shyvera

It's not difficult to plot the graph of [imath]y=||x^2-4x+3|-2|.[/imath] Start by plotting the graph of [imath]y=x^2-4x+3[/imath] and reflecting everything below the x-axis above it. Then drag it down vertically by two units and reflect everything below the x-axis above it again. While graphs themselves do not constitute proofs, they do greatly help you get started. Doing the above, I make it that the graph above intersects the line [imath]y=m[/imath] at exactly 2 points precisely when [imath]m=0[/imath] or [imath]m>2.[/imath] It remains to justify this algebraically – which can be pretty tedious, but things do become simpler once you know what you need to do.

I agree. Mathematics can be very useful as a tool for analysing theories of music, but if you want to compose decent music, you still have to understand the rules of harmony, modulation, etc. By the way, I composed a short symphony a few years ago; I call it my Short Symphony in E Minor. It's in three movements, the second of which leads without a break into the finale. I've made a MIDI version of it; see what you think. http://people.delphi...Symphony2_1.mid http://people.delphi...mphony2_2-3.mid

A moment ago, I posted something on the application of group theory to music in the thread real life applications of group theory. Here is an elaboration of that post, and another application of maths to music. Circle of fifths I mentioned in my previous post that the theory of intervals is based on the cyclic group of order 12. Here is another way of looking at musical intervals. It's a bit more complicated and less intuitive to the non-mathematical musician as it involves more mathematics – which suits people like me much better. Consider the set of all musical notes distinguished by their absolute pitch (or frequency measured in Hz). Define a relation ~ on this set by X~Y iff the notes X and Y differ by a whole number of octaves. Then it is immediately clear that ~ is an equivalence relation and there are exactly 12 equivalence classes, one for each of the 12 notes in the chromatic scale. Denote the equivalence class containing the note X by [X]. Now let us take the equivalence class [C] (containing the note middle C) as our reference. All the other equivalence classes [X] are related to [C] by the number of semitones (modulo 12) between X and C. The class [X] can therefore be defined as the interval C-X. (Of course we can choose any other class than [C] as our reference, e.g. the class [A], but as middle C is a much used in music (especially by pianists) as a reference note, we may as well choose [C] as our canonical reference.) Intervals (as equivalence classes) can be added by adding up the number of semitones (modulo 12) between the notes in each interval relative to C. For example, [E] + [F] = [A]. Note that this is relative to our reference note C; for different reference notes, the results will be different. However, given a fixed reference note, such an operation is always well defined. The set of all such intervals under this addition operation is then a cyclic group of order 12. Even temperament Temperament is a method of tuning a musical instrument by adjusting the ratios of the pitches of notes in a scale relative to a fixed note, called the tonic. Various termperaments are possible, but I am only concerned here with what is called even temperament, in which all semitone intervals have a fixed ratio, and the ratio of one note to the one exactly one octave below is 2. Let r be the ratio of the frequency of one note to that of the note exactly one semitone below. Then if we start with a fixed note, of frequency f, and ascend the chromatic scale, the frequencies of these notes are f, fr, fr2, …. The frequency of the note one octave above is 2f and that note is 12 semitones above the note we started with. This gives us [math]2f\,=\,fr^{12}[/math] or [math]r\,=\,\sqrt[12]2\,\approx\,1.059[/math] Thus in even temperament the frequency of each note is always approximately 1.059 times that of the note one semitone below. If we take the frequency of middle C as 256 Hz, then the frequency of concert A (9 semitones above middle C) is 256 × (21/12)9 or approximately 430.54 Hz. In some concerts, however, the frequency of concert A is set to 440 Hz. In this case, the frequency of middle C is 440 / (21/12)9 or approximately 261.63 Hz. Whether it is middle C that is tuned to 256 Hz or concert A that is tuned to 440 Hz depends on the performance in question.

I know of an application of group theory to music theory. The chromatic scale of Western music consists of 12 notes: C, C#, D, D#, E, F, F#, G. G#. A, A#, B. An interval is the distance from one note to the another – e.g. C–C# is an interval of a semitone, C–D is a whole-tone interval, C–D# is an interval of a minor third, etc. Note that the starting note can be any note, so F–F# is also a semitone interval. The unison interval is the interval from one to itself (e.g. C–C). All intervals that are whole octaves can be identified with the unison interval. Intervals can be “added”, the result being the number of semitones (modulo 12) from the first note the last (e.g. the sum of C–D and C–F (which is the same as D–G) is the interval C–G). It follows that the set of all intervals under this addition operation forms a group, the cyclic group of order 12. The identity element is the unision interval, and the group is generated by four intervals: semitone ( C–C# ), perfect fourth ( C–F ), perfect fifth ( C–G ), and major seventh ( C–B ). This cyclic group of order 12 is the basis on the theory of the circle of fifths. It also explains why there are only two whole-tone scales – namely, because the subgroup generated by the whole-tone interval (C–D) is a subgroup of order 6 and so has index 2.

A friend of mine suffering from depression is currently on Citalopram treatment. He is a bit uneasy about the drug and is particularly worried that it might make him suicidal. I have looked the drug on Wikipedia (http://en.wikipedia.org/wiki/Citalopram) according to which my friend should be safe as the drug only induces suicidal tendencies in patients under 24 (while my friend is over 50). Does anyone have anything about Citalopram to add to the Wikipedia article? My friend would be happy to know more about the drug.

Love waves are dangerous. They are actually earthquake waves! http://z8.invisionfr...p?showtopic=912 They have nothing to do with love, but are named after the English mathematician A.E.H. Love (1863–1940), who developed a mathematical model of these surface waves.

I’ve corrected it. By the way, ajb also left out a zero when expanding brackets: (100 + 4)(100 + 4) should be 10000 + 2(400) + 42.

By manipulating well-known algebraic formulas, you can devise your own handy arithmetic rules for mental calculation. For example, suppose you want to calculate 91 × 89 mentally. 91 × 89 = (90 + 1)(90 − 1) = 902 − 12 = 8099 Similarly, to do 122 × 118 mentally: 122 × 118 = (120 + 2)(120 − 2) = 14400 − 4 = 14396 You can also do this for division. For example, what is 22491 ÷ 153? Using the trick above, you should be able to get the answer 147 without using a calculator or long division. Now suppose you want to find the prime factors of 359999. If you start by observing that 359999 = 360000 − 1 = 6002 − 12 = (600 + 1)(600 − 1) = 601 x 599 it’s then a simple matter to verify that 601 and 599 are both primes. (This is a problem I once set for some people on another forum.)

[math]\frac{\mathrm d}{\mathrm dx}\left[\arcsin\left(\frac{x-a}{x+a}\right)\right][/math] [math]=\ \frac{\mathrm d}{\mathrm dx}\left[\arcsin\left(1-\frac{2a}{x+a}\right)\right][/math] [math]=\ \frac1{\sqrt{1-\left(1-\frac{2a}{x+a}\right)^2}}\cdot\frac{2a}{(x+a)^2}[/math] [math]=\ \frac{2a}{(x+a)^2\sqrt{\frac{4a}{x+a}-\frac{4a^2}{(x+a)^2}}}[/math] [math]=\ \frac{2a}{(x+a)^2\cdot\frac{\sqrt{4a(x+a)-4a^2}}{x+a}}[/math] [math]=\ \frac{2a}{(x+a)\sqrt{4ax}}[/math] [math]=\ \frac a{(x+a)\sqrt{ax}}[/math] [math]=\ \frac{\sqrt a\sqrt a}{(x+a)\sqrt a\sqrt x}[/math] [math]=\ \frac{\sqrt a}{(x+a)\sqrt x}[/math]

Why would you insist on integrating when you could just differentiate? The question doesn’t specifically ask you to integrate; it merely says: Show that [math]\int\mbox{LHS d}x\ =\ \mbox{RHS}.[/math] So, if you can show that [math]\frac{\mbox d}{\mbox dx}(\mbox{RHS})\ =\ \mbox{LHS},[/math] you have answered the question.

Just differentiate the right-hand side.

That won’t help. Further hint: [math]6a^4-2a^3-2a^2b+a^2-6a^2b+2ab+2b^2-b\ =\ \left(a^2-b\right)(\cdots)[/math]

I would suggesting letting H be the Sylow 11-subgroup (which is certainly characteristic since it’s unique) and K be the centre of H. K is definitely nontrivial (the centre of any p-group where p is a prime is in general nontrivial). Thus if K is a proper subgroup of H, the problem would be solved since the centre of any group is a characteristic subgroup. This leaves the case K = H (i.e. H is Abelian); again this would not be a problem unless H is an elementary Abelian group (which would not have any proper nontrivial characteristic subgroups).

This is similar to the polynomial problem you posted earlier. I get the feeling these problems are designed to test not how much you know but rather how good your observational skills are. Hint for this problem: Factorize the LHS as the product of two quadratic expressions in a.

Hint: [math]\left(x^2+ax+b\right)\left(x^2+bx+a\right)[/math]

Have you noticed that for any x such that [math]0<x<\pi[/math] you always have [math]\sec\tfrac x2>1\,?[/math] Thus, for any angles [math]A,\,B,\,C[/math] in a triangle, we always have [math]\sec\tfrac A2\sec\tfrac B2\sec\tfrac C2+\sec\tfrac A2+\sec\tfrac B2+\sec\tfrac C2-2>2[/math] and so the left-hand side of your equation can never be equal to 0. The assertion is therefore vacuously true*. *The implication [math]p\Rightarrow q[/math] is said to be vacuously true iff the statement [math]p[/math] is false.

Another method is to use the Cauchy–Schwarz inequality: [math]\color{white}......[/math][math]x^2y^2+y^2z^2+z^2x^2[/math] [math]\leqslant\ \left(x^4+y^4+z^4\right)^{\frac12}\left(y^4+z^4+x^4\right)^{\frac12}[/math] [math]=\ x^4+y^4+z^4[/math] [math]\therefore\ \frac{x^4+y^4+z^4}{x^2y^2+y^2z^2+z^2x^2}\ \geqslant\ 1\quad\mbox{if}\quad xyz\ne0[/math]

You did no such thing. You were merely stuck somewhere, and being stuck doesn’t prove or disprove anything. You are the one having problems with proofs. We also seem to be losing the plot a bit. We’re supposed to prove that [math]a^2+ab+b^2\geqslant0[/math] for all real a and b, not assume it’s true and use it to prove something else. So let’s go back a little in the thread. In post #4, I suggested attacking the problem by the determinant method. Your reply clearly showed that you didn’t know what I meant, so let me explain. Consider the following expression: [math]x^2+px+q[/math] This is a quadratic in x. The determinant of the quadratic is the quantity [math]p^2-4q.[/math] It can be shown that [math]x^2+px+q\geqslant0\ \mbox{for all}\ x\in\mathbb R[/math] [math]\Leftrightarrow[/math] [math]p^2-4q\leqslant0.[/math] Well, let’s show it, shall we? [math](\Leftarrow)[/math] Suppose [math]p^2-4q\leqslant0.[/math] Then a little rearrangement gives [math]q-\frac{p^2}4\geqslant0.[/math] Also [math]\left(x+\frac p2\right)^2\geqslant0[/math] for all real x. Adding the two inequalities gives [math]\left(x+\frac p2\right)^2+q-\frac{p^2}4\geqslant0[/math] and the left-hand side is just [math]x^2+px+q.[/math] [math](\Rightarrow)[/math] Now suppose [math]x^2+px+q\geqslant0[/math] is true for all real x. In particular, it will be true for [math]x=-\frac p2.[/math] Substituting this into the inequality and simplifying should give [math]p^2-4q\leqslant0[/math] as required. Now we return to your original problem. Treat the expression [math]a^2+ab+b^2[/math] as a quadratic in a. (Or a quadratic in b, if you like.) Now calculate the determinant. What do you find?

If x, y, z are real, then obviously x = y = z = 0 is the only solution. Suppose, to make it a little more challenging, that the variables can be complex as well as real. Then notice that [math]0=(x+y+z)^3=x^3+y^3+z^3+3(x+y+z)(xy+yz+zx)-3xyz=-3xyz[/math] Thus xyz = 0, i.e. one of x, y, z is 0. Letting one of them be 0, substituting into (1) and (2), and squaring the first equation, you should find that the product of the other two is 0 as well – which, on substituting back into the first equation again, should mean all three are equal to 0. Hence the trivial solution is again the only solution. NB: I make it that if x, y, z are, in general, elements of an integral domain whose characteristic is not 2 or 3, then the trivial solution is the only solution.

I pronounce 100.1 “one hundred point one” (which I believe is how most people in the UK say it). In fact the “a” may turn up among the positive integers much sooner than 1000, or even 101. In the past you could say “one and twenty”, “two and twenty” etc for 21, 22, etc (cf Jeremiah 52:1 of the King James Version of the Bible: “Zedekiah was one and twenty years old when he began to reign …”). This is now archaic in English, but is still used in German and Dutch: 21 is einundzwanzig in German and eenentwintig in Dutch. So, I suppose, depending on which style of English you speak (as well as your views on the numerical integrity of the word “and”), the letter “a” may appear as early as 21, or not until 1000.

If x, y, z are all real, the three equations can be reduced to just one, namely the second one.

That’s what you’re supposed to calculate! I didn’t post the full solution lest the moderators are not happy with me for doing your homework for you (and being a new member I’m trying my best not to make a bad impression). You should find that the determinant is negative if b is within a certain range of values. I make it [hide]4−2√3< b < 4+2√3[/hide] In this case, the inequality is satisfied for all real values of a. If b is outside that range of values, solve the quadratic equation for the real roots, expressed in terms of b. Then the inequality is satisfied when a is less than the lesser of the two roots or greater than the greater of the two roots.

It’s actually 101 – one hundred and one.

You are assuming [math][p\Leftrightarrow(q\Leftrightarrow r)]\Rightarrow[(p\Leftrightarrow q)\Rightarrow r][/math] to be false. This is equivalent to [math][p\Leftrightarrow(q\Leftrightarrow r)]\wedge\neg[(p\Leftrightarrow q)\Rightarrow r].[/math] What I did was to show that [math]\neg[(p\Leftrightarrow q)\Rightarrow r][/math] would lead to [math]\neg[p\Leftrightarrow(q\Leftrightarrow r)][/math] which would make a contradiction with [math]p\Leftrightarrow(q\Leftrightarrow r).[/math]