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The Math Thread

Tannhauser

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I believe that is impossible :)

It's the integral of the standard normal Gaussian density. As far as I know it cannot be expressed in terms of elementary functions.
 

Tannhauser

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An easy one:

gif.latex


Btw: site that generates formatted equations as pictures: http://www.codecogs.com/latex/eqneditor.php
 

Infinitatis

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Infinitatis

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I believe that is impossible :)

It's the integral of the standard normal Gaussian density. As far as I know it cannot be expressed in terms of elementary functions.

Exactly right. The integral can only be expressed with a non-elementary function, the error function.

1/2 erf(z/√2)
 

Tannhauser

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@Infinitatis Your integration-by-parts is correct :)

What about this one:

gif.latex
 

Cheeseumpuffs

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My notation is going to look like shit.

u=x
du=1
v=-e^(-(x^2)/2)
dv=e^(-(x^2)/2)

Integration by parts
Integral(u dv)= Integral(v du) - uv
so
Integral(xe^(-(x^2)/2))
= Integral(-e^(-(x^2)/2)) + xe^(-(x^2)/2)
= e^(-(x^2)/2) + xe^(-(x^2)/2)
evaluated from a to inf
= [e^(-inf) + inf(e^(-inf))] - [e^(-(a^2)/2) + ae^(-(a^2)/2)]
= [0 + inf(0)] - [e^(-(a^2)/2) + ae^(-(a^2)/2)]
= -e^(-(a^2)/2) - ae^(-(a^2)/2)]


I think.
I'm drunk.


Edit:This isn't right. Give me a minute
Edit2: this would be easier if I actually had pen and paper. I'll be back tomorrow
 

Tannhauser

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Hint for the integral: substitute u = x²/2
 

Tannhauser

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An easy proof: show that for all reals x, y,
gif.latex
 

Tannhauser

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No love for the integral either?

With the substitution, it becomes almost trivial:

gif.latex
 

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I remember hearing it somewhere, intuitive mathematics is my favourite so here:

What's the probability that someone lives in an even numbered house?
 

Infinitatis

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I remember hearing it somewhere, intuitive mathematics is my favourite so here:

What's the probability that someone lives in an even numbered house?

Since house numbering always begins with an odd number (1) and ends with either an odd or even number, the number of even-numbered houses is theoretically guaranteed to be less than or equal to the number of odd-numbered houses. Thus the probability that a person lives in an even-numbered house, depending on the number of houses per numbering system, is probably a little bit less than 1/2.

Here is some rough math.
Houses are numbered by the natural numbers greater than 0. Let's assume that the average number of houses in a numbering system is either 1000 or 1001. The probability that a person lives in an even-numbered house in a set of 1000 houses is 1/2, but is 500/1001 in the set of 1001 houses.

I would have to know how many houses on average are in a numbering system to calculate the probability, but considering most numbering systems have a large number of houses, I will place my guess at:

0.499±0.0005
 

Infinitatis

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Albert, Bob, and Carl have each individually scheduled to meet with Dr. Fraud, a psychologist. Dr. Fraud wants to determine whether each patient is extraverted or introverted. He has not had any sessions with any of the three patients yet, so he does not have any information to go off of, except for one thing: he knows that two of them are introverts and that one of them is an extravert.

He wants to know if they are extraverted or introverted before meeting them, so he guesses. He guesses that Albert and Bob are the introverts and that Carl is the extravert.

Albert, Bob, and Carl all know one another. They know who is the extravert, and who are the introverts. They also know that Dr. Fraud thinks the extravert is Carl. Based on Dr. Fraud's guess, they decide to let one of the introverts that is not the one who Dr. Fraud believes the extravert is to meet with him first, so they send Bob, and the other two will meet with him later.

After Dr. Fraud meets with Bob, he knows for a fact that he is an introvert.

Who is most likely the extravert?
 

The Grey Man

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Probability that the extrovert is either Alex or Carl: 2/3

Probability that the extrovert is either Alex or Carl, given that Bob is an introvert: 1

Probability that extrovert is Carl, given that it's either Alex or Carl: 1/2

Trick question.
 

Infinitatis

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Probability that the extrovert is either Alex or Carl: 2/3

Probability that the extrovert is either Alex or Carl, given that Bob is an introvert: 1

Probability that extrovert is Carl, given that it's either Alex or Carl: 1/2

Trick question.

So it would seem, but that is incorrect. Flummoxing, isn't it?
 

Infinitatis

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I fixed a subtle issue in my story problem, but it should be good now.
 

Tannhauser

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The doctor's guess is either correct for both of them or just one of them. A-priori, the probability for having both correct is 1/3. Now, if he schedules a meeting with either Albert or Bob, and picks one of them randomly, and then finds out that his guess was correct for that person, it increases the probability that both of them are introverts.

By using Bayes formula etc, the probability of the guess being correct for both of them increases to 1/2 after the meeting. When that is the case, we know that Carl is an extrovert. When it is not the case, he has a 0.5 probability of being extrovert. So in total, Carl has a 0.5 + 0.5*0.5 = 0.75 probability of being extrovert after the meeting.
 

Infinitatis

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The doctor's guess is either correct for both of them or just one of them. A-priori, the probability for having both correct is 1/3. Now, if he schedules a meeting with either Albert or Bob, and picks one of them randomly, and then finds out that his guess was correct for that person, it increases the probability that both of them are introverts.

By using Bayes formula etc, the probability of the guess being correct for both of them increases to 1/2 after the meeting. When that is the case, we know that Carl is an extrovert. When it is not the case, he has a 0.5 probability of being extrovert. So in total, Carl has a 0.5 + 0.5*0.5 = 0.75 probability of being extrovert after the meeting.

Close but not cigar. You were on the right track though.
 

Infinitatis

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I modified the wording slightly, so it makes a little bit more sense hopefully.
 

The Grey Man

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You made me incorrect with your modification :p

Anyway, probability that the first introvert is not Albert/Bob, given that it is either Albert or Bob: 1/2

Probability that Albert/Bob is the extravert, given that he is not the first introvert: 1/2

Probability that Albert/Bob is the extravert: 1/2 * 1/2 = 1/4

Probability that the extravert is either Alex or Bob: 1/4 + 1/4 = 1/2

Probability that Carl is the extravert: 1 - 1/2 = 1/2
 

Infinitatis

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Ok, I'm deeply sorry. There's still a flaw. (Don't hurt me. :p) I am almost positive I have it right this time. Here is the revised story problem:

Albert, Bob, and Carl have each individually scheduled to meet with Dr. Fraud, a psychologist. Dr. Fraud wants to determine whether each patient is extraverted or introverted. He has not had any sessions with any of the three patients yet, so he does not have any information to go off of, except for one thing: he knows that two of them are introverts and that one of them is an extravert.

He wants to know if they are extraverted or introverted before meeting them, so he guesses. He guesses that Albert and Bob are the introverts and that Carl is the extravert.

Albert, Bob, and Carl all know one another. They know who is the extravert, and who are the introverts. They also know that Dr. Fraud thinks the extravert is Carl. Based on Dr. Fraud's guess, they decide to let one of the introverts that is not the one who Dr. Fraud believes the extravert is to meet with him first, so they send Bob, and the other two will meet with him later.

After Dr. Fraud meets with Bob, he knows for a fact that he is an introvert.

Who is most likely the extravert?
I worked out the math: it works now. Sorry for the confusion.
 

Tannhauser

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If one sees it from the doctor's point of view, the only thing one can say is that he met with an introvert, so there is one introvert and one extrovert left, and there is 50% chance of either being extrovert.

His original guess does not influence the probabilities.

In light of this, this is actually the equivalent of the Monty Hall problem.
 

Infinitatis

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If one sees it from the doctor's point of view, the only thing one can say is that he met with an introvert, so there is one introvert and one extrovert left, and there is 50% chance of either being extrovert.

His original guess does not influence the probabilities.

In light of this, this is actually the equivalent of the Monty Hall problem.

You're exactly right about this being the equivalent of the Monty Hall problem. I modeled this off of it (despite a few bumps along the road).

With that said, does his original guess influence the probabilities?

Hint: Does his original guess influence whom of the three he meets with?
 

Infinitatis

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Any last guesses? I'll post the answer in a couple of hours.
 

The Grey Man

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Ok, this changes things.

Probability that Dr. Fraud guessed wrong: 2/3
...they decide to let one of the introverts that is not the one who Dr. Fraud believes the extravert is to meet with him first...
Probability that only one patient met the criteria to meet with Dr. Fraud because Dr. Fraud guessed wrong (and Bob happened to be that patient, the inevitable one introvert whose identity Dr. Fraud got right, but of course one was all that was needed*): 2/3

Probability that Carl is not the extravert: 2/3

Probability that the extravert is Albert, now the only remaining candidate: 2/3

* In earlier versions, one correctly identified introvert (which there would necessarily be because there were an odd number of categorized items and two categories, introverts being the larger category) did not need to be the first interviewee. The subject of Dr. Fraud's first interview was decided by chance, so all it told us was that he was right about Bob's introversion. It did not tell us that Bob was probably the only introvert he got right, even given that he certainly is an introvert, as the final version did. :)
 

Infinitatis

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Ok, this changes things.Probability that Dr. Fraud guessed wrong: 2/3

Probability that only one patient met the criteria to meet with Dr. Fraud because Dr. Fraud guessed wrong (and Bob happened to be that patient, the inevitable one introvert whose identity Dr. Fraud got right, but of course one was all that was needed*): 2/3

Probability that Carl is not the extravert: 2/3

Probability that the extravert is Albert, now the only remaining candidate: 2/3

* In earlier versions, one correctly identified introvert (which there would necessarily be because there are two categories, introverts being the larger one, and an odd number of categorized items) did not need to be the first interviewee; Dr. Fraud's first interviewee was decided by chance.

Correct, nice job! I realized the mistakes I made when I tried to calculate those, so I apologize for the confusion. Here are a couple of other ways that this can be solved:
Consider the events Ea, Eb, and Ec indicating that the extravert is Albert, Bob, and Carl respectively. All of these events have an equal probability of 1/3 to begin with.

Dr. Fraud initially choosing Albert, Bob, or Carl to be the extravert is described the events Xa, Xb, or Xc respectively. As Dr. Fraud's initial choice is independent of who the extravert actually is, the conditional probabilities describing who the extravert is given that he has chosen one is still 1/3 for all scenarios. Dr. Fraud meeting with Albert, Bob, or Carl is indicated by the events Ma, Mb, or Mc respectively.

P(Ea|Mb,Xc)
= P(Mb|Ea,Xc) · P(Ea|Xc) ÷ P(Mb|Xc)
= P(Mb|Ea,Xc) · P(Ea|Xc) ÷ [ P(Mb|Ea,Xc) · P(Ea|Xc) + P(Mb|Eb,Xc) · P(Eb|Xc) + P(Mb|Ec,Xc) · P(Ec|Xc) ]
= P(Mb|Ea,Xc) ÷ [ P(Mb|Ea,Xc) + P(Mb|Eb,Xc) + P(Mb|Ec,Xc) ]
= (1) ÷ [ (1) + (0) + (1/2) ] = 2/3

P(Ec|Mb,Xc)
= P(Mb|Ec,Xc) · P(Ec|Xc) ÷ P(Mb|Xc)
= P(Mb|Ec,Xc) · P(Ec|Xc) ÷ [ P(Mb|Ea,Xc) · P(Ea|Xc) + P(Mb|Eb,Xc) · P(Eb|Xc) + P(Mb|Ec,Xc) · P(Ec|Xc) ]
= P(Mb|Ec,Xc) ÷ [ P(Mb|Ea,Xc) + P(Mb|Eb,Xc) + P(Mb|Ec,Xc) ]
= (1/2) ÷ [ (1) + (0) + (1/2) ] = 1/3
Given that Carl is who he guesses the extravert is, and given that Bob is the person that he meets with first, you can use this tree diagram to determine that there exists a 2:1 ratio between Albert being the extravert and Carl being the extravert.
30jsqqr.jpg
 

Tannhauser

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A guy is lost in the forest. From where he is standing, he can take three different paths. First path takes 2 hours and will lead him home. Second path takes 4 hours and leads back to the same spot. Third path takes 5 hours and also leads back to the same spot.

He picks a path randomly always. What is the expected length of time until he gets home?
 

Tannhauser

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Hint: assume X is the time it takes for him to get home. We are looking for E[X]. If, for example, he picks the second path (call it P2), we know that E[X | P2] = 4 + E[X]. All paths have probability 1/3 of being chosen. The rest is trivial, as they say.
 

The Grey Man

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1/3 chance of taking 2 hours

(1/3 * 1/2) chance of taking 4 + 2 hours

(1/3 * 1/2) chance of taking 5 + 2 hours

(1/3 * 1/2 * 2) chance of taking 4 + 5 + 2 or 5 + 4 + 2 hours

Expected time taken = ((2 * 2) + 6 + 7 + (11 * 2))/6 = 39/6 = 6 hours, 30 minutes
 

Tannhauser

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Not quite right. It looks like you have calculated the answer for the case where he doesn't repeat his choices when he has taken the wrong paths. Here, however, his choice is always independent of the past.
 

The Grey Man

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Oh yeah, he picks "randomly always". Oops.
 

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x - x + y - y + 3z - z = 600 - 400
2z = 200
z = 100

x+y=300

tmgqkXQ.png
 

Artsu Tharaz

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Lol, me right now:

Oh, that's a cool integral, imma give it a shot!

Actually nah, I really can't be bothered.

But... integral! I gotta prove I can solve it!

Meh, I either can or I can't it don't really matter.

INTEGRAAAAAALLL!!!!!!111 :mad:
 

Ex-User (9086)

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Haha :), don't sweat this one... it's rather (relatively) difficult, it's an indefinite integral (prime antiderivative) I was going to leave it until (probably) tannhauser showed up and tried to do it, I wasn't feeling particularly creative to come up with a more widely approachable and mentally stimulating problem.
 

Artsu Tharaz

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Isn't that integral just 11th grade maths?

In Australia anyway.

Just a bit of intermediate trigonometry, and from there it's a less common but still standard integral, correct?
 

Ex-User (9086)

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Isn't that integral just 11th grade maths?

In Australia anyway.

Just a bit of intermediate trigonometry, and from there it's a less common but still standard integral, correct?
Yes, it's very simple if you were taught how to approach it. It's pure algebra, no analysis involved. There are definitely much more difficult integrals out there, though there's little ingenuity involved in solving any of them. Mathematical proofs and concrete problems worded in mathematics make for better mental conundrums.

11th graders or undergrads in engineering or maths shouldn't find it too challenging and here lies my point.

There's nothing special about most of the things you find in science, physics or maths as long as you know where to look for answers or you were taught the basic methods of that language. The barrier is one of knowledge or accessibility, not one of talent or intelligence.

Calling it an "indefinite integral" or "antiderivative" is my joke, it makes the problem seem large and serious, many people like to use big words for their status, even just to hide the fact that they were born lucky to learn about maths where others didn't have this opportunity.
 

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x= f (x)
x= 3x^2 + 4
dx= (9x + 4)dx

Taking a wild stab at it but I'm pretty out of my depth having never studied integral equations or really any form of advanced maths so I imagine I'm way off here.
 

Tannhauser

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That was a messy integral

gif.latex


I started with multiplying by the conjugate 4cos(x)-3sin(x) in the top and bottom of the fraction so that you get

gif.latex


From here one can split it up in two integrals and do the the substitution u(x) = sin(x) in the first and u(x)=cos(x) in the second. Then one makes some partial-fraction expansions and that's pretty much it.
 

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Following my suggestion of doing the substitution:

Let t = tan x/2, then without loss of generality one can describe the relations as such:

41b5acbbfd.png

After simplifying terms:

8a23126038.png

It becomes apparent that we can rely on partial fraction decomposition now:
ce9b10f59c.png

Returning the results into the equation gives:
4d395c852b.png

and multiplying both sides by 2(t+1/2)(t-2) gives:
7422162c44.png

returning to the main equation and the result:
4ad982e551.png
 

Interdimensionist

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haha wow, I'm ashamed of my feeble attempt but thanks for the links. Gonna study integrals and work through this problem backwards to see how you did it for future reference.
 

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What are we even talking about anymore?

Peace love and math bro.
 

Artsu Tharaz

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My method was completely different haha.

I can't be bothered going through all the details (I didn't even bother working them out tbh, I find arithmetic to be one of the most difficult areas of mathematics.... lol) but this is what I did:

A combination of sin and cos is just another sin/cos with a different phase, so you just have to integrate sec(x), with some constants involved.

If you don't know the integral of sec(x) of the top of your head, you can work it out using the product rule, I think it's referred to as Integration By Parts? (there aren't too many functions to choose from in doing the product rule thingy, so it's not too difficult).

Then the result will be in terms of the shifted angle trig functions, so it would be a bit tedious having to expand it all out to get it back into non-shifted functions, so hopefully that form of the answer is acceptable :P
 

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Secant and cosecant functions are 1/cos and 1/sin respectively. If you're going to merge them together you can do so describing either of them as the other with a pi/2 period added/subtracted without the need of involving inverses of them so I kind of don't follow your reasoning.
 

Artsu Tharaz

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Secant and cosecant functions are 1/cos and 1/sin respectively. If you're going to merge them together you can do so describing either of them as the other with a pi/2 period added/subtracted without the need of involving inverses of them so I kind of don't follow your reasoning.

if y = inverse tan of 4/3
then sin(y) = 4/5 and cos(y) = 3/5

so 5*sin(x+y) = 5(sin(x)cos(y) + cos(x)sin(y))
= 5(sin(x)*3/5 + cos(x)*4/5
= 3*sin(x) + 4*cos(x)

so 1/(3*sin(x) + 4*cos(x))
= 1/(5*sin(x+y))
= sec(x+y)/5, where y is just a constant, so you can just let u = x+y, then du/dx = 1

So it's 1/5 * [integral] sec(u).du

It's been like 7 years since I've done this kind of maths, so I could be off.

For the rest of it:
Using integration by parts I got that the integral of sec(x) = sin(x)*tan(x) + cos(x)

So the complete integral is (1/5)*(sin(x+y)*tan(x+y) + cos(x+y))
where y is the inverse tan of 4/3
 

Artsu Tharaz

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No one has taken on this yet. Whoever does, is my hero

Haha, I know the answer to this one already so I'll stay silent.

To anyone else who wants to have a go: go for it, it's quite simple and doesn't require anything advanced.
 
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