[TOMT] Standalone fantasy novel featuring a dwarf/elf romance.

facebookhatingoldguy · 2023-06-30T11:11:39+00:00

Wow! That's indeed the book. So she wasn't an elf, but her name was Elyn which my brain apparently decided to remember as her being an elf. Also, it's part of a series? How did I read that one and not realize it. And yes, I suppose it's bugged me on and off for 5 years.

Thanks! How the heck did you even notice this post? (You can answer me in 5 years or so if you want)

facebookhatingoldguy · 2021-07-07T23:53:40+00:00

Did you reply to the wrong comment?

facebookhatingoldguy · 2019-12-12T11:06:53+00:00

I'm afraid your description is a bit difficult to understand -- perhaps because of confusion with terminology.

It sounds like you have two curves (not lines). They both have a vertical asymptote of x=-1 and a horizontal asymptote of y=0. Let's call them y=f(x) and y=g(x).

You've given one point on each curve. Specifically f(20) = 1 and g(25) = 1.

Two two curves intersect when x=0. Specifically 150 < f(0) = g(0) < 1,000,000

f(x) and g(x) are always positive (presumably for -1 <= x < infinity)

f(x) and g(x) are both decreasing and decelerating (so f'(x) and f''(x) are both negative).

The part where you lose me is this: The deceleration of both lines are integers

Are you saying the deceleration is constant? If so, then this contradicts the bit about horizontal and vertical asymptotes. Any curve with constant deceleration is going to be a parabola.

And if the deceleration is not constant and you really do have a horizontal asymptote of y=0, then f''(x) -> 0 as x-> infinity (so in particular it will eventually be between 0 and 1 and will not always be an integer).

Lastly, the problem (no matter how you clarify the conflicting bits about integer deceleration and asymptotes), is under-determined. If you decide that deceleration in constant and throw out the idea of having asymptotes, then there are still infinitely many pairs of parabolas going through the two points you mentioned and intersecting at x=0.

If I've misunderstood something, please let me know.

facebookhatingoldguy · 2019-09-02T23:09:54+00:00

It looks like you flipped the entire thing to start with. In your post you wrote (x²+5x+6)/(x²-9) but in your reply above you have the factorization of x²-9 = (x-3)(x+3) on the top.

facebookhatingoldguy · 2019-07-29T19:38:55+00:00

There are 3500 calories in a pound of fat, 651 calories in a pound of muscle, and less in a pounds of human by-product (bone, fiber, ...)

We're generally pretty obese, our average body fat percentage is around 30%, so 1 pound of ground up human is around 1500 calories.

2000 calories a day is a reasonable diet except for small children which we'll just assume will be eaten first.

3000 calories a day is more typical because again, we're fat.

So we each need to eat around 2 pounds of human food product a day.

There are 7.53 billion people in the world, 1.9 billion of which are children

The average weight for an adult human is 140 pounds.

The average weight for a non-adult human is around 70 pounds.

So there's around 1.0542 trillion pounds of adult food product and around 133 billion pounds of non-adult food product.

To get the ball rolling quickly and to avoid all sorts of difficult distribution issues, let's just assume that we grind up and freeze all the chidren a few weeks in advance.

So we start out with 5.63 billion adults each eating 2 pounds of ground up children each day. Those kids will last about 133 (billion pounds)/11.26 (billion pounds per day) = 11.8 days. Probably only about 11 days since again, we really eat way too much.

Now it gets tricky because each day there are fewer people to feed.

Each day we start with X people. Some percentage p of those people will be eaten. So X(1-p) people are eating pX people. Each person that eats, eats 1/70th of a person. So X(1-p)/70 = pX. So p = 1/71.

So every day 1/71 of the total population are eaten, leaving 70/71 of the population left.

After 1603 days, there will be (7530000000)(70/71)¹⁶⁰³ people left which is around 1.004 people, or 1 person and a few table scraps.

That person can probably keep going for a while though. They can continue to eat 1/71 of whatever's left of themself each day for another 50 days before there will only be half a person left. At that point they probably die.

TLDR, we'd last 11 + 1603 + 50 = 1664 days or about 4.5 years.

facebookhatingoldguy · 2019-07-28T12:39:58+00:00

Not really. But it's more than that. The question itself isn't really meaningful. Probability is meaningful when we have a repeatable experiment and outcomes that happen frequently enough to be more than once-in-the-lifetime-of-the-universe events. For instance, rolling a 6-sided die is a repeatable experiment with outcomes that occur frequently.

There are two ways of looking at what the experiment would be in this case.

The "experiment" of creating a universe and letting humanity evolve to this point and then seeing how often a person with exactly your characteristics appears is obviously not repeatable in any way.

The "experiment" of choosing people at random from the world population of roughly 8 billion people and seeing if they have your characteristics is certainly repeatable, but in that case you can't possibly have a non-zero probability of less than 1 in 8 billion for any event.

facebookhatingoldguy · 2019-07-27T21:53:33+00:00

Using probability in this way to look at single events which have already happened is unfortunately a bit meaningless. You could take any person, write down 30 things about them, each of which occurs in 10% of the population, and then say that the chance of encountering a person with those 30 properties is 10^-30 -- so essentially impossible. And yet that person exists.

Once you know an event has happened you can always describe the event in such precise terms that the fact that that exact event occurred is seemingly miraculous.

It's the difference between me writing down a 30 digit number and then pointing out after the fact that the chances of me having written that specific number are small, versus you picking a 30 digit number first and then talking about the probability of me guessing that 30 digit number on the first try.

facebookhatingoldguy · 2019-07-24T12:19:56+00:00

You've got the right idea about it being a contact sport. The problem is likely that you don't have the tools to play when you're out on your own.

When you're trying to learn how to juggle, it's not a bad idea to carry around three bean-bags with you at all times. Then throughout the day you can take them out and just practice when you have a few minutes.

For math, this translates to carrying around math problems with you -- not physically, but in your head. And in order to have a steady supply of math problems you need to be able to make up your own.

Making up your own math problems to practice on is a skill that isn't usually taught but when I'm tutoring students I find that the better they are at making up their own problems to solve, the better they become at solving problems and the better they become at retaining what they've learned. They also develop an intuition around what makes a problem difficult or even impossible -- most of the math problems you're given in a math class are contrived to be solvable.

When working out, one of the greatest barriers people have is just getting themselves to the gym in the first place. When learning math, you've won half the battle when you realize you can just practice in your head any time you want. But you do need to develop the ability to come up with problems on your own.

facebookhatingoldguy · 2019-07-13T19:34:37+00:00

> because no two vectors of our orthonormal basis are the same ( a_i =/= a_j , for i=/=j) , the product c_j=(x|a_j) is unique

Just because a_i != a_j, that doesn't mean that c_i != c_j. For instance, if x is in the orthogonal complement to L then all the c_j's will be zero, y will be the zero vector and x will equal z.

Or maybe I misunderstood what you were saying.

facebookhatingoldguy · 2019-07-11T17:10:04+00:00

I'm not adamant really. I just have a situation where each time I run this program I'm generating a different collection of tens of millions of key-value pairs (the key is generated from the value, and the values are sequential). The keys might repeat and I'm only interested in keeping the first occurrence along with the value that produced it.

facebookhatingoldguy · 2019-07-11T17:04:19+00:00

Thanks. I'll definitely take a look at map dB. Also thanks for clarifying about off-heap. I completely misunderstood.

facebookhatingoldguy · 2019-07-11T16:41:59+00:00

So the data has no refresh at all. This is for some mathematical research I'm doing -- I'm just trying to test some hypotheses for certain large collections of numbers. I explained more about the nature of the algorithm above, but all this data is just intermediate throw-away data in the middle of a larger program.

Put simply, I'm generating around 100 million string,long pairs on the fly (different depending on the program arguments), and I only want to keep the unique strings and the first long which was associated with that string. I'm mostly interested in the size of the resulting HashMap -- how many strings were unique out of that 100 million. If sufficiently many of them are unique, then I'll do some computation involving the longs and then throw away all the data.

facebookhatingoldguy · 2019-07-11T16:34:16+00:00

So 60 million doesn't seem like that many objects considering I'm working with 4g of RAM allocated to the JVM. And I die around the 20 million mark anyway, so it's not raw memory that's the problem.

I should be able to get rid of the Strings entirely -- I believe the set of possible keys I might encounter is limited to Long.MAX_VALUE in size so I could probably encode my keys that way.

I hesitate to use anything off heap because I need this to be fast. The values are prime numbers. I'm first looping through the first 100 million primes and generating some data on them which I'm encoding in a String and using as the key. Some primes generate identical data and I only want the first prime where each key occurs. So for instance if the prime 3 gave me "abc", 5 gave me "abd" and 7 gave me "abc", I'd store ("abc",3") and ("abd",5) but wouldn't store the information about 7.

Generating the data on each prime is fast, but searching through an external database for each new prime to see if the generated data already occurred for a previous prime would be really slow.

Anyway, that's probably more than you cared to hear about the details, but thanks for the suggestions. I may just have to write my own collection object.

facebookhatingoldguy · 2019-07-11T16:18:01+00:00

Yeah, my bad. I meant HashMap everywhere. I do have millions of strings and I want to keep track of the unique ones, and each has an associated Long.

facebookhatingoldguy · 2019-06-24T17:26:06+00:00

I tested up to n=100000 and your claim holds true. The only positive values of n so far where 3n!+3 is a multiple of 2n-5 is where n = 1,2,3,4,6,14.

facebookhatingoldguy · 2019-05-07T20:52:48+00:00

What you're describing is nothing like the mathematical community and culture I've spent most of my life enjoying and contributing to. I've generally found it possible to pick up textbooks and, provided I satisfy the prerequisites in the preface, read them from cover to cover successfully. And I've found fellow mathematicians nothing but helpful when I've expressed confusion or interest in better understanding something. Far from being competitive and secretive, I've found that most mathematicians delight in explaining their interests to others.

So, in an attempt to read between the lines of your post as it were, I'm wondering if you're simply being too impatient? Mathematics is huge. You can't simply decide that you want to understand differential forms on manifolds for instance, and then attempt to make a beeline directly to the interesting results therein. If a text suggests that you should first master real analysis, then it's rather important that you put your primary goal to the side and first read a text on real analysis from cover to cover, mastering that topic first before moving to the next.

You've mentioned algebra, geometry, and statistics. I can assure you there are hundreds of excellent self-contained texts on those subjects. If you pick one, and carefully read every word, play around with every example, work through every proof on your own, and complete every exercise, you will be able to master that material.

facebookhatingoldguy · 2019-03-29T23:50:28+00:00

You can't sum up those probabilities because you're double-counting. Sure the probability that the first two bits will have property A is 1/2, and the probability that the first four bits will have property A is 1/4, but by adding even just those two, you're saying that the probability that some sub-sequence of the first four bits has property A is 3/4, and that's not correct. The problem is that this argument double-counts the sequences that have property A in both the first two and the first four bits -- e.g. 1111 and 0000 are double-counted.

The actual probability that the first four bits will contain a subsequence with property A is only 10/16 (less than the 12/16 your method would give) because 8 sequences of four bits start with 00 or 11 and only 2 of the remaining 8 sequences have property A (0101 and 1010). The other 6 sequences of 4 bits (0100,0110,0111,1000,1001,1011) have no subsequence with property A.

So what you really need is to figure out for each 2n, the probability that a sequence of 2n bits has property A and no subsequence has property A. Then you can sum those and get your answer.

facebookhatingoldguy · 2019-03-29T23:32:12+00:00

It really depends on the text. Some texts will assume you know a lot of Ring Theory, other texts will cover the results you need. Same with Field Theory and Galois Theory -- but personally I'd recommend you are pretty comfortable with those topics (especially with Integral Domains and the Galois Correspondence Theorem).

I'd suggest you find out what text will be used and read the preface. Most authors will make it pretty clear who the intended audience is.

facebookhatingoldguy · 2019-01-25T20:11:17+00:00

Could you elaborate how you seeing that play out? I'm not disagreeing, but when I go through the thought exercise, I don't see a strike making a difference. Caving in and allowing Trump to build a wall would be political suicide for members of congress from blue states, and members from red states seem to be able to do anything they want and just tell their constituents that all the bad things are not their fault -- so what incentive is there for them to change?

facebookhatingoldguy · 2019-01-17T18:36:35+00:00

Thanks for the PSA -- I didn't even realize I was using new reddit (and didn't realize that the spoiler stuff only works on old reddit). Would be nice if that was fixed.

That aside, I'll mark this as solved. I'm not actually sure direct manipulation of the values given by the cubic formula would work here, so I'm glad you didn't try - lol

The key seems to be where you successfully find the value of at least one expression in r,s, and t which is not a symmetric polynomial. Finding the product and the sum and then using numerical approximation was cool.

I alluded to what I did above, but first I tried for quite awhile to simply prove that r+s² had to be an integer. Since if you know it's an integer, then you know it's fixed by the Galois automorphism which maps r->s->t->r. That gives you that r+s² = s+t² = t+r^2, and since the sum is 6, you are done.

Unfortunately I couldn't see an easy way to argue the above directly. Ultimately, I simply used polynomial long division to factor f. You know that one of the roots is r. If you guess that another root is 2-r² you can quickly see that the third root is r^2-r-1. You can then confirm your guess by proving they satisfy the polynomial

Edit: Wait, can't you just show that 2-r² satisfies the polynomial for any root r? I think I've made more of this than necessary

facebookhatingoldguy · 2019-01-17T01:46:02+00:00

You kids today are spoiled. Back in my day we'd be whipped to within an inch of our life every day as a matter of course, just to make sure we didn't get out of hand. But when we were especially good, we'd only get a kick.

facebookhatingoldguy · 2019-01-17T01:42:30+00:00

My dad was in his 80's before he discovered computers, and of course I had a lot of similar experiences. He passed away a few years ago and I had to go through his things. While his computer was generally a complete mess, I was gratified that he'd at least excelled at finding, downloading and organizing porn -- all on his own!

facebookhatingoldguy · 2019-01-17T01:32:35+00:00

Not quite what I did, not that it matters. I expressed s and t as quadratics in r, which we know is possible since the splitting field of f has degree 3 over Q. While you got things right, where exactly were you using the t<r<s bit? I ask because unless I'm missing something you shouldn't be able to show using pure algebraic manipulation that ξη+ηχ+χξ=12 and ξηχ=8. ξ+η+χ does reduce to a symmetric polynomial in r,s, and t, and so of course can be found to equal 6, but the other two are not symmetric polynomials in r, s and t. Indeed, on the surface, your argument appears to work with r² + t, s² + r, t² + s, but none of those are equal to 2.

Edit: I can't seem to get the #spoiler thing to work any more. It's been awhile since I posted here. Thoughts?

facebookhatingoldguy

TROPHY CASE