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#201 Lexaeus

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Posted 19 February 2013 - 06:14 PM

I'd love to. Should I just start with polynomials or will you help me narrow this stuff down?


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#202 Ralor

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Posted 19 February 2013 - 06:35 PM

nvm sorry already got help


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#203 Lexaeus

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Posted 19 February 2013 - 07:36 PM

It was my pleasure.


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#204 Secret Felix

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Posted 25 February 2013 - 11:59 AM

So I'm working on my ODE practice test, and jeez, I have no idea what I'm doing.

 

Solve y'''(t) + 3y''(t) + 3y'(t) + y(t) = e-t ; y(0)=1 , y'(0)=0, y''(0)=0

 

D3 +3D2 + 3D + 1 = e-t

 

(D + 1)(D2 + 2D + 1) = e-t

 

(D + 1)3 = e-t

 

Now I need to make it homogeneous so I multiply both sides by the annihilator (D + 1) to get:

 

(D + 1)4 = 0

 

And this is as far as I get. I tried solving for all my cn's and I just got 0. (To be more specific I found c1=1, c2=1, c3=1/2. Which when subbed back into the original problem seemingly give y'''(t) + 3y''(t) + 3y'(t) + y(t) = 0. So yea.) Halp pls.


Edited by Artemis, 12 March 2013 - 10:52 PM.

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#205 Secret Igshar

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Posted 27 March 2013 - 09:37 AM

A question from Waker that I'll answer here for everybody.

 

There was a guy on 4chan asking for a function that would be 0 for negative numbers and x for positives. Someone gave the function:

 

[sqrt(x^2) + x]/2

 

Which nobody argued. Note that it is the POSITIVE root, only. If you look at both branches of the function, you'll find that the negative branch reverses the whole thing, putting the linear part (still positive slope of 1) to the left and the flatline to the right.

 

Someone later asked for a function that would be 0 for negative numbers and 0 to 5 and x for > 5. Someone suggested:

 

[sqrt((x-5)^2) + x-5]/2

 

And Waker then was wondering why this one works, too. There's a few simple answers to this. He was saying that it seems like it would always either be x-5 or 0, and, well, it is. If you use those same algebraic techniques on the other one, you'll end up with x or 0. So it's really no different. The ONLY difference between x and x-5 is that x-5 intersects the x-axis at x=5 instead of x=0. Note that how this was algebraically worked out to give x-5 or 0 is incorrectly handling the function.

 

They just look at the square root and are like:

if positive root: x-5

if negative root: 0

 

Which is wrong, because the function is NOT using both branches of the square root function. AND on top of that, those two cases do NOT exhaustively cover all cases, because they blatantly ignore the subtleties that come from dealing with this function on either side of x=5, which I explain below.

 

 

You need to look at the two cases. x > 5 and x < 5. x=5 is obviously 0 so we can ignore that case so long as both of the other functions are 0 at x=5.

 

For x > 5:

x-5 > 0

sqrt((x-5)^2) = x-5

So the value of the function is x-5 FOR x > 5.

 

For x < 5:

x-5 < 0

sqrt((x-5)^2) = -(x-5) [since x-5 is negative and we're dealing with the positive square root]

So the value of the function is 0 FOR x < 5.

 

Both of those are 0 at x=5, so we're good. The function is 0 until x=5, at which point it begins to rise at a rate of 1 unit per 1 unit.

 

If you were to look at the negative branch of the square root, you'd find the function to be 0 for x > 5 and x-5 for x < 5


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#206 Waker

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Posted 27 March 2013 - 07:26 PM

So, the only thing left now is to explain why the positive root is used for when x>5 while the negative root is used for X<5


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#207 Secret Igshar

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Posted 27 March 2013 - 08:04 PM

So, the only thing left now is to explain why the positive root is used for when x>5 while the negative root is used for X<5

It's not. The positive root is used every time. For x < 5, x-5 < 0, but (x-5)^2 > 0 because, well, squaring a negative gives a positive. You then just take the positive square root of that, which would be -(x-5), since x-5 < 0, and you want the positive valued root, which would be the opposite.


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#208 Waker

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Posted 27 March 2013 - 10:58 PM

I meant to say why do we use > and < for their corresponding signs before the square root. I assume the square root would have two, positive and negative. It looks like the negative of sqrt((x-5)^2) would be -(x-5) and adding that to x-5 would make it 0. And the positive of that would be 2(x-5), dividing by two making it x-5. Why are the less than and greater than signs used for the results of these sign choices, i.e. why not the other way?


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— General Kurt von Hammerstein-Equord


#209 Secret Igshar

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Posted 28 March 2013 - 12:26 AM

Because the issue is that you're missing out on something major by doing it that way.

 

If you want to do positive square root of (x-5)^2, you need to know whether or not the quantity x-5 is positive or negative. If it is positive, then the positive square root is x-5. If it's negative, then the positive square root is -(x-5). This gives two different solutions. Then you look at the negative branch of the square root function, and you're looking at that same issue in reverse (positive x-5 yields -(x-5) as your root while negative x-5 yields x-5 as your root), giving you four total cases. For the positive square root branch, you have:

 

x < 5 => f(x) = 0

x > 5 => f(x) = x-5

 

For the negative branch, you have:

 

x < 5 => f(x) = x-5

x > 5 => f(x) = 0

 

Which are two different piecewise functions. They're opposite branches of the "same" function, but they ARE two independent functions, nonetheless. Considering the fact that the first one (considering the positive roots) is the one we want, it is safe to assume that we always want to take the positive value of the square root. I hope that explains it better?

 

As an aside~

 

x < 5 <=> x-5 < 0

x > 5 <=> x-5 > 0


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#210 Waker

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Posted 31 March 2013 - 02:11 PM

Considering the fact that the first one (considering the positive roots) is the one we want, it is safe to assume that we always want to take the positive value of the square root.

Why do we take the positive root? Seems arbitrary. The negative root is the opposite of what the positive root is saying. It would be zero for all values from 5 to infinity and linearly increasing backwards from 5 to negative infinity.


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“I divide my officers into four classes; the clever, the lazy, the industrious, and the stupid. Each officer possesses at least two of these qualities. Those who are clever and industrious are fitted for the highest staff appointments. Use can be made of those who are stupid and lazy. The man who is clever and lazy however is for the very highest command; he has the temperament and nerves to deal with all situations. But whoever is stupid and industrious is a menace and must be removed immediately!”
— General Kurt von Hammerstein-Equord


#211 Secret Igshar

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Posted 31 March 2013 - 03:47 PM

We take the positive root because it gives the desired behavior. The square root relation is not a function unless you take one of the two branches (pos or neg). In this case it's the positive root. If the negative was to be considered too you'd see a plus or minus sign before it. If it were the negative branch there would be a -. It is very important to note that the square root symbol is NOT ambiguous. It ALWAYS means to get the positive root. When we put the negative before it it's to switch to the alternative branch rather than the standard one. If it were ambiguous, there would be no plus or minus in the quadratic formula. For clarification: the SYMBOL is not ambiguous. The OPERATION is.

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#212 Dion

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Posted 08 April 2013 - 09:13 AM

Having some trouble with my Discrete Structures homework. Don't quite get what this is even asking, mainly because I can't find it in the book. Any help would be appreciated.

 

> Let G be an undirected graph whose number of vertices is 1000, such that every vertex of G is pendant or isolated. Prove that the number of edges of G is less than or equal to 500.

 

and

 

>Let G be an undirected graph whose number of vertices is 1000, such that every vertex of G is pendant. Prove that the number of edges of G is equal to 500.


...V HAS COME TO...

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#213 Secret Felix

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Posted 08 April 2013 - 12:12 PM

1. Let G be an undirected graph whose number of vertices is 1000, such that every vertex of G is pendant or isolated. Prove that the number of edges of G is less than or equal to 500.

 

Proof by Cases

 

If every vertex of G is isolated, then the total number of edges is zero, and the proof holds.

 

Now recall,

 

     Σdeg(vi) = 2*E(G)

 

If every vertex of G is a pendant then, at most, each vertex of G is of degree one. Therefore,

 

     Σdeg(vi) = 1000

 

By applying our earlier relation,

 

     1000 = 2*E(G)

 

     E(G) = 500

 

Any other graph G composed of some number of isolated vertices and pendants will have less edges than a graph G composed entirely of pendants. Therefore, the number of edges of G is less than or equal to 500 for a graph composed of 1000 vertices which are either pendants or isolated

 

The second question is just the latter case in the proof.

 

Also,

 

     Discrete Math > Discrete Structures


Edited by Artemis, 08 April 2013 - 12:13 PM.

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#214 Secret Igshar

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Posted 08 April 2013 - 01:04 PM

I am crying because of how proud I am of you right now, Artemis. Crying with joy.


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#215 Dexel Hydagara

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Posted 11 April 2013 - 03:41 PM

So, some of the more simple equations on this worksheet I have, I can do fine, but some of these are really throwing me off. Help?

 

13. 1/5x+3 = (1/25)1/4-x

 

15. 37x-1 = 9-2

 

22. Cubic root of the quartic root of 5 times the quartic root of 25

 

Title of the section is Exponential and Logarithmic Equations, if that helps. I wasn't around for note-taking. I would've been instead of going to see the preview of the dance night at my school, if my teacher mentioned this shit was going to be this hard.


Edited by Dexel, 11 April 2013 - 03:49 PM.

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#216 Secret Felix

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Posted 11 April 2013 - 05:18 PM

Are we just solving for x here? If so, just remember logn(na) = a.

 

15. 37x-1 = 9-2

 

For this one, the first thing to do is to get everything in terms of base three. So we have,

 

      37x-1= (32)-2

 

      37x-1= 3-4

  

Now take the log3 of both sides,

 

      7x-1= -4

 

      7x = -3

 

      x = -3/7

 

Hopefully that will help get you started.


Edited by Artemis, 11 April 2013 - 05:19 PM.

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#217 Secret Igshar

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Posted 11 April 2013 - 07:59 PM

Things to keep in mind: The base of a logarithm must be a whole number. You can always turn a fraction into a whole number through exponential shenanigans, and for the purposes of your homework, you'll never run into situations where you need to mess with the base too much.

 

Important identities to keep in mind:

 

the "yth" root of x is x1/y

For example: sqrt(x) = x1/2

 

1/x = x-1

 

And, as Artemis said, logbbx = x

 

Those are the most basic properties of logs and exponent tricks that you'd likely ever need to worry about using in this sort of a course. Also of note is that you CAN just notice the following:

 

If xy = xz, then y=z ; x,y,z are real numbers.

(this identity doesn't hold for complex numbers since e = e3iπ = -1, but iπ =/= 3iπ, for example)



#218 Waker

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Posted 18 April 2013 - 11:50 PM

How do you find the sum of the formula/expression 50*1.9^N given a starting N and an ending N?

 

It's the formula for finding the amount of crystals needed to upgrade a stat on a monster in h-verse.

 

For example, if the STR stat is 10, with a starting stat of 5 for Humanoid monsters, the required amount of crystals to upgrade STR to level 11 would be

 

round(50*1.9^(N-5),0) where N is 10, which comes out to 1238. I want to be able to put in excel a range of levels more than just one so I don't have to calculate for each level. For level 10 to 15, for example.

 

 

And the formula for any monster class would be round(50*1.9^IF(monsterclasscell=monsterclass,N-starting stat for monster class, other monster class stat),0)


Edited by Waker, 19 April 2013 - 12:03 AM.

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"Censorship is telling a man he can't have a steak just because a baby can't chew it."
— Mark Twain

 

“I divide my officers into four classes; the clever, the lazy, the industrious, and the stupid. Each officer possesses at least two of these qualities. Those who are clever and industrious are fitted for the highest staff appointments. Use can be made of those who are stupid and lazy. The man who is clever and lazy however is for the very highest command; he has the temperament and nerves to deal with all situations. But whoever is stupid and industrious is a menace and must be removed immediately!”
— General Kurt von Hammerstein-Equord


#219 Secret Igshar

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Posted 19 April 2013 - 12:27 AM

How do you find the sum of the formula/expression 50*1.9^N given a starting N and an ending N?

 

It's the formula for finding the amount of crystals needed to upgrade a stat on a monster in h-verse.

 

For example, if the STR stat is 10, with a starting stat of 5 for Humanoid monsters, the required amount of crystals to upgrade STR to level 11 would be

 

round(50*1.9^(N-5),0) where N is 10, which comes out to 1238. I want to be able to put in excel a range of levels more than just one so I don't have to calculate for each level. For level 10 to 15, for example.

 

 

And the formula for any monster class would be round(50*1.9^IF(monsterclasscell=monsterclass,N-starting stat for monster class, other monster class stat),0)

Well I would probably use a VLOOKUP for the monster class values (just make a nice named table in another sheet with them all listed out and use smart vlookups with match functions and it should be no problem), but that's neither here nor there.

 

A good method for figuring out discrete sum formulae like this is to write it out and see how it looks. In this example, say we're looking at a sum from 5 to 10.

 

x = 50*1.9^5 + 50*1.9^6 + 50*1.9^7 + 50*1.9^8 + 50*1.9^9 + 50*1.9^10

x = (1 + 1.9 + 1.9^2 + 1.9^3 + 1.9^4 + 1.9^5)*50*1.9^5

 

And you'd try to simplify that however you'd want to. This one sort of looks like a geometric sum, so you can likely work it that way. However, in this case, all you really wanna do is use an array formula, since it's a WHOLE lot easier. Just use a clever combination of ROW and INDIRECT and it all works out really REALLY easy.

 

Variables (all as cell references):

N - starting level

M - ending level

L - initial value for the monster class

 

ROUND(SUM(50*1.9^(ROW(INDIRECT("A"&N&":A"&M-1))-L)),0)

 

And just be sure when you finish the formula you hold CTRL+SHIFT before hitting ENTER so it goes in as an Array Formula, rather than a normal one (which would return an error). And voila. Success.


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#220 Waker

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Posted 19 April 2013 - 04:05 AM

Well I would probably use a VLOOKUP for the monster class values (just make a nice named table in another sheet with them all listed out and use smart vlookups with match functions and it should be no problem), but that's neither here nor there.

 

A good method for figuring out discrete sum formulae like this is to write it out and see how it looks. In this example, say we're looking at a sum from 5 to 10.

 

x = 50*1.9^5 + 50*1.9^6 + 50*1.9^7 + 50*1.9^8 + 50*1.9^9 + 50*1.9^10

x = (1 + 1.9 + 1.9^2 + 1.9^3 + 1.9^4 + 1.9^5)*50*1.9^5

 

And you'd try to simplify that however you'd want to. This one sort of looks like a geometric sum, so you can likely work it that way. However, in this case, all you really wanna do is use an array formula, since it's a WHOLE lot easier. Just use a clever combination of ROW and INDIRECT and it all works out really REALLY easy.

 

Variables (all as cell references):

N - starting level

M - ending level

L - initial value for the monster class

 

ROUND(SUM(50*1.9^(ROW(INDIRECT("A"&N&":A"&M-1))-L)),0)

 

And just be sure when you finish the formula you hold CTRL+SHIFT before hitting ENTER so it goes in as an Array Formula, rather than a normal one (which would return an error). And voila. Success.

I dunno. I can't get your formula working. I dunno what "A" is supposed to be - http://i.imgur.com/jc26cr3.png

 

I could make mine work though - http://i.imgur.com/h6FeHYZ.png

 

WHOOPS: Changed Match to Index(Match,Match,0) got yours to work too - http://i.imgur.com/MOveUGU.png

 

Mine is shorter though... and I still don't really understand how yours works.


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"Censorship is telling a man he can't have a steak just because a baby can't chew it."
— Mark Twain

 

“I divide my officers into four classes; the clever, the lazy, the industrious, and the stupid. Each officer possesses at least two of these qualities. Those who are clever and industrious are fitted for the highest staff appointments. Use can be made of those who are stupid and lazy. The man who is clever and lazy however is for the very highest command; he has the temperament and nerves to deal with all situations. But whoever is stupid and industrious is a menace and must be removed immediately!”
— General Kurt von Hammerstein-Equord