You are not logged in.
S(x+L) is S*(x+L) and not a function, right? Using y as f(x) (as it's easier):
y = S(x+L)/x²
yx² = Sx + SL
yx² - Sx - SL = 0
Using quadratic equation:
x = (S ± √(S² - 4ySL) ) / 2y
Which is the inverse.
The title is "Proving an Idenity" so does this mean you are not allowed to use trig identities? As that would be begging the question.
Seems complicated, so let's just do it for once dice first:
The probability of rolling a 6 is 1/6. So there is a 1/6 chance that you will be done your first time:
1/6
But there is a 5/6 chance that you will have to roll again. Then there is a 1/6 what when you roll again, it will be a 6, and you are done:
1/6 + 5/6*1/6
Or an 11/36 chance that you are done after two rolls. But there is a 25/36 chance that you won't be, and then a 1/6 chance that you get a 6 when you roll again.
1/6 + 5/6*1/6 + 25/36*1/6 = 11/36 + 25/216 = 91/216
Finally, you have a 125/216 chance that you will have to roll again, and again, a 1/6 chance that you will get a 6:
1/6 + 5/6*1/6 + 25/36*1/6 + 125/216*1/6 = 91/216 + 125/1296 = 546/1296
Or a 42% chance for 1 die. You will find that using this method, the probability will approach, but never reach, 1.
But you need this for 3 dice. So that's:
546/1296 * 546/1296 * 546/1296 = 162,771,336 / 2,176,782,336
Or a 7.478% chance.
Someone please check this. At least it sounds right...
Whoops, you used a different way of finding the inverse than I'm used to, so I got mixed up.
What I normally do is:
y = x/(3x - 2)
y(3x - 2) = x
3xy - 2y = x
3xy - x = 2y
x(3y - 1) = 2y
x = 2y / (3y - 1)
What you did was pretty much the same, but you switched the x and y variables, as it is standard to want to solve for y:
x = y/(3y - 2)
x(3y - 2) = y
3xy - 2x = y
3xy - y = 2x
y(3x - 1) = 2x
y = 2x / (3x-1)
As you can see, you come up with the same answer, just different names for the variables. In mine, x is dependant on y, in yours (as it normally is), y is dependant on x.
x(3y - 2) = y
x = y / (3y - 2)
It's not quite impossible. Well, that also depends on your definition of impossible.
You can represent e^-(x^2) as an infinite series, in which case you can intergrate it. You'll still be left with an infinite series, but you can do it...
On the other hand, if you're looking for a simple function (i.e. non-infinite series), then it is impossible, at least to current knowledge.
Negative numbers are not primes, although one could certainly argue that they should be. But I'm not about to go and change the basic framework of math.
if the equation is x^2 + bx = 0, then b is the negative location of the second root (the first being 0). In other words, -b is a root of the equation.
It's not possible, and here is a simple proof to show why:
All prime numbers except 2 are odd.
Case 1: One of the prime numbers is 2:
If one of the prime numbers is 2, the other must be 9, which is not a prime number.
Case 2: Neither of the prime numbers is 2:
Then both prime numbers are odd. Let x and y be odd numbers. Then x = 2k + 1 for any integer k (ignoring the restriction for primeness) and y = 2l + 1 for any integer l.
Then x - y = 2k + 1 - (2l + 1) = 2k - 2l = 2(k - 1). Since k - l is an integer, 2(k - l) is an even integer, and thus not 7.
Therefore the are no two prime numbers such that p1 - p2 = 7. QED.
Just don't forget to also state the function is continuous from [1.1, 1.2].
I don't think I understand the question. What I think you are looking for is if the matrix formed is reflexive, I can't find a case where it's not. If they all have to have different ordering, i.e.:
1. x y z
2. x y z
3. y z x
Is invalid (the order of 1 & 2 are the same), I believe the probability is 1, although I'm not quite sure what a proof of that would look like.
If the order is completely random (and therefore two people can have the same order), then you need to find out the probability for every order to be different. This is the only time a symmetric matrix can occur.
Alabama.
900ft.
Could you post your work as well?
...have you ever heard of engineering induction? If it works 3 times, it will always work.
It's still must stronger than computer science induction: if it works once, it will work all the time. If it doesn't work all the time, it's the users fault.
krassi_holmz, I love your avatar. It is probably one of the most intriguing mathimatical relationships I have encountered so far:
e^i*pi + 1 = 0
http://aleph0.clarku.edu/~djoyce/java/trig/identities.html
Use the product identity, about 3/4th the way down the page.
If you want to get serious about the philosophy of math...
Imre Lakatos: http://en.wikipedia.org/wiki/Lakatos
Lakatos' philosophy of mathematics was inspired by both Hegel's and Marx' dialectic, Karl Popper's theory of knowledge, and the work of mathematician George Polya.
The book Proofs and Refutations is based on his doctoral thesis. It is largely taken up by a fictional dialogue set in a mathematics class. The students are attempting to prove the formula for the Euler characteristic in algebraic topology, which is a theorem about the properties of polyhedra. The dialogue is meant to represent the actual series of attempted proofs which mathematicians historically offered for the conjecture, only to be repeatedly refuted by counterexamples. Often the students 'quote' famous mathematicians such as Cauchy.
What Lakatos tried to establish was that no theorem of informal mathematics is final or perfect. This means that we should not think that a theorem is ultimately true, only that no counterexample has yet been found. Once a counterexample, i.e. an entity contradicting/not explained by the theorem is found, we adjust the theorem, possibly extending the domain of its validity. This is a continuous way our knowledge accumulates, through the logic and process of proofs and refutations. (If axioms are given for a branch of mathematics, however, Lakatos claimed that proofs from those axioms were tautological, i.e. logically true.)
Lakatos proposed an account of mathematical knowledge based on the idea of heuristics. In Proofs and Refutations the concept of 'heuristic' was not well developed, although Lakatos gave several basic rules for finding proofs and counterexamples to conjectures. He thought that mathematical 'thought experiments' are a valid way to discover mathematical conjectures and proofs, and sometimes called his philosophy 'quasi-empiricism'.
However, he also conceived of the mathematical community as carrying on a kind of dialectic to decide which mathematical proofs are valid and which are not. Therefore he fundamentally disagreed with the 'formalist' conception of proof which prevailed in Frege's and Russell's logicism, which defines proof simply in terms of formal validity.
On its publication in 1976, Proofs and Refutations became highly influential on new work in the philosophy of mathematics, although few agreed with Lakatos' strong disapproval of formal proof. Before his death he had been planning to return to the philosophy of mathematics and apply his theory of research programmes to it. One of the major problems perceived by critics is that the pattern of mathematical research depicted in Proofs and Refutations does not faithfully represent most of the actual activity of contemporary mathematicians.
Pretty interesting stuff. I've only read his stuff on the philosophy of science though, not math.
Start out with the parralle lines. What angles do you know are equal from these? And using the fact that these angles are equal, what triangles can you prove similar? And since you know that triangles are similar and the ration of one side is 2:5, what's the ration of the other sides?
"ignore air resistance and friction, my eye. Teach me something I can actually use, and then I might be interested."
Friction is related to speed. This makes things very complicated, because friction changes the speed and the speed change the friction. This lead to differential equations such as:
3y' + 5y = cos(x), solve for y, and it can't be in terms of y'.
And that's a simple one! The problem with equations like this is it takes a full understanding of integrals and derivatives to solve (basic calculus). This is why they aren't taught at a high school level. But as soon as you get the advanced math, physics becomes frighteningly accurate (as well as difficult).
Draw the triangle, and then draw it again rotated so that B is in the same corner that C was. Now since ICA < IBA and ICB < IBC, ICA + ICB < IBA + IBC. Since this is true, the angle at C is less than the angle at B. I forget the name of the proof, but if you have such a triangle, the side with a greater angle has a smaller diagonal (BB' and CC'). So Since angle C < B, CC' > BB'.
I don't understand quite understand the wording in 2.
What you're basically looking for is two pythagorean triplets that go:
b, y, x, and then y, x, a.
I don't believe any such triplets exist, although I can't prove it.
"1. If m+2=0;m=2
0x=-2
No solution"
Not true. Let m = 2, then you get the equations: 2x + y = 2, so y = -2x + 2, and y = 2x + 4. A common solution to these is the point (-1/2, 3).
"2. 2. If m+2≠0;m≠2
x=-2/(m+2)
So:
No solution-
m=2
One solution-
m≠2"
The logic here is right, if m = 2 had no solutions, than all lines without that slope intersect the line. Unforutunately, m = 2 has a solution.
"Infinite many solutions-
m∈{}"
Correct. The x-intercepts are off, and they only thing you can change is the slope.
Ah, sorry, I misread your post. But yes, that is precisely right. By multiplying the vector by -1, you are reversing it's direction, so in front becomes behind.
First, off the bat you should recognize mx+y=2 and y-2x=4 as equations of lines. Now, remember (or learn), that the "solution" to these equations is that line. That is, every (x, y) on that line will work out so that 0 = 0 when you plug it in. 1. When do two lines have no common solution? In other words, when do two lines share no points? 2. When do two lines share all points? And finally, 3. When do two lines share only one point?