And we've made use of the only property we care to remember for a group homomorphism, that is the $f(a*b)=f(a).f(b)$, so hopefully it's slightly easier to remember
Well, polynomials are defined to be expressions for which the variables can only ever have whole number exponents. So there is no way to get complex numbers. There's my answer, I guess. Nvm, then.
If you keep doing this, you will always find yourself at 4, with an infinite loop, since "four" has 4 in it.
Challenge:
Given the string length function $sl(n)$ for any english name of any mathematical object $n$. It is known that $n=4$ is a fixed point. Prove or give cnounterexample that $n=4$ is an attractive fixed point for all $n$ for this algorithm.
This might be a very stupid, and possibly philosophical question, but attempt to apply mathematics to everything plus inspired by this question caused me to ask this question
Is there any mathematical object that has been proved to exist but cannot be described in words?
If the answer is...
Well, that's true, but whether 4 is attractive or repulsive or neither makes a lot of difference. If 4 is not attractive, then there could exist $sl(n)$ that will never reach 4 after iteration
It will be very interesting if given a mathematical object with an uncountable number of words (which an example is hinted in that question above) actually do end up at 4. That will mean 4 is a fixed point of infinite "radius(?)" (forgot the correct term)
Sketch proof: Assume that whenever $sl(n)$ is unbounded (either countably infinite or uncountably infinite), we set $sl(n)=\infty$ Now $sl(\infty)=sl('infinity')=8$ Next $sl(8)=sl('eight')=5$ Next $sl(5)=sl('five')=4$
Therefore if $n$ has an unbounded number of words, the algorithm converges to 4 given the assumption.
or show that there is a counterexample that 4 is attractive fixed point for all $n$
I guess in terms of bounds as suggested by Mikemiller, one thing we can note is the trend: For all numbers $n \in \mathbb{N}$, $sl(n)$ increases much slower for each increase in $n$.
If this can be proved rigorously, then we will have the recurrence relation for large numbers $n > 100$ where $sl^{(2)}(n) < sl(n)$
and that will mean any $n > 100$ will end up in the set $\{0,1,2,\dots, 100\}$ which the results are knwon
Indeed, we can either stick to the illion system (which I think is going to run out after googol) or the english language description will be something that describes the operations e.g. "ten to the power of n"
the only thing that is certain is it is of finite length
Hello!! We have that $\mathbb{R}^{2\times 2}$ is a ring. We want to show that $K=\{\begin{pmatrix}1 & b \\ -b &a\end{pmatrix} \in \mathbb{R}^{2\times 2} \mid a,b\in \mathbb{R}\}$ with the matrix addition and multiplication is a field.
I thought to show that $K$ is a subring of $\mathbb{R}^{2\times 2}$ and it is also invertible is ths correct?
Hey everyone, just a quick question "Find the order of the cyclic subgroup of $D_{2n}$ generated by $r$", wouldn't the order of that cyclic subgroup just be $n$?
Some findings from that algorythmic stuff I was doing yesterday, it's probably simple textbook stuff but it might be interesting, it was very interesting gradually axidently discovering it:
Two sequences can be convoluted together by a simplification of pretty much literally just multiplying the sequences as if there terms were the digits of two numbers, and when you code this up short sequences can be convoluted in one instruction!
Below, the convolution of (1, 2, 3, ...) with itself is revealed to be the tetrahedral numbers. The tetrahedral numbers entry on OEIS also asserts this.
On paper, you can use a simpler version of the usual multiplication heuristic, where carrying does not occur when adding, and the adding step at the end stops early.
This also occurs when you are multiplying any two polynomials, and it is one reason why I found multiplying polynomaisl so annoying, since convolution is a nonlocal function and hence depends on every single detail in the polynomial such as their coefficients, and make it very hard to visualise what each coefficient is doing for an arbitrary polynomail
If some kind of pattern can be found in this sort of convolution function, then it will become easier to predict where the roots of a polynomials are
well, for powers of the same polynomial, by multinomial theorem you get the usual pascel triangle and its simplex generalisation on what the coefficients are. But what I am talking about is a more general case where two polynomials of different degree multiplying together
that is, all those cross terms that appeared whenever you expand something
if they actually obey some underlying rule, then predicting e.g. (a-b)(a+b) have no cross terms of the form ab or ba will become easier
Some long time ago, I asked the maths chat about a way to represent the discrete convolution as a matrix, but it does not help much because the resulting stuff is still kinda complicated
You could visualise the convolution they form like you would with a continuous convolution where it's much easier to imagine the resulting curve? Sorry if thats not that helpful.
Well, I think the gist is, since we realise multiplying any two polynomials (or as you have shown, sequences in general) is a discrete convolution, can more be learnt about this convolution function so that we can predict in advance that given two sequences with + and - terms, when you multiplying them together, which terms will vanishes
O, in that case, we will have a polynomial of degree ab. so:
e.g. $(1+2x+3x^2+4x^3)^2=1(1+2x+3x^2+4x^3)+2x(1+2x+3x^2+4x^3)+3x^2(1+2x+3x^2+4x^3)+4x^3(1+2x+3x^2+4x^3)=\textrm{a deg(9) polynomial that I am too lazy to write out}$
but for this example there are no cross terms vanishing since all are positive
What I aim for in the future is some formulae that given any $a$ and $b$, it will give me which terms will vanish in the expansion before the actual expansion is carried out
If we wrote this convolution as a matrix (before every entry is summed up), it is clear there is some kind of symmetry within, but (at least for me) it is not sure how to better characterise that symmetry and convert that into a formula
Notice the red circled entries, $8x^3$ and $2x$ will go, and only one $4x^2$ will be left behind
It seems your polynomials have terms (seperated by +) in the form of ax^b, and the convolutions have terms (I wrote them seperated by commas) in the form of just one number per term.
Yeah, if only there's some universal formula that can predict this beforehand. Everytime when I try to help myself or others to check whether some proper rational function can be simplified, this problem pops up and I often have to multiply 10+ terms and get infuriated whenever I saw terms disappear that I don't previously suspect
Solving this problem might also have implication on factorisation, in fact
So what we knew is that whatever that answer is, lies in the property of discrete convolution
hmm... the matrix is not antisymmetric, but the terms that tend to cancel are antisymmetric entries in the matrix...
If we restrict the multiplication tables to the entries that give the antisymmetric terms, then we do have submatrices that are antisymmetric
not sure what that means...
Conjecture: Given a polynomial $P(x)$ and its square is asked $[P(x)]^2$. Then for any two corresponding pair of terms in P. If they are of the form (a,b) and (-a,b), then the cross terms vanishes
e.g. $(1+2x+2x^2+\frac{4}{3}x^3+\frac{2}{3}x^4)(1-2x+2x^2-\frac{4}{3}x^3+\frac{2}{3}x^4)$
Pair 1: (1,2x) and (1,-2x). Property is fullfilled. Hence cross term of them vanishes
Pair 2: $(-2x,\frac{2}{3}x^4)$ and $(2x,\frac{2}{3}x^4)$. Property is fullfiled. Hence cross term of these vanishes
correction on typo: Conjecture: Given two polynomials $P(x)$ and $Q(x)$. For $P(x)Q(x)$, pick any two terms from each of P and Q. If they are of the form (a,b) and (-a,b), then the cross terms ab and ba vanishes
The next step is to wonder about a combinitorics problem. Given a polynomial P with n terms, the number of possible pairs of terms with no repetition can be picked from it is $\frac{{}^nC_2}{2}$ (halved to avoid double counting transpositions)
conjugation distirbutes over multiplication and addition, thus $\overline{(\overline{X1}(\overline{X2}+X3)})=X1\overline{(\overline{X2}+X3)}=X1(X2+\overline{X3})$ Work from there
(Again for the next post, please just treat 0 and 1 as just labels, I am too lazy to change them)
If you restrict your stuff to just transpositions you get some abliean group like these, which fullfill your requrement that cyclic permutation don't change them?
@AkivaW So the most obvious thing to try is to get 4 permutations $s_1, s_2, s_3, s_4$ on those 5 letters which are preserved (not individually, but as a whole) under action of $S_5$. If you could do that, you could do it all as you know how to do it on 4 letters, by the same iterative thing. I think that'd give a surjective homomorphism $S_5 \to S_4$.
Which doesn't exist (kernel is a normal subgroup of order 5; there is no such thing)
Maybe that's not entirely correct. I think that's the circle of ideas you want to get into
Actually forget surjective; it'd give a homomorphism $S_5 \to S_4$ in any case. But kernel has to be $A_5$. That means the image fits in $\Bbb Z/2 \leq A_4$. So permuting the 5 letters would just produce at most 1 conjugate of $(s_1, s_2, s_3, s_4)$... what does that mean? shrug.
Given a semigroup $S$ with possibly infinite elements , it is known that there is an element $a$ such that its left semigroup action on all other elements $x \in S$ either fixes it or map to a distinct element i.e. $ax=x$ or $ax=b$ where $b \neq x$. It is also known that a left inverse exists for $a$, thus $a$ is injective. How can I show it is surjective and hence the semigroup action $a$ and its inverse is bijective over $S$ and hence a permutation?
This might be a very stupid, and possibly philosophical question, but attempt to apply mathematics to everything plus inspired by this question caused me to ask this question
Is there any mathematical object that has been proved to exist but cannot be described in words?
If the answer is...
