Ah, right, if you pick a sequence in $\ell^{\infty}\setminus c_0$ converging to an element of $c_0$ there are issues. The ones agreeing with $1/n$ for the first $j$ terms and being constantly $1/j$ afterward for example seems problematic
The point is that if you let $x_j = (1,\frac{1}{2},\ldots,\frac{1}{j},\frac{1}{j},\ldots)$, then by your $f$, we have $f(x_j) = (\frac{1}{j},\frac{1}{2j},\ldots)$
Do try the norm idea to see if it works, though at this point I also gotta sleep
(Tbh the fact that you can have Banach spaces with non-complemented closed subspaces is probably good reason to believe the Normie Wildberger had a point...)
I guess norm idea won't work. If we let $x_j = (1/j,\frac{1}{2},\ldots,\frac{1}{j},\frac{1}{j},\ldots)$, then we have $f(x_j) = (\frac{1}{j},\frac{1}{2j},\ldots)$
A die is colored green on $4$ sides and red on $2$ sides. if the die is rolled $7$ times. Then which of following option is correct regarding occurrence of green and red faces?
(A) $2$ green and $5$ red
(B) $3$ green and $4$ red
(C) $4$ green and $3$ red
(D) $5$ green and $...
Listened to a good traditional heavy metal album called "Satan's Hallow" by Satan's Hallow. But they almost copied a Rainbow's song then stopped listening
This shows how when given a sufficiently destructive force, even some of the most numerous species (practically speaking almost infinite like) will plummet to zero
Good morning everyone! Where can I ask a logic puzzle? One of these popular puzzles with true and false statements. I am completely puzzled myself :) Thank you!
The Game is a mental game where the objective is to avoid thinking about The Game itself. Thinking about The Game constitutes a loss, which must be announced each time it occurs. It is impossible to win most versions of The Game. Depending on the variation of The Game, the whole world, or all those aware of the game, are playing it all the time. Tactics have been developed to increase the number of people aware of The Game and thereby increase the number of losses.
Though the origins of The Game are unknown, a game featuring ironic processing was played by Leo Tolstoy in 1840.
== Gameplay ==
There...
In short, similar to how it is impossible to win most of the game, it is almost impossible to stop The Plan
On a more iluminati-ish analogy: The Plan is what makes the chat flow
(Fun fact: The chemistry incident have somehow managed to cause the topology group to stop reducing weirdness (probably because with a better grasp of manifolds, I can contribute to the discussion). Currently, the weirdness reduction is only concentrated for a couple of users. You know when they happen when you saw messages that are somewhat out of context and incomprehensible)
In other news, I cannot think of any maths to discuss about yet
Herod = mix of the criminal underworld and chanting
Brando = of the 5 I like this one the most. There seemed to be a drama weaved into it since it started out lively, then go mysterious, and then the lyrics suggest some kind of struggle, then becomes hopeful, and then becomes dwelling and then more power struggle
$a$ is an integer, such write in basis $b$ (is palindrom) with only $\{0,1\}$ and the size is $n<b$. Is it true that $a^2$ is a palindrom in basis $b$ ?
I cannot think of any non number theoric ways to answer that question (and I don't know enough number theory to figure out what tools I need to answer it too)
But from a purely algebraic point of view it will probably had something to do with the following matrix:
and $a^2$ is then given by summing all the entries of the matrix
We can now observe that, for $a$ with $n$ instances of $1$s, $a^2$ will have $n^2$ instance of $1$s. Thus the algebraic way boils down to whether $n^2$ is palidrome in base $b$. That I am not very sure because I am terrible with carryovers
Doesn't this follow from just writing up the entries in the square of a polynomial and noting that the coefficients stay below $b$, so that we really can treat everything as polynomials
I am not sure whether there is an efficient way to check whether $n^2$ is palidrome in $b$ though other than writing it in base $b$ by expanding it in terms of polynomials involving $b$ and then check that the nth and n-k th matches for each k < n/2
@Dattier I tried to prove it by brute forcing the multiplication and all I get is a mess. I don't know of clever ways to deal with it since my knowledge on polynomial ideals is almost nonexistent
In generall, this is why I don't like multiplying two polynomials
P(1/X) reverses the powers of n, thus literally flip the polynomial left to right. Since P is palidrome, "filling the vacancy" by multiplying by $X^n$ gives the original polynomial, as required
@anakhronizein check the count on this page math.stackexchange.com/questions (make sure that ignored questions on the questions page are grayed out instead of hidden in your user preferences)
Now: If $P,Q$ palidrome then by the rules of degrees we get $deg(PQ) = deg(P)+deg(Q)$. Suppose this new polynomial is $R$, we want to know whether it is palidrome. Thus we compute $P(1/X)Q(1/X)=R(1/X)$ which reverses the ordering of the powers around. Now we multiply both sides by $X^{deg(P)+deg(Q)}$ to get:
Now since P,Q is palirome, by lemma 2 the LHS is P(X)Q(X)=R(X). Now using lemma 2 again, we see the equation is that of a palidrome polynomial equation, therefore $R$ must be palidrome
Now back to the original question: $a$ can be expanded as a polynomial P(b) with deg(n) and n < b. We knew that P(b) is palidrome and that all coefficients are 0 or 1. Thus when $a^2=P(b)P(b)$ is computed by lemma 2 we knew that the product is palidrome. In addition, the coefficients being 0 or 1 means their products lies in the ideal {0,1} and thus no carryovers can occur, hence $a^2$ is always palidrome
Prove that if $R$ is a commutative ring and $N=(a_1,...,a_m)$ where each $a_i$ is nilpotent., then $N$ is a nilpotent ideal.
So we need to find a $k \in \Bbb{N}$ such that $N^k = (0)$. Working through small values $m$ and $k$, I can see that $k = 1 + \max \{a_1,...,a_m\}$ seems to work. E.g...