river_rat

You cannot find the roots of a general polynomial of degree [math] \geq 5 [/math] if you limit yourself to addition, multiplication and taking square roots Atheist. I guess that is what you meant by analytical, as there are other analytical solutions if you allow other operations and functions, like the elliptic functions for the quintic case.

I'm surprised no one chastised me for my solution - oh well

Hi abskebabs Here is a big hint for you - how do you factor [math] a^n - b^n [/math]?

Hey abskebabs You can do one better then what you have posted so far: if [math] 2^n - 1[/math] is a prime number then [math]n[/math] must be prime. I'll leave the proof up to you, its not difficult

The problem is impossible to solve - Dr Math has a nice solution http://mathforum.org/dr.math/faq/faq.3utilities.html

or if you feel daring, your answer is [math] \Im \left( \int e^{(2+3i) x} dx \right) [/math] which saves you all the product rule pain PS [math] \Im [/math] denotes the imaginary part if you are wondering. The [math] dx [/math] is called a differential form (or one form in this case) and the theory here is quite interesting. It takes a surprising amount of mathematical work to get something that is more meaningful then the nonsense idea of an infinitely small but non-zero change in x.

I would suggest "Probability and Random Processes", by Grimmett I think, for a quick and easy intro to the basics here. Its nice, starting with the basics and ending with the Ito calculus.

How is the second recurrence well defined tree?

Ah, but that is a different question. I only stated i could extend the algebraic operation of addition to a larger set that includes some ideal points. To start talking about limits you must have introduced a topology. To talk about limits and addition it would be nice if addition was continuous with respect to this topology. Now if you want a "nice" topological extension of the reals i would have to suggest the Stone-Cech compactification where addition can be extended by the universal property of that compactification. Sadly I think we only get an operator which is left continuous but we are still better equipped to talk about limits here. Well share the sketch

But you have not explained why we are limited to group operations. Lets change the story a bit, we needed a way of talking about [math]\infty[/math] and addition on the naturals for this whole setup to work for the problem at hand. Now addition is not a group operation on the set of natural numbers but we have a perfectly legitimate semigroup operation which extends addition to the set [math]\mathbb{N} \cup \{ \infty \}[/math]. Just treat the added point as a zero under addition (i.e. [math] x + \infty = \infty + x = \infty[/math] [math]\forall x \in \mathbb{N} \cup \{ \infty \}[/math])

Ok, reread what you replied to again and i still can't see your connection.

Why didn't you use Ito's formula?

Hi w=f[z] I got those solutions by just kicking out the equations for Marconi, Stern and Davison and solving the resulting system and then by kicking out the equations for Marconi, Stern and Cherenkov and solving the resulting linear system.

Lol, this one is actually easy : just read the numbers out loud First Line : You have One One = Second Line Second Line : You have Two Ones = Third Line Third Line : You have One Two and One One = 4th Line 4th Line : You have One One and One Two and Two Ones = 5th Line 5th Line = You have Three Ones, Two Two's and One One so the next line is 312211

Gogo, can you show me how you got that SDE from my suggested substitution?

On the reals yes but not on the naturals so your point is?

Take [math]\mathbb{R}[/math] and adjoin two points to it, namely [math]+\infty[/math] and [math]-\infty[/math], denote this new set by [math]\left[-\infty, +\infty\right][/math]. Define addition as follows : [math] a + b = \begin{cases} a + b & \text{ if } a, b \in \mathbb{R} \\ +\infty & \text{ if } a = +\infty \text{ or } \left( a \in \mathbb{R} \text{ and } b = +\infty \right) \\ -\infty & \text{ if } a = -\infty\ \text{or } \left( a \in \mathbb{R} \text{ and } b = -\infty \right) \end{cases}[/math] This operation is closed and associative so we have a semigroup with both [math]+\infty[/math] and [math]-\infty[/math] acting as zeros on [math]\mathbb{R}[/math] under addition and is an extension of the normal addition on [math]\mathbb{R}[/math]. In fact both added points are left-zeros of the entire space The trick here is to not demand that addition be commutative for all possible values, only for real values (else define a local semigroup and work in the two point compactification of [math]\mathbb{R}[/math] if you want to keep the commutativity.) For our original problem though we only actually need to add one point to [math]\mathbb{R}[/math] so the issue is actually a moot point.

But why must the extended reals act as a group? So we declare [math]\left[-\infty, \infty\right][/math] to be a semigroup with [math]\infty[/math] seen as a zero (so it has no inverse, so what?).

Adding to what w=f[z] said, the resulting systems here are sparse so you could do them by hand if you had accountant like accuracy. Being able to accurately copy down 550 odd numbers has never been a strong point in my mathematical career though lol.

What you were looking for was "brownian motion" and not "brown's movement" which sounds like the results of a rather strong curry if you ask me. Have you tried the substitution [math]F(X) = e^{2 X(t)}[/math] ? I get a linear SDE (assuming no silly errors that is) which you can solve using the standard methods.

atheist, an addition does exist on [math]\left[-\infty, \infty\right][/math] - just treat the added points as zeros under addition. Its what the extended real system is all about

So i got Mathematica to churn through all the combinations and arrived at the following: If you remove Marconi, Stern and either Davison or Cherenkov from the list you arrive at a consistent system with an integer solution (the same solution btw) A -> 25 B -> 22 C -> 6 D -> 19 E -> 12 F -> 6 G -> 20 H -> 0 I -> 27 K -> 3 L -> 22 M -> 8 N -> 1 O -> 12 P -> 20 R -> 9 S -> 7 T -> 22 U -> 13 V -> 26 W ->4 Z -> 13 Some Trivia about each Marconi - Received prize "in recognition of their contributions to the development of wireless telegraphy" Davisson - Received prize "for their experimental discovery of the diffraction of electrons by crystals Stern - Received prize "for his contribution to the development of the molecular ray method and his discovery of the magnetic moment of the proton" Cherenkov - Received prize "for the discovery and the interpretation of the Cherenkov effect"

Ok, but the original question (which i am still concerned about if you followed my posts) was inconsistent and overdetermined. IMHO the second question was someone's cop out the original problem, so we still need to solve the problem this thread is about. For which problem, the first or second? Why is that?

You are kidding right? Excluding equations changes your solutions, and there are 2300 different solution sets for the original problem. For example, consider the system [math]x + y = 2[/math] (1) [math]x - y = 0[/math] (2) [math]x + y = 1[/math] (3) Removing equation 1 gives you the solution set [math](x, y) = (\frac{1}{2}, \frac{1}{2})[/math] while excluding equation 2 still leaves an inconsistent system. Removing equation 3 gives the solution set [math](x, y) = (1, 1)[/math]. So bar solving the 2300 systems and picking out the integer solutions and going from there as a brute force method i don't see that as a "fair" attempt at the solution. Huh? Gap in the line?

My solution [math]x=\sqrt{2}[/math] I'm too lazy to check if [math]x=-\sqrt{2}[/math] works