Support NeoGAF

[joseph1594]

All right, I need help with a word problem.

In 1990, Americans bought an average of 143 meals per year at restaurants. This phenomenon continued to grow in popularity and in the year 2000, the average reached 170 meals per year. Find a linear equation with t=0 corresponding to 1990 that makes this growth and use the equation to estimate the average number of times an American will eat at a restaurant in 2006 if the trend continues.

My first instinct is to find the slope. I did the slope formula with the two points of (1990, 143) and (2000, 170) and got a slope of 27/10. Considering t=0 corresponds to 1990, would 143 be the y-intercept? I am not sure if I am doing this right. I don't know what I need to do, am I essentially trying to find the y in the point (2006, y)?

 

[cpp_is_king]

All right, I need help with a word problem.

In 1990, Americans bought an average of 143 meals per year at restaurants. This phenomenon continued to grow in popularity and in the year 2000, the average reached 170 meals per year. Find a linear equation with t=0 corresponding to 1990 that makes this growth and use the equation to estimate the average number of times an American will eat at a restaurant in 2006 if the trend continues.

My first instinct is to find the slope. I did the slope formula with the two points of (1990, 143) and (2000, 170) and got a slope of 27/10. Considering t=0 corresponds to 1990, would 143 be the y-intercept? I am not sure if I am doing this right. I don't know what I need to do, am I essentially trying to find the y in the point (2006, y)?

Everything looks mostly correct until the end

Y = 2.7x + b

Now, two ways you can do it.

If you want 143 to be the y intercept thats fine, and then x=0 corresponds to 1990. But then you dont find y when x=2006, you fing y when x=16. If 1990 corresponds to x=0 then 2006 corresponds to x=16

If you want x to correspond directly to the year, then 143 isnt the y intercept, you would need to write 143=2.7(1990) + b and solve for b. now you can set x=2006 and find y

Either way should give the same answer

 

[SailorCosmos]

How does one find the domain of the following function?

f(x) = sqrt(log x)

 

[big ander]

How does one find the domain of the following function?

f(x) = sqrt(log x)

Best to work from the outside in with domain, I find. So what values can sqrt accept?
All values greater than or equal to zero. Thus, we know we must have:
log(x)>=0
For what values of x is log(x)>=0?

 

[Exuro]

I have another proof problem. Really terrible at these.

det(cA) = c^n*det(A)

A is a matrix of nxn and c is a scalar.

I know that when n is 1 and 2 that it's true, however I'm not sure how to describe it when it's to some n=k+1 value.


Going to try to show some sort of work.

det(cA) = c*a_11*det(c*A_1)-c*a_12*det(c*A_2)+...+c*a_1n*det(c*A_n)

How would I show det(c*A_11) is c^n*det(A_1) without doing this over and over again? Barring that, I can show that...

det(cA) = c^(n+1)*det(A_1)+c^(n+1)*det(A_2)+...+c^(n+1)*det(A_n) = c^(n+1)*det(A) for an n+1x n+1 matrix right?

 

[SailorCosmos]

Best to work from the outside in with domain, I find. So what values can sqrt accept?
All values greater than or equal to zero. Thus, we know we must have:
log(x)>=0
For what values of x is log(x)>=0?

I ended up with [1, inf). Is that correct?

I have another one?

How do I get the inverse of the following function?

f(x) = -3^(x+1)

I know I have to change variables so I get

x = -3^(y+1)

However, I don't know what to do after that. I know how to find the inverse when the variable is not an exponent but I am lost on this.

 

[ThanksVision]

I ended up with [1, inf). Is that correct?

I have another one?

How do I get the inverse of the following function?

f(x) = -3^(x+1)

I know I have to change variables so I get

x = -3^(y+1)

However, I don't know what to do after that. I know how to find the inverse when the variable is not an exponent but I am lost on this.

Logarithmic power rule. Log(3^x) can also be written as xlog(3), for example.

In this case, log both sides of the equation.

log(x) = log(-3^(y+1))

the log power rule allows you to pull the exponent out

log(x) = (y+1)log(-3)
now just solve for y

 

[rjc571]

I have another proof problem. Really terrible at these.

det(cA) = c^n*det(A)

A is a matrix of nxn and c is a scalar.

I know that when n is 1 and 2 that it's true, however I'm not sure how to describe it when it's to some n=k+1 value.


Going to try to show some sort of work.

c^n*det(A) = c^n*[a_11*det(A_11)+a_12*det(A_12)+...+a_1n*det(A_1n)]
det(cA) = c*a_11*det(c*A_11)+c*a_12*det(c*A_12)+...+c*a_1n*det(c*A_1n)

Doing it like this seems like I'd be going forever, to the point where A_xx is a 2x2 as i can easily show that the determinate with a scalar will pull out c^2. :S

Have you learned that the determinant of a product of two matrices equals the product of the determinants? Multiplying A by the scalar c is the same thing as multiplying A by the matrix c*I (where I is the identity matrix), and the determinant of cI is c^n. So there you go.

 

[Exuro]

Have you learned that the determinant of a product of two matrices equals the product of the determinants? Multiplying A by the scalar c is the same thing as multiplying A by the matrix c*I (where I is the identity matrix), and the determinant of cI is c^n. So there you go.

I know that, but it needs to be proved by induction which we haven't really learned/been taught in class. I'm just sort of confused why induction has a step of assumption for n=n.

 

[rjc571]

I know that, but it needs to be proved by induction which we haven't really learned/been taught in class. I'm just sort of confused why induction has a step of assumption for n=n.

OK, for the inductive step, if A is k+1 x k+1, then det(A) is an alternating sum of the terms in the first row times the determinant of the corresponding minor matrix (obtained by removing the 1st row and the corresponding column from A), while det(cA) is calculated in the same way. Except the terms in the first row of cA are c times the corresponding term in the first row of A, and each minor matrix for cA is c times the corresponding minor matrix for A. But the minor matrices are k x k, so by the inductive assumption, the determinant of each minor matrix of cA is c^k times the determinant of the corresponding minor matrix of A. Throw in the extra c for the first row terms, and you should get that det(cA) = c^k+1 * det(A)

To put it mathematically:
det(cA) = sum(i=1 to k+1) (-1)^i+1 * cA(1,i) * det(cA(1,i)minor)
= c*sum(i=1 to k+1) (-1)^i+1 * A(1,i) * c^k*det(A(1,i)minor)
= c^k+1 * det(A)

 

[Polydeuces]

So im doing an integration problem and I got up to here:

Integral of 1/(cosx)^4

I cant go any further right?

 

[Partial Gamification]

So im doing an integration problem and I got up to here:

Integral of 1/(cosx)^4

I cant go any further right?

Re-write it, 1/cos^4(x). = sec^2(x)*sec^2(x) = (1+tan^2x)sec^2(x)
Multiply through, now you have two integrals.

This is a quick pass, you will need; integration-by-parts, to find a recursive definition, or both. Its long but you don't need an integration table.

 

[oxrock]

I'm doing trig identities and I need some math gaf mojo to get my brain working. In theory I know how this all works but for some reason I'm just not able to do these right now. I'm hoping seeing someone else's work can get me going because I'm completely stuck even though I know it's simple.

Verify the identity:

tan x/sec x = sin x

I can only think to make it sin x/cos x divided by sec x = sin x. After that I just derp. HALP!

 

[ThanksVision]

I'm doing trig identities and I need some math gaf mojo to get my brain working. In theory I know how this all works but for some reason I'm just not able to do these right now. I'm hoping seeing someone else's work can get me going because I'm completely stuck even though I know it's simple.

Verify the identity:

tan x/sec x = sin x

I can only think to make it sin x/cos x divided by sec x = sin x. After that I just derp. HALP!

secx = 1/cosx

therefore, (sinx/cosx)/(1/cosx) = sinx

when you divide a fraction by another fraction, flip the denominator and multiply. Now see if any of the trig functions cancel.

 

[oxrock]

secx = 1/cosx

therefore, (sinx/cosx)/(1/cosx) = sinx

when you divide a fraction by another fraction, flip the denominator and multiply. Now see if any of the trig functions cancel.

Much obliged, for some reason I just need a little nudge and then all of the sudden I can do all sorts of other shit. Still have a lot more work to do on these today but I've done another 5 so far and it's been smooth sailing.

 

[oxrock]

Only been an hour but I need help again

Just need to validate the identities.

Problem: 1 - (cosx - sinx)^2 /cosx = 2 sinx

First off I foiled and I got : 1 - (cosx^2 sinx^2 -2cosx -2sinx)/cosx = 2 sinx
And I'm kind of stuck after that. I could cancel out a cosx I suppose, but then what?

 

[ActStriker]

Only been an hour but I need help again

Just need to validate the identities.

Problem: 1 - (cosx - sinx)^2 /cosx = 2 sinx

First off I foiled and I got : 1 - (cosx^2 sinx^2 -2cosx -2sinx)/cosx = 2 sinx
And I'm kind of stuck after that. I could cancel out a cosx I suppose, but then what?

I don't think you did this right.

 

[Memento_Mori15]

Only been an hour but I need help again

Just need to validate the identities.

Problem: 1 - (cosx - sinx)^2 /cosx = 2 sinx

First off I foiled and I got : 1 - (cosx^2 sinx^2 -2cosx -2sinx)/cosx = 2 sinx
And I'm kind of stuck after that. I could cancel out a cosx I suppose, but then what?

1) Make sure you factor it correctly. It should be:

Spoiler

1- (cos^2(x) + sin^2(x)-2sin(x)cos(x)) Don't distribute the negative just yet to make it easier to see what you need to the next.

2) Recall your basic trig identities. The one you'll end up using a lot in math: cos^2(x) + sin^2(x)=1. Don't we have that?
1- (cos^2(x) + sin^2(x)-2sin(x)cos(x))
1-((1)-2sin(x)cos(x))
3) Distribute the negative now
1-1+ 2sin(x)cos(x)
The ones cancel out, you're left with 2sin(x)cos(x).
4) After that, we still have that cos(x) in the denominator, cancel it out. You get 2sin(x)

I spoiled it in case you want to figure it out yourself. Basically, foil it out correctly. Use your trig identities, do algebra, and you'll get your answer.

 

[oxrock]

1) Make sure you factor it correctly. It should be:

Spoiler

1- (cos^2(x) + sin^2(x)-2sin(x)cos(x)) Don't distribute the negative just yet to make it easier to see what you need to the next.

2) Recall your basic trig identities. The one you'll end up using a lot in math: cos^2(x) + sin^2(x)=1. Don't we have that?
1- (cos^2(x) + sin^2(x)-2sin(x)cos(x))
1-((1)-2sin(x)cos(x))
3) Distribute the negative now
1-1+ 2sin(x)cos(x)
The ones cancel out, you're left with 2sin(x)cos(x).
4) After that, we still have that cos(x) in the denominator, cancel it out. You get 2sin(x)

I spoiled it in case you want to figure it out yourself. Basically, foil it out correctly. Use your trig identities, do algebra, and you'll get your answer.

I finished the problem without an issues once I copied your foil result but for the life of me I can't get that foil result. Foiling should be simple though, so I can't understand how I'm screwing it up.

(cosx - sinx)^2 "(cosx - sinx ) (cosx - sinx)"
F: cos^2
O: (cosx -sinx) O and I are essentially the same thing so -2(cosx sinx) right?
I: (-sinx cosx)
L: sinx^2

Obviously I'm doing something wrong but how am I derping? (apparently in the O and I steps) I'm just not seeing it. I blame the 5 gallons of coffee.

 

[f0rk]

What is foil?
If you add up the terms in whatever that method is you get the same as Memento did (sinx^2 - 2sinxcosx + cosx^2). Then reduce sinx^2 + cosx^2 to 1 so you are left with (1 - (1 - 2sinxcosx))/cosx

 

[Memento_Mori15]

I finished the problem without an issues once I copied your foil result but for the life of me I can't get that foil result. Foiling should be simple though, so I can't understand how I'm screwing it up.

(cosx - sinx)^2 "(cosx - sinx ) (cosx - sinx)"
F: cos^2
O: (cosx -sinx) O and I are essentially the same thing so -2(cosx sinx) right?
I: (-sinx cosx)
L: sinx^2

Obviously I'm doing something wrong but how am I derping? (apparently in the O and I steps) I'm just not seeing it. I blame the 5 gallons of coffee.

Huh, is that how people are learning it now? You'll get the same answer. Look at your O, it's cosx-sinx. Well, it's pretty much like saying cosx*-sinx, which is just -cosxsinx. Same with the I, -sinxcosx is the same as -cosxsinx. So if you add them up, you get -2sinxcosx. If you add all of them up, you get: cos^2 -2sinxcosx+sinx^2. Which is correct.

This is what I do sometimes when I foil:
Take one side, start from the left side. In this case the cos:
(cosx- sinx) (cosx- sinx)
Let's just concentrate on that for now:
Cosx (cosx-sinx)
Let's multiple:
Cosx*cosx= cos^2(x)
cosx* -sinx= -cos(x)sin(x)
Combine the two now: cos^2(x) -cos(x)sin(x)
Now let's focus on the right side
(cosx -sinx) (cosx- sinx)
-sinx (cosx-sinx)
-sinx*cosx=-cos(x)sin(x)
-sinx*-sinx= sin^2(x)
Combine the two now: -cos(x)sin(x)+sin^2(x)

From the previous: cos^2(x) -cos(x)sin(x) and what we just got: -cos(x)sin(x)+sin^2(x)
Combine them:cos^2(x) -cos(x)sin(x) -cos(x)sin(x) +sin^2(x)
cos^2(x) -2cos(x)sin(x)+sin^2(x)

Sure it's a bit long but I liked it since that's how I taught myself back in whenever I was learning that. lol Do what you feel comfortable with.

 

[oxrock]

ok guys, it's several hours later and I'm still doing this stupid math, I HATE IT! That being said, I need awesome math GAF to assist in my understanding this shit and getting a good grade.

Verify the identity:
1 - cscx\1 + cscx = sinx -1/sinx +1

AFAIK I have to pick a side to work on and that through fancy switching of shit, make them identical but this just ain't working at the moment. I have another 20 I need to hand in in the morning and I've been doing this for HOURS already. Help me out.

 

[cpp_is_king]

ok guys, it's several hours later and I'm still doing this stupid math, I HATE IT! That being said, I need awesome math GAF to assist in my understanding this shit and getting a good grade.

Verify the identity:
1 - cscx1 + cscx = sinx -1/sinx +1

AFAIK I have to pick a side to work on and that through fancy switching of shit, make them identical but this just ain't working at the moment. I have another 20 I need to hand in in the morning and I've been doing this for HOURS already. Help me out.

Can you put some parens in there? Hard to tell what the numerators and denominators of each fraction are

 

[oxrock]

1 - csc x        sin x - 1
----------   = -----------
1 + csc x       sin x + 1

I hope this makes it more clear. This is exactly as it appears in my textbook.

 

[Alaskanbullworm]

1 - cscx        sinx - 1
----------   = -----------
1 + cscx      sinx + 1

I hope this makes it more clear.

When dealing with these type of problems, I always simplify everything down to sinx and/or cosx. Since the right side is already in terms of sinx, I'd want to work with the left side to verify. Do you see now what you'd like to do to the numerator and denominator? And do you know why you can do it?

I'll spoiler what you'd have to do and why it works:

Spoiler

Notice how if you multiply the numerator by sinx you get (six - 1) and if the denominator you get (sinx + 1). Since we just have the left hand side of the equation, we are looking at a number. You can always multiply a number by 1 and get the same result. In this case our multiplying by 1 is actually multiplying the left hand fraction by (sinx)/(sinx). The numerator gets multiplied by the top sinx and the denominator gets multiplied by the bottom sinx.

 

[cpp_is_king]

1 - csc x        sin x - 1
----------   = -----------
1 + csc x       sin x + 1

I hope this makes it more clear. This is exactly as it appears in my textbook.

The above poster's solution will work, but personally I would do it a different way, which I think is a bit more straightforward. Just apply the definition of csc directly.

1 - csc x       1 - 1/sinx          (sinx - 1)/sinx          sinx - 1      sinx      sinx - 1
----------   = --------------  =   -------------------  =   ---------- x ------ =  -----------
1 + csc x       1 + 1/sinx          (sinx + 1)/sinx          sinx + 1      sinx      sinx + 1

 

[oxrock]

All the help has been greatly appreciated by the way. I've spent all day doing this crap but I think i have it... mostly
All i know is that I hate these stupid things.

 

[mattoz85]

I have a couple of questions about sequence convergence for my real analysis class.

Prove that if a converges to 2, then a^n converges to infinity.

If a converges to 1, does this imply that a^n converges to 1?

Investigate the convergence behavior of the sequence a^n. What finite values could it converge to? Could it diverge with infinite limit or with no limit? Find an example for each possibility.

I also need to prove that a=cos is divergent. I assume I can do this with a proof by contradiction?

 

[rjc571]

I have a couple of questions about sequence convergence for my real analysis class.

Prove that if a converges to 2, then a^n converges to infinity.

If a converges to 1, does this imply that a^n converges to 1?

Investigate the convergence behavior of the sequence a^n. What finite values could it converge to? Could it diverge with infinite limit or with no limit? Find an example for each possibility.

I also need to prove that a=cos is divergent. I assume I can do this with a proof by contradiction?

If a_n converges to 2, then given a real number x you can find N such that (1.5)^N > x. You can also find M such that a_n > 1.5 whenever n > M. So whenever n > max(N, M), you have (a_n)^n > x, which proves (a_n)^n diverges to infinity.

For the next part, if you fix any real number x, then the sequence a_n = (1 + x/n) converges to 1, and (a_n)^n converges to e^x. Since the exponential functions maps onto (0, infinity), the limit of (a_n)^n can be any positive real number. It can't be any negative number though, since (a_n)^n is always nonnegative and the limit of a nonnegative sequence is nonnegative.

(a_n)^n can also converge to zero, or diverge to infinity, or diverge without limit. If a_n = e^-1/sqrt, then (a_n)^n = e^-sqrt converges to zero. If a_n = e^1/sqrt, then (a_n)^n = e^sqrt diverges to infinity. If a_n = (1 + (-1)^n/n) then the terms of (a_n)^n will be bouncing back and forth between e and 1/e, so the sequence diverges without limit.

To show that cos diverges, write down the intervals on which cos(x) is greater than 1/2, and the intervals on which cos(x) is less than -1/2. There are infinitely many such intervals, and each of them has length 2pi/3, so each interval contains at least one (actually at least two) integers. So the set of integers for which the cosine function is > 1/2 is unbounded, as is the set of integers for which it is < -1/2, meaning that for any integer N, you can find an integer n > N with cos > 1/2, and another integer m > N with cos(m) < -1/2. This shows that cos diverges since it isn't Cauchy (there will always be terms of the sequence with arbitrarily large indices, which differ from each other by at least 1).

 

[Troll]

 

[Troll]

Hi,
It really depends on what theorems you already covered and if you're allowed to use them without reproving them. I'm 99% sure by the time you see Lagrange theorem you will already have covered the major theorems for cyclic groups...

We know that Z_n is a finite cyclic group, so we can apply a corollary of the fundamental theorem of cyclic groups, and I believe that's what you were referring to above.
(If you wikipedia "fundamental theorem of cyclic groups" you will see it, it's too hard to retype)

So using that theorem for Z_n, it states that if k divides n, the set <n/k> is the unique subgroup with order k of Z_n.
And this is exactly what your question is asking to prove, so that's all there is to it.

Tell me if this wasn't clear or explicit enough and I can provide more info.

Wouldn't that be of order n/k? That is the part that confuses me...

 

[mattoz85]

If a_n converges to 2, then given a real number x you can find N such that (1.5)^N > x. You can also find M such that a_n > 1.5 whenever n > M. So whenever n > max(N, M), you have (a_n)^n > x, which proves (a_n)^n diverges to infinity.

Thank you so much for your help! I understood your other explanations, but I'm still having trouble wrapping my head around this one. The TA also used 1.5 for this problem. Is that because it's greater than the square root of 2? Are you using M to represent the max of a bounded sequence? Pretty much everything we do revolves around the definition:

"For all epsilon strictly greater than 0, there exists an N such that for all natural numbers n greater than or equal to N, the absolute value of the sequence minus its limit is strictly less than epsilon."

In this case, are you making x the same thing as epsilon in the definition? I'm sorry if this sounds stupid, but if such a thing as remedial real analysis existed, I would need that.

 

[rjc571]

I'm just using the fact that the sequence {b^n} is unbounded if the base b is any number bigger than 1. (I picked b = 1.5 arbitrarily, but you can use any number between 1 and 2 as your base.) To see this, you can write b = 1+a, where a > 0. Then the binomial expansion formula tells you that (1+a)^n is bigger than n*a, and the Archimidean Principle tells you that for any real number x, there exist an integer n with n > x. So you can pick an integer n with n > x/a, so b^n = (1+a)^n > n*a > x.

(In my case, b = 1.5 and a = 0.5, so given x in R, if n > x/0.5 then b^n will be bigger than x.)

 

[mt1200]

How do you guys find Z transforms by hand?, it seems to be fucking impossible, I don't know how can books find the right sequence.

 

[MikeDip]

Wouldn't that be of order n/k? That is the part that confuses me...

No, try to find that theorem in your book. If a group is cyclic, then for every k that divides n, n the order of the group, there is a subgroup <a^(n/k)> which has order K. Since we are talking about Z_n, the groups operation is additive, not multiplicative. So the theorem becomes, if the order of Z_n is n, then there exists a unique subgroup <n/k> and it has order k.

You can see this in some examples. Z_30. This means n = 30. So if we pick a k that divides n, say k = 2, we see that the theorem tells us <30/2> which is <15>. Now we know that the order of <15> is equal to the order of the element "15". We can check this.
(since Z_n uses addition, we don't take powers to find the order)
1*15= 15
2*15= 30 = 0

So the order of 15 is 2, as expected.

Similarly, if we chose k to be 15, we see this leads to <30/15>= <2>
Checking the order of the element 2 shows it is 15.

 

[Troll]

No, try to find that theorem in your book. If a group is cyclic, then for every k that divides n, n the order of the group, there is a subgroup <a^(n/k)> which has order K. Since we are talking about Z_n, the groups operation is additive, not multiplicative. So the theorem becomes, if the order of Z_n is n, then there exists a unique subgroup <n/k> and it has order k.

You can see this in some examples. Z_30. This means n = 30. So if we pick a k that divides n, say k = 2, we see that the theorem tells us <30/2> which is <15>. Now we know that the order of <15> is equal to the order of the element "15". We can check this.
(since Z_n uses addition, we don't take powers to find the order)
1*15= 15
2*15= 30 = 0

So the order of 15 is 2, as expected.

Similarly, if we chose k to be 15, we see this leads to <30/15>= <2>
Checking the order of the element 2 shows it is 15.

Ohhhh, okay. Thank you!

 

[Troll]

Does it seem like I need to go through and prove this for each element in A or should I stick to a general proof?

 

[MikeDip]

Does it seem like I need to go through and prove this for each element in A or should I stick to a general proof?

I need to preface this by saying I hated equivalence classes and relations, so I'm not fully confident in my answer here. I'll look at b,c, and d in a minute but I think I can do a.
Also, never solve for each element here, you should find general ideas.

For a)
To show an equivalence relation you have to show Reflexivity, symmetry and transitivity, so let's do one at a time.

Reflexitivity: What we have to show is that there exists a k, such that &#963;^k(a) = a. This is the same as showing that a~a.
This is easy, since if we just take k to be 0, we see that &#963;^0=1. So we are left with 1*a=a. Good.

Symmetry: We must show that for a,y in A, that if a~b then b~a. So we must show that when is a k such that &#963;^k(a) = b then there is a j such that &#963;^j(b)=a
(which means if we have a~b, there exists a way to show b~a)
If we take (&#963;^k)^-1, then we see this equals &#963;^-k. So if &#963;^-k(b)=a. Multiply both sides by &#963;^k. This gives (&#963;^k)(&#963;^-k)b=&#963;^k(a) which gives b=&#963;^k(a) which is what we wanted.
(since (&#963;^k)(&#963;^-k) = e) We found the j that works, it was -k.

Transitivity: We need to show for every a,b,c in A that if a~b and b~c then a~c.
we have &#963;^k(a) = b and &#963;^j(b) = c, and we need to show that &#963;^k(a) = c.
Take &#963;^k(a) = b and multiply it by &#963;^j on both sides. We get:
&#963;^j&#963;^k(a)= &#963;^j(b) But &#963;^j(b) = c. So we have &#963;^j&#963;^k(a)= c. Now note &#963;^j&#963;^k = &#963;^(j+k)
This gives &#963;^(j+k)(a) = c, which is what we wanted.

So we proved that &#963; is an equivalence relation.

Do you have any questions? I know this is really messy, I'm sorry.

 

[Troll]

I need to preface this by saying I hated equivalence classes and relations, so I'm not fully confident in my answer here. I'll look at b,c, and d in a minute but I think I can do a.
Also, never solve for each element here, you should find general ideas.

For a)
To show an equivalence relation you have to show Reflexivity, symmetry and transitivity, so let's do one at a time.

Reflexitivity: What we have to show is that there exists a k, such that &#963;^k(a) = a. This is the same as showing that a~a.
This is easy, since if we just take k to be 0, we see that &#963;^0=1. So we are left with 1*a=a. Good.

Symmetry: We must show that for a,y in A, that if a~b then b~a. So we must show that when is a k such that &#963;^k(a) = b then there is a j such that &#963;^j(b)=a
(which means if we have a~b, there exists a way to show b~a)
If we take (&#963;^k)^-1, then we see this equals &#963;^-k. So if &#963;^-k(b)=a. Multiply both sides by &#963;^k. This gives (&#963;^k)(&#963;^-k)b=&#963;^k(a) which gives b=&#963;^k(a) which is what we wanted.
(since (&#963;^k)(&#963;^-k) = e) We found the j that works, it was -k.

Transitivity: We need to show for every a,b,c in A that if a~b and b~c then a~c.
we have &#963;^k(a) = b and &#963;^j(b) = c, and we need to show that &#963;^k(a) = c.
Take &#963;^k(a) = b and multiply it by &#963;^j on both sides. We get:
&#963;^j&#963;^k(a)= &#963;^j(b) But &#963;^j(b) = c. So we have &#963;^j&#963;^k(a)= c. Now note &#963;^j&#963;^k = &#963;^(j+k)
This gives &#963;^(j+k)(a) = c, which is what we wanted.

So we proved that &#963; is an equivalence relation.

Do you have any questions? I know this is really messy, I'm sorry.

Yeah, thank you. That is what I ended up doing, I don't know why but I just felt like it was wrong since he defined A, but now that I think about it he probably did that so that we could do parts b-d. For those parts I just plugged in each element in A and came up with the congruence classes to be [\sigma(a)]. Does that seem like the right idea?

 

[MikeDip]

Yeah, thank you. That is what I ended up doing, I don't know why but I just felt like it was wrong since he defined A, but now that I think about it he probably did that so that we could do parts b-d. For those parts I just plugged in each element in A and came up with the congruence classes to be [\sigma(a)]. Does that seem like the right idea?

Yeah, A is for the other parts, and what you said sounds correct.
I hate equivalence relations so much and I don't know why.
How are you finding the permutation groups? I found those to be the easiest sections in group theory.

 

[Troll]

Yeah, A is for the other parts, and what you said sounds correct.
I hate equivalence relations so much and I don't know why.
How are you finding the permutation groups? I found those to be the easiest sections in group theory.

It's funny, I was actually thinking about how easy permutations are today while sitting in class. I feel like all the permutation groups will be like little treats in the exam. And that exactly sums up how I feel about equivalence relations/classes.

 

[Glasswork]

I need a bit of help with this integral:

&#8747;3/5e^(7-2x)dx

 

[Treefingers]

I need a bit of help with this integral:

&#8747;3/5e^(7-2x)dx

Use change of variables, ie. let z = 7 - 2x, then dz = -2dx, substitute in to get it in terms of z & dz. Then it's an easy integral.

 

[Qurupeke]

lim (ax^2 +b)/(x-1)
x->1

I need to find a and b (a,b are real numbers). Well, it's not difficult, but I had a big problem. I solved it by setting f(x)=(ax^2+b)/(x-1) etc. A friend found the same a and b with Horner's method. I thought it's kinda weird. So my question is, can I apply Horner's method to this? If not, why?

 

[Partial Gamification]

lim (ax^2 +b)/(x-1)
x->1

I need to find a and b (a,b are real numbers). Well, it's not difficult, but I had a big problem. I solved it by setting f(x)=(ax^2+b)/(x-1) etc. A friend found the same a and b with Horner's method. I thought it's kinda weird. So my question is, can I apply Horner's method to this? If not, why?

Is the polynomial: ax^2+b (dividend) and: x-1 the divisor for the synthetic division using Horner?
This seems sort of familiar but to solve for a & b, are you looking for ranges that effect the limit? I mean, are you looking at cases where a=b , a > b , & a < b ? Or is there some information missing from your post, maybe I'm just a little confused about what values for a and b that you are getting. This is for a Numerical Methods class?

 

[Certinty]

Can someone give me some studying tips?

I have 3 text books to study, about 30 different topics but can't find the motivation to study as it's literally all reading examples then doing questions. Any ideas for a more interactive way of learning math?

 

[MikeDip]

Can someone give me some studying tips?

I have 3 text books to study, about 30 different topics but can't find the motivation to study as it's literally all reading examples then doing questions. Any ideas for a more interactive way of learning math?

What topics are you studying? You can probably find some interactive web aps that you can mess with. Either way though, when your studying math, there is no getting around doing sample question, and the more the better.

 

[Certinty]

What topics are you studying? You can probably find some interactive web aps that you can mess with. Either way though, when your studying math, there is no getting around doing sample question, and the more the better.

Mainly pure mathematics. Will have a look now, thanks mate.

 

[Partial Gamification]

Mainly pure mathematics. Will have a look now, thanks mate.

If you find anything interesting, please share it. I see a few apps for +16, 4-year-colleges in the US are 13-16 (after k-12), but not knowing your topics makes it a blind crap-shoot in regard to a recommendation. Is this cramming for a midterm, or developing a study plan?

 

[Certinty]

If you find anything interesting, please share it. I see a few apps for +16, 4-year-colleges in the US are 13-16 (after k-12), but not knowing your topics makes it a blind crap-shoot in regard to a recommendation. Is this cramming for a midterm, or developing a study plan?

Ah well I'm in the UK so things are a little different here. Going to get a full list from my teacher of all the topics on Tuesday, that should help. Not really sure what specifically as I'm studying Computing but most of my first year requires maths.

Study plan really. Got 2 months to learn all of it so thought I might as well start now before it's too late.