jcarlson

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Posts posted by jcarlson


"Dark matter" and "dark energy" is nonsense. Sounds like astrophysics has gone over to science fiction these days.
I have always been under the impression that dark matter was simply matter that did not emit electromagnetic radiation.... Like rock.
Without dark matter to block incoming light shouldn't we be blinded by the light of trillions of stars from incredibly far distances?
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I am not a good enough mathmatician to evaluate his math, but I do know that black holes are no longer a thoery, they are fact. This being said, I can't see how his theory could be correct.
As far as I know, there is no direct evidence of Black holes. Seeing as how no light, or any other information can escape the event horizon of a black hole, Its practically impossible to observe one directly. The closest we have come is determining several Black hole "candidates", or systems where a black hole is thought to exist based on the behavior of visible objects in the area.
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by the way dave I PM'd you a problem I think would be good for advanced high school/beginning college students.
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This guy claims to have found a small flaw in GR with the way escape velocity is described. I'm usually very skeptical about this sort of thing, and while I am of course still skeptical, all his math seems to be in order, and he claims that the expirimental evidence for General Relativity with regard to the Schwarzschild metric is valid for his idea as well. To summarize, He is basically claiming that escape velocity at a certain radius is described in the same way by both Newtonian physics and General Relativity, when it should be described relativistically in GR. The implications of this are that Escape velocity never meets or exceeds c, and thus a black hole cannot exist.
Just looking for your opinions, especially those of you knowledgeable in cosmology and Relativity.
Also, there is an ongoing discussion about the paper here, to which I have posted a few times towards the end:
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Has anybody here heard of a the GRAVITY DRIVE. What is the shortest distance betwwen any two points?
Answer = 0.
THe shortest distance between any two points is 0 ' date=' if u fold it together.
My question is , is it possible to create a black through a ship travels through, there by travelling faster than light.[/quote']
What you're referring to is known as a wormhole, and although they theoretically could exist, we don't know if they do exist, or if there is any way for man to create them if they do, or what effects they would have on anything moving through them, etc.
I think its safe to say we wont be travelling through wormholes any time soon.
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college and university are different levels? I say we start it by finding the integral of x^x from infinity to +infinity
even though i'm pretty sure there's no algebraic way to solve it
Well since its undefined at 0 you can't take the integral from infinity to +infinity
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Think of it this way. Break the graph of f(x) into 4 segments, where the segments are divided by where the graph touches the xaxis. Now, for the graph of f(x), each of those segments can either look like it does for f(x), or it can be the reflection of f(x) over the xaxis, because both will look the same when you plot f(x). Thus f(x) has 2 possible curves for each of 4 segments which are independent of the other segments. now use probablility to find the number of possible graphs of f(x) or just count them.
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you either need practice with math or with latex. (unless i helluv just read that wrong... whcih im looking at now : P)
(4! +4 *4)/4= 10
throw in some parentheses and your good to go:
((4! +4) *4)/4= 30
not sure what went wrong with this one' date=' but
(4!+4)/4 +4=11 not 31[/quote']
you missed my other factorials, its not (4! + 4*4)/4, its (4! + 4*4!)/4
4! = 24, so this is equal to 24/4 + 24 = 30.
same with the other one, missed another ! there
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I got 30
[math]
\frac{4! + 4*4!}{4}
[/math]
and 31
[math]
\frac{4! + 4}{4} + 4!
[/math]
and 32 is easy
[math]
4*4 + 4*4
[/math]
ok im gunna stop now. at least til you guys get stuck again
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Yeah I think it is easiest to simply realize that you are evaluating a constraint (sort of like a boundary condition)
f(x)=ax^2+6
g(x)=x
Such that f'(x)=g'(x)' date=' so
2ax = 1, x = 1/(2a)
This is the information we gathered from evaluating the constraint. But look we are really free to pick either a or x to suit our fancy. Nowhere does the problem stipulate that f(x) should equal g(x) only that at some point the slopes are equal.
So lets say we want a parabola that has slope one at the point x=1 then we get
1=1/2a, a = 1/2
So yes (a,x) = (1/2,1) is one solution, but for every x there is a corresponding a. In fact this is really a rather clever question. You can see that the real solution is a hyperbola by graphing x = 1/(2a). You could also think of this as a map that associates to every value x, a function f(x)=ax^2+6
Oh yeah also look at the parabola ax^2 + 6. The slope at any point is not affected by adding six (we know this from the derivative) and the parabola is still symmetric with respect to the yaxis. So if you want a parabola that has a slope of 1 at x=1 pick a =1/2. The reason this is interesting is that it flips the parabola upside down. This type of change with respect to a parameter is called a bifurcation, meaning the object is quantitatively different depending on whether a certain parameter (in this case x) has crossed a certain threshold (in this case zero). In more complicated systems of differential equations important properties of the system can change dramatically as parameters vary. I know that is far beyond the level of this question but maybe it will peak your interest.
[/quote']
This is incorrect, there is only one solution for a. You have to remember that in order for a function f(x) to be tangent to a function g(x), not only must f'(x) = g'(x) but also f(x) = g(x), or else you could end up with two lines that have the same slope, but dont touch (basically parallel at x).
if f(x) = ax^2 + 6 and g(x) = x, then we have the equations
ax^2 + 6=x AND
2ax = 1
Solving for x in the second equation, we have:
x= 1/(2a)
Now, solving the system by subsituting the second equation into the first, we have
a/(4a^2) + 6 = 1/(2a)
which simplifies to
6 = 1/2 * 1/a  1/4 * 1/a
6= 1/(4a)
1/6 = 4a
a=1/24
Now, we check our work buy subbing a into the original equation:
f(x)=1/24 * x^2 + 6
f'(x) = 1/12 * x
g(x) = x
g'(x) = 1
1 = 1/12 * x
x = 12
At x = 12, we see that f(x) = 144/24 + 6 = 12, and g(x) = 12, therefore f(x) and g(x) intersect at this point, and f'(x) = 1 and g'(x) = 1, so they are in fact, tangent at x = 12, or when a = 1/24.
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I did my utmost to elliminate ANY ref to Physicality' date=' and to keep it as Maths(ish) as I could.
the way I see the problem as boiled down to the Raw basics as possible is that which I stated before, it`s division by 2 until it`s no longer divisible (that point being the 2 minute marker).
it`s like presenting 10 / 3 =
and asking for a finite answer.
there is non. at least Not without specifying a resolution (in decimal places).[/quote']
10/3 IS a finite answer.
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Basically if the sun blinked out of existance, the earth would continue on its orbit until we couldnt see the sun anymore, that is, the 9 minutes or so it takes for the light to stop coming. This is because the change in the fabric of space time that occurs when a mass like the sun is removed occurs at the speed of light, and the spacetime around the sun would not "unwarp" itself until the light from the sun's last moment hit it. However gravity itself is simply the act of an object moving along a straight path through space time, and is therefore instantaneous.
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Media will never be pirate free for the simple fact that no matter how many protections you put on it while it is in a digital form (either as a CD or DVD or a file), at some point between the source and the output device (speakers or CRT/LCD) it has to be converted into a raw, unprotected analog signal, which, with the proper equipment, can be easily copied.
Software pirate proofing tech may have better luck in the future with advances in encryption technology, but keep in mind that those trying to crack the software will have access to the same new tech.
All in all, I'm not against companies trying to protect their work, as long as it doesn't create an overwhelming burden on the consumer. XP's authentication system is pushing it, but for other software typing in a product key isnt that much of a hassle.
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Java rules!!!!!!!!!!!!!!!!!!!!!!!!!!!
Its nice, but you can't "hack" with it. Java gives the programmer zero direct control of memory allocation and other low level things, because it has to interact throught the Virtual Machine.
If you wanna do all that crazy illegal stuff, C is the way to go.
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[hide]an onion[/hide]
a man walks out on his porch, shoots a bear, and proceeds to walk one mile south, one mile west, and one mile north back to his home. What color was the bear?
edit:
whoops... sorry
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the best i can come up with is one of the following
a b s A B S
b c t B C T
c d u C D U
...
g h y G H Y
h i z H I Z
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My favorite is notepad and cmd in windows, pico, javac, and java in unix.
those count right?
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Floppy drives are still required for loading RAID drivers when installing an OS on an unformatted pair of Hard Drives in a RAID array. Thats one of the only things I have used my Floppy drive for, now that i think about it...
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You forgot to mention the ICBMs on Subs... each sub can hold up to 24 ballistic missiles with up to 10 warheads on each, with each warhead capable of leveling a city.
Did you know that if a SSBN was its own country, it would be the third largest nuclear power in the world? (after the US and Russia)
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Now I'm having trouble with 2,2,5,5,12. Anyone got an equation for them?
does sqrt require a 2 to use? or can we just sqrt something?
if it doesnt, then sqrt(5*5) * 2 + 2 = 12
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The first task which must be done is to integrate the power with respect to time so that the situation can be analysed at a single point in time. This means that work is dealt with instead of power. The current does work on the boat. The boat is not moved because the coefficent of friction (mu, dimensionless) times the weight (mass times gravitational acceleration) of the boat is greater than the amount of force the water exerts. The current does exert a force against the boat. The bottom of the river also exerts an equal force in the opposite dirrection through friction. Because both forces cancel the boat is not moved and no work is done becuase there is no change in linear displacement (dx). In the absence of friction, the boat would move down the river regardless of the anchor. In conclusion the anchor causes a force to be exerted in the opposite dirrection of the current, but does not do work or have power (dW/dt).
In the absence of friction, the boat would remain stationary while the river flowed around it because there would be no force applied to the boat by the water.
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Assuming you haven't altered any of the methods inherited from JButton, to add text, you would invoke
cube1.setText("Text on the button");
and to display the Cube in a JPanel p1, you would call
p1.add(cube1);
p1.paint();
For all intents and purposes, you should be able to treat any instances of Cube as an instance of JButton, provided you don't alter any of the methods inhereted from JButton. The documentation for JButton is located here:
http://java.sun.com/j2se/1.5.0/docs/api/javax/swing/JButton.html
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Yah, that is interesting. It's also particularly noteworthy given the rumor flying around today that Apple is considering using Intel chips in future Macs, which would put a serious dent in IBM's chip business. A year from now the consoles may be all that's keeping them in business. Of course it's probably just a bargaining position on Apple's part, but you never know.
If apple stops using IBM chips, it won't put a dent in their chip business. IBM makes virtually ALL of its money in the server and supercomputer markets, where its chips are VERY ubiquitous.
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1. It's a known and proven fact that modern diet of processed foods' date=' et al and incidence of cancer is related.
[/quote']
correlation != causation
0
how do you proove this simple one?
in Mathematics
Posted
This is a pretty simple inductive proof for the case where a and b are both rational... I'm gunna have to think on the irrational case some.
Problem:
Assume a and b are two rational numbers, with [math]b > a[/math]. Show that there exists an infinate amount of rational numbers [math]t[/math] where [math]a < t < b[/math].
INDUCTIVE PROOF:
Let [math]f(n+1)[/math] be recursively defined as [math](f(n)+b)/2[/math] for all [math]n \geq 2[/math].
Let [math]f(1) = (a+b)/2[/math]
First we must show that every [math]f(k)[/math] where [math]1 \leq k \leq \infty [/math] is rational.
Part I:
Basis:
[math]f(1) = (a+b)/2[/math], where a and b are rational numbers.
A rational number by definition can be rewritten in the form [math]\frac{x}{y}[/math] where x and y are both integers. Thus we can rewrite a as [math]\frac{c}{d}[/math] and b as [math]\frac{e}{f}[/math], where c,d,e, and f are integers, resulting in:
[math]f(1) = \frac{\frac{c}{d}+\frac{e}{f}}{2}[/math]
This can then be written in the form
[math]f(1) = \frac{cf+ed}{2df}[/math]
Since the product of any two integers is an integer, and the sum of any two integers is also an integer, we can rewrite [math]f(1)[/math] as
[math]f(1) = \frac{g}{h}[/math]
where [math] g = cf+ed [/math] and [math] h = 2df [/math], both of which are integers. Therefore [math]f(1)[/math] is the ratio of two integers, and is by definition rational.
Inductive Hypothesis:
Assume that for some integer [math]k \geq 1[/math], [math]f(k)[/math] is rational. We need to show that [math]f(k+1)[/math] is rational.
By the definition of [math]f(n)[/math]:
[math]f(k+1) = \frac{f(k)+b}{2}[/math]
As in the case where k = 1 as shown above in the basis, since [math]f(k)[/math] (by the I.H.) and b are rational numbers, we can rewrite this as:
[math]f(k+1) = \frac{\frac{c}{d}+\frac{e}{f}}{2}[/math]
where c,d,e, and f are integers, and [math]f(k) = c/d[/math] and [math]b = e/f[/math].
This can then be written in the form
[math]f(k+1) = \frac{cf+ed}{2df}[/math]
Since the product of any two integers is an integer, and the sum of any two integers is also an integer, we can rewrite [math]f(k)[/math] as
[math]f(k+1) = \frac{g}{h}[/math]
where [math] g = cf+ed [/math] and [math] h = 2df [/math], both of which are integers. Therefore [math]f(k+1)[/math] is the ratio of two integers, and is by definition rational.
Thus we have shown [math]f(k)[/math] to be rational for every [math]k \geq 1[/math]. We must now show that [math]a < f(k) < b[/math] for all [math] k \geq 1 [/math]
Part II:
Basis:
[math]f(1) = (a+b)/2[/math]
[math]f(1) = \frac{a}{2} + \frac{b}{2}[/math]
[math]\frac{b}{2} + \frac{b}{2} = b[/math]
[math]\frac{a}{2} + \frac{a}{2} = a[/math]
since [math]a < b[/math],
[math]\frac{a}{2} < \frac{b}{2}[/math]
[math]\frac{a}{2} + \frac{b}{2} < b[/math]
[math]\frac{a}{2} + \frac{b}{2} > a[/math]
[math]a < f(1) < b[/math]
Inductive Hypothesis:
Assume for some value [math] k \geq 1 [/math], [math]a < f(k) < b[/math]. We need to show that [math]a < f(k+1) < b[/math]
By the definition of [math]f(n)[/math]:
[math]f(k+1) = \frac{f(k)+b}{2}[/math]
[math]f(k+1) = \frac{f(k)}{2} + \frac{b}{2}[/math]
[math]\frac{b}{2} + \frac{b}{2} = b[/math]
[math]\frac{a}{2} + \frac{a}{2} = a[/math]
since [math]a < f(k) < b[/math] (by I.H.),
[math]\frac{a}{2} < \frac{f(k)}{2} < \frac{b}{2}[/math]
[math]\frac{f(k)}{2} + \frac{b}{2} < b[/math]
[math]\frac{f(k)}{2} + \frac{b}{2} > a[/math]
[math]a < f(k+1) < b[/math]
Thus we have shown [math]a < f(k) < b[/math] for every [math]k \geq 1[/math].
Since there is an infinite number of values for k for which [math]k \geq 1[/math], it follows that there is an infinate number of values of [math]f(k)[/math] for which [math]k \geq 1[/math], and we have shown that every one of these values of [math]f(k)[/math] falls between the values of a and b, and that they are all rational.
EDIT: I've thought about it and the above proof could be easily extended to two irrational numbers a' and b' (and thus would be valid for all real numbers) if you show that there are always two distinct rational numbers a and b between the two distinct irrational numbers a' and b'. This can be proven by showing that there is at least one rational number b in between a' and b', and then that there is another rational number a in between a' and b. This is again, fairly simple to prove, but Its late and I'm tired, so I'm just gunna give you a link:
http://www.cuttheknot.org/do_you_know/numbers.shtml
scroll down to the section "lemma 1" and you should find the proofs.