Author Topic: The Math and Quiz Game =) (Read 36878 times)

/dev/humancontroller · « **Reply #90 on:** December 21, 2007, 01:10:29 pm »

a₁₀₁ = 1 + 0.5*(1⁴+2⁴+...+100⁴) - 40*(1³+2³+...+100³) + (1+2+...+100) - 100*(1)

The formula for (1+2+...+n) is (n*(n+1)/2), for 100 that's 5050.
The formula for (1³+2³+...+100³) is ((n*(n+1)/2)²), for 100 that's 25502500.

a₁₀₁ = 0.5*(1⁴+2⁴+...+100⁴) -1020095049

But I can't find the formula for (1⁴+2⁴+...+100⁴).
I'll google some more, or maybe you can tell me :]

TheEternalDretch · « **Reply #91 on:** December 21, 2007, 01:38:21 pm »

Why google when you can figure these things out on your own?

This problem (up to n⁹, I think) is one I played around with and found an answer to when I was 15 or 16. Of course, I was taking a very different approach, so I dunno.

I'd still like to see if someone can do it analytically. ^_^ Just one part left!

EDIT: Oh wow, I just found the general solution online. <3 Awesomeness. So never mind, I lost my curiosity in seeing how you'd do it. XD
1⁴ + 2⁴ +...+ n⁴ = (6n⁵ + 15n⁴ + 10n³ - n)/30

Your turn again!

/dev/humancontroller · « **Reply #92 on:** December 21, 2007, 02:10:10 pm »

P and T are static points in 3D.
P will throw something at T with a velocity of v.
The rock is affected by the gravity vector, G.
Where (direction: D) should P throw.. uhm, so that it hits T?

Overdose · « **Reply #93 on:** December 21, 2007, 10:04:19 pm »

D = left

TheEternalDretch · « **Reply #94 on:** December 22, 2007, 10:36:12 pm »

So, uh, is 'v' the starting velocity? And there's no friction?

Assuming that similarly to the previous problem, you're not causing any acceleration, then the only acceleration is caused by the force of gravity. I'm defining the Z axis as up.

a(t) = -G
V(t) = v - Gt
X(t) = P_x + v_xt, Y(t) = P_y + v_yt, Z(t) = P_z + v_zt - Gt²/2

Assuming that we picked the right direction, there will be a time t₀ for which our projectile hits the target. In that case,
X(t₀) = P_x + v_xt₀ = T_x
Y(t₀) = P_y + v_yt₀ = T_y
Z(t₀) = P_z + v_zt₀ - Gt₀²/2 = T_z

Of course, we also know that v_x² + v_y² + v_z² = v²

That gives us four equations, four unknowns. We'd express t₀ using v, and then find the different components of v, and thus the solution. I'd do it now, but I'd got other things to do. I'll probably get to it later.

/dev/humancontroller · « **Reply #95 on:** December 23, 2007, 09:58:33 am »

G is a vector that may or may not be pointing downwards.
Typically, it would be (0,0,-9.81), but that is not guaranteed (only that |G| > 0).
Note the conformation: the "positive direction" of the axes, the negative value in G.

TheEternalDretch · « **Reply #96 on:** December 23, 2007, 04:59:43 pm »

Quote from: /dev/humancontroller on December 23, 2007, 09:58:33 am

G is a vector that may or may not be pointing downwards.

Seeing as you put no limitations on the axes, I can simply define them so that G is indeed pointing downwards. And if that doesn't satisfy you, the four equations can easily be changed to:
X(t₀) = P_x + v_xt₀ - G_xt₀²/2 = T_x
Y(t₀) = P_y + v_yt₀ - G_yt₀²/2 = T_y
Z(t₀) = P_z + v_zt₀ - G_zt₀²/2 = T_z
v_x² + v_y² + v_z² = v²
...which is the same story. And I'm still too lazy to go through the algebra. Maybe later.

EDIT: Okay, let's do this thing.
v_x = G_xt₀/2 + (T_x - P_x)/t₀
v_y = G_yt₀/2 + (T_y - P_y)/t₀
v_z = G_zt₀/2 + (T_z - P_z)/t₀
v_x² + v_y² + v_z² = (G_xt₀/2 + (T_x - P_x)/t₀)² + (G_yt₀/2 + (T_y - P_y)/t₀)² + (G_zt₀/2 + (T_z - P_z)/t₀)² = v²
G_x²t₀²/4 + G_x(T_x - P_x) + (T_x² - 2T_xP_x + P_x²)/t₀² + G_y²t₀²/4 + G_y(T_y - P_y) + (T_y² - 2T_yP_y + P_y²)/t₀² + G_z²t₀²/4 + G_z(T_z - P_z) + (T_z² - 2T_zP_z + P_z²)/t₀² = v²
Multiply the equation by t₀²....
G_x²t₀⁴/4 + G_x(T_x - P_x)t₀² + T_x² - 2T_xP_x + P_x² + G_y²t₀⁴/4 + G_y(T_y - P_y)t₀² + T_y² - 2T_yP_y + P_y² + G_z²t₀⁴/4 + G_z(T_z - P_z)t₀² + T_z² - 2T_zP_z + P_z² = v²t₀²
Some rearranging....
t₀⁴(G_x² + G_y² + G_z²)/4 + t₀²(G_x(T_x - P_x) + G_y(T_y - P_y) + G_z(T_z - P_z) - v²) + T_x² - 2T_xP_x + P_x² + T_y² - 2T_yP_y + P_y² + T_z² - 2T_zP_z + P_z² = 0
t₀⁴(GoG)/4 + t₀²(Go(T-P) - v²) + (T-P)o(T-P) = 0
t₀² = v² - Go(T-P) + √((Go(T-P) - v²)² - (GoG)*(T-P)o(T-P)) *2/(GoG)
I'm gonna stick in L = T-P to get a cleaner formula....
t₀ = √( v² - GoL + √((GoL - v²)² - (GoG)*(LoL)) *2/(GoG) )

Now that we have t₀, we also have v_x, v_y and v_z, and therefore
D = (v_x/v, v_y/v, v_z/v) or D = Gt₀/2v + L/vt₀
Good enough? ^_^

/dev/humancontroller · « **Reply #97 on:** December 24, 2007, 02:38:07 am »

good.

TheEternalDretch · « **Reply #98 on:** December 24, 2007, 03:18:46 am »

Okay, I hope this one isn't too hard. Find a function f(x) (though general solutions are nice) for which:
f''(x)*cos(2x) + 4f(x)*cos(2x) = 1
Where f''(x) is the second derivative of f(x).

If this is too hard, I can switch it for something easier. Again, I just want to see the sort of response I'll get.

kevlarman · « **Reply #99 on:** December 24, 2007, 04:34:02 am »

Quote from: TheEternalDretch on December 24, 2007, 03:18:46 am

Okay, I hope this one isn't too hard. Find a function f(x) (though general solutions are nice) for which:
f''(x)*cos(2x) + 4f(x)*cos(2x) = 1
Where f''(x) is the second derivative of f(x).

If this is too hard, I can switch it for something easier. Again, I just want to see the sort of response I'll get.

my calculus is a little on the rusty side to be solving second order differential equations right now, but my trusty 89 says (in LaTex to avoid a mess of parentheses): f(x) = \frac{cos(2x)*ln(abs(cos(2x)))}{4} + C_{1}*cos(2x) + (\frac{x}{2} + C_{2})*sin(2x) where C₁ and C₂ are any constants.

khalsa · « **Reply #100 on:** December 24, 2007, 07:41:07 am »

I dunno why, but reading latex is one of my pet peeves:

TheEternalDretch · « **Reply #101 on:** December 24, 2007, 01:32:34 pm »

That is very correct. o_o What is this trusty 89, eh, precious? What is it, eh?

kevlarman · « **Reply #102 on:** December 24, 2007, 09:32:06 pm »

ti-89, the motorola 68K calculators from can solve many first and second order differential equations (but you have to be careful because they don't tell you when they are wrong, it's easy to check though, since they always give the correct answer when differentiating)

TheEternalDretch · « **Reply #103 on:** December 25, 2007, 12:31:27 am »

Okay, then next time I should give a third order equation.

Calculators are for the weak, anyway.

kevlarman · « **Reply #104 on:** December 25, 2007, 01:15:52 am »

MATLAB > 3rd order differential equations i think
a calculator is useless for this one (this is pretty unreadable but i'll let khalsa fix it up again):

det \left( \left[ \begin{array}{cccccc}
-3 & -3 & -5 & -4 & 6 & -6 \\
7 & -4 & 3 & 5 & 6 & -9 \\
-3 & 6 & -1 & -2 & -1 & -2 \\
9 & 9 & 4 & 2 & -6 & 2 \\
-1 & 2 & 9 & -8 & -6 & -3 \\
1 & -8 & -5 & -7 & 3 & 9
\end{array} \right] - \lambda I_{6} \right)
where lambda is any eigenvalue of that 6x6 matrix.
[edit] bonus question:
list all values of eigenvalues of the matrix, i will accept answers accurate to 32 bits or better[/edit]

khalsa · « **Reply #105 on:** December 25, 2007, 04:59:37 am »

Why must you abuse my pet peeves?

Also, matlab solved this in under a second >_>

But I'll leave these ones for the plebs.

Khalsa

* khalsa tries to ignore this thread.

TheEternalDretch · « **Reply #106 on:** December 25, 2007, 04:12:31 pm »

...this is just raw calculating. How boring. If I was going to solve this, I would also have used matlab, so khalsa might as well post the solution and give us a new problem. ^_^

(Since when can matlab solve differential equations? That's new to me. How do you do it?)

khalsa · « **Reply #107 on:** December 25, 2007, 05:02:53 pm »

Just need to break down the equation a bit:

http://www.mathworks.com/products/pde/description3.html

http://web.mit.edu/answers/matlab/matlab_ode.html

Khalsa

kevlarman · « **Reply #108 on:** December 25, 2007, 05:40:41 pm »

if you think the original problem (not the bonus question) requires a lot of calculating then you aren't thinking hard enough
(also wtf partial differential equations have nothing to do with this)

TheEternalDretch · « **Reply #109 on:** December 25, 2007, 09:15:54 pm »

Quote from: kevlarman on December 25, 2007, 05:40:41 pm

if you think the original problem (not the bonus question) requires a lot of calculating then you aren't thinking hard enough

As far as I can currently tell, both problems require the same effort. So sure, I'm missing something. *shrug* Thanks, Khalsa.

kevlarman · « **Reply #110 on:** December 25, 2007, 10:21:00 pm »

hint: attempting to solve the second problem by hand should reveal a trivial solution to the first.
edit: in case i wasn't completely clear in my original post, i want the value of the whole expression, not what lambda has to be to satisfy the conditions (that's the bonus question)

khalsa · « **Reply #111 on:** December 26, 2007, 04:47:32 am »

Just cause kevlarman was being an annoying brat:

Answer is:
0
Proof:

Quote

"the solutions to det(A - \lambda I)=0 are the eigenvalues of A" ;
"the eigenvalues of that matrix btw are roughly 1.355 and 16.982"

Khalsa

Edit: My actual proof was "I'm retarded"

kevlarman · « **Reply #112 on:** December 26, 2007, 05:08:23 am »

at least take the effort to copy the program i gave you into your 92 and give me eigenvalues to (almost) 64 bits.

khalsa · « **Reply #113 on:** December 26, 2007, 05:23:54 am »

Actually cause I'm a lazy bastard, let's just call your bonus question the question I pose for the people

.

It's easy as pie folks, finish it quick and shut kevlarman up.

Khalsa

TheEternalDretch · « **Reply #114 on:** December 26, 2007, 06:51:02 pm »

....duh. >_< Way to forget what eigenvalues are, Eternal! For some reason, I thought you wanted the polynomial for an unknown lambda. Bah.

But I'm still not planning on doing that determinant by hand, since there would be 720 products. And I only have matlab at school.

kevlarman · « **Reply #115 on:** December 26, 2007, 07:18:37 pm »

Quote from: TheEternalDretch on December 26, 2007, 06:51:02 pm

....duh. >_< Way to forget what eigenvalues are, Eternal! For some reason, I thought you wanted the polynomial for an unknown lambda. Bah.

But I'm still not planning on doing that determinant by hand, since there would be 720 products. And I only have matlab at school.

i chose a large matrix for a reason, most calculators (once again 89/92 are smarter than that) choose to calculate a determinant naively. doing this properly should take roughly 30 additions and 36 multiplications.
(the determinant of a triangular matrix is simply the product of the diagonal, swapping 2 rows negates the determinant, multiplying a row by 1/c will multiply the determinant by c, adding another row times a constant to a row does not change the determinant), the only difficulty is the 6th order polynomial that results from this (ti-89s can't find exact solutions unless the polynomial is trivial to factor).

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Author Topic: The Math and Quiz Game =) (Read 36878 times)