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The Math Help Thread
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Boogie9IGN
Member
I'm having some trouble in statistics. This girl in our group was supposed to send us her answers for our lab but the class is in 1.5hours and still no email. I HATE GROUP PROJECTS
Anyway, I did the empirical probabilities easily. The theoretical ones are what's causing me trouble. It's the usual M&M lab :lol
Here's the data:
Population
Color Quantity
Yellow (Y) 6
Green (G) 3
Blue (BL) 3
Brown (BR) 3
Orange (O) 6
Red (R) 3
Theoretical Probabilities
With Replacement Without Replacement
P (2 reds) 6/24 5/24
P (R1BR2 OR BR1R2) 6/24
P (R1 AND G2) 6/24
P (G2 | R1)
P (no yellows) 18/24
P (doubles)
P (no doubles)
Am I on the right track or am I completely lost? I'd like help with figuring out the rest (if I'm right) or everything (if I'm not)
Anyway, I did the empirical probabilities easily. The theoretical ones are what's causing me trouble. It's the usual M&M lab :lol
Here's the data:
Population
Color Quantity
Yellow (Y) 6
Green (G) 3
Blue (BL) 3
Brown (BR) 3
Orange (O) 6
Red (R) 3
Theoretical Probabilities
With Replacement Without Replacement
P (2 reds) 6/24 5/24
P (R1BR2 OR BR1R2) 6/24
P (R1 AND G2) 6/24
P (G2 | R1)
P (no yellows) 18/24
P (doubles)
P (no doubles)
Am I on the right track or am I completely lost? I'd like help with figuring out the rest (if I'm right) or everything (if I'm not)
Subliminal
Member
Boogie9IGN said:
P(2 Reds) w/replacement: 3/24*3/24 = 1/63 w/o replacement = 3/24 * 2/24 = 1/96
i think
Boogie9IGN
Member
Subliminal said:
Wow I was hella off then :lol
so going by what you did, I have now:
With Replacement Without Replacement
P (2 reds) 1/64 1/96
P (R1BR2 OR BR1R2) 1/64 5/96
P (R1 AND G2) 1/64
P (G2 | R1) 1/64
P (no yellows) 9/16 17/32
P (doubles)
P (no doubles)
I can't get the blanks though and I'm thinking if all four are 1/64 I must've screwed up again
Boogie9IGN said:
i dont have a calculator that can transform decimals into fractions, so ill let you do the final work yourself, but...
Theoretical Probabilities
With Replacement Without Replacement
P (2 reds) 6/24 117/552
remember that if you arent replacing them, then the total number goes doesn. so for two reds, its 3/24+2/23=117/552
P (R1BR2 OR BR1R2) 9/24 117/552
6/24+3/24=9/24 with replacement, 6/24+3/23= ?? without replacement
P (R1 AND G2) 6/24
3/24+3/24=6/24 with replacement, 3/24+3/23= ?? without replacement
P (G2 | R1)
i dont know what this notation means
P (no yellows) 12/24
1- (6/24+6/24) = 12/24 with replacement, 1-(6/24+5/23)= ?? without replacement
P (doubles)
P (no doubles)
not sure on the last two, gotta think about it :/
Subliminal
Member
Boogie9IGN said:
Have you tried drawing out a tree diagram for this, it always helps.
With the doubles its the probability of multiple branches so P(YY)+P(RR)+...
Subliminal said:
doubles is basically asking what the probability is of getting YELLOWYELLOW OR ORANGE ORANGE OR GREEN GREEN OR BLUE BLUE OR BROWNBROWN OR REDRED... right?
i was thinking it would be
YELLOW:6/24*6/24
ORANGE: 6/24*6/24
GREEN: 3/24*3/24
BLUE: 3/24*3/24
BROWN: 3/24*3/24
RED: 3/24*3/24
with replacement, chances are .25
YELLOW:6/24*5/23
ORANGE: 6/24*5/23
GREEN: 3/24*2/23
BLUE: 3/24*2/23
BROWN: 3/24*2/23
RED: 3/24*2/23
without replacement, chances are 0.1521739
edit: this answer doesnt jive with my last post, i obviously forget too much about stats to remember when to add and when to multiply. sorry boogie
What are you talking about? Of course you can use L'Hôpital. Plugging in x = 1 gives 0/0, making it a valid situation.7h1ag0 said:help!
I know the answer is 3, and I can't use L'Hôpital's rule
Derivative of the top is a little tedious, but it's ((1/2) * (x^2 - 3x + 3) ^ (-1/2) * (2x - 3) - (1/2) * (x^2 + 3x - 3) ^ (-1/2) * (2x + 3))
Derivative of the bottom is 2x - 3
Subbing in x = 1 now will yield -3 / -1, or 3.
Unless you meant that your teacher will not allow you to use L'Hôpital's, in which case, fuck him or her, that's how this problem is done.
â Narayan
Member
Prove: (sinx+cscx)(sin^2x+csc^2x-1)=sin^3x+csc^3x
LOL. Nevermind.
Looking at basic differential equations in Calc I. I am a bit confused over this basic differential equation and its solution: dy/dx = ky & y = Ae^kx
|y| = e^c * e^kx
y = +/- e^c * e^kx
y = Ae^kx
I am completely fine with the formula and everything calculus about the darn thing... but I cannot understand how the absolute value "goes away" and the constant term becomes +/- Could someone explain this concept a bit more? Thanks If you know of any videos about this particular type of differential equation that'd be great too.
|y| = e^c * e^kx
y = +/- e^c * e^kx
y = Ae^kx
I am completely fine with the formula and everything calculus about the darn thing... but I cannot understand how the absolute value "goes away" and the constant term becomes +/- Could someone explain this concept a bit more? Thanks
If you know of any videos about this particular type of differential equation that'd be great too.Pyke Presco
Member
Not actually that hard, just multiply it out and expand, it all simplifies:
(sinx+cscx)(sin^2x+csc^2x-1)
=sin^3x + sinxcsc^2x - sinx + cscxsin^2x + csc^3x - cscx (this is just simple expansion)
Notice that cscx = 1/sinx
Then the term sinxcsc^2x = sinx*1/sin^2x=1/sinx=cscx
Similarly, sin^2xcscx = sin^2x*1/sinx=sinx
So we get
sin^3x + sinxcsc^2x [=cscx] -sinx + cscxsin^2x [=sinx] + csc^3x - cscx (bolded in brackets is just the simplification of the term)
=sin^3x + csc^3x + cscx - cscx + sinx - sinx
=sin^3x + csc^3x
EDIT: No problem, good luck with the rest
â Narayan
Member
Haha. Yeah. I just solved it. Brain fart. Thanks though!Pyke Presco said:
Pyke Presco
Member
barnone said:Looking at basic differential equations in Calc I. I am a bit confused over this basic differential equation and its solution: dy/dx = ky & y = Ae^kx
|y| = e^c * e^kx
y = +/- e^c * e^kx
y = Ae^kx
I am completely fine with the formula and everything calculus about the darn thing... but I cannot understand how the absolute value "goes away" and the constant term becomes +/- Could someone explain this concept a bit more? Thanks
Taking the absolute value of a number simply gives you its magnitude, ie the distance that number is away from zero, which is always a positive value.
Say that y=-2
The absolute value of this is |y|= 2 (since -2 is 2 units away from zero)
Now say that y=2
The absolute value of this is also |y|=2 (same reasoning)
Then clearly, if |y|=2, then y can be either +2 or -2
Since e^c is just a constant, we can think of it as a simple number, like 2 (or 10, or Pi, or e^c, whatever floats your boat). Therefore, just like above with |y|=2
in your case we would have:
|y|=e^c(constant)* e^kx(variable) and so
y=+/- e^c(constant) * e^kx(variable)
Sums and series suck.
So I'm trying to solve a limit.
lim of ((x^2) +(y^2) - 2y +1)/(xy-x)
(x,y)->(0,0)
I know that it doesn't exist, but I have to prove it. Is this a correct demonstration?
Since I know the function always has to approach the limit as I reach (0,0) from any path, I can do:
lim of ((x^2) +(y^2) - 2y +1)/(xy-x)
(x,y)->(0,y)
(the limit as I approach (0,0) moving on the y axis)
which has different one sided limits, which proves the last limit doesn't exist, and thus the first limit doesn't exist either.
Is that correct, GAF?
So I'm trying to solve a limit.
lim of ((x^2) +(y^2) - 2y +1)/(xy-x)
(x,y)->(0,0)
I know that it doesn't exist, but I have to prove it. Is this a correct demonstration?
Since I know the function always has to approach the limit as I reach (0,0) from any path, I can do:
lim of ((x^2) +(y^2) - 2y +1)/(xy-x)
(x,y)->(0,y)
(the limit as I approach (0,0) moving on the y axis)
which has different one sided limits, which proves the last limit doesn't exist, and thus the first limit doesn't exist either.
Is that correct, GAF?
Rich Uncle Skeleton
Member
Feep said:
Maybe they haven't learned it in the class yet. Maybe they haven't even learned derivatives yet. You can take that limit without L'hopital by multiplying the numerator and denominator by the conjugate of the numerator (i.e. the result of replacing the - in the middle by a +).
That gives (-6x + 6)/(x-2)(x-1)(conjugate). You can cancel x-1 to get -6/(x-2)(conj.).
Plugging in x=1 gives -6/(-1)(1+1), which is 3.
-COOLIO- said:
I don't remember the formula. But I can give you a geometicral construction to find the result. Would that work?
find the normal vector to the directional vector of the line. Use that vector as a directional vector of a line that passes through the point and find the intersection point of the two lines.
The distance between the intersection point and your original point is your distance.
Because it is not the 'length' between a point and a line, it is the 'distance'.-COOLIO- said:
And what type of point do you want to get? 2 dimensional or 3 dimensional? Correlation error (residual) or just the distance? You have to figure it out yourself lol.
Even though I suck at math, I will give you a good advise. Always remember the purpose of what you are trying to find, what it is exactly called and how it is interpreted.
Edit: this is a calculator if you are too lazy to calculate it. Online calculators FTW.
close to the edge
Member
Not sure if this has been posted already but this site here is a lifesaver if you have problems with math:
http://www.khanacademy.org/
He explains stuff pretty slowly and very thoroughly, so even for me who usually learns math in German (my native language) it's easy to understand. Thanks to him I was able to calculate the determinant of a 4x4-matrix and the eigenvectors and eigenvalues of a 3x3 matrix. (still a pain in the ass) By the way, Eigenvalue is a funny word.. the German word is Eigenwert, where Wert means value, so it's only half translated.
http://www.khanacademy.org/
He explains stuff pretty slowly and very thoroughly, so even for me who usually learns math in German (my native language) it's easy to understand. Thanks to him I was able to calculate the determinant of a 4x4-matrix and the eigenvectors and eigenvalues of a 3x3 matrix. (still a pain in the ass) By the way, Eigenvalue is a funny word.. the German word is Eigenwert, where Wert means value, so it's only half translated.
-COOLIO- said:ive been stuck on this for days. is there a solution or does the limit not exist?
So, your problem is that both the numerator and the denominator are zero when x = -1?
What you want to use is L'Hopital's rule. Basically, since they both go to zero (it also works if they both go to infinity), you can simply use the derivative of both the numerator and denominator instead.
So your problem goes:
lim(x->-1) (x^(1/3) + 1)/(x^(1/5) + 1)
=>
lim(x->-1) ((1/3)x^(-2/3))/((1/5)x^(-4/5))
So then, subbing in x=-1, you get
(1/3)/(1/5) = 5/3
Ventrue said:
Technically, the equation x^3 + 1 = 0 has 3 roots. Two are complex, one is -1. Typically, ^(1/3) denotes the "principle" cube root, which would be the real one, -1. It is similar to why sqrt(16) = 4, even though x^2 = 16 has solutions of 4 and -4.
So your line of thinking is completely fine.
Typically for Calc 1/2 stuff, you completely ignore imaginary numbers. If you have a sqrt(negative number), you just say its undefined and leave it as that. You are technically studying functions f: R --> R, i.e., real domain and real range.
Adding complex numbers leads to a whole mess of things, and you have to treat it much differently than you would in Calc 1.
womfalcs3
Banned
Ventrue said:
That's certainly a solution to z^3=-1, but graph the above function that Coolio posted. It's not defined in x<0. So how can the limit be 5/3 as x approaches -1 from either side?
I'm fairly sure the limit has an imaginary component.
Like bachikarn said, complex limits aren't dealt with in early calculus courses, so I don't know. I've never dealt with them myself.
EDIT: He can give the response in expression form?
(-1)^0.5 is i.
Rich Uncle Skeleton
Member
womfalcs3 said:
Yeah, I think he meant to say odd, not even, which is what we have in the problem. If x is a positive number then (-x)^3 = -x^3, so it makes sense to say that -x is a cube root of -x^3. Similarly for fifth roots.
womfalcs3
Banned
tsef said:
It's defined in x<0 in the reals? You sure? I get complex numbers. There's a real component, but there's also an imaginary one.
I just used Wolfram Alpha to check, and it shows a complex limit. I've tried several other functions to which I know the solution. It worked fine for them.
Typing... limit of (x^(1/3) + 1)/(x^(1/5) + 1) as x approaches negative 1
If I'm wrong, I'd like to learn from this.
Yeah, I just tested it in maple and I get a complex solution. But the thing is that x^(1/3) can be defined on R by using this : If x < 0 then x^(1/3) = -((-x)^(1/3)).
When I plot the function using Grapher i get this :
And there you can see that lim(f,x=-1)=5/3
*Edit* It looks like Maple doesn't use the real solution but the complex solution to (-1)^(1/3)
When I plot the function using Grapher i get this :
And there you can see that lim(f,x=-1)=5/3
*Edit* It looks like Maple doesn't use the real solution but the complex solution to (-1)^(1/3)
womfalcs3 said:
It is defined for x < 0. As tsef said, these programs aren't taking the real solution as the principle value. Matlab is doing it too.
The reason is because you can find those roots by De Moivre's theorem. It uses a polar representation for the complex numbers. The principle root can be defined as the root with the smallest angle in polar form.
So say x^3 - 1 = 0. It has solutions as shown in the following graph:
Clearly for 1^(1/3), you would think the answer is 1. As shown in the graph, 1 corresponds to the root with the smallest angle. This isn't the case for (-1)^(1/3).
The solutions to x^3 + 1 = 0 are shown:
We see from this one of the imaginary solutions has an angle that is the smallest. So these math programs (Wolfram Alpha, Matlab, Maple, etc) are outputting that as the "principle" value instead of the real one.
So back to the limit question, the function is defined for x < 0, and if you can set your math program to output the real values as tsef did, you can graph it and see that the limit would be 5/3.
womfalcs3 said:
Not really. It is all how you define your function. A function cannot have multiple outputs. So if you define x^(1/3) to output a complex number when x < 0, then it will still only have one limit. Complex functions tend to be "multi-valued functions," which is technically not a function. So you can have things where x^(1/3) will spit out 3 different values, but limits stop making sense in that context. What they do is restrict the range (output) to a principle "branch," so it can be treated as a function. Then you can start doing stuff like limits. This is analogous to having x^(1/3) only spitting out real numbers. You learn all about this stuff in a Complex Variables class.
womfalcs3 said:
As Bachikarn showed in his post when you look at x^(1/3) : R->C for each value of x you have 3 solutions so it's not a function, there lies the problem.
A function must have only one answer for each value you input. And in C and in R if a function has a limit it is unique.
Edit : Ok Bachi answered it more properly than I did ^_^
The limit does exist.-COOLIO- said:ive been stuck on this for days. is there a solution or does the limit not exist?
http://www07.wolframalpha.com/input/?i=lim+x-%3E-1+((x^1/3)+%2B+1)/((x^1/5)+%2B+1)
And it's pretty obviously continuous on both sides of the gap if you plot it.
http://www07.wolframalpha.com/input/?i=plot+x=[-2..0]+((x^1/3)+++1)/((x^1/5)+++1)
So GAF, I'm needing a considerable amount of help here.
I could do a couple of those, but in general I have some doubts (mostly on the sums and the limits, the others not so much). I could really use your help. Thanks in advance.
EDIT: Edited and left the limits I have trouble with.
I could do a couple of those, but in general I have some doubts (mostly on the sums and the limits, the others not so much). I could really use your help. Thanks in advance.
EDIT: Edited and left the limits I have trouble with.
harriet the spy
Member
BlueMagic said:So GAF, I'm needing a considerable amount of help here.
I could do a couple of those, but in general I have some doubts (mostly on the sums and the limits, the others not so much). I could really use your help. Thanks in advance.
EDIT: Edited and left the limits I have trouble with.
Because I am bored...
For 1), the integral of f(t) from a to b is equal to F(b)-F(a), where F is an antiderivative of f.
In your case, call F the antiderivative of f(ln(t)). You need F(exp(t))-F(1) = 1/2 t^2.
Say F(1) is 0 for kicks (antiderivatives are defined up to additive constants).
So you need F(exp(t))=1/2 t^2.
What is the F which satisfies this equation? (hint, change the variable t)
Once you find F, you can find f fairly easily.
For 2 and 3, every answer is easy once you remember the following fact (i'll let you prove them, since i don't know what tools yo have to prove the results)
- exponential of large positive numbers (exp
) are really fucking big.- exponential of large negative numbers (exp(-n)) are really fucking small (close to zero).
- a term of the term a^n can be rewritten as exp(n log (a)). In other words, if a>1, it's the exponential of a large positive number (if n is large), and a large negative number (if n is large).
- series converge when their terms go to zero really fast.
Rich Uncle Skeleton
Member
BlueMagic said:So GAF, I'm needing a considerable amount of help here.
I could do a couple of those, but in general I have some doubts (mostly on the sums and the limits, the others not so much). I could really use your help. Thanks in advance.
EDIT: Edited and left the limits I have trouble with.
Harriet explained the first one well, so I won't add anything. For 2ii you should use L'hopital. For the sums, in two of the three cases the terms don't even tend to zero, so the sum definitely won't converge. For the other, the ratio test is probably the easiest way to go (actually, the ratio test works for all three if you can't see why the terms get large in the other two).
^^^^^^^
If I were you, I would go to the school and ask around. First, go the tutoring lab and see if anyone there knows. Then try to find some math professors or even the dept. head.
Now, once you get in your class, GO TO DA TUTORING!! My math muscle was flabby and weak, but after going to the tutoring lab all de time, I got A! Hiya!!
If I were you, I would go to the school and ask around. First, go the tutoring lab and see if anyone there knows. Then try to find some math professors or even the dept. head.
Now, once you get in your class, GO TO DA TUTORING!! My math muscle was flabby and weak, but after going to the tutoring lab all de time, I got A! Hiya!!
Help guys.
Exam in 2 hours and I'm having trouble finding the second derivative of a quotient. I can use the quotient rule to find the first derivative easily but I get so confused when trying to derive the second.
For example, I know that for something like f(x)= x^2+1/x the derivative would be f'(x)=x^2-1/x^2 but I can't find the second derivative. Answers say it's f''(x)= 2/x^3
Exam in 2 hours and I'm having trouble finding the second derivative of a quotient. I can use the quotient rule to find the first derivative easily but I get so confused when trying to derive the second.
For example, I know that for something like f(x)= x^2+1/x the derivative would be f'(x)=x^2-1/x^2 but I can't find the second derivative. Answers say it's f''(x)= 2/x^3
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