The Math and Quiz Game =)

[TheEternalDretch]

Sorry, but no. The answer I'm looking for is independent of x.

[/dev/humancontroller]

using an Easy 3rd Degree Equation Solver, I got x = 1.3247.
...then I picked up a calculator and got 2.0

[TheEternalDretch]

Like I said, it's a very easy problem to solve with a calculator. But if you guys don't want to delve into the algebra, that's fine by me. Go ahead, your turn.

[/dev/humancontroller]

I was just kidding xD

Base equations used:
x³-x-1 = 0
x = x³-1
x³ = x+1
1 = x(x²-1)

The path:
( 3x²-4x )^1/3 + x*( 2x²+3x+2 )^1/4
( 3(x³-1)²-4x )^1/3 + x*( 3x²+3x+2x(x²-1) )^1/4
( 3x⁶-6³+3-4x )^1/3 + x*( x(3x+3+2x²-2) )^1/4
( 3x⁶-6³+3-4(x³-1) )^1/3 + x*( x(2x²+2x+1) )^1/4
( 3x⁶-10x³+7 )^1/3 + x*( x(2x²+2x+x(x²-1)) )^1/4
( 3(x³)²-10x³+7 )^1/3 + x*( x²(2x+2+x²-1) )^1/4
( 3(x+1)²-10x³+7 )^1/3 + x*( x²(x²+2x+1) )^1/4
( 3x²+6x+3-10x³+7 )^1/3 + x*( x²(x²+1)² )^1/4
( -10x³+3x²+6+10 )^1/3 + x*( x(x²+1) )^1/2
( -x³-9x³+3x²+6+10 )^1/3 + x*( x(x²+x(x²-1) )^1/2
( -x³-9(x+1)+3x²+6+10 )^1/3 + x*( x²(1+x²-1) )^1/2
( -x³+3x²-3x+1 )^1/3 + x*( x²(x²) )^1/2
( 1-3x+3x²-x³ )^1/3 + x*( x⁴ )^1/2
( (1-x)³ )^1/3 + x*x²
1-x+x³
(x³-x-1)+2
2

PS1: btw do u know how much it took me to get this through algebra? 4 fucking hours

PS2: Troy, I posted correct answers to your question before anyone else. Why the silent ignore? You could have just said "I need details".

[TheEternalDretch]

Yaaaay! =D

PS1: btw do u know how much it took me to get this through algebra? 4 fucking hours

PS2: Troy, I posted correct answers to your question before anyone else. Why the silent ignore? You could have just said "I need details".

Sorry! o_o It took me a while too, but I thought it was fun...

And I hadn't noticed that you had reached the correct answer, since you wrote it in decimal form, so maybe Troy didn't either?

[/dev/humancontroller]

Yaaaay! =D

PS1: btw do u know how much it took me to get this through algebra? 4 fucking hours

Sorry! o_o It took me a while too, but I thought it was fun...

Yeah it was! (Did I sound angry? xD)
Could you plz name the source (book,etc.)?

I'll think of a challenge tomorrow. // ZzZ...

[/dev/humancontroller]

I am in Newton's 3D space.
I am at location P, and moving in direction&speed S.
My target is at location T, and moving in direction&speed V.
I am to shoot a projectile of a constant speed f in a direction D, so that it hits my target.
The projectile picks up my current speed, as in Newtonian physics.
Nothing is accelerating/decelerating.

Where should I shoot? (Create a formula to get D!)

[==Troy==]

I am in Newton's 3D space.
I am at location P, and moving in direction&speed S.
My target is at location T, and moving in direction&speed V.
I am to shoot a projectile of a constant speed f in a direction D, so that it hits my target.
The projectile picks up my current speed, as in Newtonian physics.
Nothing is accelerating/decelerating.

Where should I shoot? (Create a formula to get D!)

Can I just quote the particle prediction algorithm used for the trapper?

Working on 3D model, got the 2D equations, now its just a hassle of maths going from 2D to 3D

[TheEternalDretch]

Sorry! o_o It took me a while too, but I thought it was fun...

Yeah it was! (Did I sound angry? xD)

Well, your use of the word "fucking" gave me that idea.

Could you plz name the source (book,etc.)?

Yup - here it is. (TamTam is me)

Now for the new problem. Lessee....
Since you want to negate your current speed, and since you want to catch up to the second particle, I'll start with a speed and direction of (V-S). Using only this speed, the projectile would end up moving parallel to the target, matching its speed. So now the problem can be simplified to two static objects. The direction from one object to the other is (T-P), so we'll need to add some speed E in that direction. The magnitude of E doesn't matter terribly, as long as it's greater than zero.
So I'd say f = (V-S) + E*(T-P), and the direction is already included in there, since we're dealing with vectors.
I know you asked specifically for D, but can I get away with D = f/|f|? XD

[/dev/humancontroller]

f is a linear speed value, like 800m/s. The direction is unknown. Combining speed and direction, the raw trajectory would be f*D+S.

For example:
Alert! Enemy starship approaching P(5,4,3)km, S(1,2,3)km/s from T(11,-2,-3)km, V(3,2,4)km/s! We are shooting a linear rocket moving at f=20km/s. But uhm, where to? A moving target is hard to hit... or is it?

OK let's rename!
First, v := f. Second, we are relatively stationary, and are at the relative origin. The target is at location L (= T-P), and heading in direction&speed H (= V-S).

[TheEternalDretch]

Okay, now I have absolutely no idea what you're talking about. XD Wait, nevermind. I got it.

I still think you should fire the rocket at H + E*L, where E > 0. You want that in specifics?
H = (2,0,1), L = (6,-6,-6). V_final = (2+6E,-6E, 1-6E)
f = |V_final| = √((2+6E)² + (-6E)² + (1-6E)²) = √(4+24E+36E²+36E²+1-12E+36E²) = √(108E²+12E+5)
D = V_final/f = ( (2+6E)/f, -6E/f, (1-6E)/f )

Why do I feel like I'm making a silly mistake? 

[Nux]

NitPicking time!

The projectile picks up my current speed, as in Newtonian physics.
Nothing is accelerating/decelerating.

Kudos for launching a projectile without acceleration. (OMG a joke! Yes I know it's not important to the problem/accelerated yet)

For example:
Alert! Enemy starship approaching P(5,4,3)km, S(1,2,3)km/s from T(11,-2,-3)km, V(3,2,4)km/s! We are shooting a linear rocket moving at f=20km/s. But uhm, where to? A moving target is hard to hit... or is it?

Where are these values relative to? Surely we only need the velocity and position relative to us.

Also, why are we firing a rocket if we're in space and it has a constant velocity? Are we using Hollywood physics? I want to fire a hammer instead!

[TheEternalDretch]

Haha, Nux! XD But you know, in physics we use infinitesimal acceleration all the time....I've gotten used to it already. ^_^

And who said we're in space, huh? What if we're in starships in a non-vacuum, and our rocket is fired with continuing acceleration that exactly negates the friction?

Addition to my solution - the projectile will take 1/E seconds to reach the target.

[Nux]

And who said we're in space, huh?

Alert! Enemy starship approaching...

Perhaps the starship is approaching on wheels? Maybe it's lost in deep-sea and we're an angry submarine?

[TheEternalDretch]

Many sci-fi starships can handle themselves just fine in an atmosphere, and even underwater. ^_^

[Nux]

I would -have -said -that -be-fore -you -if -it -weren-'t -for t-his -damn -min-us -key sub-tracting- -everything- -I -say ->

[TheEternalDretch]

Aww. *pat pat* Try shifting that evil minus into an underscore! Maybe that'll help.

Anyway, I want to know if I'm right so I can post another problem. x_x Come on, humancontroller! I don't have all day!

[/dev/humancontroller]

Well maybe I don't have all day too... to press the refresh webpage button every minute.

I'm asking for a formula, like D = bla(bla/sqrt(bla+bla)-bla)

Why is your answer/start_of_answer weird? well,
f*D = H + E*L
f*D = H + 1/t*L
t*f*D = H*t + L
L = H*t+L
H*t = 0
wtf, t = 0, or |H| = 0?


TH3 SYST3M: http://drop.io/7ovjgpj


We are not in water, not air, not is vacuum, but the world of pure Newtonian space. That's something that doesn't exist. Just like geometry. Humans invented, in their mind, both.

Troy, the blob algorithm is, well, OMFGWTFNOOBHAX?!

[TheEternalDretch]

.....except t*f*D != L. I still think I'm right.

*sigh*

D = (H + E*L)/√((H_x+EL_x)² + (H_y+EL_y)² + (H_z+EL_z)²)

I'm almost certain that's not what you want....but I can't find a problem with it, nor can I find a better way to write it. :| I phail.

And by the way, I was kidding about not having all day.... XD

[/dev/humancontroller]

OK I've slept on this one. Yeah you're right, it was just weird for me at a first glance. Still, I'm telling you: |D| = 1
We still don't know the value of E, but keep going



Here's an example:
Target: T(-2,4), V(7,6)
Me:     P(-3,-5), S(3,3)
Deltas: L(1,9), H(4,3)
rocket speed: v = 13

results of calculation:
direction: D(5/13,12/13) (|D| = 1)
time to hit: t = 1 // but this isn't the question
rocket's relative trajectory: v*D = (5,12) (|v*D| = 13 = v)
rocket's absolute trajectory: S+v*D = (8,15) (absolute speed = 17 !)

target's relative coordinates after t time: L + t*H = (5,12)
rocket's relative coordinates after t time: t*v*d = (5,12)
target's absolute coordinates after t time: T + t*V = (5,10)
rocket's absolute coordinates after t time: P + t*(S+v*D) = (5,10)
// bullseye in both cases

[TheEternalDretch]

My |D| = 1, thankyouverymuch.

Your example uses my formula with E = 1, and therefore t = 1/E = 1. If, for example, you had used E = 2, you would have gotten:

Rocket speed = f = √477
D = (6/√477,21/√477)
|D| = 1
t = 1/E = 1/2

target's relative coordinates after t time: L + t*H = (3,10.5)
rocket's relative coordinates after t time: t*f*d = (3,10.5)
target's absolute coordinates after t time: T + t*V = (1.5,7)
rocket's absolute coordinates after t time: P + t*(S+f*D) = (1.5,7)

Still bullseyeing. ^_^ Any E works, I'm telling ya.

[/dev/humancontroller]

But where do u get E from, and why do you say "you would have gotten f = √477", when f is a given value?

Listen. Here's the given data:
Dretch: T(5,-1), V(7,6)
Me: P(-13,-3), S(3,3)
Lucifer cannon's speed: v (aka. f) = 13
(and although the lucifer cannon doesn't pick up my current speed, and the luci's speed isn't actually "13", but it doesn't matter, it's only for the sake of example)

gimme D, so that i can hit that dretch

[Bullislander05]

But wait.  Why are you giving examples in 2-space?  Isn't this 3-space?

I attempted the problem on my own using vector analysis (Velocity is a vector quantity that can be broken up into 3 subsections, x, y, and z by using trigonometry and two angles.  Yes, two angles in 3-space.  One to determine how far above the ground and how far in the x-direction and another to determine how far off of the xy plane it is (z).)

I got extremely complicated formulas that did not effectively isolate theta and alpha (my two angles).  I had 3 equations with 3 unknowns, though.  Here's what I had, maybe someone else can extrapolate a bit more:

T = Target
R = projectile
V = Target velocity
S = projectile velocity
[] = anything within there is a subscript.
t = time
Target directions are given in the problem statement, so the only unknowns are time, and theta and alpha.

T

• = T[x[0]] + Vcos(theta[Target direction])t

T[y] = T[y[0]] + Vsin(theta[Target direction])t
T[z] = T[z[0]] + Vcos(alpha[Target direction])t

R

• = 0 + Scos(theta)t

R[y] = 0 + Ssin(theta)t
R[z] = 0 + Scos(alpha)t

In my analysis, since there is no acceleration, basic kinematic equations can be used.  The equation I used is x[t] = x[0] + vt + 1/2at^2.  a = 0 so it's simply x[t] = x[0] + vt.  Since V is a velocity magnitude, in order to find the direction in a certain axis direction basic trig must be used to find exactly how much of V works in that direction.

Also, I "moved" my axis system so that the origin was located at where the projectile was fired.  It's a simple calculation, you only subtract the particles initial position from the targets initial position and set the particle at the origin.  It simplifies the algebra a bit, yet does not yield a fully simplified answer using the tools that I currently know.

So, after that little note you can set those six equations equal (In their respective axes), solve for t, substitute, and you will have 2 equations with only theta and alpha as unknowns.  It's only complicated because simplifying trig functions isn't as easy as factoring out common factors in algebraic functions.

[/dev/humancontroller]

But wait.  Why are you giving examples in 2-space?  Isn't this 3-space?

An example is just an example. The perfect solution would be for N dimensions, not just 2 or 3.

I attempted the problem on my own using vector analysis (Velocity is a vector quantity that can be broken up into 3 subsections, x, y, and z by using trigonometry and two angles.  Yes, two angles in 3-space.  One to determine how far above the ground and how far in the x-direction and another to determine how far off of the xy plane it is (z).)

I got extremely complicated formulas that did not effectively isolate theta and alpha (my two angles).  I had 3 equations with 3 unknowns, though.  Here's what I had, maybe someone else can extrapolate a bit more:

T = Target
R = projectile
V = Target velocity
S = projectile velocity
[] = anything within there is a subscript.
t = time
Target directions are given in the problem statement, so the only unknowns are time, and theta and alpha.

I would say "not only the time, but also the angles are unknown".

T

• = T[x[0]] + Vcos(theta[Target direction])t

T[y] = T[y[0]] + Vsin(theta[Target direction])t
T[z] = T[z[0]] + Vcos(alpha[Target direction])t

R

• = 0 + Scos(theta)t

R[y] = 0 + Ssin(theta)t
R[z] = 0 + Scos(alpha)t

In my analysis, since there is no acceleration, basic kinematic equations can be used.  The equation I used is x[t] = x[0] + vt + 1/2at^2.  a = 0 so it's simply x[t] = x[0] + vt.  Since V is a velocity magnitude, in order to find the direction in a certain axis direction basic trig must be used to find exactly how much of V works in that direction.

Also, I "moved" my axis system so that the origin was located at where the projectile was fired.  It's a simple calculation, you only subtract the particles initial position from the targets initial position and set the particle at the origin.  It simplifies the algebra a bit, yet does not yield a fully simplified answer using the tools that I currently know.

So, after that little note you can set those six equations equal (In their respective axes), solve for t, substitute, and you will have 2 equations with only theta and alpha as unknowns.  It's only complicated because simplifying trig functions isn't as easy as factoring out common factors in algebraic functions.

Here's a reference manual to help you with simplifications:
http://www.alcyone.com/max/reference/maths/trigonometry.html

If you can do this with trigonometry, then that's fine by me.
But for a general, N-dimensional solution, you would have to use vectors.

[TheEternalDretch]

But where do u get E from, and why do you say "you would have gotten f = √477", when f is a given value?

Ohhhhh. Now I understand why we weren't understanding each other - I thought I was supposed to find f as well as D.

Well then. Since |H + E*L| = f, we can find E:
√((H_x+EL_x)² + (H_y+EL_y)² + (H_z+EL_z)²) = f
H_x² + 2EH_xL_x + E²L_x² + H_y² + 2EH_yL_y + E²L_y² + H_z² + 2EH_zL_z + E²L_z² = f²
E²(L_x² + L_y² + L_z²) + E(2H_xL_x + 2H_yL_y + 2H_zL_z) + H_x² + H_y² + H_z² - f² = 0
Slightly simplified, E²*|L|² + E(2H_xL_x + 2H_yL_y + 2H_zL_z) + |H|² - f² = 0

E = -(2H_xL_x + 2H_yL_y + 2H_zL_z) + √( (2H_xL_x + 2H_yL_y + 2H_zL_z)² - 4*|L|²*|H|² ) * 1/(2|L|²)

And once you have E, D = (H + E*L)/√((H_x+EL_x)² + (H_y+EL_y)² + (H_z+EL_z)²)

Is that too complicated? 

[Bullislander05]

But wait.  Why are you giving examples in 2-space?  Isn't this 3-space?

An example is just an example. The perfect solution would be for N dimensions, not just 2 or 3.

I attempted the problem on my own using vector analysis (Velocity is a vector quantity that can be broken up into 3 subsections, x, y, and z by using trigonometry and two angles.  Yes, two angles in 3-space.  One to determine how far above the ground and how far in the x-direction and another to determine how far off of the xy plane it is (z).)

I got extremely complicated formulas that did not effectively isolate theta and alpha (my two angles).  I had 3 equations with 3 unknowns, though.  Here's what I had, maybe someone else can extrapolate a bit more:

T = Target
R = projectile
V = Target velocity
S = projectile velocity
[] = anything within there is a subscript.
t = time
Target directions are given in the problem statement, so the only unknowns are time, and theta and alpha.

I would say "not only the time, but also the angles are unknown".

T

• = T[x[0]] + Vcos(theta[Target direction])t

T[y] = T[y[0]] + Vsin(theta[Target direction])t
T[z] = T[z[0]] + Vcos(alpha[Target direction])t

R

• = 0 + Scos(theta)t

R[y] = 0 + Ssin(theta)t
R[z] = 0 + Scos(alpha)t

In my analysis, since there is no acceleration, basic kinematic equations can be used.  The equation I used is x[t] = x[0] + vt + 1/2at^2.  a = 0 so it's simply x[t] = x[0] + vt.  Since V is a velocity magnitude, in order to find the direction in a certain axis direction basic trig must be used to find exactly how much of V works in that direction.

Also, I "moved" my axis system so that the origin was located at where the projectile was fired.  It's a simple calculation, you only subtract the particles initial position from the targets initial position and set the particle at the origin.  It simplifies the algebra a bit, yet does not yield a fully simplified answer using the tools that I currently know.

So, after that little note you can set those six equations equal (In their respective axes), solve for t, substitute, and you will have 2 equations with only theta and alpha as unknowns.  It's only complicated because simplifying trig functions isn't as easy as factoring out common factors in algebraic functions.

Here's a reference manual to help you with simplifications:
http://www.alcyone.com/max/reference/maths/trigonometry.html

If you can do this with trigonometry, then that's fine by me.
But for a general, N-dimensional solution, you would have to use vectors.

=\.  None of the formulas on that site you gave me simplify my trig.  Oh well.  I've given the problem enough thought for now.  Unless I come across a new idea in the future, I'll most likely just wait for someone to solve the problem.

But, just to clarify myself, I am using vectors in my analysis.  If I truly wanted, I could express position and velocity both in terms of vectors and do some analysis that way, but I doubt I'll get too far by doing it, so I'm not going to even bother.  Kudos to making a fantastic problem.

-Bull

[/dev/humancontroller]

Well then. Since |H + E*L| = f, we can find E:
√((H_x+EL_x)² + (H_y+EL_y)² + (H_z+EL_z)²) = f
H_x² + 2EH_xL_x + E²L_x² + H_y² + 2EH_yL_y + E²L_y² + H_z² + 2EH_zL_z + E²L_z² = f²
E²(L_x² + L_y² + L_z²) + E(2H_xL_x + 2H_yL_y + 2H_zL_z) + H_x² + H_y² + H_z² - f² = 0
Slightly simplified, E²*|L|² + E(2H_xL_x + 2H_yL_y + 2H_zL_z) + |H|² - f² = 0

E = -(2H_xL_x + 2H_yL_y + 2H_zL_z) + √( (2H_xL_x + 2H_yL_y + 2H_zL_z)² - 4*|L|²*|H|² ) * 1/(2|L|²)

And once you have E, D = (H + E*L)/√((H_x+EL_x)² + (H_y+EL_y)² + (H_z+EL_z)²)

almost Correct! You forgot f out of the equation

To show what I mean about a general vector formula, I'll just rewrite this:
E = (-HoL +- √(HoL² + (LoL)(f² - HoH))) / (LoL)   // o signifies a dot product
D = (H+E*L)/f   // if E > 0

OK, Eternal, your turn.

[TheEternalDretch]

*cough cough* Yeah, I fail. I forgot the f and I missed a much simpler way of writing the answer. >_< But I'm out of practice! Haven't done any math in a while.

Mmm, my turn.....I don't know. Oh, y'know what? I'd like to see how you guys would solve this:

a_n+1 = a_n + 0.5n⁴ - 40n³ + n - 1
a₁ = 1
a₁₀₁ = ?

I never learned a formal way to solve this, though the solution is simple enough. I'm just wondering how widespread it is.

[/dev/humancontroller]

an instant c program shows 5071616, now let's see if i can do this analytically xD

[TheEternalDretch]

Yep, that's right. If nobody succeeds in doing it analytically, you take the next one.