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#1 Re: Help Me ! » Skippy the kangaroo » Today 22:12:55

Okay, so I have discovered that my code for the optional backwards step is not correct, despite giving the correct result (it fails on lower numbers, etc.).

I /have/ written new code for this one, which is the same principle, basically...

#2 Re: Help Me ! » Why Rationalize Infinite Limits » Today 20:59:14

I notice Mathematica can't actually expand around negative infinity.

#3 Re: Help Me ! » Why Rationalize Infinite Limits » Today 19:08:29

I am not sure what you even tried to do.

#6 Re: Help Me ! » Why Rationalize Infinite Limits » Today 16:55:13

Taylor does not exactly make sense at infinities...

And, before we go any further, you'll have to agree that that would lead to the incorrect answer of 2.

#7 Re: Computer Math » How to do it. » Today 16:32:41

No, of course. There was mention of him as well.

I particularly liked Borcherds's proof of Jacobi's triple product identity. Very elegant!

#8 Re: Help Me ! » Why Rationalize Infinite Limits » Today 16:26:11

Relevant line:

anonimnystefy wrote:

Here, I applied Taylor (or more specifically Maclaurin), I just didn't state it explicitly.

#10 Re: Computer Math » How to do it. » Today 16:19:27

Yeah, there is some of his stuff and also some of Ramanujan.

#11 Re: Help Me ! » Why Rationalize Infinite Limits » Today 16:17:02

... Which is basically the same thing I did above. smile

#12 Re: Help Me ! » Mysterius semiprime fact in other bases » Today 16:16:01

Here is a way of looking at it:


If
, with
, and
has a hundreds place equal to 1, that means that
and
. So, for some
,
, i.e.,
, or further,
, and the representation of
in base
will be exactly
, from which the representation of
will be
.

#13 Re: Computer Math » How to do it. » Today 16:04:42

Ooh, I miss these!

I think I might post a similar thing (or two), but maybe in Euler's Avenue; it's not really Computer Math. I had a Descrete Mathematics course this semester, and the prof showed us some very cool proofs involving partitions, mostly combinatorial!

#14 Re: Help Me ! » Skippy the kangaroo » Today 15:47:33

Yeah, basically what I did is partition the length into 1's and 2's, then find the number of ways to permute each one and to insert the one step backwards. (I.e., the same thing as you, phro, except I did not actually permute anything, just counted the permutations.)

#15 Re: Help Me ! » Skippy the kangaroo » Today 15:26:19

I will have to agree on 167.

#16 Re: Help Me ! » Why Rationalize Infinite Limits » Today 15:05:50

Nehushtan wrote:
Numerical wrote:

Hello MathisFun forummers,

I stumbles accross this excercise:

and I was thinking why should I rationalize it?

You are NOT rationalizing it; you are doing the reverse!

To rationalize is to eliminate the square-root signs in the denominator. Here you want to put the square-root signs back in the denominator!

And why do you that? Because it's the only way to solve limit problems of this sort!

Not exactly true...


Maybe a bit more complicated than the anti-rationalization method, but much more generalizable. smile

#17 Re: Help Me ! » Simplifying this function for partial differentiation » 2015-12-14 06:02:11

Hi all,

Technically, that simplification is correct only if y is not negative. Otherwise, you'd need to have |y| instead of y in the numerator there. smile

#18 Re: Help Me ! » Min of the largest sum of adjacent numbers in a 2nx2n array » 2015-09-22 17:32:26

Well, their proof shows that the minumum is 19 or greater. After that, we only one example of the sum being 19 to prove that is the minimum.

#21 Re: Help Me ! » The definition of ''or'' when sketching a set » 2015-09-20 15:19:10

In all of mathematics, "or" usually signifies what would be often denoted as "and/or", so in this case z is something that satisfies at least one of those constraints.

#22 Re: Help Me ! » Min of the largest sum of adjacent numbers in a 2nx2n array » 2015-09-20 09:49:40

Hej,

You are usign the highest sum you can find in the grid. In the 2x2 case, that is 6. If you arrange the numbers any other way, you'll get a higher maximum sum. For example:


Here, the highest sum is 7, which is higher than the 6 for the previous grid, so we do not want that.

For the second, 4x4 grid, the highest sum inside it is 19. To prove that this is the minimal highest sum, we would need to prove that for all other arrangements of numbers from 1 to 16 into the grid, there will be two adjacent numbers which sum to 19 or higher.

#23 Re: Help Me ! » Min of the largest sum of adjacent numbers in a 2nx2n array » 2015-09-20 05:38:11

Hej bobbym

You look at all possible fillings of the grid and for each filling you see what the maximum sum of adjacent ellements is. Then you find the minimum of those numbers.

For example, for n=1, you are filling a 2x2 grid with numbers 1,2,3,4. The lowest adjacent sum, which is 6, can be obtained with the filling:


or any equivalent filling, (obtained by rotation or symmetrical mapping, t.ex.)

For n=2, a filling with S=19 *can* be found, but I am not sure if there is a lower one (probably not):

#24 Re: Help Me ! » Min of the largest sum of adjacent numbers in a 2nx2n array » 2015-09-20 02:42:32

The minimum of S across all possible ways to fill the grid up, I am guessing?

#25 Re: Maths Teaching Resources » The Nine Point Circle » 2015-09-19 08:13:05

Well, t.ex. and osv. are abbreviations from Swedish, not English, so that is not too surprising. big_smile

I shall take a look at that page.

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