The Geek Culture Forums
Ask a Geek! give me an algerbra problem

UBBFriend: Email This Page to Someone!  next newest topic  next oldest topic 
Author  Topic: give me an algerbra problem 
geekwannabebritt Geek Apprentice Posts: 41 
posted August 12, 2002 23:15
i want to see if i can solve it IP: Logged 
Drazgal GeekinTraining Posts: 35 
posted August 13, 2002 04:03
Heres a fun one from my A Level revision notes Solve the differential eq^n y * cotx * dy/dx = 2 at x = Pi/4 and y = 0 IP: Logged 
Alien Investor Assimilated Posts: 488 
posted August 13, 2002 07:03
Drazgal, your problem is 2nd year calculus, not algebra. (1) A Manhattan resident commutes back and forth to work for a total of two trips per day. An ordinary commuting week is five days. It costs $1.50 to ride the subway or $6.00 to take a taxicab. The Manhattanite spent $28.50 on commuting this week. How many subway trips and how many taxicab trips did they take? To solve this properly, use variables (x = # of subway trips, y = # of taxi trips) and set up two equations in two unknowns. Then reduce to 1 equation in one unknown. (2) Find the sum of the integers from 1 to 100, inclusive (this is not quite algebra). Then find a more general formula for the sum of the integers from 1 to N (this is algebra). (3) Find a solution to the general quadratic equation, a x^2 + b x + c = 0. It's okay to use books and the Internet to search. Study the solution until you can do it yourself from memory. Then try it with different letters and different signs all the way through, such as k y^2  m y + n = 0. This will give you plenty of practice in pushing equations around.
IP: Logged 
nekomatic Assimilated Posts: 452 
posted August 13, 2002 07:25
(4) Let x be 2. If it tries to be anything else, prevent it. Use force if necessary. (5) If m is the codependant of n, n is fixated upon half of y and z is a member of a triangle involving m and q, what is the solution and should y be held responsible? (6) Find the cube root of b. I can't remember where I put it and I really need it for tomorrow.
IP: Logged 
spungo Highlie Posts: 749 
posted August 13, 2002 07:30
quote: Err... that's high school calculus. Oh  and the answer is y = i*SQRT(2*ln(2*(cos^2(x)))) (sorry... the geek in me can never resist!) 
IP: Logged 
maxomai Super Geek Posts: 248 
posted August 13, 2002 07:51
Easy: Show that Rubik's Cube is abelian. (This is an edit; the original was wrong) Medium: Prove that it is impossible to square the circle (that is, given a circle of a given area, to produce a square of exactly that area) using a straightedge and a compass. Hard: Find a general algorithm for generating the Galois groups for C^n as an extension field of R^n. (Yes, these are algebra problems, and they are solvable. And yes, I'm a sadistic jerk.)
IP: Logged 
+Andrew Alpha Geek Posts: 275 
posted August 13, 2002 09:02
(1) 7 subway, 3 taxi (2) 5050. Formula: (1+N)*(N/2) (A famous problem) IP: Logged 
Alien Investor Assimilated Posts: 488 
posted August 13, 2002 09:58
quote: This looks like it came out of a program, what with the redundant instances of 2 in "2 * ln(2 * ...)", and especially multiplying i by the square root of the logarithm of a number in the range [0..1]. That is to say: cos^2(x) is in [0,1]; so ln cos^2(x) is negative; so sqrt(2*ln(2*cos^2(x))) is pure imaginary, and i times that mess is real again. Try solving by hand and the redundant 2's come out and a "4 ln cos x" is really "4 ln sec x" and stays purely real. Deep breath ... y * cot x * dy/dx = 2 y dy = 2 tan x dx 1/2 y^2 = 2 ln sec x + C1 y^2 = 4 ln sec X + C2 # Constantofintegration fiddling y = +2 sqrt ln sec X + C3 Plug the initial conditions and solve for C3, C4: 0 = +2 sqrt ln sec pi/4 + C3 Similarly, 0 = 2 sqrt ln sec pi/4 + C4 (Note I am using sqrt here to mean "positive square root" always, consistently.) So:
IP: Logged 
Alien Investor Assimilated Posts: 488 
posted August 13, 2002 10:08
quote: I think the Cube is nonAbelian. For example, turn the top face one quarter turn to the right, then turn the right face one quarter turn back. Those two operations do not commute. (Follow the top left front corner cubie). I understand the Medium problem, but I can't solve the whole problem. First, show that the compass and straightedge can perform a subset of the rational functions (addition, subtraction, multiplication, division, square roots, nth roots). That's not hard. Then, prove that pi is transcendental. That's beyond me, but I remember where to find the proof in Spivak. I don't even understand the Hard problem, my group theory is gone with the wind.
IP: Logged 
maxomai Super Geek Posts: 248 
posted August 13, 2002 10:27
quote: I should have listened to my first instinct. You're right, of course.
quote: That's about the gist of it although if I recall correctly cube roots are impossible .. that would be the doublingthecube problem.
quote: It's pretty easy as it turns out, but it took me forever to figure it out when I was a grad student. Chalk it up to my being an idiot. Anyways: Suppose n = 1. The Galois group Gal(C,R) consists of identity (a+bi => a+bi) and conjugate (a+bi => abi). This makes sense since [C:R] = 2. Note that there are no intermediate splitting fields since [C:R] is prime. Suppose n = N > 1. C^n is an extension field of C^(n1) * R (up to isomorphism). Furthermore Gal (C^n, C^(n1) * R) = {phi1, phi2} such that: phi1 is identity; Note that Gal(C^(n1) * R, C^(n2) * R^2) is isomorphic to Gal (C^(n1), C^(n2) * R). Thus the induction hypothesis applies. I admit to some hand waving and detail skipping here. This should help you follow along: http://www.math.niu.edu/~beachy/aaol/galois.html
IP: Logged 
Xanthine Uber Geek Posts: 815 
posted August 13, 2002 10:29
quote: I agree. After a year and a half of college calculus I don't even understand what the easy one is asking. Here's a relatively easy one for her.  IP: Logged 
maxomai Super Geek Posts: 248 
posted August 13, 2002 11:06
quote: If you're a math major you might understand all three questions by your Senior year. (Yes, I'm sick .. sick sick sick.)
IP: Logged 
SpikeSpiegel Highlie Posts: 739 
posted August 13, 2002 11:08
quote: oh so thats why i couldnt do it, here i thought i was just losing my mind  IP: Logged 
neotatsu Uber Geek Posts: 839 
posted August 13, 2002 15:53
And to think, this thread was started by a high school FRESHMAN to see if she could do SIMPLE ALGEBRA, and because I'm in a pissy mood, I'm not going to post any either(though I'm seriously tempted to write a few and send them out in an email, though I don't think that'd be a good idea, Twinkie only has access to a public terminal I believe..) IP: Logged 
skylar Alpha Geek Posts: 342 
posted August 13, 2002 17:50
I was just thinking it's a shame Twink's not here to read this. This is definitely her kind of thread... sky IP: Logged 
Alien Investor Assimilated Posts: 488 
posted August 13, 2002 23:22
Just to let Britt in on the joke ... There are two different branches of math, both named algebra. The first "algebra" starts with simple word problems like "how often does the commuter take the subway" and goes up to, more or less, solving polynomial equations. Americans usually learn this algebra in 7th or 8th or 9th or 10th grade. I can just feel the geek oneupmanship reflex kicking in among all the posters here, so I'll say that I started learning this style of "algebra" in the summer between 2nd and 3rd grade. The second "algebra" starts with generalizations over all the possible varieties of the first "algebra". It starts with matrices (linear algebra) and includes group theory (abstract algebra) and plenty of other stuff that I've either forgotten or never knew. American geeks usually learn this algebra starting around the second year of college. There's a large gap between "algebra" (type 1) and "algebra" (type 2). I'm assuming from Britt's school level (starting high school) that she is interested in the first type of algebra, and the joke is that people are posting type2 algebra problems. Which is fine with me, I love math. And it would be even funnier if Britt were the second incarnation of Ramanujan and she really is looking for problems like the Goldbach conjecture (every even number > 2 is the sum of two primes), which is my favorite unsolved problem. IP: Logged 
greycat Assimilated Posts: 374 
posted August 14, 2002 13:13
Here's a nice, easy one that a high school student should be able to handle: Pick any number, N. Square it and subtract 1. Remember the answer. Now take N+1 and N1 and multiply them. Does the answer look familiar? Here's a gruntwork one (not a lot of thought involved, just technique): Find the solutions to the equation x^3 + x^2  4x  24 = 0. (Hint: one is real, and two are complex. Once you find the real root, you can get the other two with the quadratic formula [Alien Investor problem #3].) Here's a much harder one: Prove that there are infinitely many primes. (There are many famous proofs of this; if you get one from a book or the Internet, you need to make sure you understand it completely and can reproduce it from memory.) IP: Logged 
spungo Highlie Posts: 749 
posted August 15, 2002 02:01
quote: Err... no, it was done by hand  you may dislike the style, but I thought it was a more compact expression. I always thought if one can be express it conveniently in one term (instead of two) than one should do so. 
IP: Logged 
snupy Highlie Posts: 763 
posted August 15, 2002 03:10
quote: **whew** lucky meI learned my lesson last time and brought a change of underpants. IP: Logged 
Alien Investor Assimilated Posts: 488 
posted August 15, 2002 06:53
quote: There's an advantage to 1 term instead of 2, but you still missed the point about generating imaginary intermediate results. If nothing else, on your terms, that's some redundant multiplication that can be folded together. Viz: y = i sqrt (2 * ln (2 * cos^2(x))) Versus: y = sqrt (2 * ln (2 * cos^2(x))) And then: y = sqrt (2 * ln (1/2 * sec^2(x))) y = sqrt (ln (1/4 * sec^4(x))) Cheers.
IP: Logged 
spungo Highlie Posts: 749 
posted August 15, 2002 07:16
Ok, fair enough  but I disagree that leaving the 'i' on the outside is implicitly wrong  it's not as if the argument to the sqrt will always be negative.
quote: Not true! 
IP: Logged 
maxomai Super Geek Posts: 248 
posted August 15, 2002 08:38
quote: If you like that one: A Smith number is a positive composite integer, such that the sum of its digits, is equal to the sum of the digits of its prime factors. Examples: 4 is a Smith number (prime factors are 2 and 2. 2+2 = 4) 23 is not a Smith number because it is prime. 666 is a Smith number. (prime factors are 2, 3, 3, 37. 2+3+3+3+7 = 18; 6+6+6 = 18) Prove that there are infinitely many Smith numbers. (This has been done before) Bonus question: prove that for each K, there are infinitely many positive composite integers so that K divides both the sum of the digits and the sum of the digits of the prime factors. (This has been done before.) Super duper bonus question: prove that there is a prime number between every two squares (e.g., 2 is between 1 and 4; 5 is between 4 and 9; 11 is between 9 and 16; etc.) This is called Sierpinski's Conjecture and it hasn't been proven yet, despite the efforts of a bunch of math geeks one weekend at Purdue University. The smartest of the bunch thinks the solution is algebraic (that's "type 2" algebra). IP: Logged 
Alien Investor Assimilated Posts: 488 
posted August 15, 2002 08:41
The cool thing about Geek Culture is that nobody will dare yell at us for being too geeky in a public place. You are right! 2 cos^2(x) is in the range [0,2], not [0,1]. I dropped the "2 *" part in my head when I was analyzing it. Oops. Just looking at the equation again: y * cot(x) * dy/dx = 2 The "y * dy/dx" term implies that there will be two mirrorimage solutions, with y = f(x) and y = f(x).
IP: Logged 
spungo Highlie Posts: 749 
posted August 15, 2002 08:47
quote: I know! It's a place we can all be ourselves  we need to arrange some Geek Pride marches, or something!  IP: Logged 
snupy Highlie Posts: 763 
posted August 15, 2002 09:40
It's kinda like online porn for me, given what geekiness does to me... IP: Logged 
spungo Highlie Posts: 749 
posted August 15, 2002 09:51
quote: Online porn for me would be... I suppose... a beer forum... or a beeranddoughnut forum.... (mmmmm.... doughnuts!...) **tremble**  IP: Logged 
Alien Investor Assimilated Posts: 488 
posted August 15, 2002 11:18
quote: New York Apartment Porn: "Now I'm walking into your ENORMOUS living room ..." I swear, there are days I would pay $2 a minute to hear something like that. IP: Logged 
snupy Highlie Posts: 763 
posted August 15, 2002 11:41
quote: ROTFLOL!!!!!!!!! IP: Logged 
maxomai Super Geek Posts: 248 
posted August 15, 2002 12:40
quote: Which, by coincidence, is almost as much as it costs to rent in Lower Manhattan IP: Logged 
All times are Pacific Time  next newest topic  next oldest topic 
© 2002 Geek Culture® All Rights Reserved.
Powered by Infopop www.infopop.com © 2000
Ultimate Bulletin Board 5.47e