 posts by 5 people.  
Answer Snippets (Read the full thread at omgili):
Re(z) = z.r(z)/q(z) On Wed, 5 Aug.
Let r(z) = p(z)/z.
0 = p(0)/q(0).
Hence p(z) = 0.
roots.
Imaginary Numbers have always confused me. I get that their squares are negative. And I understand a good way to picture them is on the Y axis if the X axis is Real Numbers. But beyond that I'm completely baffled. What are their practical uses? Do they...
Started by on , 20 posts by 10 people.  
This mathematical object has shown us so such failures come down... .
Real represented in the x axis and imaginary in the y axis.
It is plotted using both real and imaginary numbers.
In particular, it is the Mandelbrot Set.
It is a fractal.
On Wed, 25 Feb 2009 01:03:09 -0800 (PST), Bennett Haselton <bennett@peacefire.org Something's bugged me about imaginary exponents ever since learning in high school that e^ix = cos(x) + isin(x). Now that I am passing on my alleged wisdom to high...
Started by on , 17 posts by 12 people.  
Answer Snippets (Read the full thread at omgili):
Then how do ....
The mathematics is sound: root.) When I teach about exponents, I usually take the following route (or should that be root roots.
Of fractional exponents, irrational exponents and even imaginary exponents.
On Tue, 23 Jun 2009 06:08:24 -0700 (PDT), pavel <pavelmakedonski@yahoo.com Suggestion of the Roots of Macedonians into Antiquity is a Total Nonsense Skopje, June 23rd, 2009, Utrinski daily http://www.utrinski.com.mk/?ItemID=45ED08226B240E4E82838...
Started by on , 23 posts by 10 people.  
Answer Snippets (Read the full thread at omgili):
Of the Macedonian Academy) for "creating (Macedonian) history without roots" ("history" beginning to create it and that is, the imaginary and fictional connection between your South Slavonian people it and that is, the imaginary....
On Mon, 16 Mar 2009 09:34:54 GMT, Albert <albert.xtheunknown0@gmail.com Hello: I have a problem that I'm not able to completely solve yet. It reads: 'Show that the roots of the equation 4(m + 1)x^2 - 4(m - 1)x - 3 = 0 are real for all real m....
Started by on , 7 posts by 5 people.  
Answer Snippets (Read the full thread at omgili):
Use the fact that all the roots of a second degree equation are real if and only if the discriminant Santos <jcsan......
On Mon, 16 Mar 2009 12:25:56 +0100, "Philippe 92" <nospam@free.invalid José it .
With imaginary roots.
On Thu, 12 Feb 2009 04:28:31 EST, Zak Seidov <zakseidov@yahoo.com Simple/silly Q(?) Take y(x)= a x^2+ b x + x. If discriminant d = b^2 - 4*a*c isn't negative, then x_1,2 = (-/+ b - Sqrt[b^2 - 4*a*c])/(2*a) have clear geometric meaning: the...
Started by on , 4 posts by 2 people.  
Answer Snippets (Read the full thread at omgili):
Geometric meaning of imaginary roots?" at <http://groups.google.com/group/sci.math/browse_thread (in the school textbook! while i'm now 70!) the next "geometrical meaning" of the roots x_1,2 if d<0.
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