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#1 Re: Help Me ! » group with union or intersection as the operation » 2016-10-24 11:11:21

Symmetric Difference should be doable, right? A o B = B o A, ∅ is the neutral element and A^(-1) = A.
Closure should be achievable, too, right? Let's say A o B = C, then A o C = B -> Whatever you do, you still land in the group. It's late at night here, so please someone correct me if I'm wrong, but it sounds plausbile in my head.

#2 Re: Help Me ! » projectile » 2016-10-24 01:18:39

Just a suggestion, but v and u are nearly impossible to distinguish in LaTeX, I'd usually avoid using them both at the same time.

#3 Re: Help Me ! » group with union or intersection as the operation » 2016-10-23 11:16:45

I could only think of the complement, but that doesn't fulfill the associativity requirement.
You could argue that in modular arithmetic, every element (e.g. of (Z_7, +)) is actually a set, but the operation wouldn't be one of the usual set operations.

#4 Re: Help Me ! » projectile » 2016-10-23 10:12:41

You made two mistakes in your formula: a) You used 21 instead of 28 and b) you used 100 instead of 10 (you can use 100 if you leave out the square root).

I assume that 28 is the vertical velocity (it looks like it in your approach).

Then we get: v(t) = 8i+(28-gt)
speed(t) = |v(t)| = sqrt(8²+(28-gt)²) = 10
8²+(28-gt)² = 100
(28-gt)² = 36
784-56gt+g²t² = 36
748-56gt+g²t² = 0

Just plug that into any formula for solving quadratics of your liking smile

If we set g as 9.8 you get exactly the answers given.

#5 Re: Help Me ! » Rearranging students in a class » 2016-10-23 02:09:47

That's what the factor 2 is accounting for (cases a) and b) in my post).

#6 Re: Help Me ! » Rearranging students in a class » 2016-10-23 00:37:14

Thickhead is right. Let's first seat all the "noname-students" randomly, there are 6! ways to do that. Now we choose a random spot for our Frankenstein's Monster (aka both of them tied together), which can be in one of 7 places, i.e. 7C1 (which is just 7), so we get 6!*7 = 7!. Now we can double that count for a) Gina-Christina and b) Christina-Gina.

So we get 7!*2, which is obviously the same as 7!*2!.

Edit: Just to clarify why the opening post is wrong: You assume there are 8 different spots for Gina to be seated, but you can't sit Gina to the right (or left) of Christina, so there are actually only 7.

#7 Re: Help Me ! » Lagrange Interpolation in a finite Field » 2016-10-22 13:02:18

/u/ImYourWenis was able to decipher the problem: The notation used in the paper is just really unintuitive!

This really comes from mixing up the polynomial and elements of GF(2³) (To any aspiring authors out there: Don't use x for both!).
Anyway, I'm glad it's clearing up, because that was really giving me a bad headache.

#8 Re: Help Me ! » Lagrange Interpolation in a finite Field » 2016-10-22 10:25:36

Note my edit.

I assume SE means stackexchange? I have not, but I will for sure.

Even if they are reduced, I don't think that should change anything (after all, the congruence is what makes modular arithmetic so useful).

#9 Re: Help Me ! » Lagrange Interpolation in a finite Field » 2016-10-22 10:19:16

bobbym wrote:

I do not see the relationship either...

Do you have any example of how those interpolating polynomials were formed? What was the data?

Well, the data is right there in the table?

(The first value corresponds to x=0, the second one to x=1, ...)

Given that this is GF(2³), there are no more possible data points, as x is inbetween 0 and 7.

Edit: For k=3 and k=4 

yet the polynomials evaluate to 20 and 18 respectively... Something fishy seems to be going on here.

#10 Re: Help Me ! » solve » 2016-10-22 09:57:53

Primenumbers wrote:

7 is prime so only factorable by 1 or 7.

the only squares less than 6 are 1 and 4 therefore a=1, r=2

a=7, r=0
73 is prime so only factorable by 1 or 73. 73 too big so must be 1. Only
smaller than 73 is r= 1 or 2. 1 is too small so a=1, r=2.

You seem to implicitly assume that a is an integer, because for a,r€R there are a lot more possiblities than your first few lines mention.

#11 Re: Help Me ! » Lagrange Interpolation in a finite Field » 2016-10-22 09:52:26

It shouldn't have anything to do with the question though, which I think is a much more fundamental misunderstanding on my part.

#12 Re: This is Cool » Fast way to find primes. » 2016-10-22 09:49:22

Primenumbers wrote:

I'm not sure but I'm guessing this would be a faster way to compute finding primes. I.e. using Euclid's algorithm.

Primality Tests are usually done on numbers in the range of 2^1024 to 2^2048 (or much, much bigger numbers, when it's simply about "finding a prime", not crypto). Enumerating (and multiplying!) all the primes below those numbers is an insurmountable task.

If enumerating all primes was easy, RSA would be useless :-p

Edit: I just saw this thread is kinda old (still first page), I hope this doesn't count as necromancy smile

#13 Help Me ! » Lagrange Interpolation in a finite Field » 2016-10-22 09:39:04

Replies: 9

As part of my bachelor, I'm currently working on a presentation on this paper:

You probably don't need to look at the paper, because the important part (the one I don't quite understand) is here:

If I'm not mistaken, the interpolation polynomial is a closed form for

(choosing the one corresponding to k). Note: For some reason LaTeX doesn't like my apostrophe, so I will just leave it out.

Now, when I actually try to compute the corresponding values using the polynomials, I don't get the correct results. The first column is obviously correct - all these polynomials map 0 to 0. Let's look at k=0, so the first row.
1 is, even in this representation, simply 1, so the polynomial evaluates to 15, right?
I'm not quite sure how to interpret that number, but any way I've tried has lead to a solution that does not equal 5. If I interpret this as a member of GF(2³), I get

Which obviously doesn't equal 5.

I'm really at a loss here, as these polynomials don't seem to form any relation with the given table?

Help will be appreciated,

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