Vector Signal
2009
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Vector Signal
The Best Car Radar Detector: Beltronics V940 Vector /Laser Detector
You will be able to travel safely from one place to other places. You do not have to worry about exceeding speed limit in a certain area. You can buy car radar detector to make you have more comfortable trip.
Sometimes you will find that you have driven your car faster than usual. This situation may make you get ticket since driving car over speed limit is against the law. You will need to have car radar detector to reduce the risk of getting ticket.
A smart choice to avoid getting ticket is this device. You will be able to have comfortable trip on your own. The device will tell you if you have exceeded the speed limit. You need to know that having this device in some region is against the law. Thus, you need to find the proper device.
This posting reviews about Beltronics V940 Vector radar/Laser detector. This device will help you to limit your own speed. This device is available in black and silver color. It can display text to tell you about your speed. The most beneficial feature of this device is the immune ability. The police or other law enforcement will not be able to detect this device.
Beltronics V940 Vector radar/Laser detector is one of the best devices that will be good for travelers. This device will help all people that want to have save trip in all time. You will be able to avoid ticket and will move in safe speed. The other features of this detector are full laser alert, X, super-wide ka, Ku and K.
The screen will display the message to alert and remind you about your speed in driving you car. This easy to use laser and radar detector will give you the best performance of car radar detector. You can easily read the massage on its ultra bright 280-LED alphanumeric display. The other feature that you need to know is about then DSP or digital signal processing technology and multiple laser sensors that are able to give you long-range performance. These features are also able to reduce the number of having false alarm.
You do not need to worry about the price of this excellent device. This device is sold in very affordable price. You will find that the features and the ability of this device is much beneficial than the other device.
The manufacturer of this car radar detector gives one year limited warranty. You can hide this device from the law enforcement since this device has been equipped with travel case so that, you will be able to bring this detector with you in all time. The design and the model of this device are compatible and quite small so that you can bring and put it in small space. You do not need to think twice about this device. You will get laser and radar detector in one device in very affordable price.
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Avoid getting ticket by having
car radar detector
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Linear Algebra NEED HELP !!!!?
1/ Prove that the set V=R+ ( the set of all positive real numbers) is a vector space with the following nonstandard operations: for any x,y belong to R+ & for any scalar c belong to R:
x O+ ( +signal into circle) y=x.y (definition of vector addition) & c O ( dot signal into circle) x = x^c (definition of scalar multiplcation) ( must verify that all 10 axioms defining a vectorspace are satisfied).
2/ Consider the vector space V = C (-infinite,infinite)= all fumctions f(x) which are continuous everywhere. Show that the following subset H of C (-infinite, infinite) is in fact a subspace of C (-infinite,infinite):
H= {all functions f(x) satisfying the differential equation f " (x) +25 f(x)=0}
(need to verify all 3 subspace requirements)
1) for ease of notation, let me write the non-standard vector sum as x†y, and the non-standard scalar multiplication as c*x.
so x†y = xy, and c*x = x^c.
first we need to show that (R+,†) is an abelian group.
R+ is closed under †: if x,y are real numbers greater than 0, so is xy.
† is associative: for x,y,z in R+, x†(y†z) = x(yz) = (xy)z = (x†y)†z
† posseses an identity in R+, namely 1: x†1 = x1 = x, 1†x = 1x = x.
(although it sounds wierd, the real number 1 will be our "zero-vector").
for every x in R+, there is a unique inverse x^-1 (a "negative x" for †) for x under †:
we can take x^-1 to be the real number 1/x, which is in R+ whenever x is, and
x†(1/x) = x(1/x) = x/x = 1, (1/x)†x = (1/x)x = x/x = 1.
finally, † is commutative: for any x,y in R+, x†y = xy = yx = y†x.
now it should be clear that c*x = x^c is a positive real number whenever x is, so
*:R x R+ ---> R+ (that is, c*x is in R+). we also have to verify that * is compatible with the vector addition †:
so suppose x,y are arbitrary positive real numbers, and that a,b are arbitrary real numbers.
then (a+b)*x = x^(a+b) = (x^a)(x^b) = (a*x)†(b*x).
a*(x†y) = (xy)^a = (x^a)(y^a) = (a*x)†(a*y).
(ab)*x = x^(ab) = x^(ba) = (x^b)^a = a*(x^b) = a*(b*x)
finally, we need to show that 1*x = x:
1*x = x^1 = x.
2) given: f and g both satisfy the differential equation, so
f "(x) + 25f(x) = 0, g"(x) + 25g(x) = 0.
so (f+g)"(x) + 25(f+g)(x) = [(f+g)'(x)]' + 25(f(x) + g(x))
= [f '(x) + g'(x)]' + 25f(x) + 25(g)x) =
f "(x) + g"(x) + 25f(x) + 25g(x) =
f "(x) + 25f(x) + g"(x) + 25g(x) = 0 + 0 = 0
thus f+g satisfies the differential equation as well.
does af given by (af)(x) = a(f(x)) satisfy it as well?
(af)"(x) + 25(af)(x) = [(af)'(x)]' + 25(a(f(x)) =
[a(f '(x)]' + (25a)f(x) = a[f '(x)]' + a(25f(x)) =
af "(x) + a(25f(x)) = a(f "(x) + 25f(x)) = a0 = 0, so yes.
(i leave it to you to show that f+g and af are continuous for all real x if
f and g are).
finally, if we consider the 0-function 0(x) = 0 (for all real x),
we see that this function (being constant) is continuous everywhere,
and since 0'(x) = 0, for all x, we must have 0"(x) = 0 for all x as well,
thus 0"(x) + 25(0(x)) = 0 + 25*0 = 0 + 0 = 0, that is, the 0-function belongs to H.
hence H is indeed a subspace of V.
Agilent HP 89410A Vector Signal Analyzer, for sale
