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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...

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

I am not sure what you even tried to do.

See ya around! And thanks!

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.

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

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

Relevant line:

anonimnystefy wrote:

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

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

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

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 .

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!

I will have to agree on 167.

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.

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.

"No problem".

Hej bobbym

Inget problem!

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.

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):

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

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

I shall take a look at that page.