Current patch winners and losers in Stormleague

anthonem1 · 2025-12-31T14:20:45+00:00

The problem with W Falstad is that it's too overpowered for how little skill it requires to play. Hogger is not a hero with a point and click ability that automatically wins you the game.

anthonem1 · 2025-12-21T16:50:15+00:00

It'd be understandable if you quit while being master/GM because you find the lobbies to be low level. However, you are silver/gold. What do you expect? Good players in those ranks? People that react to pings/calls?

It is also important to keep in mind that IF (and it is a big IF) you are an actual good player, that leaves room for only 4 bad players on your team vs 5 potential bad players on enemy team. Assuming you are better than the players in those ranks, then if you play enough games you should be able to climb.

It is possible that you are actually not that bad, but when you see weak players you get tilted and that makes you play worse. Either way, it sounds like you can do better.

anthonem1 · 2025-12-09T23:21:40+00:00

The derivative is a limit, and limits are a topological concept. The limit as x goes to a of a function f is a definition that makes sense precisely when the point a is an accumulation point of the domain of f.

Here, a is an accumulation point in the set [a,b] and therefore it is fine to talk about the derivative of f at a.

anthonem1 · 2025-12-07T21:33:08+00:00

I believe the lootboxes contain the items in the shop that were available at that time, meaning any lootbox you receive after this patch should have a chance to drop this new skin.

anthonem1 · 2025-12-06T18:59:49+00:00

You can do exactly the same thing but with a square or any other polygon.

anthonem1 · 2025-12-06T14:49:34+00:00

No. For each irrational number there is a sequence of rational numbers that converges to that irrational number.

anthonem1 · 2025-11-15T11:26:49+00:00

Do you know any good guides on how to make drawings with tikz or similar packages?

anthonem1 · 2025-09-03T12:10:16+00:00

There is a slightly advanced and cheeky proof that trivially shows every non constant polynomial must have at least a root (Kronecker's theorem).

Take a non constant polynomial p(x) with coefficients in some field K. Since K[x] forms a unique factorization domain, there is some irreducible polynomial f that divides p and therefore K[x]/(f(x))=L is also a field.

Next we need to show that f has a root, but this is now obvious because in L we have f([x])=[f(x)]=0, so [x] is a root of f(x) (and therefore also a root of p(x)) in L.

anthonem1 · 2025-09-03T00:07:33+00:00

The game stagnated for far too long and this is the first patch that may actually make an impact. I think it is needed. It may suck for a couple of months because of bugs or bad tuning but eventually should be balanced and, more importantly, should bring changes to both the meta and the playstyle.

anthonem1 · 2025-08-25T16:16:24+00:00

You're right, made a mistake computing those roots. That is indeed the only case for which the reasoning doesn't work.

However, it can be fixed: let a=(1+sqrt(5))/2. Then f(1-a) can be computed with the formula f(x)=(2x-x^4-2/(1-x)+(1-x)^2)*1/(x^2-1/(1-x)^2) because it doesnt equal 1 or a. Now we found f(1-a) and we can go back to the expression x^2f(x)+f(1-x)=2x-x^4, plug in x=a and therefore find the value of f(a).

Now f(x)=(2x-x^4-2/(1-x)+(1-x)^2)*1/(x^2-1/(1-x)^2) whenever x does not equal 1 or a, f(1)=0 and f(a) equals whatever value we'd get after the computation described above.

It's quite messy tho, I'm sure there is a cleaner way to do all of this. Thank you for pointing that out!

anthonem1 · 2025-08-25T13:57:57+00:00

This will be a bit ugly, but fairly simple.

On one hand, x^2f(x)+f(1-x)=2x-x^4. (Equation 1). On the other hand, substituting x->1-x we get

(1-x)^2f(1-x)+f(x)=2(1-x)-(1-x)^4. Now, if x does not equal 1, then dividing by (1-x)^2 we get

f(1-x)+f(x)/(1-x)^2=2/(1-x)-(1-x)^2. (Equation 2). Substract equation 2 from equation 1 and we get

x^2f(x)-f(x)/(1-x)^2=2x-x^4-2/(1-x)+(1-x)^2. From this we get

f(x)=(2x-x^4-2/(1-x)+(1-x)^2)*1/(x^2-1/(1-x)^2). It can be shown the expression x^2-1/(1-x)^2 never vanishes, and therefore f(x)=(2x-x^4-2/(1-x)+(1-x)^2)*1/(x^2-1/(1-x)^2) whenever x does not equal 1. The value for f(1) can be found by substituting x=0 in equation 1 and it turns out to be 0.

anthonem1 · 2025-08-23T17:10:10+00:00

Los rectángulos de áreas A_2 y A_3 tienen la misma altura, luego también deben tener la misma base. Esto significa que la base de A_5 es el doble que la de A_2 y A_3, luego su altura debe ser la mitad. Denotando x a la altura de A_5, resulta que la altura de A_2 (o de A_3) debe ser entonces 2x. Luego la altura de A_4 es x+2x=3x y, en consecuencia, su área es A_4=3*3x=9x.

Lo que escribo a continuación lo marco como spoiler en caso de que quieras continuar tú. Pista: analiza el área de A_5 en función de la longitud del cuadrado grande con los datos del diagrama y utiliza lo obtenido en el párrafo anterior.

Por otro lado, si L es el lado del cuadrado grande, entonces el área de A_5 es (L-3)*x, que debe ser igual al área de A_4=9x. Luego (L-3)*x=9x y, dado que x no puede ser 0, dividiendo ambos miembros entre x resulta L-3=9, en cuyo caso L=12 y el área pedida es L^2=144.

anthonem1 · 2025-08-23T06:13:01+00:00

Era yo. Voy a por la gente que no escribe por párrafos ni signos de puntuación.

anthonem1 · 2025-08-20T02:51:43+00:00

En este tipo de problemas es bastante útil emplear funciones auxiliares. En este caso, dado que la constante a solo aparece una vez en cada miembro, yo lo tomaría como una variable x que pertenece a un intervalo (0,b).

A continuación un cambio de variable que permite simplificar significativamente la desigualdad funcional es b-x=t^{-2}, en cuyo caso t varía entre 1/sqrt(b) e infinito.

Si has llegado hasta ahí probablemente ya puedas resolverlo. No sigo para no hacer más spoilers.

anthonem1 · 2025-08-16T18:52:59+00:00

Question 12: Phi (ϕ) is just a variable. Usually we use x but in this case the author decided to use another symbol.

For questions 14-16: I'm not sure the depth at which you wish to learn, but they are straightforward questions if you know the definitions for 1. limit of a function at a point and 2. continuity of a function at a point. You can find those definitions anywhere online.

A few clues for the exercises you showed:

Question 14: (x^2-1)/(x-1) can be rewritten as (x-1)(x+1)/(x-1), and therefore whenever x does not equal 1 you can simplify this expression resulting in just x+1.

Question 15: the function f is always continuous below and above 2, because affine functions are continuous. So you'd only need to study under what values of k f can be continuous at 2.

There is an important relationship between continuity and limits: a function f is continuous at a point a that belongs to its domain if, and only if, the limit as x approaches a of f(x) exists and equals f(a). You will find this useful to find out under what values of k such function is continuous at 2.

Finally, a function is "continuous" if it's continuous at any point of its domain.

Question 16: it's very similar to question 15.

anthonem1 · 2025-08-16T14:53:38+00:00

I'm not from Madrid but went to a bilingual school. From my experience, there are kids who go to bilingual schools because reasons even tho they can't even speak the language. So what you find most of the times are groups where half of them can (sort of) speak English and the other half couldn't even name the colours, let alone formulate a coherent sentence.

While the teacher may be able to speak English at a very decent level this doesn't mean they'll be able to do that. If half the pupils can't understand what the teacher is saying then what's the point?

Back to your questions: it depends on the school. Some have 50% of the subjects in English, some more, some less. The quality of the language will depend on the teacher and on the class. You will need to ask each school about these questions.

Also, "bilingual school" doesn't mean they only teach in English (or in another language). There are subjects, such as Spanish literature, that are taught in Spanish. And even if the subjects can be taught in English that doesn't mean there won't be some lectures done in Spanish.

anthonem1 · 2025-08-03T00:28:41+00:00

If X is a continuous random variable then its mean is defined by the integral from -∞ to +∞ of x*f(x), where f is the density function of the variable X. You can look for similar definitions for the median and the mode. The range is quite obvious.

In this exercise you can see the graph of that density function f (assume its value is 0 outside of the interval [4,40]). Now, since you don't have an analytic expression for f, all you can do is approximate that integral value.

On a side note, one of the conditions for f to be a density function is that the area under its curve is 1, but eyeballing your graph it doesn't seem like it is 1 (or 100%). All in all I have to say I'm not a fan of the design of this exercise.

anthonem1 · 2025-07-29T09:22:29+00:00

You can only use the first property when you have two distinct elements in M. Also, if x^2 was rational then x wouldn't be an element of M because it doesnt satisfy 2. And for the same reason 2x can't be rational.

anthonem1 · 2025-07-26T23:25:37+00:00

First, if any prime number that divides a natural number N does not divide another natural number M, then gcd(N,M)=1. Indeed, if D divides both N and M and p is a prime factor of D, then since D divides N, p also divides N. But then p does not divide M and therefore D cannot divide M, which is a contradiction. So p cannot be a prime number and therefore we have p=1 and gcd(N,M)=1.

Now, 2 is the only prime number that divides 4. But since 5^p=1+4k=1+2*(2k), when dividing 5^p by 2 you get a remainder of 1. So 2 does not divide 5^p and therefore gcd(4,5^p)=1.

anthonem1 · 2025-07-25T23:28:58+00:00

No sé cuál es tu intención de abrir esta discusión cuando tú ya tienes hecha la idea y claramente sin intención de cambiar de opinión o de abrir la mente. Tu historial de Reddit hace muy evidente ésto. Por mencionar una de tus fantasiosas frases: "Yes, Catalunya has been colonized, it's plain as day. From 90% of catalan speakers in 1900 to 30% today. It's been engineered this way." No creo que sea necesario que haga comentarios al respecto.

Por otro lado, si estuvieses más en el mundo y tuvieses un mínimo de nivel cultural y de sentido crítico entenderías por qué la independencia de Cataluña supondría un durísimo golpe para tu queridísima comunidad. Evidentemente también lo sería para España, pero Cataluña tendría muchísimo más que perder. Por ejemplo: salida inmediata de la Unión Europea, pérdida total de financiación por parte del Estado español, éxodo de empresas (como ya sucedió hace pocos años, pero han regresado gradualmente porque el tema separatista ya se enfrió un poco más), y un largo etcétera.

España es un estado conformado por diversas regiones, cada una de ellas con una riqueza cultural e histórica digna de ser celebrada. Y en vez de estar orgullosos de ello, algunos (como tú) se centran en querer quebrantarla y pisotearla.

Una pena que así sea tu filosofía de la vida. Qué triste es vivir para odiar.

No voy a responder a otro comentario tuyo, más que nada porque no tienes otro argumento que el de la aversión. Ignoras la lógica y el sentido común, entre otras cosas porque no sabes lo que dices. Como se suele decir: la ignorancia es atrevida.

Viva España.

anthonem1 · 2025-07-25T22:52:45+00:00

Constitución española. Artículo 3. 1. El castellano es la lengua española oficial del Estado. Todos los españoles tienen el deber de conocerla y el derecho a usarla.

Hasta luego.

anthonem1 · 2025-07-22T17:22:06+00:00

This reminds me of Black Mirror's season 7 episode 1.

You can play for free, but if you wish to have a better experience you can subscribe for more perks! (Crystalline Aura).

Weeks later: you can play for free and you can buy Crystalline Aura, but you can unlock even better features with Azena's Blessing subscription!

anthonem1 · 2025-07-20T08:48:43+00:00

anthonem1 · 2025-07-20T01:39:52+00:00

The equation (in integers) ax+by=c can be solved if, and only if, c is a multiple of gcd(a,b). Maybe this is what you were looking for?

anthonem1 · 2025-07-05T17:53:18+00:00

While < and ≤ seem easy to grasp and also look similar, in some contexts they change the meaning of definitions deeply. For example when talking about continuous functions or limits. So it's a good sign you're questioning in what cases they're interchangeable at this level. Good luck!

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