Solve the differential eq^n

(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.

(5) If m is the co-dependant 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.

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Originally posted by Alien Investor:
Drazgal, your problem is 2nd year calculus, not algebra.

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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!)

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"No longer empty and frantic like a cat tied to a stick, that's driven into frozen winter shit (the ability to laugh at weakness), calm, fitter, healthier and more productive, a pig in a cage on antibiotics."

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.)

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Originally posted by spungo:
y = i*SQRT(2*ln(2*(cos^2(x))))

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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 # Constant-of-integration fiddling

y = +2 sqrt ln sec X + C3
y = -2 sqrt ln sec X + C4
# twin solutions

Plug the initial conditions and solve for C3, C4:

0 = +2 sqrt ln sec pi/4 + C3
0 = +2 sqrt ln sqrt 2 + C3
0 = +2 sqrt ((ln 2)/2) + C3
0 = sqrt(ln 8) + C3
C3 = - sqrt ln 4

Similarly,

0 = -2 sqrt ln sec pi/4 + C4
0 = -2 sqrt ln sqrt 2 + C4
0 = -2 sqrt ((ln 2/2) + C4
0 = -sqrt(ln 4) + C4
c4 = sqrt ln 4
[/i]

(Note I am using sqrt here to mean "positive square root" always, consistently.)

So:

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Originally posted by maxomai:
Easy: Show that Rubik's Cube is abelian. ([b]This is an edit; the original was wrong)
[/B]

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I think the Cube is non-Abelian. 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 straight-edge 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.

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Originally posted by Alien Investor:
I think the Cube is non-Abelian. 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).

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I should have listened to my first instinct. You're right, of course.

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I understand the Medium problem, but I can't solve the whole problem. First, show that the compass and straight-edge 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.

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That's about the gist of it although if I recall correctly cube roots are impossible .. that would be the doubling-the-cube problem.

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I don't even understand the Hard problem, my group theory is gone with the wind.

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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 => a-bi). 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^(n-1) * R (up to isomorphism). Furthermore Gal (C^n, C^(n-1) * R) = {phi1, phi2} such that:

phi1 is identity;
phi2 is identity on all but one dimension of C^n, and is conjugate for the one exception.

Note that Gal(C^(n-1) * R, C^(n-2) * R^2) is isomorphic to Gal (C^(n-1), C^(n-2) * 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

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Originally posted by maxomai:
Easy: Show that Rubik's Cube is abelian. ([b]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.)

[/B]

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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.
This is the concentration:volume relationship used for diluting solutions in chemistry: initial concentration(initial volume) = final concentration(final volume). You've got a stock solution of 12 M hydrochloric acid (pour carefully). You want to make a 50 mL, 2 M solution (M represents the units of concentration). How much 12 M stock and how much water are you going to need?

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Originally posted by Xanthine:
I agree. After a year and a half of college calculus I don't even understand what the easy one is asking.

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If you're a math major you might understand all three questions by your Senior year.

(Yes, I'm sick .. sick sick sick.)

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Originally posted by Alien Investor:
Drazgal, your problem is 2nd year calculus, not algebra.

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oh so thats why i couldnt do it, here i thought i was just losing my mind

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 one-upmanship 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 type-2 algebra problems. Which is fine with me, I love math.

Pick any number, N. Square it and subtract 1. Remember the answer. Now take N+1 and N-1 and multiply them. Does the answer look familiar?
Why? Either prove that the two answers are always the same for all N, or find a counterexample N for which they are different.

Here's a grunt-work 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:

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Originally posted by Alien Investor:
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].
[/i]

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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.

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"No longer empty and frantic like a cat tied to a stick, that's driven into frozen winter shit (the ability to laugh at weakness), calm, fitter, healthier and more productive, a pig in a cage on antibiotics."

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Originally posted by spungo:
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!)

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Originally posted by spungo:
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.

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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.

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From Alien Investor:
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,

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Not true!
sqrt(2*ln(2*cos^2(x))) is NOT pure imaginary! 2*cos^2(x) is NOT always between 0 and 1.

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"No longer empty and frantic like a cat tied to a stick, that's driven into frozen winter shit (the ability to laugh at weakness), calm, fitter, healthier and more productive, a pig in a cage on antibiotics."

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Originally posted by greycat:
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.)

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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)
22 is a Smith number (prime factors are 2 and 11. 2+1+1 = 4)

23 is not a Smith number because it is prime.
26 is not a Smith number. (prime factors are 13 and 2. 1+3+2 = 6; 2+6 = 8)

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.)

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 mirror-image solutions, with y = f(x) and y = -f(x).

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Originally posted by Alien Investor:
The cool thing about Geek Culture is that nobody will dare yell at us for being too geeky in a public place.

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I know! It's a place we can all be ourselves - we need to arrange some Geek Pride marches, or something!

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Originally posted by snupy:
It's kinda like online porn for me, given what geekiness does to me...

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Online porn for me would be... I suppose... a beer forum... or a beer-and-doughnut forum.... (mmmmm.... doughnuts!...) **tremble**

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Originally posted by snupy:
It's kinda like online porn for me, given what geekiness does to me...

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New York Apartment Porn:

"Now I'm walking into your ENORMOUS living room ..."

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Originally posted by Alien Investor:
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.

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Originally posted by Alien Investor:
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.

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