TVTropes Now available in the app store!
Open

Follow TV Tropes

Following

Let's make a homework help thread!

Go To

randomtropeloser Since: Jan, 2001
#576: Aug 22nd 2010 at 9:23:12 PM

You know, I've been considering making a follow-up thread since school is almost back in and this one seems to be getting a bit cluttered. I wouldn't want to make two of the same thread though, and this one still pops up on the front page sometimes.

Madrugada Since: Jan, 2001
#577: Aug 22nd 2010 at 10:17:35 PM

No need to make a new one. This one is working fine.

Tzetze DUMB from a converted church in Venice, Italy Since: Jan, 2001
DUMB
#578: Aug 26th 2010 at 9:20:14 AM

This isn't homework, but it's complicated and I'm completely stumped.

It has to do with signals. There's a system with three nodes: xj(n), x'j(n), and yk(n). xj(n) has no input and outputs to x'j(n). x'j(n) gets input from xj(n) and yk(n), and outputs to yk(n). yk(n) gets input from x'j(n) and outputs to x'j(n). A simple feedback loop.

The connection from x'j(n) to yk(n) is considered as an operator, A. That is, yk(n) = A[x'j(n)]. The connection from yk(n) back to x'j(n) is similarly considered B.

Here's the bit I don't get. From the conception above:

  1. yk(n) = A[x'j(n)]
  2. x'j(n) = xj(n) + B[yk(n)]

According to the thing I'm reading, from this it can be found that yk(n) = (A/(1-AB))[xj(n)].

Now this makes sense if A and B are considered as numbers, but they're described as "operators". What am I missing here?

[1] This facsimile operated in part by synAC.
Hydrall Since: Jun, 2009
#579: Aug 26th 2010 at 7:08:42 PM

^... Now I feel dumb in comparison. sad

Hey, how does one solve an equation with variables (the same variable) in both the sides exponents?

edited 26th Aug '10 7:08:48 PM by Hydrall

Tzetze DUMB from a converted church in Venice, Italy Since: Jan, 2001
Hydrall Since: Jun, 2009
#581: Aug 26th 2010 at 7:10:53 PM

Well, I still have to do it on my own, but...

2^(3x+2)=4^(2x)

I don't need the answer, just an idea of how to get to it.

Tzetze DUMB from a converted church in Venice, Italy Since: Jan, 2001
DUMB
#582: Aug 26th 2010 at 7:11:47 PM

Oh, those problems are a bitch. First you have to make sure that the bases are the same number  hint

, and then you take the logarithm of both sides.

[1] This facsimile operated in part by synAC.
Taelor Don't Forget To Smile from The Paths of Spite Since: Jul, 2009
Don't Forget To Smile
#583: Aug 26th 2010 at 11:05:00 PM

edited 26th Aug '10 11:06:07 PM by Taelor

The Philosopher-King Paradox
DaeBrayk PI Since: Aug, 2009
PI
#584: Aug 29th 2010 at 12:27:30 PM

So, this is kind of lame following up all that crazy calculus shit, but I have to make graphs in Excel, and the instructions are for a newer version than I have. Most things translate easily enough, but does anyone know how to get the equation shown with the graph in the 2002 version?

Nvm, found it. =) It's an option when you add a trendline.

edited 29th Aug '10 12:33:18 PM by DaeBrayk

TsundeRay HOORAY! from Santa Clara, California Since: May, 2009
HOORAY!
#585: Aug 31st 2010 at 12:30:17 AM

Anyone have some good articles on "Cogito ergo sum"? I looked up stuff on Google and all I got was high-brow stuff that I have trouble understanding.

http://twitter.com/raydere | http://raydere.tumblr.com
Taelor Don't Forget To Smile from The Paths of Spite Since: Jul, 2009
Don't Forget To Smile
#586: Aug 31st 2010 at 8:17:25 AM

Basically, Descartes was reacting to the general climate of philosophical skepticism that was brewing in France by attempting to find at least one thing that is absolutely certain; he went about doing this by systematically doubting everything until he found something that could not be doubted; what he eventually settled on was that his own existence: according to Descartes, there is no way for him to doubt his own existence, because that would require him to be able to think, which in tun requires that he exists (this strategy of combating doubt by claiming that the doubt itself presupposes the doubted thing is called a transcendental argument, btw) . Thus, he thinks, therefore he knows that he exists. He later went on the use some truly idiotic reasoning to prove that God and everything else exist too, but that is tangential to the Cogito Ergo Sum thing. What's ionic about the whole thing is that because of this, he's often misunderstood to have been a solipsist, when really, that's the sort of thing he was trying to discredit. Don't know any good articles about it, though. Hope that helps.

edited 31st Aug '10 8:25:16 AM by Taelor

The Philosopher-King Paradox
Mapi "keionbu ni yokusou, nyan?" from Sakurakou Keionbu Since: Aug, 2011
"keionbu ni yokusou, nyan?"
#587: Aug 31st 2010 at 8:21:12 AM

Tzetze: Are these logical operators? (AND, OR, NOR, NAND, XOR, XNOR, etc...)

I'm not sure I can help though...

My FF.net account
Tzetze DUMB from a converted church in Venice, Italy Since: Jan, 2001
DUMB
#588: Aug 31st 2010 at 8:24:52 AM

No, they're multiplication operators. Or linear operators. Or something.

[1] This facsimile operated in part by synAC.
Deboss I see the Awesomeness. from Awesomeville Texas Since: Aug, 2009
I see the Awesomeness.
#589: Aug 31st 2010 at 11:14:57 AM

We called those logical operators.

Fight smart, not fair.
melloncollie Since: Feb, 2012
#590: Sep 4th 2010 at 12:30:06 PM

Does anyone know what happens when you do a definite integral using integration by parts? I've forgotten.

Carciofus Is that cake frosting? from Alpha Tucanae I Since: May, 2010
Is that cake frosting?
#591: Sep 4th 2010 at 3:10:28 PM

According to the thing I'm reading, from this it can be found that yk(n) = (A/(1-AB))[xj(n)].

Now this makes sense if A and B are considered as numbers, but they're described as "operators". What am I missing here?

Unless I am mistaken, they are assuming that A and B are linear operators (or matrices, if you will) and that 1-AB is invertible, where 1 is the identity operator over the range of x^'j(n) and AB is the operator defined by AB(x) = A(B(x)).

Then you can work almost as if A and B were numbers: in brief  *

, since

y = A(x') = A(x + B(y)) = A(x) + AB(y)

it holds that

(1 - AB)(y) = y - AB(y) = A(x)

and from this and the invertibility of 1-AB it follows that

y = (A/(1-AB))(x).

However, the notation A/(1-AB) is kind of ambiguous here, since this "product" is not commutative: the intended meaning is the inverse of (1-AB) applied to A, that is,

(A/(1-AB))(x) = (1-AB)^{-1}(A(x))

and not the other way around.

If I can handwave things a little, when you calculate AB(y) you "assume that x=0", compute the value of x' corresponding to y under this hypothesis, and then use this new value of x' to compute the new value of y - if x were 0 we would have that y = AB(y), of course; more in general, you assume that the "control dial" x is set to 0 and you go through the feedback loop once more under this hypothesis.

Hence, (1-AB)(y) can be seen as the difference between the current value of y and the one that it would result after going through the loop once more if it held that x=0.

Furthermore, A(x) can be interpreted as follows: you take the value of x (the setting of the "control dial", so to say), assume that x' = x (that is, that no feedback is coming from y to x', for example because the system has just now been turned on) and you compute the corresponding value for y.

Thus, the formula

(1-AB)y = Ax

has a rather natural meaning: the difference between the value of the output y and the one that it would take after one cycle of feedback if the dial were set to 0 is precisely the value that the output y would take if no feedback existed from y to x' .

But in general, (1-AB) is not necessarily invertible, I think, unless there is some further hypothesis of the problem...

There are ways of finding approximate inverses of a non-invertible operator if need be, of course, but I am not sure if I remember enough of them to really go into it now (also, that's not really related directly enough with the problem at hand).

Hope this helps!

edited 4th Sep '10 3:19:04 PM by Carciofus

But they seem to know where they are going, the ones who walk away from Omelas.
Taelor Don't Forget To Smile from The Paths of Spite Since: Jul, 2009
Don't Forget To Smile
#592: Sep 5th 2010 at 2:46:27 PM

Does anyone know what happens when you do a definite integral using integration by parts? I've forgotten.
This video may be helpful.

The Philosopher-King Paradox
ColorPrinter Since: Dec, 2011
#593: Sep 12th 2010 at 12:27:20 PM

ARG, this problem has been driving me mad for HOURS.

n^3 = n(n-1)(n-2) + 3n(n-1) + n

We're supposed to do a combinatorial proof. Only problem is I have no earthly idea what in the world the right side is supposed to be.

Zizoz Since: Feb, 2010
#594: Sep 12th 2010 at 4:13:29 PM

If I understand what a combinatorial proof is, n(n-1)(n-2) would be the number of ways to choose an order of three objects out of n objects. (I didn't word that very well...)

ColorPrinter Since: Dec, 2011
#595: Sep 12th 2010 at 4:16:04 PM

I just figured it out, thanks to my textbook. It's the number of lists with 3 elements you can make.

The first part is where they're all different, and "n" is where they're all the same.

The second part must be where two are the same, but I have to figure out how that works.

I should really read the "suggested textbook problems" section on my syllabus.

Hydrall Since: Jun, 2009
#596: Sep 23rd 2010 at 5:02:20 PM

I need help. I need to find stuff about the genre Magic Realism and how it relates to the book we've been reading, but so far every single source I've visited has either given a different definition or no definition at all. What is Magic Realism?

Taelor Don't Forget To Smile from The Paths of Spite Since: Jul, 2009
Don't Forget To Smile
#597: Sep 23rd 2010 at 5:19:09 PM

Magical realism is a literary genre in which supernatural things happen in our world, but are not really explained. If a story provides a framework for why supernatural things happen (even a Fridge Logicy one), then it isn't Magical Realism. If it takes place in a world where such things are normal, then it is also not Magical Realism. Or at least, that's how it was explained to me when I was in high school, like, 4-5 years ago. I may be misremembering part or all of it.

The Philosopher-King Paradox
melloncollie Since: Feb, 2012
#598: Sep 25th 2010 at 4:04:44 PM


This post was thumped by the Shillelagh of Whackingness

Deboss I see the Awesomeness. from Awesomeville Texas Since: Aug, 2009
I see the Awesomeness.
#599: Sep 25th 2010 at 4:21:37 PM

en-1/en-> en*(en-1/en)

edited 25th Sep '10 4:22:09 PM by Deboss

Fight smart, not fair.
JuiceBoxHero from the butthole of Texas Since: Aug, 2009
#600: Oct 5th 2010 at 9:08:02 AM

Can I necro this?

If any of you tropers are from a country where gays serve openly in the military, how is that received by the general public and how comfortable are military personnel with it?

/needs to finish this paper on her country's embarrassingly quaint military laws.


Total posts: 1,508
Top