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[jepense]

I'm having a hard time figuring out how this:

(log is the natural logarithm (ln))

is equal to this:

Because it's not?

Try ln(x(sqrt(1+exp(x))-sqrt(exp(x)))) + ln(sqrt(1+exp(x))+1) instead.

Hints:
1. ln a + ln b = ln(a*b)
2. -> you get ln(x*f(x)) = ln(x) + ln(f(x)), where f(x) contains all the sqrts and exps.
3. c ln d = ln(d^c), try ln(f(x)) = 1/2 * ln(f(x)^2)
4. f(x)^2 turns out to be exp(-x) which is almost what you want. You may need e^(2x)+2e^x+1 = (e^x+1)^2.

 

[survivor]

I have some quick algebra question. How would go around to find if 2 lines intersect if they are in R4 or more. Like each line has (x, y, z, w) for coordinates and direction vectors. Would you just make x=x, y=y and so on like I would do if it was on a (x, y, z) plane? Oh also I don't think I can use matrices which I assume can be used for solving the question.

 

[bachikarn]

I have some quick algebra question. How would go around to find if 2 lines intersect if they are in R4 or more. Like each line has (x, y, z, w) for coordinates and direction vectors. Would you just make x=x, y=y and so on like I would do if it was on a (x, y, z) plane? Oh also I don't think I can use matrices which I assume can be used for solving the question.

yeah, sounds like you have the right idea


Line1: l1(x,y,z,w) = (x1_0,y1_0,z1_0,w1_0) + t(v1,v2,v3,v4)

Line2: l2(x,y,w,z) = (x2_0,y2_0,z2_0,w2_0) + s(u1,u2,u3,u4)

Where the v, and u are the direction vectors. So just set them equal to each other and see if there exists a pair (t,s) that satisfies that equality.

 

[Ave22]

Damn it. I feel I'm going to be using this thread weekly now.

Let p and q be real numbers such that p - q > 1. Prove there is an integer n such that q < n < p.

Fuck, man. I can explain it in words but not in math. I can't use the well ordering principle, because that assumes an integer exists. I'm trying to use the theorem stating something like for every real number x, there exists integers n and n+1 such that n < x < n+1, but this is leading me to make some odd reasoning:

q < n + q < p
q <= n < n+q < p
q <= n < p.

This doesn't feel right though.

 

[Rich Uncle Skeleton]

Damn it. I feel I'm going to be using this thread weekly now.

Let p and q be real numbers such that p - q > 1. Prove there is an integer n such that q < n < p.

Fuck, man. I can explain it in words but not in math. I can't use the well ordering principle, because that assumes an integer exists. I'm trying to use the theorem stating something like for every real number x, there exists integers n and n+1 such that n < x < n+1, but this is leading me to make some odd reasoning:

q < n + q < p
q <= n < n+q < p
q <= n < p.

This doesn't feel right though.

Does the bold mean you aren't allowed to use it or that you can't see how to use it? If you're allowed to use it, you can just let A be the set of all integers greater than p. This has a smallest element n, and n-1 will be smaller than p but larger than q.

Edit: Technically, to use the well-ordering principle you need to assume p is positive, but if p and q are both negative then negate them, apply the above technique to their negatives, then negate whatever integer you find between them.

 

[Ave22]

I can't see how to use it. Let me think about what you're saying. Thanks for the response.

 

[harriet the spy]

Damn it. I feel I'm going to be using this thread weekly now.

Let p and q be real numbers such that p - q > 1. Prove there is an integer n such that q < n < p.

Fuck, man. I can explain it in words but not in math. I can't use the well ordering principle, because that assumes an integer exists. I'm trying to use the theorem stating something like for every real number x, there exists integers n and n+1 such that n < x < n+1, but this is leading me to make some odd reasoning:

q < n + q < p
q <= n < n+q < p
q <= n < p.

This doesn't feel right though.

I don't understand - what are you allowed to use?
If p is integer, than let n=p-1. Clearly, n>q, otherwise you'd have p-1 <=q or p-q <=1, which you know is not true.

If p is not an integer, let n=E[p] (i.e. the integer n such that n<p<n+1).
Then, from p<n+1, you have p-n<1, and 1<p-q, so you get p-n<p-q, or q<n as desired.

You can prove the existence of E[p] using the well-ordering principle, so if you have that, you're golden.

 

[Ave22]

Thanks a bunch. I finally understand it.

 

[Lonely1]

Dam, finally an Algebra question and I missed it.

 

[Government-man]

Because it's not?

Try ln(x(sqrt(1+exp(x))-sqrt(exp(x)))) + ln(sqrt(1+exp(x))+1) instead.

Hints:
1. ln a + ln b = ln(a*b)
2. -> you get ln(x*f(x)) = ln(x) + ln(f(x)), where f(x) contains all the sqrts and exps.
3. c ln d = ln(d^c), try ln(f(x)) = 1/2 * ln(f(x)^2)
4. f(x)^2 turns out to be exp(-x) which is almost what you want. You may need e^(2x)+2e^x+1 = (e^x+1)^2.

Thanks for the answer, but I stated my question wrong. What I meant was not the two above expressions added together would equal the one below, rather they both by themselves equal the one below. My bad.

Though now a have a trig question which probably is really easy if you know how to approach it but I'm seriously stuck...

cos(2x) + 3cos(x) - 1 = 0

How do I solve x?

 

[Leezard]

Thanks for the answer, but I stated my question wrong. What I meant was not the two above expressions added together would equal the one below, rather they both by themselves equal the one below. My bad.

Though now a have a trig question which probably is really easy if you know how to approach it but I'm seriously stuck...

cos(2x) + 3cos(x) - 1 = 0

How do I solve x?

expand cos(2x) into (cos^2(x) - sin^2(x)) and then you should be able to easily work it out, assuming you have done some trig before.

 

[innervision961]

So I'm taking pre-algebra at the community college this year before jumping straight into college algebra because it has been years since since I've done any math and I never was that good to begin with... I never really "got it" when I was in school but I desperately want to do good in my math classes, can you guys help me, or point me to some help, in these simple areas?

1) how do you solve a simple linear equation in two variables? 2x+3y=12 (completely made this problem up, but it is one similar to a problem I got wrong recently on a test)

2) simplifying/FOIL I start to get really really confused when I am foiling something with exponents and coefficients...

I can multiply x and x^2 and get x^3 right? What about 2x and x^2? This is the kind of stuff I keep getting wrong... I don't know why, I just can't get a simple intuition for what I can and can't do.

thanks guys, I'll be back, surely.

 

[SmokeMaxX]

So I'm taking pre-algebra at the community college this year before jumping straight into college algebra because it has been years since since I've done any math and I never was that good to begin with... I never really "got it" when I was in school but I desperately want to do good in my math classes, can you guys help me, or point me to some help, in these simple areas?

1) how do you solve a simple linear equation in two variables? 2x+3y=12 (completely made this problem up, but it is one similar to a problem I got wrong recently on a test)

2) simplifying/FOIL I start to get really really confused when I am foiling something with exponents and coefficients...

I can multiply x and x^2 and get x^3 right? What about 2x and x^2? This is the kind of stuff I keep getting wrong... I don't know why, I just can't get a simple intuition for what I can and can't do.

thanks guys, I'll be back, surely.

1) With two variables in a pre-alg course, they're either asking for you to substitute in which they'll give you values for x/y that you just plug in OR more likely, they're asking you to isolate a variable (most likely Y, but maybe both).

So for your example, Y = -2/3 X + 4; X = - 3/2 Y + 6 if you move everything to the other side of the equation and divide by whatever's in front of the variable you're looking for. If they gave you a value like Y = 2, then you could plug it in to those isolation equations and see that X = -3 + 6 = 3. Or if they said X = 3, then you'd see that Y = -2 + 4 = 2.

2) I believe simplifying is taking common denominators out of an expression? So simplifying 3x^2 + 9X = 3X(X+3). You're just finding the biggest thing you can divide the components of your expression by and putting it outside the parenthesis. For FOIL, you don't really have to do it First Outer Inner Last like the expression says, you just have to remember to get everything in the expression.

So, (3x + 3) (4X + 2) --> First is 3x * 4X = 12X^2. Outer is 3X * 2 = 6X. Inner is 3 * 4X = 12X. Inner is 3 * 2 = 6. You multiply for all those steps. THEN at the end, you add all your products together to get... 12X^2 + 6X + 12X + 6 = 12X^2 + 18X + 6... which you can simplify to 6(2X^2 + 3X + 1).

X * X^2 = X^3.
Likewise, 2X * X^2 = 2X^3.

 

[Ventrue]

1) I'm not sure if you meant to solve just one linear equation or a series of linear equations. Since this might be helpful: You need to have an equation for each variable. Otherwise you can only come up with a solution like x = something y.

For example, you might get a problem like:

5x =3y
2y = 5 + x

The idea is to solve one of them as above, to get x = something y.

e.g.

5x = 3y
x = 3/5y

Then you put that value in the second equation, so you only have one variable.

2y= 5 + x
2y = 5 + (3/5y)
10/5y = 5 + 3/5y
7/5y = 5
7y = 25
y = 25/7

Then use that solution to get a value for the other variable.

x = 3/5y
x = 3/5(25/7)
x = 75/35
x = 15/7

So your solution is y = 25/7, x = 15/7.

If you have three equations and three variables, then you can do the same thing, except you'll have x = something y something z, and then in the second equation you do z = something y, etc.

 

[MYE]

fuck math

 

[Door2Dawn]

Math is the devil.

 

[poweld]

1) I'm not sure if you meant to solve just one linear equation or a series of linear equations. Since this might be helpful: You need to have an equation for each variable. Otherwise you can only come up with a solution like x = something y.

For example, you might get a problem like:

5x =3y
2y = 5 + x

The idea is to solve one of them as above, to get x = something y.

e.g.

5x = 3y
x = (3/5)y

Then you put that value in the second equation, so you only have one variable.

2y= 5 + x
2y = 5 + (3/5)y
(10/5)y = 5 + (3/5)y
(7/5)y = 5
7y = 25
y = 25/7

Then use that solution to get a value for the other variable.

x = 3/5y
x = (3/5)(25/7)
x = 75/35
x = 15/7

So your solution is y = 25/7, x = 15/7.

If you have three equations and three variables, then you can do the same thing, except you'll have x = something y something z, and then in the second equation you do z = something y, etc.

Please be careful when showing your work. Remember PEMDAS.

So I'm taking pre-algebra at the community college this year before jumping straight into college algebra because it has been years since since I've done any math and I never was that good to begin with... I never really "got it" when I was in school but I desperately want to do good in my math classes, can you guys help me, or point me to some help, in these simple areas?

1) how do you solve a simple linear equation in two variables? 2x+3y=12 (completely made this problem up, but it is one similar to a problem I got wrong recently on a test)

2) simplifying/FOIL I start to get really really confused when I am foiling something with exponents and coefficients...

I can multiply x and x^2 and get x^3 right? What about 2x and x^2? This is the kind of stuff I keep getting wrong... I don't know why, I just can't get a simple intuition for what I can and can't do.

thanks guys, I'll be back, surely.

To answer your second question, I always found it helpful when I'm not sure of the rules to break things down by more basic rules. For example, a series of multiplication operations can be broken up as much as you like.

You asked how to multiply 2x and x^2, so let's break it down.

2x = 2 * x
x^2 = x * x

So now we have: 2 * x * x * x

or

2 * x^3

or

2x^3

 

[TL4E]

1) how do you solve a simple linear equation in two variables? 2x+3y=12 (completely made this problem up, but it is one similar to a problem I got wrong recently on a test)

I'm guessing "solve" here means write the equation in terms of x and then find the zeros?

You'd write the problem in y= form, so in this case you subtract 2x from both sides and then divide both sides by 3, yielding:

y = 4 - (2/3)x

Then have y=0

0 = 4 - (2/3)x

And solve it like a regular equation

-4 = -(2/3)x

-4(-3/2) = x

x = 6


If they let you use a graphing calculator, you can confirm your answer while taking the test by graphing the equation and checking where the equation you graphed crosses the y-axis at y=0. "Finding the zeros" is a fancy way of saying for what values x is y = 0.

ps: I whole-heartedly recommend khanacademy.org for help in math -- he does an excellent job giving you the intuition on solving problems rather than teaching how to mechanically solve problems, which will get you nowhere in the long run.

 

[ChocolateCupcakes]

I like this thread. Subbed.

 

[sestrugen]

I will be taking Calculus I next semester at college, I have never really grasped math but I am very interested in changing that. I want to start from the basics since my foundations are not very good but I am at a loss on where to begin, what areas will I need to focus a bit more on (Pre-Algebra, Algebra, Trig)?
I am alredy watching the information on the Khan Academy website which has helped me to remember a few things but I want to focus on specifics and do a lot of problems for those subjects.

Can anyone recommend and outline of what should I be focussing to get a better foundation on my math skills?

 

[RustyNails]

Yeah you don't solve linear equations unless you want to plot them, or you solve an equation until it becomes linear or quadratic.

I would stop at y = 4 - (2x)/3, unless you're plotting.

x is only 6 if y is 0.

 

[Tornado Condor]

I will be taking Calculus I next semester at college, I have never really grasped math but I am very interested in changing that. I want to start from the basics since my foundations are not very good but I am at a loss on where to begin, what areas will I need to focus a bit more on (Pre-Algebra, Algebra, Trig)?
I am alredy watching the information on the Khan Academy website which has helped me to remember a few things but I want to focus on specifics and do a lot of problems for those subjects.

Can anyone recommend and outline of what should I be focussing to get a better foundation on my math skills?

I'd suggest you remember your trigonometry and basic algebra skills (mostly manipulation). If you get advanced (ala Calculus 2 and 3) you will definitely need to know how to properly manipulate numbers around.

 

[RustyNails]

I will be taking Calculus I next semester at college, I have never really grasped math but I am very interested in changing that. I want to start from the basics since my foundations are not very good but I am at a loss on where to begin, what areas will I need to focus a bit more on (Pre-Algebra, Algebra, Trig)?
I am alredy watching the information on the Khan Academy website which has helped me to remember a few things but I want to focus on specifics and do a lot of problems for those subjects.

Can anyone recommend and outline of what should I be focussing to get a better foundation on my math skills?

I went till Calc 4 and slightly beyond at college. Here's the breakdown for Calculus at my school:

Math 148: Algebra, linear/quadratic equations, Trig, graphs, etc.
Math 150: Elementary functions.
Calc 1: Limits and Derivatives
Calc 2: Integration
Calc 3: Series and sequences
Calc 4: Differential equations and multivariable calculus

You basically need the algebra and Trig down pat before starting Calc I. See if you can solve this Math 148 Midterm from my school. If you can, you should be well equipped in algebra department. Also make sure that you really understand graphs.

I'd suggest you remember your trigonometry and basic algebra skills (mostly manipulation). If you get advanced (ala Calculus 2 and 3) you will definitely need to know how to properly manipulate numbers around.

Yep. Its funny how as more you progress in Math, the less numbers you use.

 

[TL4E]

I will be taking Calculus I next semester at college, I have never really grasped math but I am very interested in changing that. I want to start from the basics since my foundations are not very good but I am at a loss on where to begin, what areas will I need to focus a bit more on (Pre-Algebra, Algebra, Trig)?
I am alredy watching the information on the Khan Academy website which has helped me to remember a few things but I want to focus on specifics and do a lot of problems for those subjects.

Can anyone recommend and outline of what should I be focussing to get a better foundation on my math skills?

I found being comfortable with algebraic notaion and manipulation immensly useful in calculus, e.g. knowing (a/b)/(c/b) = (a/c), understanding composition notation f(g(x)), being familar with trig identities such as (sinx)^2+(cosx)^2=1 etc.

It's hard to pinpoint where exactly to start. Imo, watching khan's videos on elementary math topics you're unfamiliar with should be good enough.

@ RustyNails: well, it depends on what the question wants you to do with the equation. Often times I would be asked to find the zeros in equations of that form when I took basic algebra.

 

[RedShift]

I'm confused by the prep material my soon to be new University sent me. One of the arithmetic exercises had me giving the prime factorisations of numbers (without a calculator), starting easy with 42, 63, 385, slightly harder with 1,188, and then 132,461.

So when I got to 132,461 It started alright, I found 7 as a factor, divided through to get 132,461 = 7 x 18,923. So then I start trying to factorise 18,923, checking if it divided by 7, 11, 13 and then by 17 got bored and checked WolframAlpha, which told me 132,461 = 7 x 127 x 149. Am I missing something obvious which would make this easier than it seems? (No calculator allowed)

EDT: The next number was 9009, which wasn't hard to reduce down to 3 x 3 x 7 x 11 x 13

 

[poweld]

I'm confused by the prep material my soon to be new University sent me. One of the arithmetic exercises had me giving the prime factorisations of numbers (without a calculator), starting easy with 42, 63, 385, slightly harder with 1,188, and then 132,461.

So when I got to 132,461 It started alright, I found 7 as a factor, divided through to get 132,461 = 7 x 18,923. So then I start trying to factorise 18,923, checking if it divided by 7, 11, 13 and then by 17 got bored and checked WolframAlpha, which told me 132,461 = 7 x 127 x 149. Am I missing something obvious which would make this easier than it seems? (No calculator allowed)

EDT: The next number was 9009, which wasn't hard to reduce down to 3 x 3 x 7 x 11 x 13

As far as I know, the only way to find this out is to do what you were probably already doing: divide by a small prime and then get the prime factors of the result. 127 and 149 though... that would take a long while, especially if you're not using a calculator.

Seems like a test of patience or something.

 

[Lonely1]

Um RedShift, that problem sucks. Yeah, there's no other way. Maybe someone screwed up and was supposed to be a different number?

 

[1138]

I've been stuck on this problem for a while now, and unable to see how i can solve this:

f(x,y) has continous partial derivatives, and the maximum directional derivative of f(0,0) is 4 along the vector <1,3>. Find the gradient of f(0,0).

I've calculated the unit vector, and if im not wrong it is <(1/&#8730;10), (3/&#8730;10)> But i'm not able to get any further from this point, so any help is appreciated.

 

[Dkong]

I've got a quick question for the math guys or even better for the physics guys at GAF. I'm currently finishing my thesis for the dutch equivalent of a college information technology bachelor. I want to go to a university next to study physics and they've told me that the only thing I need to work on before applying is math (I need to master the highest level math given at dutch high schools). I was wondering if you guys could recommend a couple of categories that will be of special importance to master if I want to aqcuire a physics bachelor or higher. I've already started and became addicted to the Khan Academy site since I have nothing else to do at this moment. I also plan to arrange some kind of tutorage or private teacher or evening course for math but I have a feeling that any help is welcome when you prepare for university physics with a background with little math.

 

[Vaporak]

I've got a quick question for the math guys or even better for the physics guys at GAF. I'm currently finishing my thesis for the dutch equivalent of a college information technology bachelor. I want to go to a university next to study physics and they've told me that the only thing I need to work on before applying is math (I need to master the highest level math given at dutch high schools). I was wondering if you guys could recommend a couple of categories that will be of special importance to master if I want to aqcuire a physics bachelor or higher. I've already started and became addicted to the Khan Academy site since I have nothing else to do at this moment. I also plan to arrange some kind of tutorage or private teacher or evening course for math but I have a feeling that any help is welcome when you prepare for university physics with a background with little math.

I don't know what the highest level of mathematics at dutch high schools is, but the important math for physics will be what you learn while doing the degree really. If you haven't started calculus yet I'd suggest just knowing the basic algebra well, too many people try and get into calculus without good enough algebra skills imo. (Mathematics major here)

 

[Lonely1]

I've got a quick question for the math guys or even better for the physics guys at GAF. I'm currently finishing my thesis for the dutch equivalent of a college information technology bachelor. I want to go to a university next to study physics and they've told me that the only thing I need to work on before applying is math (I need to master the highest level math given at dutch high schools). I was wondering if you guys could recommend a couple of categories that will be of special importance to master if I want to aqcuire a physics bachelor or higher. I've already started and became addicted to the Khan Academy site since I have nothing else to do at this moment. I also plan to arrange some kind of tutorage or private teacher or evening course for math but I have a feeling that any help is welcome when you prepare for university physics with a background with little math.

Calculus and Analytical Geometry. If your pre-calculus algebra is lacking, it will show while studying those.

 

[jepense]

Calculus and Analytical Geometry. If your pre-calculus algebra is lacking, it will show while studying those.

Pretty much. Physics is full of partial differential equations, so you must know calculus. Things usually also happen in space, so you need to know geometry to handle the relevant coordinate systems. You'll also need linear algebra to solve anything for real, but that's probably not high school level stuff anymore.

 

[jepense]

I've been stuck on this problem for a while now, and unable to see how i can solve this:

f(x,y) has continous partial derivatives, and the maximum directional derivative of f(0,0) is 4 along the vector <1,3>. Find the gradient of f(0,0).

I've calculated the unit vector, and if im not wrong it is <(1/&#8730;10), (3/&#8730;10)> But i'm not able to get any further from this point, so any help is appreciated.

Gradient is a vector with the length of the maximum directional derivative, in the direction of fastest change. So, you just need to multiply your unit vector by 4, the value of the derivative.

 

[1138]

Gradient is a vector with the length of the maximum directional derivative, in the direction of fastest change. So, you just need to multiply your unit vector by 4, the value of the derivative.

I don't have any more info than what i listed in my previous post. The answer is supposed to be &#8711;f(x,y)= < (2&#8730;10)/5 , (6&#8730;10)/5 > but i still can't see how i get there.

 

[BlueMagic]

I don't have any more info than what i listed in my previous post. The answer is supposed to be &#8711;f(x,y)= < (2&#8730;10)/5 , (6&#8730;10)/5 > but i still can't see how i get there.

As jepense said, you have to multiply the unit vector (which you calculated) by 4. Then, take the square root out of the denominator (multiplying by (&#8730;10/&#8730;10)) and that's it.

 

[1138]

As jepense said, you have to multiply the unit vector (which you calculated) by 4. Then, take the square root out of the denominator (multiplying by (&#8730;10/&#8730;10)) and that's it.

I feel kinda dumb that i didn't notice that before, but at least i get it now. Thanks for the help.

 

[RSTEIN]

Can anyone please explain what a moment generating function is? I know that it's

In probability theory and statistics, the moment-generating function of any random variable is an alternative definition of its probability distribution.

but I think I need an example to fully comprehend it.

For a binomial distribution, the moment generating function is (1-p+pe^t)^n. For the normal distribution it's e^(ut+.5sd^2t^2).

So... how do I use this?

 

[Ave22]

I'm really struggling with this Real Analysis class. So much of the work we're given seems like it's so easy, but it ends up seeming impossible to explain.

Suppose the sequence {An} converges to A, {Bn} converges to B, and An <= Bn for all n natural numbers. Show that A <= B.

I have tried messing around with many things, but I can't see how to show that A <=B. I tried assuming A > B to show a contradiction, but I have no idea what that implies.

 

[Prophet Steve]

I'm really struggling with this Real Analysis class. So much of the work we're given seems like it's so easy, but it ends up seeming impossible to explain.

Suppose the sequence {An} converges to A, {Bn} converges to B, and An <= Bn for all n natural numbers. Show that A <= B.

I have tried messing around with many things, but I can't see how to show that A <=B. I tried assuming A > B to show a contradiction, but I have no idea what that implies.

Not sure if I understand this right. But won't divide both sides by n works?

 

[Ave22]

Sorry, I should have said n is just a subscript, the nth term of the sequence. I miss the good old days when it was that easy.

 

[zoku88]

EDIT: I think this only works if both sequences were positive to begin with. (or, if I put absolute values around each sequence, it would need to be that the sets would need to be absolute convergent.)

Maybe you could be a new sequence, Cn, which is An-Bn. Cn will always be greater than 0.

Cn converges to some C. If B were greater than A, this C would be negative. That's impossible. There doesn't exist a series with all positive terms that converges to a negative number.

I'm not sure how to do it when An and Bn are allowed to be negative, though.

EDIT2:&#12288;I think I mixed up your A and B. What I just did was when An > Bn for all natural numbers.

EDIT3: I'm sorry if I'm not much help. I'm sure if I had my analysis book on me, I could actually do something. I think I may have done this problem before :lol :lol

 

[Therion]

I'm not sure how to do it when An and Bn are allowed to be negative, though.

I'm fairly certain that your method is correct whether the sequences are negative or not. The difference will still be positive, so C_n->C is positive, and C=A-B=>A=C+B=>A>B (using the reversed A and B from your answer).

 

[Rich Uncle Skeleton]

I'm really struggling with this Real Analysis class. So much of the work we're given seems like it's so easy, but it ends up seeming impossible to explain.

Suppose the sequence {An} converges to A, {Bn} converges to B, and An <= Bn for all n natural numbers. Show that A <= B.

I have tried messing around with many things, but I can't see how to show that A <=B. I tried assuming A > B to show a contradiction, but I have no idea what that implies.

Proof by contradiction seems like a good approach to me. What it implies is that there is some epsilon > 0 such that

(1) A = B + epsilon.

But, going to the definition of limits, there's some N such that for n > N,

(2) A_n > A - epsilon/3

and

(3) B_n < B + epsilon/3.

If you combine (1), (2), (3), and the assumption that A_n <= B_n, you can derive a contradiction. Hope that helps; let me know if you need me to clarify anything.

 

[innervision961]

I just wanted to let you guys know, I read and re read your replies a few times and BAM! it finally hit me... Now I feel like an imbecile for having trouble with something so simple :lol

It turns out that, yes, the two variable equation was just to help solve for x and y intercepts and slope.

So if I had 2x + 3y = 12
3y = -2x + 12
(for instance)


I took my final today, and I got a 97! I passed the class with an A

 

[poweld]

I took my final today, and I got a 97! I passed the class with an A

 

[Ave22]

EDIT: I think this only works if both sequences were positive to begin with. (or, if I put absolute values around each sequence, it would need to be that the sets would need to be absolute convergent.)

Maybe you could be a new sequence, Cn, which is An-Bn. Cn will always be greater than 0.

Cn converges to some C. If B were greater than A, this C would be negative. That's impossible. There doesn't exist a series with all positive terms that converges to a negative number.

I'm not sure how to do it when An and Bn are allowed to be negative, though.

EDIT2:&#12288;I think I mixed up your A and B. What I just did was when An > Bn for all natural numbers.

EDIT3: I'm sorry if I'm not much help. I'm sure if I had my analysis book on me, I could actually do something. I think I may have done this problem before :lol :lol

This makes sense, but I don't know how to explain that a series of positive terms must have a positive limit.

Proof by contradiction seems like a good approach to me. What it implies is that there is some epsilon > 0 such that

(1) A = B + epsilon.

But, going to the definition of limits, there's some N such that for n > N,

(2) A_n > A - epsilon/3

and

(3) B_n < B + epsilon/3.

If you combine (1), (2), (3), and the assumption that A_n <= B_n, you can derive a contradiction. Hope that helps; let me know if you need me to clarify anything.

This one feels like juvenile trickery showing that epsilon < 2epsilon/3 since we can pick epsilon to be whatever we want. But I suppose it works out, because it doesn't work when A > B, and it does when A = B and A < B.

Thanks for the responses.

 

[Lonely1]

This makes sense, but I don't know how to explain that a series of positive terms must have a positive limit.



This one feels like juvenile trickery showing that epsilon < 2epsilon/3 since we can pick epsilon to be whatever we want. But I suppose it works out, because it doesn't work when A > B, and it does when A = B and A < B.

Thanks for the responses.

No. Epsilon is not that you want, but rather what you need. You can define Epsilon to be a dependent of A and B if you wanna be more formal, but using it as it is accepted.

 

[Therion]

This makes sense, but I don't know how to explain that a series of positive terms must have a positive limit.

Contradiction. Suppose C_n->C, where C<0. Choose epsilon=|L/2|. Then |C_n-C|>epsilon for all n, since C_n is nonnegative.

 

[Undubbed]

Better Explained is a supurb site that tries to break out of the rote bullshit and actually explains what's going on.

Favorite one is the one about complex numbers, for example.

And I found this excellent article on that site.

 

[nincompoop]

Can anyone please explain what a moment generating function is? I know that it's



but I think I need an example to fully comprehend it.

For a binomial distribution, the moment generating function is (1-p+pe^t)^n. For the normal distribution it's e^(ut+.5sd^2t^2).

So... how do I use this?

Given a random variable x and a constant t you can apply the function e^(t*x) to give you a new random variable y = e^(t*x). You can then compute the expected value of y which gives you a new function (called the moment generating function of x) which is a function of t because it varies as you choose different values for t. The significance of the moment generating function is that you can use it to find the nth moment of x by differentiating it n times and evaluating it at t=0. In particular, the expected value of x is M_x'(0) and the variance of x is M_x''(0).