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The Math Help Thread

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Partial Gamification said:
It seems to me like its more along the lines of considering its a bijective function. Something simple:
setq7k2d.png

C union D could be empty, or not. The inverse function of C could be A, and inverse function of D could be B.

inverseF(C - D) = inverse F(C) - inverse F(D)
Assume RHS then prove LHS, then reverse.
A not in B = A not in B

Maybe I'm confusing the use of the "-" operator.
Click to expand...

That could work too, considering for a function to have an inverse it needs to be bijective, and it would be best to assume finite sets to help further the point.
Thinking about it a bit more, you could just use a proof by contradiction. That if F: X -> Y then F^-1: Y -> X. So, F(C -D) must be an element of X. Although I can't recall if a function has an inverse, whether it is a linear transform or not.

Edit: Yeah, just saying it's a bijective function is basically enough to prove it all.
 
Thanks so much for your feedback, Dropped_Off and Partial Gamification, but I'm still a little confused. Aren't we already assuming the function F is bijective from the get go?

The way I was thinking of tackling the problem goes something like this:

Let x be an element of C - D and let y be an element of F(C - D) such that F(x) = y. Then since x is an element of C - D, by the set difference law, x is an element of C intersect complement of D (I think this is where Partial Gamification was confused with the "-" operator. It just means C intersect complement of D). Then since x is an element of the intersection, it is an element of C individually and an element of the complement of D individually. Thus y must be an element of F(C) individually and an element of the complement of F(D) individually. Using the set difference law, I can now say y is an element of F(C) - F(D). Then somehow I have to use inverses which is where I hit the wall... Any ideas?

Edit: Unless I made some mistakes, I think I proved F(C - D) = F(C) - F(D). Should be trivial to prove the same for the inverses shouldn't it? Can I just say that y is an element of F(C - D) and also an element of F(C) - F(D), then by definition of an inverse function in a one-to-one correspondence, x is an element of inverse F(C - D) and also an element of inverse F(C) - inverse F(D), thereby completing the proof?

Double edit: Just thought I'd clarify one part of my proof. If x is an element of the complement of D, it is not an element of D. Therefore y is not an element of F(D), so it is an element of the complement of F(D). I think then you can use the set difference law to say y is an element of F(C) - F(D).
 
Roflobear said:
Edit: Unless I made some mistakes, I think I proved F(C - D) = F(C) - F(D). Should be trivial to prove the same for the inverses shouldn't it? Can I just say that y is an element of F(C - D) and also an element of F(C) - F(D), then by definition of an inverse function in a one-to-one correspondence, x is an element of inverse F(C - D) and also an element of inverse F(C) - inverse F(D), thereby completing the proof?
Click to expand...

I would think so, although my set theory is a tad neutered in some areas. But, by the definition of an inverse function your proof makes sense.
 
campfireweekend said:
I am completely stumped by the second part of this problem.
Click to expand...
Does it have to do with the circulation line (normal to the plane of circulation) [yes but not confusing with what the line integral is]? Look at the Fundamental Theorem of Line Integrals and Stoke's form for counterclockwise circulation (viewed from above), the integrands are the same. Its kind of funny, not referencing the given information directly and so one could (lose credit) making up their own givens.
 
Roflobear said:
Thanks so much for your feedback, Dropped_Off and Partial Gamification, but I'm still a little confused. Aren't we already assuming the function F is bijective from the get go?
Click to expand...

Might be a little late on this one, but the statement is true even if F is not a bijection, assuming that by inverse F(C) you mean the set of all elements x in X such that F(x) is in C. But even if you mean a strict inverse function, only an injection is needed, not a bijection. Anyway, my point is that you may not want to assume F is a bijection unless the problem actually says that it is one.

Edit: Also note that you will not have F(C-D)=F(C)-F(D) without assuming that F is at least an injection (consider the constant function as a counterexample).
 
I have a question about notation.

so theres factorial

!5 = 5 x 4 x 3 x 2 x 1

how do you notate a factorial but only up to a certain number for example...

5 x 4 x 3

just the first 3 numbers of a factorial.
 
HoodWinked said:
I have a question about notation.

so theres factorial

!5 = 5 x 4 x 3 x 2 x 1

how do you notate a factorial but only up to a certain number for example...

5 x 4 x 3

just the first 3 numbers of a factorial.
Click to expand...

First of all, the ! comes after the number. It's not !5, it's 5!.

Second, what you're looking for is called a Falling Factorial, and in your example, 5 x 4 x 3 would be written (5)_3 (the _ indicates a subscript). that said, this is not common notation, and I wouldn't expect someone other than a mathematician or student of higher math to recognize it. Because of that, if you use it in some paper or while showing your work for some problem you're solving, you should make a note before you use it along the lines of "the notation (x)_n denotes the falling factorial, i.e. x*(x-1)*...*(x-(n-1))"
 
I'd never heard of that notation. Interesting.

If you're just looking for a simple way to get it across, something like 5!/2! is pretty intuitive.
 
HoodWinked said:
I have a question about notation.

so theres factorial

!5 = 5 x 4 x 3 x 2 x 1

how do you notate a factorial but only up to a certain number for example...

5 x 4 x 3

just the first 3 numbers of a factorial.
Click to expand...

An easy way to do it is just to divide by another factorial. So the one in your example would be 5!/2!. If you want the first k terms of n! you'd write n!/(n-k)!

That seems to be how it's most commonly done anyway.
 
Hey I've got a graph theory question.

So the question is:

If a connected planar graph with n vertices all of degree 4 has 10 regions, determine n.

I understand we have to use Euler's formula

r = e - v + 2

10 = (2e = 4*n) - n + 2
10 = 2n - n + 2
10 = n+2
n = 8

So there are 8 vertices, 16 edges, and 10 regions.

However, the second part is to draw the graph. I have no idea how to do this. I assume it has to do with Euler's rule and the proof that if an edge connects 2 vertices, it creates a new region, but I have no idea where to begin drawing this. It's very daunting.
 
Differential equation help needed.

y(x)=e^x is a solution of the diff. eq.

(x+1)y'' - (x+2)y' + y = 0

Q: Determine the general solution of the equation

What now? I can verify that indeed y(x) is a solution to the equation but (i believe) i need a 2nd solution to determine its wronskian and the general solution in the form of y(x) = C1Y1(x) + C2Y2(x)

right?
 
mooooose said:
Hey I've got a graph theory question.

So the question is:

If a connected planar graph with n vertices all of degree 4 has 10 regions, determine n.

I understand we have to use Euler's formula

r = e - v + 2

10 = (2e = 4*n) - n + 2
10 = 2n - n + 2
10 = n+2
n = 8

So there are 8 vertices, 16 edges, and 10 regions.

However, the second part is to draw the graph. I have no idea how to do this. I assume it has to do with Euler's rule and the proof that if an edge connects 2 vertices, it creates a new region, but I have no idea where to begin drawing this. It's very daunting.
Click to expand...

Did you actually try to do it? I just threw down 8 vertices and started connecting the adjacent ones, making sure I never went over degree 4, and got an appropriate graph on the first try.
 
Therion said:
Did you actually try to do it? I just threw down 8 vertices and started connecting the adjacent ones, making sure I never went over degree 4, and got an appropriate graph on the first try.
Click to expand...
I had been trying, and not getting anywhere, but I tried again and got it.
 
Hey, I need a little more help with something.

How many different numbers can be formed by various arrangements of the six digits 1,1,1,1,2,3

I tried breaking this into cases, but it clearly is more trouble than it should be. Can anyone guide me towards the easiest way of figuring this out?
 
mooooose said:
Hey, I need a little more help with something.

How many different numbers can be formed by various arrangements of the six digits 1,1,1,1,2,3

I tried breaking this into cases, but it clearly is more trouble than it should be. Can anyone guide me towards the easiest way of figuring this out?
Click to expand...

6! / 4! = 30 I think...it's been so long since I've done this.
 
Buggy Loop said:
Differential equation help needed.

y(x)=e^x is a solution of the diff. eq.

(x+1)y'' - (x+2)y' + y = 0

Q: Determine the general solution of the equation

What now? I can verify that indeed y(x) is a solution to the equation but (i believe) i need a 2nd solution to determine its wronskian and the general solution in the form of y(x) = C1Y1(x) + C2Y2(x)

right?
Click to expand...

You can use the characteristic equation to find the second one.
 
I can't remember any math besides dividing, subtracting and that basic stuff. It's been 10 years since I left high school, and now I need to take math and advanced math over the next year at my community college.


I feel like a lost cause, because I never learned equations. I tried to, but I never understood how, or why I had to do it or what it can help me with. I am not really sure exactly how much I need to learn, but as far as equations goes I need to understand quadratic equations and be able to do them with no help and calculator for an exam.

So I have a year. fuck. Formulas is another one. I tried using Khan Academy but even that is too advanced for me. I can't even remember how to use a calculator. I forgot how to use SIN and all that jazz.



Anyone have some ideas, even if it's time consuming, I really need to learn this shit right.
 
Vigilant Walrus said:
I can't remember any math besides dividing, subtracting and that basic stuff. It's been 10 years since I left high school, and now I need to take math and advanced math over the next year at my community college.


I feel like a lost cause, because I never learned equations. I tried to, but I never understood how, or why I had to do it or what it can help me with. I am not really sure exactly how much I need to learn, but as far as equations goes I need to understand quadratic equations and be able to do them with no help and calculator for an exam.

So I have a year. fuck. Formulas is another one. I tried using Khan Academy but even that is too advanced for me. I can't even remember how to use a calculator. I forgot how to use SIN and all that jazz.



Anyone have some ideas, even if it's time consuming, I really need to learn this shit right.
Click to expand...

Attend all your lectures, ask questions, and see the math tutor at school.

To prepare, figure out what the course structure is via the community college's website and take a crack at whats on the outline/syllabus.
 
Vigilant Walrus said:
I can't remember any math besides dividing, subtracting and that basic stuff. It's been 10 years since I left high school, and now I need to take math and advanced math over the next year at my community college.


I feel like a lost cause, because I never learned equations. I tried to, but I never understood how, or why I had to do it or what it can help me with. I am not really sure exactly how much I need to learn, but as far as equations goes I need to understand quadratic equations and be able to do them with no help and calculator for an exam.

So I have a year. fuck. Formulas is another one. I tried using Khan Academy but even that is too advanced for me. I can't even remember how to use a calculator. I forgot how to use SIN and all that jazz.



Anyone have some ideas, even if it's time consuming, I really need to learn this shit right.
Click to expand...

Hey, don't feel alone in this matter. I can relate.
 
Ok so this is somehow related to math but is some sort of puzzle as well. The information:

Constellation within...

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
Click to expand...

The answer is a 40 character hexadecimal. I am completely stuck. I tried converting the above to ascii to see if a message was hidden inside, looking for patterns, etc. No luck.
 
I'm not sure if this is the right thread to be asking for help but...

lets say (hypothetically speaking, of course) that I have an job interview for a machine learning entry level position at a startup on Monday. anyone know any good resources that could get me up to beginner level knowledge on the subject? thanks
 
I have no idea how to factor 5y^4 - 18y^2 - 8
Maybe it's obvious, but I've been trying to do this for the past 20 minutes and still have no clue.
 
dorfism said:
I have no idea how to factor 5y^4 - 18y^2 - 8
Maybe it's obvious, but I've been trying to do this for the past 20 minutes and still have no clue.
Click to expand...

Typically an easy trinomial to factor would be in the form such as y^2 - y - 8

If you notice it's always a variable squared and then the same variable for the second term. So for your problem you want to get it in that form as well. You want to write the variable in the first term as (y^2)^2 and in the second term you hope the variable remaining is the same as in the parenthesis from the first term which is (y^2)

So, rewrite the trinomial as 5(y^2)^2 - 18(y^2) - 8 and factor like normal, just treat the y^2 in the parenthesis as you would any other single variable.

Oh and always check your factors after you do that to see if you can take the factor another step
 
Hello. Precalc here.

I'm having trouble with graphing this particular rational function: f(x) = 2/(x^2+x-2)

My problem is finding the x-intercepts. I know there aren't any.

I would normally set the numerator to zero and solve for x but the numerator is just 2. The graph of this function does not hit 2. It isn't supposed to, as there is a horizontal asymptote at y=0. But I am looking for a systematic way of telling that there are no x-intercepts here.
 
For a rational function to be 0, the numerator must be 0. It's impossible for the numerator to be 0, as you've observed, so f(x) is never 0. Therefore it has no x-intercepts.
 
I guess the "systematic" way is to set f(x) = 0 = 2/(x^2+x-2)
Then multiply both sides with x^2+x-2
So you get: 2 = 0 which is a contradiction.
Therefore the function never takes the value 0
 
Thanks, guys.

I really needed a word from others of how to look at this kind of problem and now I can feel sure of what to do next time I come across something like this. (:
 
Kimosabae said:
Should I just assume this was too elementary for someone to answer or....
Click to expand...

I took a look at it and it did seem wrong. I didn't see any obvious mistake that could produce the answer he gave (if only I could see what was typed into that calculator...). There were comments below that also said the answer was incorrect.
 
There are several groups, each with different colored things.
Code: 
-Group 1-Group 2-Group 3-
| red   |       | red   |
|       | green |       |
|       | blue  | blue  |
If I randomly grab one thing from each group, how many things of each color should I expect to pull?

Do I just find out the color probabilities per group, then add them? It seems too simple! Going on what I think is right, I should expect 1.5 reds, 0.5 greens, 1 blue.

I'm also programming this, so a follow up question is if there's a better way to determine this "expected outcome" beyond calculating the averages per group and summing them.
 
Differential equation : αF′′(x) − (λ + β^2)F(x) = 0

Im not sure if i can do it this way but...

Caracteristic equation : α^2R^2 - (λ + β^2) = 0

Delta = -4α^2(-(λ + β^2)) = 4α^2(λ + β^2)

And then should i do the 3 cases of delta > 0 , delta = 0 and delta < 0 ? But how do i go with that, like, &#955; > -&#946;^2 ???

Some fucked up shit.
 
eot said:
I've been trying to prove this identity:
d/dx ( J_n(x) / x^n ) = - J_(n+1)(x) / x^n
Where J is the bessel function.

I decided to look it up and found a proof here (page 4)
http://homepage.tudelft.nl/11r49/documents/wi4006/bessel.pdf
But I don't get the second to last step, that was the step I couldn't get right myself.

I think I need to eat.
Click to expand...

Since you have a factor k in the sum, the first term (k = 0) will be zero. They want the sum to be from 0 to inf, so they change the indexing inside the part that is summed such that they can remove the first, zero, part.
edit: you change the starting index from 1 to 0, thus you must change all k to (k+1)
 
Wanted to say thanks to all the people who helped me out last semester. I got an A- in Calc 2. :D Now i just need Physics 1 & 2.
 
Leezard said:
Since you have a factor k in the sum, the first term (k = 0) will be zero. They want the sum to be from 0 to inf, so they change the indexing inside the part that is summed such that they can remove the first, zero, part.
edit: you change the starting index from 1 to 0, thus you must change all k to (k+1)
Click to expand...

Jesus christ.
That teaches me to not be so sloppy about writing out indices.

Thank you very much.
 
My brother tells me that he needs help writing a function with points passing through (2,3)(3,7)(4,11). I'm usually good with this stuff, but I'm honestly stumped.

So far, we got the easy ones:

f(x)=4x-5
f(x)=4|x|-5

Any function at all (that isn't piece-wise, ie 3 different functions). Help?
 
EightBitNate said:
My brother tells me that he needs help writing a function with points passing through (2,3)(3,7)(4,11). I'm usually good with this stuff, but I'm honestly stumped.

So far, we got the easy ones:

f(x)=4x-5
f(x)=4|x|-5

Any function at all (that isn't piece-wise, ie 3 different functions). Help?
Click to expand...

By inspection the points exist on a straight line. Constant slope.
 
EightBitNate said:
So then we could only use functions of straight lines? So linear, abs value.... Any others?
Click to expand...
Well, if you plot the three points in question, I am sure you could draw (freehand) all sorts of functions that go through them all. So there are infinitely many other choices.

"Ah," you will probably object, "but I wanted a function that has a 'formula', i.e., that is expressible in terms of elementary functions!" So we will have to put in a little extra effort. Consider the linear function f(x) = 4x-5 that you found already, which passes through the three given points. Let g(x) be one of these mysterious other functions that also passes through the given points. Let h(x) be the difference f(x) - g(x). What special properties must such a function h(x) have? Is there an easy way to cook up functions that have these properties? How does that help?
 
boviscopophobic said:
Well, if you plot the three points in question, I am sure you could draw (freehand) all sorts of functions that go through them all. So there are infinitely many other choices.

"Ah," you will probably object, "but I wanted a function that has a 'formula', i.e., that is expressible in terms of elementary functions!" So we will have to put in a little extra effort. Consider the linear function f(x) = 4x-5 that you found already, which passes through the three given points. Let g(x) be one of these mysterious other functions that also passes through the given points. Let h(x) be the difference f(x) - g(x). What special properties must such a function h(x) have? Is there an easy way to cook up functions that have these properties? How does that help?
Click to expand...

I'm a little hazy on subtracting functions. Let's see.

If (f-g)x=(h)x then g(x)=(h+f)x? That's the only conclusion I can draw, unless I'm missing something. But I doubt that's what you're suggesting.
 
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