Algebracus

You can show this as follows: [math]cos u = cos v \Leftrightarrow cos u - cos v = -2 sin\frac{u - v}{2} sin \frac{u+v}{2} = 0[/MATH]. [MATH]sin\frac{u - v}{2} = 0 \Leftrightarrow u - v = 2\pi n[/MATH] [MATH]sin\frac{u + v}{2} = 0 \Leftrightarrow u + v = 2\pi n[/MATH].

http://mathworld.wolfram.com/Circumradius.html The formula can easily be combined to a shorter formula by using Heron's formula: http://mathworld.wolfram.com/HeronsFormula.html [MATH]R = \frac{abc}{4A}[/MATH]

[MATH]\sqrt{0.9}=\sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}}[/MATH], which is irrational, so you have to stick with the approximations. An elementary method is to start with some intervall [MATH]a < \sqrt{0.9} < b[/MATH], and then to decrease the width of the intervall by picking [MATH]c, a < c < b[/MATH] and checking whether [MATH]c^2 < 0.9[/MATH] or [MATH]c^2 > 0.9[/MATH]. In either case, we have reduced the intervall to either [MATH]a < \sqrt{0.9} < c[/MATH] or [MATH] c < \sqrt{0.9} < b[/MATH]. Then you pick another number [MATH]d[/MATH] and do it all over again to make your new intervall even smaller. The method is very easy to understand, although there are stronger methods out there.

I don't agree with you when you are saying that my replies don't relate to your problem the way you formulated it firstly. Just look at your first post in the thread: I have shown that you were wrong and that the actual number is 5, not 3. Since the result is wrong, your problem is not interesting, unless you are talking about representing parabolas as functions of x. If you assumed this restriction, you should have mentioned it in your first post! On the other hand, circles doesn't satisfy this assumption; they can't be represented by a single function f(x). A parabola satisfying [math]y^2 = x[/math] isn't worse than a circle satisfying [math]y^2 = 1 - x^2[/math]. So, why did you mention circles in your first post?

Now, I am going to show that [math]N = 5[/math] Let some arbitrary parabola be given. We use a coordinate system having the y-axis as the symmetric axis of the parabola and origo as the intersection point of the y-axis and the parabola; the formula for the parabola can now be settled as [math]y = ax^2[/math]. The general formula for a parabola is [math]Ax^2 + By^2 + Cxy + Dx + Ey + F = 0[/math] , with some restrictions that is not problematic here. We are going to show that such a parabola can not have more than four points in common with the given parabola. Points satisfying both parabolas are points satisfying the equation as follows: [math]Ax^2 + Ba^2x^4 + Cax^3 + Dx + Eax^2 + F = 0[/math]. This is a polynomial equation of degree 4, and it has at most 4 different real solutions, that is, the two parabolas have at most 4 points in common; [math]N \leq 5[/math]. Since [math]N > 4[/math], [math]N = 5[/math]. I think a modified procedure can be used to prove that [math]N \leq 9[/math] for all conics. Then, we can show by a simple example that [math]N = 9[/math] for hyperbolas, and it is also easily proved that [math]N = 3[/math] for circles, [math]N = 2[/math] for lines and [math]N = 1[/math] for points. I think [math]N = 5[/math] for ellipses, but have not proved it yet. The proof is probably rather trivial.

The natural question to ask is: What is the smallest number [math]n[/math] so that given [math]n[/math] points in the plane, there is at most one parabola which satisfy the points? Name this smallest number [math]N[/math]. I have already shown that [math] N > 3[/math], and it doesn't take much effort to show that [math]N > 4[/math]: The two parabolas [math]y = x^2[/math] and [math]x = (y - 2)^2 - 2 = y^2 - 4y + 2[/math] have four points in common. This follows because the equation [math]x^4 - 4x^2 - x + 2 = 0 = (x + 1)(x - 2)(x^2 + x - 1)[/math] has four real solutions.

The formula [math]y = ax^2 + bx + c[/math] is not a general formula for a parabola. Another formula is for instance [math]x = ay^2 + by + c[/math], and I will use these two formulas to show that given three points [math](x_0, y_0), (x_1, y_1), (x_2, y_2)[/math], there may be more than two parabolas which satisfy the points: Let the points be (0,0), (-1,1) and (1,2). y(x) = ax^2 + bx + c y(0) = c = 0 y(1) = a + b + c = a + b = 2 y(-1) = a - b + c = a - b = 1 That is, [math]y = 1.5x^2 + 0.5x[/math] x(y) = ay^2 + by + y x(0) = c = 0 x(2) = 4a + 2b = 1 = 2(a + b) + 2a = -2 + 2a x(1) = a + b = -1 That is, [math]x = 1.5y^2 - 2.5y[/math]. Needless to say, there may be even more parabolas satisfying these three points, that is, som skew parabolas.

I used the formula for a geometric series to show the result in most cases, except those cases where the formula implie a division by 0. Then I checked the cases which where left. Although you by some reason say the opposite, I really think a/b * b/c = a/c is obvious! It doesn't seem like you have heard about the formula for geometric series, but here it comes (with a huge hint for how to derive it): S = 1 + a + a^2 + ... + a^n aS = a + a^2 + ... + a^(n + 1) S = (a^(n+1) - 1)/(a - 1). "You are starting to prove theorems like an Englishman". What do you mean by this??!

You just check the axioms. For the existence of an inverse, we use the fact that G is abelian: a^n * (a^-1)^n = (a * a^-1)^n = ...

At the age of digging into modern algebra, geometry kind of vanishes.

Challenge 1: In most cases we can rewrite this as (a^(n + 1) - 1)/(a - 1) = (a^(m + 1)(s + 1) - 1)/(a^(m + 1) - 1) * (a^(m + 1) - 1)/(a - 1), which surely is true. There are two cases left: a - 1 = 0 and/or a^(m + 1) = 1. Case I: a = 1. The expression is simply (n + 1) = (s + 1)(m + 1). Case II: a = -1 and m is odd. Now n also is odd, so both 1 + (-1) + (-1) + ... + (-1)^n and 1 + (-1) + (-1) + ... + (-1)^m are zero, and the result follows.

I think Psion meant that Pluto may be a satellite to the new-found planet, not a satellite to the Earth. Anyway, this does not make the hypothesis more likely; Pluto is moving in a manner that does not exactly fit the idea.

Are you sure you have not assumed what you were going to prove? Anyway, a method that could be used is induction: First show that formula is correct for n = 1 and for n = 2, then show that if the formula is true for n = k and for n = k + 1, then it also is true for n = k + 2. A more direct approach which does not assume that you already know or have guessed the formula for a_n is to use some knowledge of difference equations: a_(n + 1) = 5a_n - 6a_(n - 1) has the characteristic equation x^2 - 5x + 6 = (x - 2)(x - 3), so therefore a_n = C*2^N + d*3^N for alle n. We find C and D by using a_1 = 2C + 3D = 0 and a_2 = 4C + 9D = 2(2C + 3D) + 3D = 6. This gives D = 2 and C = -3, or a_n = 2*3^n - 3*2^n.

You say the Aboriginals are apelike? I have never seen anything else than pictures (including movies) of Aborginials, but they does not look more apelike to me than for instance New Guineans (which are more or less closely related to the Aboriginals) or some people from the country of mine (except for the skin colour). I surely know the difference between a chimpanzee and an Aborignal. You say the Aboriginals are descendent of Homo erectus? We are all descendents from Homo erectus! The only question is how it happened; multiregional or out-of-Africa (for the second time, as should not be that impossible - if Homo erectus made the way to Australia, why not Homo sapiens too? Aren't the Homo sapiens said to be less and not more primitive?) How do you know Indonesia and Australia was not connected by land for some thousand years ago (and they were, or the seadistance was at least way shorter)? You say Aboriginals are primitive? That's may be true, but are you yourself less primitive? What have you invented? You say Aboriginals did not invent farming, but how many times has farming been invented and under what circumstances? Deserts aren't exactly the right place to begin, and one need also need suitable crops and animals (a puzzle that one). And if one cannot farm, how can one then get the population density needed for going into metallurgy or such things? Aboriginals did not invent farming, the did not start metallurgy, but they survived in their surroundings. In Eurasia the competition was higher - those who started farming and those who organized themself into states got a higher chance of surviving, but that cultural development may have nothing do to with the physcial or psychical capabilities: To organize oneself into states isn't necessary in a low-density area, so why should one when one have the nicely fitted tribe-structure?

Johnny5: I didn't ask for the meaning of x^y because I don't know what it mean myself, but because you don't seem to have a total understanding of the expression. Your attempt of proving that 0^0 = 1 if 0! = 1: You started by stating that [MATH]e^x = \sum_{k = 1}^{\infty} x^n/n![/MATH] for all x, but do you know whether this expression is so easily written because of a convention of letting 0^0 = 1 or because of something else? (And this is another rethorical question, asked because I hope you will see the systematic approach of starting at the definitions, which are needed (to start with everything you find mathematically true is really to let "human whim" take over a bit too much), and then go upwards in a rigourous manner. A little mistake, and you will lead yourself to contradictions, possibly because the starting foundation is wrong (like in naive set theory), but more likely because you broke a rule.)