There’s no need to lie milsy.
Your best subject was always detention
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Making a video game has so far taught me more math than any formal class.
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Mental
Abuse
To
Humans
M.A.T.H
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I dislike maths.
It makes me feel physically ill…
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I think I’m hitting the upper limit to what I can do here, differential geometry for general relativity as well as Feynman propagators and their associated rules are just reading as an incomprehensible mess every time I look in the textbook.
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Been doin’ a course in Abstract Algebra this semester, and I love it. The proofs are relatively simple to understand and its been really interesting to see how some of math is constructed; with groups, rings, etc.
There’s this great series of videos on the subject by the YouTube channel Socratica that I’d highly recommend. I think Abstract Algebra is a great place to start with higher math since the concepts are pretty intuitive, and it gives a broader sense as to what math is all about:
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Taking abstract algebra next semester I believe, can’t wait
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I’m so happy I’ll never take a mathclass again. Literally I’ll never use division again for the rest of my life.
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So apparently the golden ratio is approached by any sequence where each element is the sum of the past two elements, sans the first two, not just the Fibonacci sequence
also I know it was a year ago but [quote=“Maphrox, post:253, topic:4630”]
I think I’m hitting the upper limit to what I can do here
[/quote]
is this a pun
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I have a habit of doing math stuff the hard way, and then realizing afterword that there was a much easier way to do it. And I think there’s no better example of that than this:
If you’re trying to calculate how many possible orderings of 4 items you can get out of n items, it’s really easy – you just do n(n-1)(n-2)(n-3). So, for example, if you have 8 items, you would do 8x7x6x5, or 1680.
Willess now knows this. The Willess of 6 years ago, however, did not know this, and when pressed to find an equation for the number of possible orderings of 4 items, he came up with this:
(n4 + (n x (n - 1) x 3)) - n - ((n x (n - 1)) x 4) - ((n x (n - 1)) x 6) - (n x ((((n - 1) / 2) x n) - (n - 1)) x 12)
Where the Karzahni did I even get this from? I can’t even figure out how I got most of this insanity – there is one part I recognize here, but the rest is unfamiliar.
This monstrosity of an equation is actually correct, too. If you distribute stuff through all the parentheses and simplify it down, you can get n(n-1)(n-2)(n-3).
I got the answer I was looking for, but in quite possibly the hardest way possible.
~W12~
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First, Willess, thanks for one of the best laughs I’ve had in a while. XD
I think I figured out your thought process. Let me know if this rings a bell:
First, I’m going to clean up the formula a bit, and remove some of the unnecessary brackets, while still keeping the “structure”, so that it’s easier to refer back to:
n^4 + 3n(n-1) - n - 4n(n-1) - 6n(n-1) - 12n(n(n-1)/2 - (n-1))
You start out by allowing anything, including repeated elements, i.e. strings like “1 1 1 1”. That’s the n^4. Then you need to subtract all the different ways you can make strings with such repetition.
The first case is easy. There are n ways to make strings where all 4 characters are the same. “1 1 1 1”, “2 2 2 2”, …, “n n n n”. Hence, we have the “-n” in the formula above.
Then you subtract the number of ways you can make strings with exactly 3 repetitions, e.g. “7 7 5 7”. This is the “-4n(n-1)”; n ways to choose the repeating element, n-1 ways to choose the single element, 4 ways to place the single element into the string.
Then, the number of ways to make exactly one pair. I.e. strings like “1 1 2 3”, but not “1 1 2 2”. There are choose(4,2)*n(n-1)(n-2) = 6n(n-1)(n-2) ways to do this, which ends up being equivalent to the 12n(n(n-1)/2 - (n-1)) in the formula. So, how did you come up with this 12n(…) form? My best guess is the following:
- You knew there were 6 ways to choose the location for the “pair” elements.
- 2 ways to order the remaining elements
- n ways to choose the repeated element
- choose(n-1,2) ways to pick the last element. How did choose(n-1,2) end up being n(n-1)/2 - (n-1)? My best guess: You pretended you had n elements, and chose 2 from those, then subtracted the (n-1) possible cases that would have been invalidated by picking the repeated element as one of your choices.
Anyway, subtract this 12n(…) thing you and you remove the cases with exactly 1 pair.
That leaves the last case of repetition: two pairs, e.g. “1 1 2 2”. This is the guess I feel less comfortable about, since the two portions are so spread apart in your formula. But if you combine them, the 3n(n-1) - 6n(n-1) = -3n(n-1) represents removing the cases with exactly 2 pairs; n ways to pick the first element, n-1 for the second, and choose(4,2)/2 = 6/2 = 3 ways to order the elements.
Having removed all of the invalid cases, you have your pick(n,4). Let me know what you think.
If so, may I suggest the slight modification: “I think I’m hitting the supremum of what I can do here” 
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n(n-1)/2 is actually the equation I memorized for the number of possible combinations of two, so I know where that came from at least. That’s the one part I did recognize before.
Knowing the way I do math, that’s probably because I wanted to group up all the stuff I was adding first, and then subtract everything else. Made it a bit easier to condense down the equation afterward.
This looks a lot like the way I would’ve reasoned through this problem, actually (I’ve done similar combination/ordering problems with methods like this). Though maybe a bit more “manual” (more focus on what I was doing and less on why I was doing it) but basically the same thought process. Thank you for helping me figure this out, I appreciate it!
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This semester showed me that no math class can ever prepare me to use it for physics.
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Yes. It was also an inaccurate statement. Turns out I have ascended and now I’m solving divergent integrals and getting finite results.
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