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The Math Help Thread
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Biggest-Geek-Ever
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Your limit will equal |x/7| < 1 --> |x| < 7^1/3, making your interval of convergence of (-7^1/3, 7^1/3). Testing the first end point:
series 0 to infinity (-1)^n (2(-7^1/3)^3n+1)/7^n+1
= series 0 to infinity (-1)^(4n+1)*(2*7^n*7^1/3)/(7^n*7)
This does not converge by the p-series test as (7/7)^n = 1^n. Testing the second endpoint is similar.
r^2*r^2 = r^4, 4r^2 - r^2 = 3r^2, and 4*-1 = -4, so:
r^4 + 3r^2 - 4 = (r^2+4)(r^2-1) = (r^2+4)(r+1)(r-1)
Looking at that, one that immediately jumps out at me is (r - 1), since 1 + 3 - 4 = 0. (r + 1) works as well, because (-1)^4 + (-1)^3 - 4 is also 0. So already we have:
(r - 1)(r + 1)(ar^2 + br + c).
Comparing this to the original expression, a = 1 and c = 4 (-1 * 1 * c has to equal -4, and 1 * 1 * a has to equal 1):
(r - 1)(r + 1)(r^2 + br + 4)
Expand:
r^4 + br^3 + 3r^2 - br - 4
If we compare to the original expression, we see that b has to be 0.
Hence, (r - 1)(r + 1)(r^2 + 4) = (r - 1)(r + 1)(r - 2i)(r + 2i)
(Two real roots, two imaginary roots)
Edit: ninja'd. Guess that happens when I post on mobile!
Orbis Tabula
Member
Arguing with a friend over a probability problem. We're getting different answers and aren't sure who is right. I thought I'd throw it up here and see if I can get a third opinion to help us because we're not sure where the other one fucked up.
You have an urn with 8 amber balls, 8 blue balls, and 2 green balls. You draw 7 balls from it. What is the probability of getting at least one amber ball, at least one blue ball, and no green balls?
You have an urn with 8 amber balls, 8 blue balls, and 2 green balls. You draw 7 balls from it. What is the probability of getting at least one amber ball, at least one blue ball, and no green balls?
Can I have your results?
I got a quick and dirty result that probably is wrong:
(c(16,7)-2*c(8,7))/c(18,7) = 14/39.
Plz, tell me if I'm wrong since I'm studying this area that I have neglected. :S
My reasoning:
c(18,7) is the total number of possible draws.
c(16,7) is the number of possible draws without green balls
c(8,7) is the number of possible draws with only either amber or blue balls.
c(16,7)- 2*c(8,7) is the total number of possible draws without green balls and with at least one of each of the other.
By diving by the total number of combinations, we have the probability.
Orbis Tabula
Member
Can I have your results?
I got a quick and dirty result that probably is wrong:
(c(16,7)-2*c(8,7))/c(18,7) = 14/39.
Plz, tell me if I'm wrong since I'm studying this area that I have neglected. :S
My reasoning:
c(18,7) is the total number of possible draws.
c(16,7) is the number of possible draws without green balls
c(8,7) is the number of possible draws with only either amber or blue balls.
c(16,7)- 2*c(8,7) is the total number of possible draws without green balls and with at least one of each of the other.
By diving by the total number of combinations, we have the probability.
Looks very convincing to me. Also looks exactly like how my friend did it, which is discouraging to me. Hahaha. But it makes sense. I'm curious to see if everyone will get the same answer. The issue was that, like most probability problems, there are multiple ways of going at it. I chose a different way and got a different answer, but odds are you and my friend are both right.
Can you share how you did it?
Orbis Tabula
Member
Yeah, sure. Let me preface this by saying that I prefer doing all probability problems the way that you did this one. Find the size of the solution set, find the size of the set of all outcomes, and divide. I think it's the best way to do all probability problems, given a few rare exceptions. However, when I was given the problem on an exam, the professor specified that we should do it by counting each drawing as individual, independent events, and do the probabilities by multiplying each event. So tell me where I went wrong, if you can.
First you pick the first ball. Either amber or blue is good for this one, so the probability of getting a success on the first draw is (16/18). Next, you draw ball 2. WLOG you can make this draw the one where you draw the other color you need. So the probability of success is (8/17). Now you have 5 more balls to draw, and they can be amber or blue. It doesn't matter. So the third ball you have probability of (14/16). Then for the fourth ball, you have (13/15). Then fifth, (12/14). Then (11/13). And finally (10/12) for the last ball. All the drawings are independent, so you just multiply them all together.
(16/18)(8/17)(14/16)(13/15)(12/14)(11/13)(10/12)=.192
So, can anyone find my incorrect assumption?
Imo, this part is wrong. You don't necessary need that specific draw. What do you mean by WLOG?
Orbis Tabula
Member
Yeah, I'm sure I'm under counting something. I mean "without loss of generality" and I'm pretty sure that's where I'm making my mistake.
yeah, You can't assume WLOG there. There is indeed a loss of generality.
battousai11
Member
Do you know what the correct answer is? I ended up with .28
Care yo share how you did it?
Orbis Tabula
Member
I don't know yet but I'll post it when I find out. Could you show what you did?
I Stalk Alone
Member
Hi,
I have a math problem that I cannot figure out.
y = 791.19e^-0.03x
firstly how do i rearrange this equation to determine x?
I have a math problem that I cannot figure out.
y = 791.19e^-0.03x
firstly how do i rearrange this equation to determine x?
= ln(791.19e^-0.03x) => lnbattousai11
Member
I have to redo my math; I don't know that I did it correctly anymore.
I'll get back to you
Okay EskimoJoe. Got I just spent a little more time on your problem and I did it like your professor told you:
(16/18)(15/17)(14/16)(13/15)(12/14)(11/13)(10/12)-2(8/18)(7/17)(6/16)(5/15)(4/14)(3/13)(2/12) = 14/39
The first expression is the probability of getting no green balls, lets call it P1. The second is the probability of getting all amber or all blues, lets call it P2.
So, P = P1-P2.
There you go.
Again, I would really appreciate if someone could verify this results.
(16/18)(15/17)(14/16)(13/15)(12/14)(11/13)(10/12)-2(8/18)(7/17)(6/16)(5/15)(4/14)(3/13)(2/12) = 14/39
The first expression is the probability of getting no green balls, lets call it P1. The second is the probability of getting all amber or all blues, lets call it P2.
So, P = P1-P2.
There you go.
Again, I would really appreciate if someone could verify this results.
I get the same result with another method.
Let A be the number of amber balls, B the number of blue balls and G the number of green balls. You want P(A >= 1, B >= 1, G = 0).
Now, let's consider the possible cases:
If A = 1, then B = 6 and the amount of ways to draw 1 amber ball and 6 blue balls are C(8,1)*C(8,6).
If A = 2 then B = 5 and the amount of ways for this to happen is C(8,2)*C(8,5).
If A = 3 then B = 4 and the amount of ways is C(8,3)*C(8,4)
Same for A = 4, 5, 6 (note that A = 4 is the same than A = 3, A = 5 is the same than A = 2 and A = 6 is the same than A = 1).
Anyway, summing those cases gives you 11424.
The total amount of ways of extracting 7 balls is C(18,7) = 31824.
Finally, the probability you're searching for is 11424/31824 = 14/39.
-------------------------
Another similar way is calculating the probability of the negation, which means that either you get no amber balls or you get no blue balls or you get at least 1 green ball.
Once again, you consider 3 cases.
If G = 0 (no green balls), the only possible case is drawing either 7 amber balls or 7 blue balls: 2*C(8;7) cases.
If G = 1, it doesn't matter what else you draw: C(2,1) * C(16, 6) cases.
If G = 2, repeat: C(2, 2) * C(16, 5) cases.
Then, the probability of the negation is the sum of the cases divided by the total amount of cases, which is: 20400/31824 = 25/39.
Hence, the probability you were originally searching for is 1 - 25/39 = 14/39.
There are many ways of solving these problems!
Orbis Tabula
Member
Thanks for all the responses guys! I wish the professor hadn't specified for me to solve it in the way he did as I much prefer the other approach. Regardless, it was my mistake but I'm glad to get that cleared up.
Piercedveil
Member
EDIT: Think I got it now!
Piercedveil
Member
spelltropy
Member
Oswald fired a rifle from approximately 90 metres away.If the bullet travelled in a straight
line after leaving the rifle (at a velocity of ~700m/s) then the sight
aimed about 10 cm too high at a target 90 metres away. How much would the
drop in the trajectory due to gravity compensate for this? (The initial vertical
velocity is zero, and satisfies dv/dt = −g, while the horizontal velocity is constant
if we neglect air resistance.)
Can anyone push me in the right direction for this? Its a Diff eqns class but I can't figure out how to do it... Just looking for a push, not an answer, if possible ^^
line after leaving the rifle (at a velocity of ~700m/s) then the sight
aimed about 10 cm too high at a target 90 metres away. How much would the
drop in the trajectory due to gravity compensate for this? (The initial vertical
velocity is zero, and satisfies dv/dt = −g, while the horizontal velocity is constant
if we neglect air resistance.)
Can anyone push me in the right direction for this? Its a Diff eqns class but I can't figure out how to do it... Just looking for a push, not an answer, if possible ^^
Remember projectile motion from high school physics? You can basically derive those constant acceleration equations on your own now.
Position (s)
Velocity (v) = ds/dt
Acceleration (a) = dv/dt
In the vertical direction, dv/dt = -g
In the horizontal direction, dv/dt = 0
Integrate twice to get the equations for velocity and position. The constants that come out are the initial conditions.
It didn't like my answer. The y^2/16 was a pretty big hint to square everything, so after that I just solved for theta in terms of x. This involved taking the arcsin.
I can't tell if there is something wrong with my formatting or if I have no idea what I'm doing. I did the other ones fine but the trig functions in this one give me a bad feeling that I didn't do it right. Am I making a mistake?
It didn't like my answer. The y^2/16 was a pretty big hint to square everything, so after that I just solved for theta in terms of x. This involved taking the arcsin.
I can't tell if there is something wrong with my formatting or if I have no idea what I'm doing. I did the other ones fine but the trig functions in this one give me a bad feeling that I didn't do it right. Am I making a mistake?
You are making it more complicated than it needs to be. Just use the trigonometric identity sin^2(theta) + cos^2(theta) = 1
Needing more math help with parametric curves.
It's probably simple but I'm not seeing it.
x = e^t
y = (t-6)^2
Find an equation y = mx + b of the tangent to the curve at (1, 36)
So my equation of the line is y - y1 = m(x - x1)
or
y - 36 = m(x - 1)
All I need to find is the slope.
I take the derivatives of x and y with respect to t.
x' = e^t
y' = 2t - 12
And according to the formula dy/dx is dy/dt divided by dx/dt
Which just gives me (2t - 12)/e^t
Now I have a slope in terms of t that I'm not sure what to do with. Usually they give you a t-value instead of an (x,y) coordinate, and I would have used the t-value to find my (x,y) from the initial equations, and at this point I would also use that same t-value to find my slope.
But I can't do that since I don't know t.
And even more confusing, if I plug in the x and y values into the equations and solve for t, I get two different values for t.
(1, 32)
1 = e^t
t = 0
32 = (t-6)^2
t = not zero
So I'm stuck because I don't know how to find the slope.
Any help?
It's probably simple but I'm not seeing it.
x = e^t
y = (t-6)^2
Find an equation y = mx + b of the tangent to the curve at (1, 36)
So my equation of the line is y - y1 = m(x - x1)
or
y - 36 = m(x - 1)
All I need to find is the slope.
I take the derivatives of x and y with respect to t.
x' = e^t
y' = 2t - 12
And according to the formula dy/dx is dy/dt divided by dx/dt
Which just gives me (2t - 12)/e^t
Now I have a slope in terms of t that I'm not sure what to do with. Usually they give you a t-value instead of an (x,y) coordinate, and I would have used the t-value to find my (x,y) from the initial equations, and at this point I would also use that same t-value to find my slope.
But I can't do that since I don't know t.
And even more confusing, if I plug in the x and y values into the equations and solve for t, I get two different values for t.
(1, 32)
1 = e^t
t = 0
32 = (t-6)^2
t = not zero
So I'm stuck because I don't know how to find the slope.
Any help?
Perfect Cha0s
Member
You can also put t in terms of x.
x=e^t -> ln(x) = t
y = [ln(x)-6]^2
dy/dx = 2[ln(x)-6] * (1/x)
x=e^t -> ln(x) = t
y = [ln(x)-6]^2
dy/dx = 2[ln(x)-6] * (1/x)
Permanently A
Junior Member
Having trouble understanding implicit differentiation. I watched Khan Academy's video on it, and I understand it in the context of x^2 + y^2 = 0 since the value of y is a function of x so you use chain rule.
However, I don't understand it in this problem:
Why does x have a dx/dt? Isn't x the independent variable, not the dependent?
However, I don't understand it in this problem:
Why does x have a dx/dt? Isn't x the independent variable, not the dependent?
Permanently A
Junior Member
Oh! Okay, I get it. The derivative is with respect to t, not x. Thanks!
An ecologist wishes to mark off a circular sampling region having radius 10 m. However, the radius of the resulting region is actually a random variable R with the following pdf.
What is the expected area of the resulting circular region?
m2
Anyone know how I would go about answering this in R? I can't for the life of me figure out how to write the code.
What is the expected area of the resulting circular region?
m2
Anyone know how I would go about answering this in R? I can't for the life of me figure out how to write the code.
boviscopophobic
Member
Are you supposed to give an approximate answer via sampling, or an exact answer? If you're supposed to sample, you could use either CDF inversion or rejection sampling to generate draws from the specified PDF. Another strategy could be to use the integrate function to numerically compute the required integral. Either way, being forced to "use R" for this problem is pretty silly since a) it's easy to solve analytically without R; b) R doesn't seem to offer any special support for this problem.
Also, if the ecologist actually wants a radius of 10 meters, he is pretty incompetent at his job.
Also, if the ecologist actually wants a radius of 10 meters, he is pretty incompetent at his job.
boviscopophobic
Member
Hmm. Well, R doesn't have any built-in symbolic integration capabilities (though there are third-party packages to provide this), so it's not really a good choice if you want an exact result. It seems designed more for dealing with empirical data, which points to an approximate approach via simulation.
spelltropy
Member
How would I go about proving that if a sequence tends to infinity, its reciprocal tends to 0?
Here's a rough outline: Pick an epsilon > 0, then compare the sequence with 1/epsilon, applying the definition of a sequence tending to infinity. Then take another reciprocal so you are comparing with epsilon, and apply the definition of a sequence tending to zero.
In response to concerns about nutritional contents of fast foods, McDonald's has announced that it will use a new cooking oil for its french fries that will decrease substantially trans fatty acid levels and increase the amount of more beneficial polyunsaturated fat. The company claims that 97 out of 100 people cannot detect a difference in taste between the new and old oils. Assuming that this figure is correct (as a long-run proportion), what is the approximate probability that in a random sample of 1000 individuals who have purchased fries at McDonald's,
(a) At least 32 can taste the difference between the two oils?
The binomial distribution I used pbinom(32,1000,0.03) which gave me 0.6865675.
1-(32-0.5-1000*0.03)/sqrt(1000*0.03*0.6865675) = 0.3305134
Which gives me 0.3305134. And yet I can't find the prob on the table of standard normal curve areas. Anyone know where I went wrong?
https://www.stat.tamu.edu/~twehrly/651/ztable.pdf
(a) At least 32 can taste the difference between the two oils?
The binomial distribution I used pbinom(32,1000,0.03) which gave me 0.6865675.
1-(32-0.5-1000*0.03)/sqrt(1000*0.03*0.6865675) = 0.3305134
Which gives me 0.3305134. And yet I can't find the prob on the table of standard normal curve areas. Anyone know where I went wrong?
https://www.stat.tamu.edu/~twehrly/651/ztable.pdf
Biggest-Geek-Ever
Member
0.3305134 is only the resultant of (32-0.5-1000*0.03)/sqrt(1000*0.03*0.6865675). Subtract that from 1 and you will get a probability of 0.6694.
ItWasMeantToBe19
Banned
Starting Analysis next semester, how should I prepare for it?
Permanently A
Junior Member
This is probably a pretty simple question that I'm just not getting.
Why is the derivative of xa^22 just a^22?
Why don't you use the product rule and get (22a^21)x + 1(a^22) ?
Why is the derivative of xa^22 just a^22?
Why don't you use the product rule and get (22a^21)x + 1(a^22) ?
ItWasMeantToBe19
Banned
Is "a" a constant in this equation?
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