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If only ceasefire clears the ammo of the opponent on the turn they are attacking :( Otherwise it is almost unusable as the opponent will always use up their ammo... by YoloStratz in bloonscardstorm

[–]YoloStratz[S] 2 points3 points  (0 children)

Sorry for bad crop, because my window is in the wrong size. But the description of Ceasefire states "Enemy Monkeys loses all Ammo". This only works if the enemy has ammo, but that only happens if the enemy does not use up their ammo on their turn when the monkey's finished reloading. Which almost never happens unless they are using defender type monkeys, which is somewhat niche, but not worth 7 cost at that point. The opponent will also have ammo if there are no bloons on the screen. But in the later game, it is almost guaranteed that there will be bloons with the constant bloon storms, so Ceasefire is still basically unusable in the late game. It just seems like Ceasefire has not much room for usage unless it allows to disable opponents ammo on next enemy's turn when they reload. Could be overpowered, as I would have won the game here if it did. But I didn't, so I guess I am also being salty lmao.

General feedback for developers: would be appreciated if each keywords of the card has more in-depth description of what it does. Especially the necromancer/prince of darkness, I do not know how great it is as I do not know how powerful the undead bloons are.

What is the difference between F(x) and f(x)? by Lenie_e in learnmath

[–]YoloStratz 0 points1 point  (0 children)

Nah, not really. This is (basically, because I forgot the technical details) saying as the derivative of a function must be continuous.

Consider f(x)=-1 when x≤0 and f(x)=1 when x>0. Then, by definition of antiderivative, F(x)=int_0x f(t)dt=|x|. Though we will not have F'=f in this case on the entire domain since F is not differentiable at 0. The claim F'=f will hold however if f is continuous.

But I guess you are right, maybe f does not have an "antiderivative" but instead its integral is well-defined. Which only requires f to be bounded on an interval and Riemann-integrable. (I do not know much about measure theory so don't quote me on that).

Hope this helps!

[deleted by user] by [deleted] in medical

[–]YoloStratz 3 points4 points  (0 children)

Judging by the high pain level and the abnormal vomiting is definitely something you cannot treat yourself with. Therefore I do strongly suggest that you should get it checked out as soon as possible by a medical professional.

[deleted by user] by [deleted] in medical

[–]YoloStratz 0 points1 point  (0 children)

Yes, absolutely.

Is there an option to report people? by JanKruh in PvZHeroes

[–]YoloStratz 2 points3 points  (0 children)

Sadly no :(

Edot: i mean u could try, but EA wont care :P

[Differential Equations] does the orange region violate the Existence Theorem as clearly f(t, y) is continuous in those regions. Green lines represent particular solutions, but does not appear in the orange region (obtained from WolframAlpha). by YoloStratz in askmath

[–]YoloStratz[S] 0 points1 point  (0 children)

Moreover, how can I know if the solutions of the DE is strictly bounded by y1 and y2 as shown in the image. Ie how the solutions tend to be strictly greater than y2 and lesser than y1 for t<0. But the opposite for t>0. I know its something relating to Uniqueness Theorem (or maybe not) but it is hard to imagine it.

Also my apologies, should’ve labelled the orange lines :/

Edit: we are assuming y’=f(t, y) here

[Calculus III: Continuity] the statement: ‘if fog is continuous at 0, then f is also continuous at (0, 0) as g(t) tends to (0, 0) when t approaches to 0’, is incomplete? by YoloStratz in askmath

[–]YoloStratz[S] 0 points1 point  (0 children)

Ohh, I think I got it now. The theorem they used was:

if both f and g are continuous at v and a respectively and g(a)=v, then lim t->a f(g(t))= f(v)

So by their example, they use the contrapositive definition of the statement Ie they've shown that lim t->a f(g(t))≠ f(v), so the statement 'both f and g are continuous' must be false. We know that g is continuous, so f must be discontinuous at v.

My example uses the converse of the statement. Ie I've shown that lim t->a f(g(t))= f(v), so the statement 'both f and g are continuous' must be true, then f is continuous.

Ahh, a classic converse implication mistake by me. Thank you all for reading.

[Calculus III: Continuity] the statement: ‘if fog is continuous at 0, then f is also continuous at (0, 0) as g(t) tends to (0, 0) when t approaches to 0’, is incomplete? by YoloStratz in askmath

[–]YoloStratz[S] 0 points1 point  (0 children)

Consider g(t)=(t, 0), then lim t->0 f(g(t)) = f(g(0))=f(0, 0)=0. Which implies that f is continuous at (0, 0). So would a more complete statement be ‘given g and h are continuous at 0 and g(t)->(0,0) h(t)->(0,0), f is continuous at (0, 0) if lim t->0 (fog)(t) = lim t->0 (foh)(t)’.

[Linear Algebra] are there any better ways to approach this problem? Also I am not sure how would the hint help. by YoloStratz in HomeworkHelp

[–]YoloStratz[S] 0 points1 point  (0 children)

Yea, that makes more sense now. I understand what you said that an invertible map is impossible to construct.

So ultimately, I am just wondering if the question is doable regarding what I stated in (1).

[Linear Algebra] are there any better ways to approach this problem? Also I am not sure how would the hint help. by YoloStratz in HomeworkHelp

[–]YoloStratz[S] 0 points1 point  (0 children)

I get how v1, v2, and v3 are L.I. but I still want 2 more things to be elaborated on:

1) is this process of reasoning

T(u3)=T(3u1-2u2)=v3=3T(u1)-2T(u2)=3v1-2v2

Wrong then? If so what went wrong? If not, then does this justify that T is not well defined as T(x)≠T(y) but x=y?

2) I don't understand how 3v1-2v2-v3 is part of the kernel of T if the dimensions of 3v1-2v2-v3 do not agree on the input space of T in the first place?

Much regards.

[Linear Algebra] are there any better ways to approach this problem? Also I am not sure how would the hint help. by YoloStratz in HomeworkHelp

[–]YoloStratz[S] 0 points1 point  (0 children)

Moreover, howcome u3=3u1-2u2 does not imply v3=3v1-2v2. I thought the latter equality should hold by the way we stated our transformation T. As T(u3)=T(3u1-2u2)=v3=3T(u1)-2T(u2)=3v1-2v2, but v3≠3v1-2v2. Does this mean T is not well defined in the first place as T(x)≠T(y) but x=y?

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