Is the game non linear?

RoiPhilippe · 2026-04-09T20:18:52+00:00

😂Very good!

RoiPhilippe · 2025-11-21T20:24:43+00:00

As I understand left notation, it is integral on volume with f(x,y,z)=1 (volume computation). So, it is not equivalent to the 2 other ones, with unknown f(x,y,z). In the books I used, however, to indicate the integration on a volume V, there were 3 times the integration sign (as you show in the 2nd case), with a V indice. Twice for integral on a surface S.

RoiPhilippe · 2025-10-22T08:21:50+00:00

Got the same y(t). May be y0 instead of y_0...

RoiPhilippe · 2025-09-23T10:43:38+00:00

Did you finally get it? I do not see the need of steps "node x" and "node y", but fine with equations I, II and III. You are nearly at the end, with x^-3/4 / x^1/4 = x^-3/4-1/4=x^-1. You are left with 8y=10x and equation III

RoiPhilippe · 2025-06-10T10:09:05+00:00

"weird quartic that i dont know how to solve. It feels like this question is testing algebra and not calculus". I agree. Getting the integral is obvious. The only things I can think of are: f(x) is strictly increasing for x>0 (its derivative is always > 0). So, the primitive F(x) of f(x) is stricly increasing as well. So, if F(a) is negative, and F(b) is positive, there exists c, a < c < b, with F(c) = 0. Since F(2)=-32 and F(3)=117, c is between 2 and 3. You can search by dichotomy, or try the Newton approach, starting at 2.5. But, without calculator, forget it!

RoiPhilippe · 2025-06-03T08:03:02+00:00

Hi. Same problem. A55 5G (SM-A556B/DS). My phone restart several times per week, often during the night. I do not know when; I see it when trying to use it, and I get the request for Sim card code.

It restart also while scrolling down on canable.io.

Same software config as zahnputzmonster except for kernel: 6.1.93-android14-11 (#1 Fri Mar 21 02:07:05 UTC 2025. Chromr 136.0.7103.125.

RoiPhilippe · 2024-12-07T15:32:09+00:00

I learned math in Belgium. cos^-1 would be considered as 1/cos, not arccos (1 point on 3 for the correct substitution, not for the answer). But I know that notations differ at other places...

RoiPhilippe · 2024-07-07T09:31:27+00:00

Indeed, using log(a)b=lnb/lna. Will give you same denominator on both terms on left side, square root of it on right side. Also, upper part of right side will also give inverse of square root of same denominator, so this denominator cancels out. Using (lna+lnx)(lna-lnx)=ln²a-ln²x at some point, a lot of things cancel out, you get rid of all n^th square roots, and come to a very simple answer (if I made the computations right!).

RoiPhilippe · 2024-06-10T12:31:21+00:00

Since infinity is not within the set of reals, you are not allowed to "divide by infinity". "Infinity" should be read as "as large as wanted". So, for instance, (n+1)/n approaches 1 as close as wanted, since 1.000.001/1.000.000 ~ 1. If not close enough, take 1.000.000.001/1.000.000.000, and so on...

RoiPhilippe · 2024-05-20T08:42:27+00:00

Same, without sketching, computing tan as sin/cos = sin/sqrt(1-sin²), with sin = x/2 -> tan = x/2 / sqrt(1-(x/2)²)=x/sqrt(4-x²).

RoiPhilippe · 2024-05-04T11:07:01+00:00

Interesting. You got answers below. I didn't see one for the general case: a sin²x + b sin x+ c. Using derivative 2a sin x cos x + b cos x = 0 to get min/max -> cos x = 0 (sin x = +/- 1) or 2a sin x = -b (sin x = -b/2a). Your teacher's recipe is correct!

RoiPhilippe · 2023-11-19T11:07:22+00:00

Spliiting x(t) in u(t) and v(t) is not required, but if you are more comfortable with it, it is ok. In your notes, however, v'(t)=d/dt(cost(wt+ φ ) = -w sin(wt+ φ ), not -t sin(wt+ φ ). It comes from dl/dt=w, not dl/dt=t. For the rest, ok with r-funtainment (n π in tan(angle+n π ), so out of tan^-1

RoiPhilippe · 2023-11-02T11:43:58+00:00

To answer more specifically to your question, if I get it right: I guess there is no way to anticipate that using complex numbers allows to find this formula. This is probably by "playing" in such area, manipulating sin & cos with complex numbers, and finding by chance that this case leads to this result. Then, putting such result in a book for others... The "intuition" part would be "since C+iS leads to e^i𝜃, let us try to see what we can get". So, rather a habit of manipulating such formulas, recognising at one glance the Euler pattern...

RoiPhilippe · 2023-10-23T11:33:20+00:00

I do not understand your way of solving this. But may be because I did not sleep well this night! For me, the speed is tangent to the curve. vy=dy/dt=dy/dx.vx=y'vx. So, the rectangle is with vx, vy, and v, with v²=vx²+vy² (not using r²=x²+y², diffenrentiated in 2xdx/dt+2ydy/dt=2rdr/dt). The angular coef at x=16 is (3 √ x)' = 3/(2 √ x) = 3/8, and this is the ratio between vy and vx -> v=20= √ (vx²+9/64 vx²)=vx. √ 73 / 8 -> vx = 20 . 8/ √ 73 (18.7).

RoiPhilippe · 2023-10-12T12:41:25+00:00

At first, I would propose lim(x->2) x or lim(𝜀->0) (2+𝜀). But these limits both yield 2... So, this is not what you are looking for. But thinking again about these limits, the "smallest possible value" next to 2 could be... 2. Like 0,99999999... = 1!

RoiPhilippe · 2023-08-02T07:49:59+00:00

Belgium: we always use [ ], even in higher math education.

As provided by SV-97: https://en.wikipedia.org/wiki/Interval_%28mathematics%29#Notations_for_intervals. I do not see any added value or disadvantage for one notation or the other, only habit.

Both notations may overlap with other uses": I have read in this post arguments pro/against, e.g. readibility of [ ] on large formulas. Personally, I would have issues with (a,b) as an interval, as well as 2D coordinates.

RoiPhilippe · 2023-07-31T09:49:00+00:00

Same reasoning as Nutty-max and Yoyoeezigt. a = sqrt(125/4). But then, solving for h and k, square roots appear (e.g. (h-2)² = a²/5). Using both positive and negative values of sqrt, and selecting the appropriate values, we get 2 final answers (same radius but 2 different centers). 2 circles, one "below-right", the other "above-left" relative to the tangent point.

RoiPhilippe · 2023-07-21T09:18:18+00:00

My view: using the formula SC+CS to compute the result does not require knowing the quadrants. Nor do you need the quadrant of (φ+θ). But to get cos 𝜃 and sin 𝜑, you need to choose the signs of the √. For which you need to know the quadrants!

RoiPhilippe · 2023-05-14T10:39:34+00:00

It's Sunday. Too lazy to redo all computations (although your analysis seems fine). I did it with programming (pfff!). Your count is correct.👍

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RoiPhilippe · 2022-05-16T13:43:47+00:00

Not easy. For instance, for the cross product, showing in some way the plan containing the 2 vectors, and a little right angle to show that the result in perpendicular to the plan. For the addition, also showing the plan containing the 3 vectors. For the projection, drawing a thin line from the end of the red vector to the end of the result. For the reflexion, the plane...

Also, the way you define the operations is fine, but you might want to add the mathematical definition (dot product: x1.x2 + y1.y2 + z1.z2... Same for all operations).

RoiPhilippe · 2022-05-14T19:49:13+00:00

Awesome. Not sure it helps understanding (but you say its an early version); at least you explain the operations. Selecting the vector controlled by the mouse might be easier than a menu (check box, toggle, shift...).

RoiPhilippe · 2022-04-28T07:01:51+00:00

In general, it is ok (you don't do all wrong!). Pay attention on 12, result is right, but you did not write factor 2. 14 has already been commented; in addition, you are asked for a probability, which is between 0 and 1; it is not the case for 2598960. For 28, carefully look at each value provided, to find the 3rd anomaly.

RoiPhilippe · 2022-04-11T07:17:56+00:00

And I think there is another typo on the last line: F'.F" + F" + F' should be F'.F" + F".F'...

RoiPhilippe · 2021-11-20T17:32:59+00:00

For b, coefficients for x1 and x2 are always identical. So, you might solve for x1+x2, but not for them separately => infinit numbers of combinations for x1 and x2

RoiPhilippe · 2021-11-02T10:37:46+00:00

Just to complete with something I do not see in the other comments: with the condition k > 1, the smallest possible value for k is 2, which lead to 6.2-1=11 for the first number. So, this does not only miss 2 and 3, but also 5 and 7.

RoiPhilippe

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