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itxaka · Dec 12, 2010

poweld said:
Because the function is highly sensitive to its initial conditions, we can't reliably determine where the ball will fall. For example, the minute scoring on the ball may affect the way it rolls slightly, which may vastly change the outcome.

Thanks, I thought that it was almost impossible to calculate it but I wasn't sure.

Le-mo · Dec 14, 2010

I'm taking precalculus right now and I can not remember how to do this problem.

x^6 + 7x^3-8=0

Find all the real solutions

Einherjar · Dec 14, 2010

Le-mo said:
I'm taking precalculus right now and I can not remember how to do this problem.

x^6 + 7x^3-8=0

Find all the real solutions

It's been a long time since I've done math, but isn't it just:

let y = x^3

y^2 + 7y - 8 = 0
Quadratic equation
solve for x?

Lonely1 · Dec 14, 2010

Yes, you are right Einherjar.

Lagspike_exe · Dec 16, 2010

I have an issue with this problem:

Define "x" so the numbers log2, log (2^x -1), log (2^x +3) are consecutive.

I know that log (2^x -1) = log2 +d ; log (2^x +3) = log2 + 2d

... but no idea what to do next. Help, GAF?

Lagspike_exe · Dec 16, 2010

Lagspike_exe said:
I have an issue with this problem:

Define "x" so the numbers log2, log (2^x -1), log (2^x +3) are consecutive.

I know that log (2^x -1) = log2 +d ; log (2^x +3) = log2 + 2d

... but no idea what to do next. Help, GAF?

Anyone?

RedShift · Dec 16, 2010

Lagspike_exe said:
Anyone?

Only had a quick go but use the exponential function on each equation you have with d in? Then you end up with simultaneous equations with e^d and 2^x.

Lagspike_exe · Dec 16, 2010

RedShift said:
Only had a quick go but use the exponential function on each equation you have with d in? Then you end up with simultaneous equations with e^d and 2^x.

The issue here is that I've learned log last year and I've completely forgoten how to use them. :lol

-COOLIO- · Dec 27, 2010

math is SOOOOO much easier to follow when youre hot for teacher:

http://www.youtube.com/watch?v=qFoa3jbVG3U

dinazimmerman · Jan 3, 2011

If I divide the D into two sets, one where x_1>=0 and the other where x_1<0, then I can easily find two orders of integration (integrate first with respect to x_3, then x_2 or x_1, then which ever variable is left).

However, I can't seem to identify a third order of integration. I need to figure out how to integrate first with respect to x_1 or x_2 and then x_3 second, but this is hard since the limits of x_3 depend on both x_1 and x_2, since -sqrt(5 - x_1 - x_2) < x_3 < sqrt(5 - x_1 - x_2).

dinazimmerman · Jan 4, 2011

EDIT: nvm figured out the problem myself

Meatpuppet · Jan 4, 2011

Can anyone help with some differential equations?

I have data along two straight lines, which are oriented such that one is parallel to the x axis, and the other is at omega degrees to the y axis (lets call the quantity u = u(x,y)). I know u(x,y) along each line, and can calculate u_x(x,y0), u_y(x0,y0), u_xx(x,y0), and the combined quantity [sin2(omega)u_xx(x0,y0) + 2sin(omega)cos(omega)u_xy(x0,y0) + cos2(omega)u_yy(x0,y0)]

I am trying to derive either u_xy(x0,y0) or u_yy(x0,y0) from the data I have. I'm not sure if this is possible. I can set up a PDE, which from responses in this thread can be rephrased as an ODE and solved accordingly... however I'm a bit lost trying to follow the explanations. Its been a long time since I played with ODEs and PDEs etc

http://www.physicsforums.com/showthread.php?t=460600

Ducarmel · Jan 4, 2011

I'm tutoring this high school kid in geometry anybody has a good internet resource with good high school level geometry examples he can look at on his own. I am aware of Khan Academy but their geometry section is kind of poor.

Feep · Jan 13, 2011

Why do I feel like both A and C are true? (This is a Data Structures and Algorithm question)

Suppose T1

= O(g

) and also T2

= O(g

). Which of the following are true?

A) T1

+ T2

= O(g

)
B) T1

- T2

= o(g

)
C) T1

/T2

= O(1)
D) T1

= O(T2

)

-COOLIO- · Jan 13, 2011

ive just started linear algebra and OOOHH MYY GOOOOD IMM INTIMIDATED

-COOLIO- · Jan 13, 2011

Feep said:
Why do I feel like both A and C are true? (This is a Data Structures and Algorithm question)

Suppose T1 = O(g) and also T2 = O(g). Which of the following are true?

A) T1 + T2 = O(g)
B) T1 - T2 = o(g)
C) T1/T2 = O(1)
D) T1 = O(T2)

i would think that b is definitely true. the Ts cant be negative right?

Feep · Jan 13, 2011

-COOLIO- said:
i would think that b and c are definitely true

A is definitely true.

Keep in mind, Part B is talking about Little O notation, not Big O. It is my understanding that f

= 2n is NOT o

, so if T1

= 3n and T2

= n, T1

- T2

would be 2n and thus not o

.

-COOLIO- · Jan 13, 2011

Feep said:
A is definitely true.

Keep in mind, Part B is talking about Little O notation, not Big O. It is my understanding that f = 2n is NOT o, so if T1 = 3n and T2 = n, T1 - T2 would be 2n and thus not o.

oh whoops.

ok, ya just A then. at first i thought i wouldnt be a if one T grew just as fast as g, but then i remembered that doesnt matter

Feep · Jan 13, 2011

-COOLIO- said:
oh whoops.

ok, ya just A then. at first i thought i wouldnt be a if one T grew just as fast as g, but then i remembered that doesnt matter

What about C? Isn't that still true?

-COOLIO- · Jan 13, 2011

Feep said:
What about C? Isn't that still true?

i dont think so because t1 could be like n^2, t2 could be like n^-2, and then t1/t2 would be n^4 which would grow faster than n^3 which could have been g

but then aaaggain, i dont think n^-2 would be allowed as a T

as a T it has to grow, so any T1 divided by another T2 should shrink T1

so actually i would say that c has to work.

but wait, this is all in O(1), and i was thinking it was in o of g, so once again i misinterpreted a question. : [

listen to the guy below me.

it's just a)

Rich Uncle Skeleton · Jan 13, 2011

Feep said:
A is definitely true.

Keep in mind, Part B is talking about Little O notation, not Big O. It is my understanding that f = 2n is NOT o, so if T1 = 3n and T2 = n, T1 - T2 would be 2n and thus not o.

Yeah, I agree B false.

C is also false. We know T2

is O(g

), but it could be o(g

). That is, it could grow much slower than g. For example, take g

= n^2, T1

= n^2, and T2

= n. Then T1 and T2 are both O(g

), but T1/T2 = n, which is not bounded.

D is false for basically the same reason.

Feep · Jan 13, 2011

Rich Uncle Skeleton said:
Yeah, I agree B false.

C is also false. We know T2 is O(g), but it could be o(g). That is, it could grow much slower than g. For example, take g = n^2, T1 = n^2, and T2 = n. Then T1 and T2 are both O(g), but T1/T2 = n, which is not bounded.

D is false for basically the same reason.

Oh, damn, I forgot, only Omega notation implies an exact match. In other words, f

= n is technically O(n^4), right?

-COOLIO- · Jan 13, 2011

Feep said:
Oh, damn, I forgot, only Omega notation implies an exact match. In other words, f = n is technically O(n^4), right?

ya

Rich Uncle Skeleton · Jan 13, 2011

Feep said:
Oh, damn, I forgot, only Omega notation implies an exact match. In other words, f = n is technically O(n^4), right?

Yes, f

= n is O(n^4). I wasn't familiar with omega notation, but according to wikipedia it doesn't mean exact match, but rather it's just an analog of big-O with a ≥ rather than a ≤. But if f is both O(g) and Omega(g), then it's an exact match, in the sense that for large enough n, mg

≤ f

≤ Mg

, so that C would be true.

-COOLIO- · Jan 13, 2011

ifffff i recall

f in O(g) implies that f grows at a smaller or the same rate as g

f in theta(g) implies that f grows at the same rate as g

f in omega(g) implies that f grows faster than or at the same rate as g

Feep · Jan 13, 2011

-COOLIO- said:
ifffff i recall

f in O(g) implies that f grows at a smaller or the same rate as g

f in theta(g) implies that f grows at the same rate as g

f in omega(g) implies that f grows faster than or at the same rate as g

Yeah I meant Theta. Good call.

Gouty · Jan 15, 2011

Write an equation of the line satisfying the given conditions. Write the answer in slope-intercept form. The line passes through (9, -5) and (9, 2).

I got y = -5 but I have no idea what to do when you end up with a slope of 0.

hemtae · Jan 15, 2011

Gouty said:
Write an equation of the line satisfying the given conditions. Write the answer in slope-intercept form. The line passes through (9, -5) and (9, 2).

I got y = -5 but I have no idea what to do when you end up with a slope of 0.

The slope isn't zero

Try visualizing a line that goes through those points

x = 9

Gouty · Jan 15, 2011

I thought the formula to find the slope is

Gouty · Jan 15, 2011

fuck this shit, my teacher is awful.

-COOLIO- · Jan 15, 2011

Gouty said:
I thought the formula to find the slope is

it is

so the slope is 7/0. which is undefined, so the line is a straight, non angled, vertical line.

TL4E · Jan 15, 2011

Gouty said:
I thought the formula to find the slope is

The slope of a vertical line (x=9) doesn't exist.
That formula is correct.

Gouty · Jan 15, 2011

ok so I wind up with 7/0 for my slope, but how does that work within y - y1 = m (x - x1)? I can't plug that number in for M so where do I go?

Gouty · Jan 15, 2011

This is why I hate this fucking teacher. Every single problem she assigns never works out, you constantly wind up with weird ass shit that I have no idea what to do with. For instance, I'm working on a similar problem and just got a slope of negative zero over eleven. WTF am I suppose to do with that? Can't she come up with at least one fucking problem where the slope is 6 or half or something that I can work with? lksdzv
lksdfglkfg
olbfg

FUCK

Kalnos · Jan 15, 2011

Gouty said:
This is why I hate this fucking teacher. Every single problem she assigns never works out, you constantly wind up with weird ass shit that I have no idea what to do with. For instance, I'm working on a similar problem and just got a slope of negative zero over eleven. WTF am I suppose to do with that? Can't she come up with at least one fucking problem where the slope is 6 or half or something that I can work with? lksdzv
lksdfglkfg
olbfg

FUCK

0 over anything is 0.

TL4E · Jan 15, 2011

Gouty said:
ok so I wind up with 7/0 for my slope, but how does that work within y - y1 = m (x - x1)? I can't plug that number in for M so where do I go?

That's point-slope form, not slope-intercept form (as you said in your last post).

I'm not sure what your question is. You can't put the line x=9 into slope intercept form (y=mx+b) because there's no slope and no

intercept.

Gouty · Jan 15, 2011

I already posted the question.

dinazimmerman · Jan 19, 2011

Kalnos said:
0 over anything is 0.

0/0 is indeterminate

zoku88 · Jan 19, 2011

Since it's a vertical line, that means the equation for the line is "x=?"

This is because y can be anything but x can only be one number.

You just have to find out what ? is, which you can find by looking at the points you were given.

EDIT: Oh, someone already gave you the answer. Yea, slope intercept form is impossible.

Kalnos said:
0 over anything is 0.

Oh god.....

poweld · Jan 19, 2011

zoku88 said:
Oh god.....

Oh c'mon, he was right for ∞-1 cases.

malsumis · Jan 19, 2011

poweld said:
Oh c'mon, he was right for ∞-1 cases.

well, technically ∞-1 is still ∞, and technically he was right for an ∞ cases

poweld · Jan 19, 2011

malsumis said:
well, technically ∞-1 is still ∞, and technically he was right for an ∞ cases

Ass.

You're so right

close to the edge · Jan 21, 2011

I'm trying to do this problem, but I have no idea how to do it:

Find the arc length of the curve c:

Is this something you need a contour integral for?

edit: Okay, no contour integral from what I've gathered, just a curve length (length = integral(c'(t)^2) dt where c(t) is a parametrized curve). So I differentiate both parts of the vector and plug them into the formula, which, according to Wolfram Alpha, results in:

I don't get the simplifying the integrand part. How the hell does sin^2(2t) become sqrt(5)?

Mudkips · Jan 21, 2011

Goya said:
0/0 is indeterminate

Goya
Incurious Bastardv

Mecha_Infantry · Jan 22, 2011

Hey y'all...DOing some cryptography based stuff and I'm trying to work out the easiest way of knowing the total numbers in an n-bit number

FOr example I think I read it goes like this:

32bit number = 32x31x30....x2x1= x

Also is this called factorising? Cheers

zoku88 · Jan 22, 2011

wait. Are you trying to find the max number n bits can store? Or the amount of numbers an n bit sequence can possibly store.

The answer to the first is 2^n - 1. The second is 2^n.

What you were doing is a factorial (n!) Very different.

close to the edge said:
I don't get the simplifying the integrand part. How the hell does sin^2(2t) become sqrt(5)?

It doesn't. They applied the double angle formula and factored.

You should try to re-memorize these http://en.wikipedia.org/wiki/List_of_trigonometric_identities

Please respond if you need to see how they applied it and factored the root(5) out.

peakish · Jan 22, 2011

close to the edge said:
I'm trying to do this problem, but I have no idea how to do it:

Find the arc length of the curve c:

Is this something you need a contour integral for?

Wolfram seem to do it a bit harder than necessary. You have a line s(t) (or, rather, s(x(t),y(t)). An easy way to find out how it changes with x and t is to use the Pythagorean theorem, (ds/dt)^2 = (dx/dt)^2 + (dy/dt)^2. So, ds = dt*sqrt( (dx/dt)^2 + (dy/dt)^2 ). Then, integrating over all t:

s = int_0^pi/2 dt*sqrt( (dx/dt)^2 + (dy/dt)^2 )

You'll need to use some trigonometric properties on the way.

f0rk · Jan 22, 2011

close to the edge said:
I don't get the simplifying the integrand part. How the hell does sin^2(2t) become sqrt(5)?

Try using sin(2t) = 2cos(t)sin(t).

Mecha_Infantry · Jan 22, 2011

zoku88 said:
wait. Are you trying to find the max number n bits can store? Or the amount of numbers an n bit sequence can possibly store.

The answer to the first is 2^n - 1. The second is 2^n.

What you were doing is a factorial (n!) Very different.

I'm trying to figure this out:

Note that TCP ISNs are 32-bit numbers. This makes for 4,294,967,296 possibilities for an ISN.

Thanks

Mecha_Infantry · Jan 22, 2011

Man, maths is hard

Learning modular arithmetic makes me feel incredibly dense!

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