Not enough real math on this supposedly math board so here's one for you: solve the exercise then submit another one, gradually increasing difficulty. I'll start not too hard:
For n>=2, does *pic related* have any roots of multiplicity >1 in the complex field?
数学が嫌いよ!!
Lol that's just e^x-(some constant)
So probably not.
it is only e^x if you take the limit n-> infinity which we do not. These are polynomes of rank n which have n roots in the complex field
answer is no?
It's math, so you need a proof. Test it for n=2 and n=3 and you'll be able to conjecture what to prove.
P_n has a nonsimple root if and only if P_n and P_n' share a root. But P_n' = P_{n-1}, so we need P_n = P_{n-1} = 0. But P_n = P_{n-1} + \frac{X^n}{n!}. This forces \frac{X^n}{n!} = 0 which occurs iff X=0. But X=0 is not a root of P_n, so all of the roots of P_n are simple.
got you now
Proof:
A polynomial of the nth degree must have exactly n solutions in the complex plane. Pn is a nth degree polynomial (due to the X^k term in the summation) therefore must have n solutions in the complex plane/field.
therefore, for n>=2, Pn has >= 2/>1 solutions in the complex field. QED.
(Do I need to prove the statement "A polynomial of the nth degree must have exactly n solutions in the complex field"?)
Was just writing something like this after getting out D&F for that differential theorem, nice. A lot easier problem than I expected.
not OP but that doesn't say anything about the multiplicity of the roots
Misread the question.
Sorry.
The answer is no. For n=1 this is obviously true. If you write any polynomial with a root of multiplicity >1 as the product of (x-root[i]) and take the derivative it is clear that it should vanish at this root as well. Now take the derivative of the given formula P[n] and see that it is just P[n-1]. That means that if P[n] has roots with multiplicity >1, P[n] and P[n-1] share a root.
Now we only have to proof that this is impossible. Obviously P[n](0) = 1, so 0 is not a root of P[n]. Let's assume x[k] is a root of P[n].
P[n] - P[n-1] = x^n/n!.
-P[n-1](x[k]) = x[k]^n/n!
We have already seen that x[k] cannot be 0 since 0 is not a root of P[n] => P[n-1] and P[n] do not share a root what completes the proof.
heres how the roots look like btw
i plot it for some other question on sci long ago
As for the next puzzle:
You have a polygon of n sides. You start as one side and go one step to the next sides. for the next step you go two sides further. for the kth step you go k sides further.
Give me all possible n so that you visited each side at least one time for k -> infinity.
Nice.
Anyone care to submit another problem?
Any triangle number.
Distance moved after k steps = k+(k-1)+(k-2)...+2+1 = T(k) = k(k+1)/2
oh, didn't see someone already submitted a proof. nice work, I hope you don't mind that I submitted another puzzle
But this doesn't prove you visit each side does it? It only shows that you moved past k(k+1)/2 sides after k steps. Try doing this with a triangle as a simple example. You will only ever land on two sides (unless I misunderstand the problem or didn't let k go high enough)
That's not really the answer to tat question. take for example a triangle. you start with side 0, after the first step you are at side 1, after the second step you go two sides hich brings you back to side 0. Taking step 3 (3 sides) is equivalent to no step at all and it starts again since a four side step equals a one side step. You will never reach side 2 even if you take infinitely many steps.
6 ÷ 2(1+2)
I think I remember that problem. Did it have anything to do with the trajectories of Mario fireballs? I think the solution ended up being a cycloid.
Exactly. it works for a square though, so that makes me think it only works for even n. You need to make sure that k mod n is *something* I just haven't figured out that something part yet.
>two is a triangle number
you would go:
side 0 -> side 1 (1)
side 1 -> side 1 (3)
side 1 -> side 1 (6)
side 1 -> side 2 (10)
thus reaching every side.
on a n sided polygon starting at side 0, after k triangle number steps, the next step would land you on the( (T(k-1)+ T(k) )mod n) th side
If you try it until n=16 it seems that n must be of the form 2^m with m a natural number. I could not find a proof yet.
I think you're misunderstanding. The first step you move 1 side over. The second step you move two sides over. The third step you move three sides over. That is a set part of the problem. You cannot choose to move a triangle number of sides. The question is pertaining to the polygon itself. What number of sides does a polygon need to be in order for this weird condition to satisfy? You have no freedom in choosing how many sides you move, it's set. It cannot be triangle numbers because it doesn't work for a triangle and does work for a square.
I'll check for correct solutions when I wake up.
Even the simplest of household tasks can present complicated problems in operational research. Consider the preparation of three slices of hot buttered toast. The toaster is the old-fashioned type, with hinged doors on its two sides. It holds two pieces of bread at once but toasts each of them on one side only. To toast both sides it is necessary to open the doors and reverse the slices.
It takes three seconds to put a slice of bread into the toaster, three seconds to take it out, and three seconds to reverse a slice without removing it. both hands are required for each of these operations, which means that is is not possible to put in, take out, or turn two slices simultaneously. Nor is it possible to butter a slice while another slice is being put into the toaster, turned, or taken out. The toasting time for one side of a piece of bread is 30 seconds. It takes 12 seconds to butter a slice.
Each slice is buttered on one side only. No side may be buttered until it has been toasted. A slice toasted and buttered on one side may be returned to the toaster for toasting on its other side. The toaster is warmed up at the start. In how short a time can three slices of bread be toasted on both sides and buttered?
Bonus points for PDDL code
I will admit that I don't know how to solve the problem, but If you would be kind enough to tell me what to go learn to be able to solve it then I will try and solve it later.
Here is the solution. Have fun.
If you want to understand these answers Study complex analysis.
you don't even need much of complex analysis, only the derivative rules of a polynomial and that you can expand a polynomial as a*sum[(x-x_k)] with x_k being roots of the polynomial and a being a constant.
I've been working on that. Also do I need to study all the theorems in this book or just the fundamental principles. Some of it seems not very important.
nice try, you can't divide by zero
Oh I got that one, it's the first book I've ever bought. It's neat but the "direction" formalism for limits is unusual (it's not the usual one and it's not really a filter, it's weird), not sure you'll encounter it in your class.
Beyond that, it's all important. You probably won't remember all of it after a while, but it's good to try to do everything
0! = 1 dumbass