Math Is Fun Forum

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.

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".

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.

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:

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

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.

Well, "ex-" sure sounds much better there. "e-" does not seem like a productive suffix.

And besides, you cannot expect a suffix to recognize mathematical terminology, can you? I doubt it would recpgnize "homomorphism" and the like. xD

Hej Bob!

These threads are a fantastic idea! Nice work.

An intersting fact about the Nine Point Circle (I also know it under the name of Euler's Circle) is that when you apply inversion around the Nine Point Circle, the circumcircle gets mapped into the incircle and vice versa!

Does entering an empty math tag somewhere before help?

Hi mrpace

You can prove that there is at least one number c in (a,b) such that f(c)=C by the Intermediate Value Theorem. To prove that there is only one such number, we will take any two numbers, c1 and c2, such that f(c1)=f(c2)=C. If c1 were less than c2, than f(c1) would be less than f(c2) which is not possible. Also, if c1 were greater than c2, then f(c1) would be greater than f(c2), which is impossible as well. Thus, c1 must be equal to c2, which proves that there is exactly one such number.