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The Official Math/Physics/Whatever Homework questions thread The Official Math/Physics/Whatever Homework questions thread

10-27-2009 , 08:51 AM
Quote:
Originally Posted by neurogenesis
Yes you are correct. I think this works:

x + (-x) = 0

let x = 1, so 1 + (-1) = 0, however x+y=max(x,y), so 1 + (-1) = max(1,-1) = 1, therefore is is not a vector space.

Thanks
No No No No No.

1 + (-1) = 1. Because "+" is the max operator now. You don't have regular addition anymore. You can't "add" things. You can multiply by a scalar, and you can take a max. That's it.

You're not contradicting anything with what you wrote.

edit: you need to find a vector space axiom that these fail OR show that all are satisfied.

edit2: e.g. See if it's true that c(x+y) = cx+cy.
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10-27-2009 , 09:00 AM
Quote:
Originally Posted by Wyman
No No No No No.

1 + (-1) = 1. Because "+" is the max operator now. You don't have regular addition anymore. You can't "add" things. You can multiply by a scalar, and you can take a max. That's it.

You're not contradicting anything with what you wrote.

edit: you need to find a vector space axiom that these fail OR show that all are satisfied.

edit2: e.g. See if it's true that c(x+y) = cx+cy.
Wow I suck. OK last try before I need a pointer. There should be a zero vector x + 0 = x for all x. So can I use this to show that if x = -1, -1 + 0 = 0, not x? I am using the + operator to mean max(x,0) which = 0 in this case.

If this is not correct, do you have a suggestion on which rule to try?
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10-27-2009 , 06:54 PM
Showing work is helpful but if not its fine too
Questions I don't know how to do:
Bank lends two people money for a total of $b. First person pays x% interest and second person pays y% interest. The amount of interest received by the bank is the same for reach loan. How much was loaned at x%??

a) (b+y) / xy
b) bx /(x+y)
c) by /(x+y)
d) (x+y) / by
e) bxy / (x+y)

-------------------

Questions that I might know how to do but would take a while:

If the solutions to the two equations are the same, find k
(x+7)^2 + (y-3)^2 = 0
-2x + 3y = k

Choices are:
23
-5
5
-23
18

------------------------------------------

(x^3 + 14) (x^3 - 51) = 1014
has two different real roots, find the difference between the roots.
- I know you prime factor 1014 (2 x 3 x 13 x 13) but I can't get past there
Choices are
4, 5, 6, 7, 8

------------------------------

One day visibility was limited to 1300m. Ship Blue was travelling west on a parallel course to Ship White, travelling east. The courses are 500m apart. Blue ship's velocity is 6 km/h. If the ships were in sight of each other for 15 mins, what's the veolcity in km/h of white ship?
-I think you're supposed to use pythagorous (5-12-13 or in this case 500-1200-1300 triangle) but am not sure what to do after that.
Choices: 4.5, 4.2, 3.8, 3.6 and 3.2
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10-27-2009 , 07:17 PM
Quote:
Originally Posted by thylacine
fyp

also solve the recurrence relation and that'll be your formula

or get the generating function if you like

I tried both of these ways, but I didn't get anything that could easily be solved by hand.

I'm thinking maybe that it is in the form 2^n - (number of 111s in an n digit binary number), but the only way I can see doing this is by inclusion/exclusion.
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10-27-2009 , 08:10 PM
Quote:
Originally Posted by neurogenesis
Wow I suck. OK last try before I need a pointer. There should be a zero vector x + 0 = x for all x. So can I use this to show that if x = -1, -1 + 0 = 0, not x? I am using the + operator to mean max(x,0) which = 0 in this case.
The additive identity is denoted by "0" or "0v" in a vector space but that doesn't imply that it is the real number zero. Of course, with the *normal* definition of addition, zero is the additive identity. In this case, you have correctly shown that zero is *not* the additive identity of this vector space.

I think you are on the right lines now, see what happens if you try to find an additive identity "0" with x+"0" = x for all x. Remember that this means that max(x,"0") = x for all x.

Last edited by Pyromantha; 10-27-2009 at 08:22 PM.
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10-27-2009 , 08:13 PM
Quote:
Originally Posted by neurogenesis
Wow I suck. OK last try before I need a pointer. There should be a zero vector x + 0 = x for all x. So can I use this to show that if x = -1, -1 + 0 = 0, not x? I am using the + operator to mean max(x,0) which = 0 in this case.

If this is not correct, do you have a suggestion on which rule to try?
That's a good idea. However you've only shown that 0 does not work as a zero vector. You need to show that there is no z in R that acts as a zero vector. To do this you might notice:

z+x = max(z,x) = x
if and only if
z <= x


PairTheBoard
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10-27-2009 , 08:47 PM
Quote:
Originally Posted by Myrmidon7328
I tried both of these ways, but I didn't get anything that could easily be solved by hand.

I'm thinking maybe that it is in the form 2^n - (number of 111s in an n digit binary number), but the only way I can see doing this is by inclusion/exclusion.
You've set up the recurrence relation. Do you know how to solve recurrence relations?
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10-27-2009 , 09:54 PM
Quote:
Originally Posted by thylacine
You've set up the recurrence relation. Do you know how to solve recurrence relations?
I tried the r^n method, but that yields a cubic that I can't do by hand.

I also tried the generating function method, and got a fraction like this for G(x):

G(x) = (1+x+x^2)/(1-x-x^2-x^3)

Which I don't know how to solve without doing some pretty nasty partial fractions.
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10-27-2009 , 11:52 PM
Quote:
Originally Posted by Myrmidon7328
I tried the r^n method, but that yields a cubic that I can't do by hand.

I also tried the generating function method, and got a fraction like this for G(x):

G(x) = (1+x+x^2)/(1-x-x^2-x^3)

Which I don't know how to solve without doing some pretty nasty partial fractions.
Yes, factorizing that cubic is an obstacle. (And I can see you know what you're doing.) I'm not sure what you're expected to do at this point. You'll have to find out what the expectations are for you solving this.
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10-28-2009 , 03:58 PM
Quote:
Originally Posted by DoctorFarha
Showing work is helpful but if not its fine too
Questions I don't know how to do:
Bank lends two people money for a total of $b. First person pays x% interest and second person pays y% interest. The amount of interest received by the bank is the same for reach loan. How much was loaned at x%??

a) (b+y) / xy
b) bx /(x+y)
c) by /(x+y)
d) (x+y) / by
e) bxy / (x+y)
Set up 2 equations in 2 variables, where the variables are the amounts loaned to each person.


Quote:
(x+7)^2 + (y-3)^2 = 0
-2x + 3y = k

Choices are:
23
-5
5
-23
18
Hint1: What would equation 1 be a graph of if the right side was r^2 instead of 0?
Hint 2: (x+7)^2 = -(y-3)^2. The left side is >= 0, and the right side is <= 0.


Quote:
(x^3 + 14) (x^3 - 51) = 1014
has two different real roots, find the difference between the roots.
- I know you prime factor 1014 (2 x 3 x 13 x 13) but I can't get past there
Choices are
4, 5, 6, 7, 8
If you multiply this out and set equal to zero, then the constant is -1728. The only prime factors of 1728 are 2 and 3. Integer solutions must be positive or negative factors (not necessarily prime factors) of 1728 which you get from the prime factors. Try the lowest ones first, and you should find a couple solutions fairly quickly.

Another way without trial and error is to notice that this is a product of 2 numbers that differ by 14 - (-51) = 65. This means there is some number, call it y, that is 65/2 = 32.5 from each of these numbers, and we can write

(y + 32.5)(y-32.5) = 1014

This is a difference of 2 squares:

y^2 - (32.5)^2 = 1014

Now solve for y (both solutions), then solve

x^3 + 14 = y

for x, using both values of y to get 2 values for x. Be sure to check in original equation.


Quote:
One day visibility was limited to 1300m. Ship Blue was traveling west on a parallel course to Ship White, traveling east. The courses are 500m apart. Blue ship's velocity is 6 km/h. If the ships were in sight of each other for 15 mins, what's the velocity in km/h of white ship?
-I think you're supposed to use pythagorous (5-12-13 or in this case 500-1200-1300 triangle) but am not sure what to do after that.
Choices: 4.5, 4.2, 3.8, 3.6 and 3.2
Since 1200m is the East-West separation when they are 1300m apart as the crow flies, they must be within 1200m of each other in either direction for 15 minutes. Calculate how fast they are moving with respect to each other, and subtract the blue ship's contribution to the speed to get the white ship's speed.

Last edited by BruceZ; 10-28-2009 at 04:27 PM.
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10-28-2009 , 04:46 PM
Quote:
(x^3 + 14) (x^3 - 51) = 1014
has two different real roots, find the difference between the roots.
- I know you prime factor 1014 (2 x 3 x 13 x 13) but I can't get past there
Choices are
4, 5, 6, 7, 8
Perhaps the easiest way to do this problem is to write it as a quadratic by a change of variables.

Let y = x^3

(y + 14)(y - 51) = 1014

y^2 - 37y -1728 = 0

Now solve this for y with the quadratic formula, or by factoring. Then use the values of y to compute the values for x.


Also, in my second solution,

Quote:
x^3 + 14 = y
should be

x^3 + 14 = y + 32.5

Last edited by BruceZ; 10-28-2009 at 05:05 PM.
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10-28-2009 , 05:24 PM
thanks =D
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10-28-2009 , 11:31 PM
Sighs, this is a genetics problem and most of the problems I see here are math. I guess it actually is just a logic problem but I'm a nl50 grinder and haven't attained this level of thinking yet, I don't think. I know this is a containment thread for all homework questions but if no one gets it would it be ok for me to make a new thread?

This chromosome mapping crap is killing me. The combos were limited when it was just mapping three traits. But four and five are killing me. For three traits I know the isolated gene in the double crossover is the gene in the middle of the parental chromosome but I don't know what too look for when mapping four or five or even more traits. Here's a problem from my homework.

You are trying to map 5 traits, a, b, c, d, and e relative to each other. You take a heterozygous animal and testcross it. The following 10,000 progeny were found:

2300 ade
45 wild type
105 d
3 a
2500 bce
2400 ad
95 abc
90 de
55 abcd
4 bcd
1 ae
2210 bc
50 e
40 abcde
100 abce
2 bcde

a) What is the genotype of the starting heterozygous parent?
b) Map the loci (i.e., determine the position and map distances among the genes).

From this I can tell the chromosome is trans. Whats confusing me is the four main groups in the f1 generation. excluding any crossovers, the ratio of genotypes is 1:1:1:1. It seems there's an extra pair of genotypes in there and I don't know how to account for it. Is one them a really high frequency crossover? Is there only partial linkage here and some traits are on another chromosome or something. I know this is jargon to anyone not familiar with this but I'm hoping someone can help me.
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10-29-2009 , 12:49 PM
Let H be the subset of M_2(R) consiting of all matrices of the form

[a -b]
[b a]

for a,b is an element of R.

show that <C,+> is isomorphic to <H,+>

Show that <C,*> is isomorphic to <H,*> * equals multiplication
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10-29-2009 , 04:34 PM
Quote:
Originally Posted by smcdonn2
Let H be the subset of M_2(R) consiting of all matrices of the form

[a -b]
[b a]

for a,b is an element of R.

show that <C,+> is isomorphic to <H,+>

Show that <C,*> is isomorphic to <H,*> * equals multiplication
Write down the map, and show that it's an isomorphism. [Right? I mean what else could your strategy be?]
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10-30-2009 , 02:07 AM
hopefully this is physics enough to belong in this forum / thread. the problem is a bit longer but I have everything worked out to this point.

(quick back up...assume I have 2 parallel wires, one of infinite length and one of finite length. the one of infinite length has an AC current running through it, creating a magnetic field that varies with time. I need to figure what that magnetic field needs to be in order to induce a certain voltage / current in the other wire.)

I started with finding the magnetic field produced by the AC current carrying wire. then, I know that I need to find the change of the magnetic flux through the second wire with respect to time in order to find the induced emf. what I cannot figure, however, is over what surface I should be integrating the magnetic field in order to find the magnetic flux and then how to integrate it. a rectangle (simplified rectangular prism) would certainly be easy to integrate but I think that would be oversimplifying.

can anyone help? many thanks in advance.
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11-02-2009 , 02:04 PM
One to One and onto when describing groups

Can someone please help me with this concept

I know one to one and onto is called a bijective function. And I know what the two mean. But when I try to bring it into group theory I get a little lost.

So if I am asked to give an example of a non abelian group G and an abelian group G' with an onto homomorphism phi:G-->G'

or

an example of a non abelian group G and an abelian group G' and a one to one homomorphism phi:G-->G'

I know an example of the first is

phi:S_3-->Z_2

and the second doens not exist.

So What am I looking at here? The homomorphism function with manipulation?

Like we know if it is a homomorphism then:
phi(xy)=phi(x)phi(y)

but if I want to prove its onto then does it need to be

phi(xy)=(phi(x)phi(y))^-1???

Let me ask it this way, other than the example above, if I was given the same question what is my strategy to start?
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11-02-2009 , 02:38 PM
Quote:
Originally Posted by smcdonn2
So if I am asked to give an example of a non abelian group G and an abelian group G' with an onto homomorphism phi:G-->G'
Let G' = {e}, and map every element of G to {e}. This is the simplest example.

Quote:
or

an example of a non abelian group G and an abelian group G' and a one to one homomorphism phi:G-->G'
Take arbitrary x,y in G.

phi(xy) = phi(x)phi(y) =* phi(y)phi(x) = phi(yx), with the * equality because G' is abelian.

But since phi is 1:1, this implies that xy=yx. Since x,y were arbitrary, G must also be abelian.


You just figure out what you're trying to prove (or disprove) and use properties of the homomorphism to do it.
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11-02-2009 , 03:49 PM
Anyone have any hints for my P-Chem test on Wednesday? Any useful things to write down on our 1 sheet of notes? Our professor emphasizes that we will not have time to read our notes while taking the test, so add only important details/information about quantum mechanical systems.

We have covered particle in 1d box, particle in a 2d box, harmonic oscillator, rigid rotor, photoelectric effect and blackbody radiation. We also covered commuting and noncommuting operators, normalization, probability (of finding a particle in the nth state), and boltzmann's distribution.

Thanks if anyone decides to respond!
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11-02-2009 , 03:59 PM
Rethorik question

2+2 = ?
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11-02-2009 , 09:52 PM
Problem from my Quantum Mechanics HW:



so I started by finding the equation for the stationary states of an infinite square well and finding the fourier transform of it. two problems i'm having:

1. as the problem suggests i re-wrote sine in the complex exponential form. after i took the integral for the fourier transform i can't write the expression I have in terms of sine.

2. i have no idea what it means when it asks me to compute the wave function in the p-eigenbasis.

problem 1 is not too much of an issue i don't think. if someone could just explain the p-eigenbasis thing it would help a lot. thanx.
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11-02-2009 , 10:55 PM
Quote:
Originally Posted by solsek
Anyone have any hints for my P-Chem test on Wednesday? Any useful things to write down on our 1 sheet of notes? Our professor emphasizes that we will not have time to read our notes while taking the test, so add only important details/information about quantum mechanical systems.

We have covered particle in 1d box, particle in a 2d box, harmonic oscillator, rigid rotor, photoelectric effect and blackbody radiation. We also covered commuting and noncommuting operators, normalization, probability (of finding a particle in the nth state), and boltzmann's distribution.

Thanks if anyone decides to respond!
hey finally a non math question

i'm feeling too lazy to type out all this stuff so here are a couple pictures of old cheat sheets from the quantum mechanics part of my p-chem class, looks like it covers most of the stuff you mentioned except for the statistical mechanics stuff which we had already covered the previous semester

honestly this was a few years ago so i don't remember what all the letters in the equations mean but i'm sure you'll be able to figure most of the out, but you can ask and i might be able to decipher it





o and don't laugh at my simple trig identities please
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11-03-2009 , 01:33 AM
I got a few problems I need help on that are gonna make you guys think I'm an idiot but w/e

Problem 1:

Simplify


4 + 6/y
_______

2 + 3/x^2

Need to simplify that problem. From what I can remember you add a denominator to the 4 and 2 and then flip to get:

4/1 + 6/y x 2/1 + 3/x^2

Where I go from there I have no clue.

Problem 2:

Solve:

t+12/t^2-t-6 + 4/t-3 = 1/t+2

My guess for this one is to subtract 1/t+2 to get everything on one side to get everything equal to 0. After that simplify the first denominator and multiply the missing half of the denominators to the 2nd and third fractions to get:

t+12/(t-3)(t+2) + 4t+8/(t-3)(t+2) -t+3/(t-3)(t+2) = 0

Add and subtract the like terms to get:

4t+23/(t-3)(t+2)

Problem 3:

Divide

x^3 - 4x^2 + 6x -10/x-3

This one I thought to simplify first but that accomplishes nothing and I'm just lost on where to even start.
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11-03-2009 , 01:37 AM
Problem 1 you have the right idea w/ re-writing 4 as a fraction but to add two fractions you need them to have the same denominator.

problem 2 you seem to be on the right track.

problem 3 you just need to do long division. its an algorithm. you can look it up online.
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11-03-2009 , 04:04 AM
I've got a few more linear algebra questions. I just can't wrap my head around this yet.

1. Let S = {v1, ... , vn} bet a set of vectors in vector space V. Prove that span S is a subspace of V.

2. Suppose that S = {v1, ... , vn} is a linearly independent set of vectors in a vector space V. Show that any non-empty subset of S is linearly independent.

Thanks.
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