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).