12 Difficult Mathematical Questions for 12 Days of Christmas?

1. There is actually some debate about how many of Euclid's elements were Euclid and how many he "borrowed" from other mathematicians. Now I have one for you. Why is Euler pronounced Oiler while Euclid is pronounced U-klid?

2. chaos was on trial

3. But division is just multiplication by the inverse.

4. Nothing. Topologically it isn't a knot.

5. Who'd want to make a building with Euly bricks?

6. Abstract numbers. Cotton candy for mathematical minds.

7. Something to do with radicals trying to legalize homomorphisms.

8. Any crops. But they can't be more than four levels deep.

9. Steve Jobs tried to sue him but the lawyers are still fighting it out.

10. Actually it was what he called his Chihuahua. I have been reminded that he really called his dog Cauchy because it left a residue at every pole. I call mine (see avatar) Liberace because he is the peeanist dog I've ever seen.

11. No. But it will help you find them. Maybe someone was sitting in a tree and dropped it on Newton's head when he got too excited playing angry birds. I go to Fry's to pick my Apples.

12. It helps a lot if you use the identity (n+1)! / n! = n!/(n-1)! + 1.

13. Merry Christmas.

1) How many are Euclid's elements?

How many would you like them to be? And how much are you willing to pay for them to be that many?

2) What crime was prosecuted and punished in Bernoulli Trial?

I don't know, but I think that someone was just accused at random.

3) Why did Lagrange use Multipliers only, but neglected Divisors?

You know that bright minds are usually very distracted. It happens to me all the time! hehehe.

4) What can be wrapped up in the Folium of Descartes?

I think we could try to wrap our next burrito.

5) Why Euler Bricks are not used in the construction?

I depends on what you are trying to build.

6) What material are Dedekind, Artinian and Noetherian Rings made of?

Dedekind said: some kind of material.

Noether said: neither.

Artinian said: This question doesn't ring any bells.

7) What is the difference between Right Ideal and Wrong Ideal?

Ideally, I should know what you are talking about.

8) What crop grows best on the Galois Fields?

I think that Galois and Gaulois are similar words. And, since he was French, they are probably related.

So, you should ask Asterisk and Obelik, don't you think so?

9) In the good old days of DOS there was a mathematical software 'Eureka - the Solver'. Did Archimedes (as far as I know he has lived also some time ago) use 'Eureka' to discover the Law of Archimedes?

10) French mathematician Fermat is known for his 'Fermat's Little Theorem' - does it mean he has not been capable of proving something big?

In Spanish, as far as I know, we call that the Last Fermant Theorem. Means that little are always last? Maybe size matters, after all.

11) The most famous apple in the history of science has fallen on Sir Isaac's Newton's head. These days I heard about Newton's Method - is it a method for painless picking of apples?

Maybe his method was to just wear a good helmet?

12) I like that approach of one of the answerers. Here are my 2 cents

1!/0! + 2!/1! + 3!/2! + . . + 99!/98! + 100!/99! =

= 1!/1! + 2!/0! + 3!/2! + . . + 99!/98! + 100!/99! =

Can we call this commutative property?

I thought of this one too:

1!/0! + 2!/1! + 3!/2! + . . + 99!/98! + 100!/99! =

= 0!/0! + 1!/1! + 2!/2 + 3!/2! + . . + 99!/98! + 100!/99!

Now, seriously. I realised some time ago that

sum [1 to n] i = C(n+1,2)

and I found a few more similar relations (see below)

Now, coming back to what you posted,

Sum [1 to n] i!/(i-1)! = Sum [1 to n] C(i,i-1) = sum [1 to n] C(i,1) = C(n+1,2)

We can go from there, and combine these to what Madhukar discovered.

----

Notation remark: Instead of sum [1 to n] (i^2), I willl just write sum i^2.

I didnt really recall the formula, this was the reason why I had to find a way to deduce it.

I first proved (using induction) that (sum x_i)^2 = sum (x_i)^2 + 2sum [(x_i)*(x_j)], (i different from j)

Then I expressed sum i^2 this way:

1 + (1+1)^2 + (1+1+1)^2 +(1+1+1+1)^2+... +(1+1+...+1)^2 (In this last terms there are n ones)

This is equals to

1 + (1+1+2) + (1+1+1+2+2) + ... + (1+1+1+...+1+ 2+2...+2)

since any of the products (x_i)*(x_j) are 1.

In the last term, there are n ones and C(n,2) twos

1 + (2 + 2*1) + [3+ 2(1+1)] + [4 + 2 (1+1+1+...+1)] + ... + [n + 2 (1+1+1+...+1)]=

=1 + [2 + 2 C(2,2)] + [3 + 2 C(3,2)] + [4 + 2 C(4,2)] + ... + [n + 2 C(n,2)]=

= (1+2+3+4+...+n) + 2 [C(2,2)+ C(3,2)+ C(4,2)+... + C(n,2)]

The first term is n(n+1)/2

So, I just had to calculate this:

C(2,2)+ C(3,2)+ C(4,2)+... + C(n,2)

In order to use the Stieffel theorem [C(m,n) + C(m,n+1) = C(m+1,n+1)], I changed C(2,2) by C(3,3). Both combinatory numbers are equals to 1.

C(3,3) + C(3,2) + C(4,2) + C(n,5) + ... + C(n,2) =

= C(4,3) + C(4,2) + C(5,2)+ ... + C(n,2) =

= C(5,3) + C(5,2)+...+ C(n,2) =

.......

= C(n,3) + C (n,2) = C(n+1, 3)

Hence sum i^2 = n(n+1)/2 + 2 C(n+1,3)

sum i^2 = n(n+1)/2 + 2(n+1)n(n-1)/ 6

sum i^2 = n(n+1) [3+2(n-1)]/6

sum i^2 = n(n+1)(2n+1)/6

BTW, I noticed that sum i = C(n+1,2)

And sum i^2 = 2 C(n+1,3) + C (n+1,2) or

sum i^2 = C(n+1,3)+ C(n+2,3)

Funny, isnt it?

1. Nothing is really known of the physical form of Euclid, but if he was human, and

not some remarkable, alien computing machine, then his (essential) elements would

have been 24 (O, C, H, N, Ca, P, S, K, Na, Cl, Mg, Si, Fe, F, Zn, Cu, Mn, Sn, I, Ni,

Mo, V, Cr and Co). It's not known how many others were incorporated during his life.

4. Just in the loop, I can wrap up the largest circle, of radius √(√3 - 3/2).

EDIT: I should add that the equation used was x^3 + y^3 = 3xy.

11. Newton's method is to choose a spot under an apple tree and stand there to see

if the first apple that drops, falls on your head. If not, then at intervals, shake the tree

a little. The longer you stand there and shake the tree, the greater the chance you

will be bopped. If you were unsuccessful, try another tree that's on a different slope.

Failing that, the method then requires you to dig up the ground underneath your spot

and search around until you find a root. This is almost always successful.

I can answer a couple:

10) Fermat is a mathematical genius, and his theorem is so non-trivial that it has still not been proven. FLT is important. (all f them)

11) Newtons method is a method used in calculus to approximate functions.

12) let the number you want to add up to be 99. So I take 99+1, 98+2... which is equal to 100*50. Now I add 100, and I get 100*50 + 100 which is 5100.

Your question 12) gives me an idea

(n+2)!/(n-1)! = n*(n+1)*(n+2) = n^3 + 3n^2 + 2n

=> 3!/0! + 4!/1! + 5!/2! + .... + (101)!/(98)!

= Σ n^3 + 3 Σ n^2 + 2 Σ n ... (n = 1 to 100)

= [n(n+1)/2]^2 + n(n+1)(2n+1)/2 + n(n+1) ... n=100

= n(n+1) [n(n+1)/4 + (2n+1)/2 + 1] ... n=100

= n(n+1) [n(n+1) + 4n+2 + 4]/4 ... n=100

= n(n+1)(n^2+5n+6)/4 ... n=100

= n(n+1)(n+2)(n+3)/4 ... n=100

= 100 * 101 * 102 * 103 / 4

= 26527650.

Edit:

There is an interesting symmetry for series of the above types as under:

1) Σ n! / (n-1)! = n(n+1)/2

2) Σ (n+1)! / (n-1)! = n(n+1)(n+2)/3

3) Σ (n+2)! / (n-1)! = n(n+1)(n+2)(n+3)/4

4) Σ (n+3)! / (n-1)! = n(n+1)(n+2)(n+3)(n+4)/5

5) Σ (n+4)! / (n-1)! = n(n+1)(n+2)(n+3)(n+4)(n+5)/6

I have checked the first three manually and last two using Wolfram Alpha.

Intuitively, the following result must be true.

Σ (n+k)! / (n-1)! = n(n+1)(n+2)(n+3) ... (n+k+1)/(k+2)

I Like it!

You've never been on a date, have you?