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.)

There is sort of a blind alley with 0^0, sort of not-defined-thing. I must ask another time: What does the expression x^y mean? Let us define the function f(x,y) = x^y, at every position (x,y) where x^y is normally defined, with a possible exception for (0,0) (that is, we don't bother about whether 0^0 is defined or not). Then it is natural to say that for specific numbers a and b, a^b is only defined if the limit for f(x,y) when x and y gets close to a and b respectively exists. What do we then find for the case of a = 0, b = 0? That the limit does not exist, that is, 0^0 is not defined. You say mathematics does not have many conventions? Are you sure about that? You know, we cannot just use pure logic without having anything to start with, so I would rather say mathematics is full of conventions, but, and this is important, not more nor less conventions than logic could survive. For instance, is the definition of a group, or the name "group", purely based upon pure logic? No, it is led to by cause of human mind, but in mathematics it is now stated completely by means of logic. "Human whim" indeed has much to say for mathematics; we are the ones exploring mathematics (or possibly creating it), and our way of thinking settles the way of exploring. A little tip: Start looking at the definitions (of for instance n!) before you go into something, and then you possibly would avoid some of the worst flaws (that 1 = 1/0 if 1 = 0!, for instance; n! =1 * 2 * 3 * ... * n only works for natural numbers n, not every single integer).

I will ask a simple question, Johnny5: What does the product symbol tell? And the answer: It is just a product symbol. By definition, this symbol can only be used under the following circumstance: Suppose n and m are integers, and that [MATH]n \leq m[/MATH]. Also, suppose f is a function such that [MATH]f(i)[/MATH] is defined for all natural numbers i that satisfy [MATH]n \leq i \leq m[/MATH], and suppose we have defined some multiplication on the set in which all [MATH]f(i)[/MATH] are members. Then we can use the product symbol, by the mean that [MATH]\prod_{i = n}^{m} f(i) := f(n) \cdot f(n + 1) \cdot ... \cdot f(m)[/MATH]. (I have possibly not given all restrictions, which would be a result of lack of knowledge.) The product symbol cannot be used under any other circumstances, at least as long as we use the given convention, and that is indeed what we do. So to write that [math]\prod_{k=a}^{b} f(k) = \prod_{k=b}^{a} f(k)[/math] is in fact completely meaningless, unless [MATH]a = b[/MATH]. On the other side, the expression [math]\prod_{k=a}^{b} f(k) = \prod_{k=a}^{b} f(a + b - k)[/math] in some situations make sence (and there you got your commutative law). I have to ask another question: Why do you (Johnny5) have so severe problems with conventions? When people in some cases use the convention 0^0 = 1, this should not be a problem at all. Firstly, 0^0 is not defined, so the convention would not work against anything, and secondly, as Matt Grime already has written, people only use the convention when they say they use it, and not without saying. It seems that for you, it is the not-defined part that is the problem. But have you then asked yourself what the expression x^y means? (A little regression: The product symbol can be used in expressions such as [MATH]\lim_{m\to\infty}\prod_{i = n}^{m} f(i)[/MATH].)

1: I was assuming positive divisors, but the wording "divisors" does not prohibit negative divisors. The reason of my assumption were merely based upon the fact that if we allow for negative divisors, the answers to the problem become very trivial; 2005N surely has the divisors -1, 1, 5, -5, 2005, - 2005, 401 and -401, 8 there is, and if N is not +/- 1, then the number possibly have more. So the answers are: (a) No such N exists, (b) M = 1 og M = -1. 2: Let's go on with positive divisors, and the formula for the numbers of positive divisors of a given natural number. There are several ways of finding the formula, I would better suppose a combinatorical argument and/or induction. Let d(N) be the number of positive divisors of N. Then: N = 1 has exactly 1 positive divisor. N = p^f, p a prime and f a natural number, has exactly (f + 1) positive divisors; 1, p, p^2, ..., p^f. N = p^f*M, gcd(M,p^f) = 1, has exactly (f + 1)d(M) divisors: A divisor of N is p^g*M', M' a divisor of N. We can choose this in (f + 1)d(M) different ways; (f + 1) choices of g, d(M) choices of M', and no two choices give the same divisor, since p^g and M' are relatively prime. The formula then must be [MATH]d(\prod_{i = 1}^{n} p^{e_i}_i) = \prod_{i = 1}^{n} (e_i + 1)[/MATH] How I got the answers? By finding the exponentials of 2005N and 2005M, respectively (what was the options??). Let's look at the latter case of finding M: We have [MATH]\prod_{i = 1}^{n} (e_i + 1) = 8 = 2^3[/MATH]. Now, 8 = 2*2*2 = 4*2 = 8*1, that is, e_1 = e_2 = e_3 = 1 or e_1 = 3, e_2 = 1 or e_1 = 7, all other exponents being zero. The first one gives threee single prime factors, but then M is a prime, that is, not composite. The third one gives p^7 = 2005M, surely not true for any prime p. So the second is the only one working, namely, 2005M = p^3*q for some primes p and q. Now, 2005 = 5*401, so p and q must be 5 and 401, that is 2005M = 5^3*401 or 2005M = 401^3*5, that is, M = 5^2 or 401^2. Got it? I better give some hints for the problem I posed, since noone seems to bother about it right now: Find all (I originally wrote 'the', but it possibly is two) integer M with lowest absolute value |M| such that there exists a positive integer N so that 2005N has exactly N + M divisors, and for such M, find the lowest possible N + M. Hint 1: What about the case M = 1, N = 5? Hint 2: For M = 0, does such an N exist. Suppose it does, in which cases do you get a contradiction, and in which cases does such an N exist (if in any)? Mere to say, a start would be: Suppose p is a prime and e a non-negative integer. Then (i) [MATH]p^e \geq e + 1[/MATH] with equality for p = 2, e = 1 or for e = 0. (ii) [MATH]p^e \geq e + 2[/MATH] if p > 2 and e > 0, with equality for p = 3, e = 1. (iii) [MATH]p^e \leq 2(e + 2)[/MATH] for p > 4 only if p = 5 and e = 0 or 1. (iv) [MATH]p^e \leq 4(e + 1)[/MATH] only if p = 2, 0 < e < 5 or if p = 3, 0 < e < 4. I will not give any proofs; they are easy and only based upon the simple fact that the left hand side grows faster than the right hand side. And, for the case of M = 0, (a) N = 5^f * 401^g * M, gcd(M,2005) = 1, f > 0, g > 0. Then d(2005N) = (f + 2)(g + 2)d(M) = N = 5^f * 401^g * M > (f + 2)(g + 2)d(M), a contradiction. (b) N = 5^f * M, gcd(M,2005) = 1, f > 1. Then d(2005N) = 2(f + 2)d(M) = N = 5^f * M > (f + 2)d(M), a contradiction. © N = 5M, gcd(M,2005) = 1. Then d(2005N) = 6d(M) = 5M. What next? (d) N = 401^f * M, gcd(M,2005) = 1, f > 0. Then d(2005N) = 2(f + 2)d(M) = N = 401^f * M > (f + 2)d(M), a contradiction. (e) gcd(N,2005) = 1. Then d(2005N) = 4d(N) = N. What next? For which N does we surely get a contradiction? What N are the "left-overs". I must admit, making a computer program would be the simplest way of solving the problem; everything I have written it typically brute-force-methodic.

2005 = 5*401, both 5 and 401 being primes. A number p^v*q^b*r^n*... (standard prime factorization with uncommon notation) have (v + 1)(b + 1)(n + 1) divisors. We are seeking 6 = 2*3 (a*b*..., a,b,... > 1) divisors, so N = 5 or 401. (2005 itself has 4 divisors, 1, 5, 401 and 2005, and we shall only add two new divisors; in the case N = 5 we add 10025 and 25.) 8 = 4*2 = 2*2*2, so M can be 5^2 or 401^2 (all primes would also work, if it was not for the restriction of M being composite). Here is a similar one: Find the integer M with lowest abosulte value |M| such that there exists a positive integer N so that 2005N has exactly N + M divisors and for the given M, find the lowest N so that 2005N has exactly N + M divisors.

The fundamental theorem of algebra: Every polynomial p(x) with complex coefficients of degree >= 1 has a complex root. The statement that a complex polynomial with degree n >= 1 has n roots is in fact a statement where multiplicity is encounted. The "real" theorem is the one stated above (if you allow multiple roots, then the second statement can easily be shown by induction). Other formulations of the theorem is: - Every complex polynomial have a factorization where no polynomial factor is of degree greater than 1. - C splits over C. - The field C of complex numbers is algebraically closed. I suggest you should allow yourself to look into some books introducing you to the field of abstract algebra. Then you hopefully would understand the concept of i; it is more to it than lower calculus is telling you.

æ, ø and å are three vowels that are used in the Norwegian and the Danish writing systems. In Norwegian, 'æ' equals the vowel in 'cat' or 'plait', the 'ø' is more or less equal to the vowel in 'rough' or 'tough', and the 'å' equals the 'aw' in 'drawer' or the 'o' in organ. We have got 02 29 02 30 02 48 01 97 01 98 02 16, or possibly (by 0 being dividers) 229 230 248 197 198 216. But you have already seen that. Makes Σµ° ┼╞ ╪ any sense to you?

[MATH]f(f(x))=\frac{f(x)-a}{bf(x)-c}=\frac{\frac{x-a}{bx-c}-a}{b\frac{x-a}{bx-c}-c} = 1[/MATH] for all x (except those few that does not define f(x) or f(f(x))) if and only if [MATH]\frac{x-a}{bx-c}-a=b\frac{x-a}{bx-c}-c[/MATH] for all x, that is, [MATH]x - a - abx + ac = bx - ba - cbx + c^2[/MATH], that is, [MATH]x(1 - ab + bc - b) = c^2 + a - ac - ba[/MATH]. If this is going to work for every x, then we need both [MATH]1 - ab + bc - b = 0[/MATH] and [MATH]c^2 + a - ac - bc = 0[/MATH]. By checking, we find that [MATH]b=1, a=c[/MATH] is a solution to these equations. We can now look for other solutions, and begin by solving the second equation for b: c = 0 or [MATH]b=c-a+a/c[/MATH]. The first eqation gives a+1=c or [MATH]b=1/(a+1-c)[/MATH]. By this, we check the cases: c = 0 gives 1 = b(a + 1) and a = 0, that is, b = 1. c = a + 1 gives 1 - ab + ab + b - b = 1 = 0, which cannot be true. [MATH]b = c - a + a/c = 1/(a + 1 - c)[/MATH] gives [MATH]c = (a + 1 - c)(c^2 - ac + a)[/MATH], or [MATH]c^3 - (2a + 1)c^2 + (a+1)^2c - (a^2+a) = 0[/MATH]. We already know the solution c = a, so we can reduce this to [MATH]c^2 + (a+1)c + (a + 1) = 0[/MATH] by dividing with c - a. This new equation has the solutions [MATH]c_1(a), c_2(a)[/MATH], and from this we can find all triplets (a,b,c) so that f(x) is self-inverse. An example is for instance (3, -6.5, - 2). As a paranthesis, c = 1 gives ab = 1 and 1 = b, that is, a = b = c = 1, a trivial case in which f(x) = 1 for alle x not being 1.

When the task is to find the center of the circle using only a compass and an unmarked ruler, is it then really allowed to base oneself upon approximations?? A method not based on approximations is to draw two lines, both dividing the circle in two parts. Then you find the midpoints of the two circles, and construct a right angle there. The two new lines cross eachothers in the center of the circle, as can easily be proved.

If you do not want to use sine and cosine, then you can use similarities instead, and if you do not want to use the Heron formula, you can always work hard with Pythagoras formula. I do not know why that would be better - it is easier just learning the stuff, or possibly even better: how to derive it. The point in the specific 7.5-10-12.5- exercise is that right triangles are inscribed in a circle with the hypotenus as a diameter (Thales theorem, a special case of the "periferivinkelsetninga" - I don't know the english name - that the center angle is twice the angle at the circumference. Funny enough, this sentence is not taught at Norwegian schools anymore, although it is very simple and something everyone that know a bit about geometry should know. Last year I was talking to a person who now studies physics and electronics at Princeton, and he had never heard about the theorem, nor did he manage to derive some simple consequences of it. So much for geometry in Norway!)

Another approach is to try to derive the digits by the following method: 54321*abcde=.....12345. e must be 5, since 1*e should end in 5. Now, we have 54321*abcd5 ........271605 ...... Next, d should be 4, since 1*d should end in 4: 54321*abc45 ........271605 ......217284 or in abbreviate form 54321*abc45 ......2444445 Next, 1*c+4 should end in 3, giving c = 9, and then 54321*ab945 ......2444445 ....488889 or 54321*ab945 ....51333345 We now need another 9, so that b + 3 ends in 2, and get 54321*a9945 ....51333345 ..488889 or 54321*a9945 ..540222345 And at last, a = 9: 54321*99945 ..540222345 488889 5429112345 The method will work in all cases (if you substitute 12345 and 54321 with whatever you want), but in this case the method is especially well-applicable, since 54321 ends in 1, and we therefore don't need to check multiple possibilities.