Partial Gamification
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Use the unit vector. Normalize like simplayer said, same direction length one.
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The Math Help Thread
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campfireweekend
Member
Sorry for being dense, but am I just normalizing r(t)? Because that won't give me a numerical answer will it?
Maybe its because its not in conventional vector form. The 1st question was quite easy because it gave me another point and thus an apparent direction, but I only have one point in this problem.
So my direction unit vector would be 1/|r(t)|(5sin(2pit), 0, 5/3cos(2pit)?
Partial Gamification
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Yep, r(t)/|r(t)| ; "multiply that with your differentiated T function."
Partial Gamification
Banned
Explain what you know, what you need to know for the test (the material you have covered in class-briefly), and any questions you have about that.
Not to just throw a question back at you (and it might sound stupid) but it seems like a good study chore to have you type that up. Get it in your head, even if its a lot to summarize, don't remember it all at once and type up what you know about polar equations.
Basically I have to know the back and forth between polar and cartesian, and finding solutions for each.
Partial Gamification
Banned
The Pythagorean theorem is your friend, sorry if this is late.
Think of the x and y coords from Cartesian as the sides of a right triangle.
r is the hypotenuse of that triangle, so: r^2 = x^2 + y^2 [edit or:] sin^2(θ
+cos^2(θ = r^2From the trig identities theta can be found with: θ = tan^-1 (y/x)
The direct conversion (to memorize for the test) is:
x = r*cos(θ
y = r*sin(θ
The graph helps me. If you have a polar equation and are unsure, pointplot from zero to 2*pi, every pi/6 or pi/4 to get an idea of the graph.
There are some common ones, r=1 is the same as x^2 + y^2 = 1
What would θ = 0 look like?
The Pythagorean theorem is your friend, sorry if this is late.
Think of the x and y coords from Cartesian as the sides of a right triangle.
r is the hypotenuse of that triangle, so: r^2 = x^2 + y^2 [edit or:] sin^2(θ
From the trig identities theta can be found with: θ = tan^-1 (y/x)
The direct conversion (to memorize for the test) is:
x = r*cos(θ
y = r*sin(θ
The graph helps me. If you have a polar equation and are unsure, pointplot from zero to 2*pi, every pi/6 or pi/4 to get an idea of the graph.
There are some common ones, r=1 is the same as x^2 + y^2 = 1
What would θ = 0 look like?
Thanks for the help. I think I did alright. It mostly depends on how insane my professor's grading will be this time around.
I have never really been any good at maths, and this introductory calculus class is killing me. I'm trying to solve this problem:
I get that it's true from substituting random values into it, but have no idea how to show it's true for all values of a≥1. I am thinking it has something to do with 3! being equal to 3x2x1 which are the same values in the parentheses? Is that on the right track at all?
I get that it's true from substituting random values into it, but have no idea how to show it's true for all values of a≥1. I am thinking it has something to do with 3! being equal to 3x2x1 which are the same values in the parentheses? Is that on the right track at all?
toddhunter
Member
Whenever you see a≥1 in a question it is a fair bet the method needed will be proof by mathematical induction. It shouldn't be too hard using that approach.
Partial Gamification
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I'm going to work toddhunter's suggestion, just for the enjoyment of a little coffee and induction.
Proof by induction:
base step:
n = 1, n^3 + 6n^2 +11n + 6 = 1 + 6 + 11 + 6, 24/6 = 4
n = 2, 8 + 24 + 22 + 6 = 60, 60/6 = 10
inductive step:
assume: (n^3 + 6n^2 +11n + 6)/6 is a Natural Number.
step (n+1): (n+1)^3 + 6(n+1)^2 + 11(n+1) + 6 = ...
..= (n^3 + 3n^2 + 3n + 1) + (6n^2 +12n +6) + (11n +11) + 6
reduces to: n^3 + 9n^2 + 26n + 24
Getting nth step out of (n+1)th step:
n^3 + 6n^2 + 3n^2 + 11n + 15n + 6 + 18 = ...
...= (n^3 + 6n^2 + 11n + 6) + (3n^2 + 15n + 18),
3n^2 + 15n + 18 = 3( n+2 )( n+3 ),
3(n+2)(n+3)/6 = (n+2)(n+3)/2
If (n+2)(n+3) is even then the inductive step holds.
(n+2)(n+3) = n^2 + 5n + 6,
if n=even, then: even^2 + even + even = even,
if n=odd, then: odd^2 + odd + even = odd + odd + even,
odd + odd = even, and thus the inductive step holds.
Therefore, the given polynomial is divisible by 6 for all Natural Numbers.
It might be easier not expanding the given.
Could someone please explain to me how to calculate th probability of this situation?
A test consists of 25 multiple-choice questions. Assume that a test taker randomly guesses a choice for each and every question, and each correct answer is worth 4 points. Clearly state what probability model(s) you use and what value each parameter is equal to.
If 22 questions has three choices each (one correct, two incorrect) and three has true-false questions (one correct, one incorrect), what is the probability that the test taker will get a B or better?
Thanks
A test consists of 25 multiple-choice questions. Assume that a test taker randomly guesses a choice for each and every question, and each correct answer is worth 4 points. Clearly state what probability model(s) you use and what value each parameter is equal to.
If 22 questions has three choices each (one correct, two incorrect) and three has true-false questions (one correct, one incorrect), what is the probability that the test taker will get a B or better?
Thanks
There are 100 points maximum, and you want to know what the probability is that a person scores at least 80. That means the person can miss 0, 1, 2, 3, 4, or 5 questions.
Do you know how you'd go about doing this if all 25 of the questions had three choices?
It's possible to get a B just by getting 20-22 of the 22 three-choice questions right. In that case it doesn't matter how many of the true/false questions are answered correctly. But if only 19 of the multiple choice questions are answered correctly, at least one of the true/false questions has to be answered correctly.
Does that help?
Yes, I think I know how to solve it if all 25 had three problems. P(x<=5) binomcdf(25,2/3,5)= 2.269E-6
Here goes.
I am using this format. P(x=5) means the probability of exactly 5 wrong answers.
p=2/3 for the first 22 and p=1/2 for the last 3.
For example, if I only want to have exactly 80 then the probability of getting 5 wrong out of 22 (p=2/3) is multiplied with the probability of getting 0 wrong out of 3 (p=1/2) and I add down until I have 2 out off 22 wrong multiplied by 3 out 3 wrong, correct?
P(x=5)
+(22 nCr 5)*(2/3)^5*(1/3)^17 * (3 nCr 0)*(1/2)^0*(1/2)^3
+(22 nCr 4)*(2/3)^4*(1/3)^18 * (3 nCr 1)*(1/2)^1*(1/2)^2
+(22 nCr 3)*(2/3)^3*(1/3)^19 * (3 nCr 2)*(1/2)^2*(1/2)^1
+(22 nCr 2)*(2/3)^2*(1/3)^20 * (3 nCr 3)*(1/2)^3*(1/2)^0
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Then add those up. and I get P(x=5)= 4.90619E-06, which means the probability of getting exactly 5 wrong. That should be the answer if I want exactly 80. But the question is asking for at least 80. So I keep doing this down to P(x=0)
P(x=4)
+(22 nCr 4)*(2/3)^4*(1/3)^18 * (3 nCr 0)*(1/2)^0*(1/2)^3
+(22 nCr 3)*(2/3)^3*(1/3)^19 * (3 nCr 1)*(1/2)^1*(1/2)^2
+(22 nCr 2)*(2/3)^2*(1/3)^20 * (3 nCr 2)*(1/2)^2*(1/2)^1
+(22 nCr 1)*(2/3)^1*(1/3)^21 * (3 nCr 3)*(1/2)^3*(1/2)^0
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Then I do the same for P(x=3), P(x=2), P(x=1), and(P(x=0).
And finally add them all up to get the final probability. P(Atleast 80)= P(x=5)+P(x=4)+...+P(x=0).
Is this correct?
Yeah, I think that's right. A quick monte carlo simulation suggests the answer is about double what it was for all 25 being multiple choice.
Dropped_Off
Neo Member
This is a metric space question, I can generally see where the answer is going, but I'm not too sure how to start it, since this is for uni work just an example of a similar question or the theory would be helpful.
Let p be a prime number. In the field of rational numbers Q, let:
Prove that:
a) ||r1 + r2||p <= max{ ||r1||p, ||r2||p }
For clarification the p's at the end of the absolute value markers are meant to be subscript.
Any help would be great, thanks.
Let p be a prime number. In the field of rational numbers Q, let:
||r||p = p^(-k), if r = p^k * (m/n)
where m, n are co-primes with p. Then define:d(x, y) = ||x - y||p
Prove that:
a) ||r1 + r2||p <= max{ ||r1||p, ||r2||p }
For clarification the p's at the end of the absolute value markers are meant to be subscript.
Any help would be great, thanks.
Partial Gamification
Banned
I'm thinking (and could be wrong about setup) the LHS of the inequality can be represented as the distance, d(r1, -r2) or |r1 - (-1)*r2|ₐ.
Use the definition of |r|ₐ to both sides. (I couldn't find subscript p)
How is k defined? As a→∞ the |r|ₐ get smaller?
The only other thing coming to mind is that the absolute values and the whole of the inequality can be squared. These are just positive values, the associates aren't being considered are they? I'm not 100% on the best route to a solution here, it looks like the inequality needs to be in another form and them a little algebra needs to be used; my apologies if that's just a "no shit sherlock" sort of statement. Maybe a three parter with r1<r2, r1=r2, and r2<r1 ?
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Is a HTML entity (decimal) the best way to input special characters? Aside from cutting and pasting? Is there a 'p' subscript?
I'm having trouble telling which parameters are necessarily the same for r1 and r2 and which can differ. r1 = p^k1 * (m1/n1) and r2 = p^k2 * (m2/n2)? Or do the m and n have to be the same for both - are m and n supposed to be given in the metric in the way that p probably is?
Suppose they are. Then r1 = p^k1 * (m/n), r2 = p^k2 * (m/n), and r1+r2 = (p^k1 + p^k2) * m/n. The thing you're trying to prove is that if we let p^k3 = p^k1 + p^k2, then k3 >= max(k1, k2), because k is the only free parameter when evaluating the metric. Intuitively this is pretty straightforward, since p is positive, p^a is positive, and the sum of two positive numbers is at least as great as either of them.
Partial Gamification
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From |r|_p = p^(-k), if r = p^k * (m/n)
can't it be said that:
|(p^k)*(m/n)| = p^(-k)
so that;
m/n = (+/-)p^(-2k)
For the bolded, isn't that the reverse of the inequality:
The positive sum is less than or equal to the largest of the summands.
Could you let: |r1+r2|_p = |r3|_p
and: |r2+r3|_p = |r4|_p
... |r_{n-2} + r_{n-1}|_p = |r_n|_p
Maybe I'm reading the question all wrong or don't know enough about this topic but it seems like the only way to prove this is to have an alternating sequence (+/-) for
{r1, r2, ..., r_n}. [
edit: I don't think its possible to have an alternating sequence;
with: m/n = p^(-2k)
then: r = p^(-k) -or- r = |r|
A math major friend of mine points out that k needs to be an integer and m and n need to depend on r in order for r1 + r2 to always be rational.
In that case you start by setting up r1+r2 = p^(k2) *(p^(k1-k2)m1/n1 +m2/n2)), for k1 >= k2. The goal is to show that that big thing on the right works out to be write-able as m3/n3, where both are coprime with p. She worked that out and seemed convinced the proof held (prove by assuming that the numerator is not co-prime with p).
http://en.wikipedia.org/wiki/P-adic_number#Analytic_approach
Partial Gamification
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Dropped_Off
Neo Member
Thanks for all the help, got a fairly good idea how to handle this question now.
Hey GAF I need help with a probability question. I'm not sure how to do it, but I have the feeling once someone explains it to me I'll be frustrated I didn't know how to do it... anyways here it is:
Milk Chocolate (aka plain) M&Ms come in six different colors. According to the manufacturer, 23% of these M&Ms are blue, 14% are brown, 16% are green, 20% are orange, 13% are red, and 14% are yellow. Suppose you open up a pack of M&Ms and pour out two candies. What is the probability that these two are the same color?
Milk Chocolate (aka plain) M&Ms come in six different colors. According to the manufacturer, 23% of these M&Ms are blue, 14% are brown, 16% are green, 20% are orange, 13% are red, and 14% are yellow. Suppose you open up a pack of M&Ms and pour out two candies. What is the probability that these two are the same color?
The probability of pouring out two blues is 0.23^2. The probability of pouring out two browns is 0.14^2. And so on. Add up all the possibilities and that should be the answer.
So all I would do is 0.23^2 + 0.14^2 + 0.16^2 etc for each color?
Have a rather simple question that I wanted to get checked. A random variable X has a probability density function fx(x) = Ax where 2<= x <= 8. The question asks to determine A, and I believe I just need to integrate the pdf and set it equal to 1 to get A=1/30. Can someone confirm/deny this? Thanks.
EDIT: Also, if I have two random independent variables how do I find the join pdf of them? I can figure out each's pdf as I'm given a range but I don't know what to do to make a join pdf from them.
EDIT: Also, if I have two random independent variables how do I find the join pdf of them? I can figure out each's pdf as I'm given a range but I don't know what to do to make a join pdf from them.
velociraptor
Junior Member
Note: / = square root
/49,000 = /4.9 x l0^4 = 2.1 x l0^2
I'm currently preparing for the MCAT and the book shows how to 'calculate' square roots without a calculator.
Now my question is... how exactly does one calculate the square root of 49,000 using scientific notation, like the above, without a calculator?
I cannot figure out intermediary steps. Is there something I am missing?
/49,000 = /4.9 x l0^4 = 2.1 x l0^2
I'm currently preparing for the MCAT and the book shows how to 'calculate' square roots without a calculator.
Now my question is... how exactly does one calculate the square root of 49,000 using scientific notation, like the above, without a calculator?
I cannot figure out intermediary steps. Is there something I am missing?
First thing first. /49,000 =221.359436212 = 2.2x10^2, (not sure what does this say about our doctors...)
If you only need the answer to the first decimal place, just do "trial and error", /49,000 = /4.9 x l0^4 , sqrt of 10^4 is 100, and
2*2=4, 2.1*2.1 = 4.41, 2.2*2.2 = 4.84, good enough. Almost as fast as pressing keys on a calculator.
velociraptor
Junior Member
Yeah I had copied that straight from the book lol.
I still don't see how I can do the trial and error in my head like that. I suppose there is no other way to work it out then?
You are not allowed even a piece of blank paper?
Buggy Loop
Gold Member
Finding the error with the approximation of taylor with 2 variables is killing me. Any good sites that explain this?
I have for example,
f(x,y) = sin(x) + sin
I have to find the error of approximating it with a Taylor polynomial of degree 1.
B (the disk circling the error) is
B(pi/4, pi/4) with distance d=1
Now
fx' = cos(x)
fy' = cos
The formula for linear error is |El(x,y)| <= 2 Ml d^2
d^2 = 1, no problems.
Ml though, i know that max value of sin(x), sin, cos(x) or cos would be 1
and the answer is |El(x,y)| <= 2
But i dont know how he got there, what happened to the (pi/4, pi/4) ? or what do i do with the derivatives?
B(pi/4,pi/4) = {(x,y) | (x-(pi/4))^2 + (y-(pi/4))^2 <= 1 }
Sorry if its a not using typical USA terminology, sometimes maths from french to english can make things incomprehensible.
I have for example,
f(x,y) = sin(x) + sin
I have to find the error of approximating it with a Taylor polynomial of degree 1.
B (the disk circling the error) is
B(pi/4, pi/4) with distance d=1
Now
fx' = cos(x)
fy' = cos
The formula for linear error is |El(x,y)| <= 2 Ml d^2
d^2 = 1, no problems.
Ml though, i know that max value of sin(x), sin
, cos(x) or cos would be 1and the answer is |El(x,y)| <= 2
But i dont know how he got there, what happened to the (pi/4, pi/4) ? or what do i do with the derivatives?
B(pi/4,pi/4) = {(x,y) | (x-(pi/4))^2 + (y-(pi/4))^2 <= 1 }
Sorry if its a not using typical USA terminology, sometimes maths from french to english can make things incomprehensible.
Start by attaching probabilities to the three cases in the question.
The probability that Lee wins immediately is 1/6.
The probability that Lee rolls something other than a 4, then Sharland rolls something other than a 4, then Lee rolls a 4 is (5/6)^2 * 1/6.
The probablity that four non-4 rolls occur followed by a roll of 4 is (5/6)^4 * 1/6.
So for these three cases, the total probability that Lee wins is 1/6 * (1 + (5/6)^2 + (5/6)^4). See where that's going?
You could start by writing out the probabilities for the three situations described.
Professor Lee wins if he throws a 4 immediately
p_0 = (1/6)
the results are non-4 for Lee, non-4 for Sharland, 4 for Lee
p_1 = (5/6)*(5/6)*(1/6)
non-4 for Lee, non-4 for Sharland, non-4 for Lee, non-4 for Sharland, 4 for Lee
p_2 = (5/6)*(5/6)*(5/6)*(5/6)*(1/6)
There's a pattern here. 5/6 multiplied together an even number of times, and then multiplied by 1/6. You could write in general, the probability of a situation in which Lee wins is:
p_n = (1/6)*(5/6)^(2n)
Now you just need to sum up the probabilities of all the situations in which Lee wins.
SUM (n=0->infinity) (1/6)*(5/6)^(2n)
I think that's right ... not completely sure. Hope it helps!
Professor Lee wins if he throws a 4 immediately
p_0 = (1/6)
the results are non-4 for Lee, non-4 for Sharland, 4 for Lee
p_1 = (5/6)*(5/6)*(1/6)
non-4 for Lee, non-4 for Sharland, non-4 for Lee, non-4 for Sharland, 4 for Lee
p_2 = (5/6)*(5/6)*(5/6)*(5/6)*(1/6)
There's a pattern here. 5/6 multiplied together an even number of times, and then multiplied by 1/6. You could write in general, the probability of a situation in which Lee wins is:
p_n = (1/6)*(5/6)^(2n)
Now you just need to sum up the probabilities of all the situations in which Lee wins.
SUM (n=0->infinity) (1/6)*(5/6)^(2n)
I think that's right ... not completely sure. Hope it helps!
p = success probability = 1/6
q = failure probability = 5/6
x = max number of attempts = infinite
X = finite number of attempts (where prof Lee's attempts will be odd numbered since he goes first)
P(X < x) = P(X=1) + P(X=3) + P(X=5) + ... = p + q^2p + q^4p + ...
using a geometric series where a (the first term) is p and r (the common ratio) is q^2 you get:
P = a/(1-r)
= p/(1 - q^2)
= (1/6)/(1 - (5/6)^2)
= (1/6)/(1 - 25/36)
= (1/6)/(11/36)
= 36/66
= 6/11
Wow, you guys all did a great job of explaining it! I got it Gotchaye explained it but I appreciate everyone's help a lot. I may be coming to you guys when I get stuck more often, thank you all so much for being so prompt and helpful, I really appreciate it. Definitely subscribing to this thread 
Unless I'm misinterpreting the question, the number of entries in a joint probability distribution should just be the total number of possible system states (a probability distribution just says what the probability of each state is).
If you have a system with two binary nodes and one ternary node, you have 2*2*3 = 12 possible states (000,001,002,010,011,012,100,101,102,110,111,112).
If you have a system with four binary nodes and twelve decimal nodes, you have (2^4)*(10^12) = 1.6*10^13 possible states.
Etc.
thequickandthedead
Member
How do i simplify this?:
6(3^(k)-1) + (4×3^(k+1))
6(3^(k)-1) + (4×3^(k+1))
Partial Gamification
Banned
(2*3)*(-1+3^k) + 4*3^(k+1)
2*(-3+3^(k+1)) + 4*3^(k+1)
-6 + 2*3^(k+1) + 4*3^(k+1)
-6 + 6*3^(k+1)
6*(3^(k+1) - 1)
2*(3^(k+2) - 3)
second to last line is probably as reduced as it gets.
thequickandthedead
Member
Thanks. I hadn't thought of splitting the 6 into 2*3.
This is a binomial problem. I haven't done them in a long time, so I need a little help. We did a problem in class, but I can't remember how we got to the next step.
Problem: (((2x) / 3) - (3 / (4 * √x ) )) ^ (-2/3)
This is the next step:
((2x)/3)^(-2/3) * (1 - 9 / (8x * √x )) ^ (-2/3)
How did we get (1 - 9 / (8x * √x ))?
Problem: (((2x) / 3) - (3 / (4 * √x ) )) ^ (-2/3)
This is the next step:
((2x)/3)^(-2/3) * (1 - 9 / (8x * √x )) ^ (-2/3)
How did we get (1 - 9 / (8x * √x ))?
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