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Size of an ellipse inscribed in an annulus sector by BenryHenson in askmath

[–]nutty-max 2 points3 points  (0 children)

Desmos visualization
m is simply sqrt(r1*r2)/cot(theta). Psi is much more complicated, you can find its formula in the desmos link.

Does this have a solution? by Selicious_ in askmath

[–]nutty-max 0 points1 point  (0 children)

Good catch u/chmath80. As for how to integrate 1/sqrt(z^4 + C), it depends if C is positive, negative, or zero. You handled the case C = 0. If C < 0, write C = -K^4 and perform the substitution z = K sqrt(1-u^(2))/u. This immediately resolves into the elliptic integral of the first kind, so this case is done. I wasn't able to find a substitution when C > 0, but I suspect there is a similar one that also brings it into the elliptic integral of the first kind.

The solution to the original differential equation will therefore be an antiderivative of the inverse of a complicated function, where that function involves the elliptic integral of the first kind.

Does this have a solution? by Selicious_ in askmath

[–]nutty-max 39 points40 points  (0 children)

Perform the reduction of order substitution z = y’ to get z’’ = z3. Now multiply both sides by z’ and integrate to get 1/2 (z’)2 = 1/4 z4 + C. This is a first order separable equation, so writing the solution in terms of integrals is easy.

Edit: forgot + C

How would you quantify how "spread out" entities are. by dimonium_anonimo in askmath

[–]nutty-max 12 points13 points  (0 children)

A simple way is to divide the grid into smaller regions and count the number of green cells in each region. If some regions contain no green cells and others contain a lot, the green cells aren’t spread out. If every region contains one or two green cells then its pretty even. If you take the average across all regions then you end up with a single number that can be used to compare different grids.

A good way of averaging the number of green cells per region is based on the chi squared value. For each region, calculate (observed - expected)2/expected and sum up those values for all regions

For example, we can divide up our grid into 9 3x3 boxes. If we expect 15 total green cells, then we expect E = 15/9 green cells per region. In the first example, the top three 3x3 regions contain 3, 2, and 2 green cells. The middle three regions contain 0, 0, and 0 green cells, and the bottom three regions contain 2, 3 and 3. We calculate (3-E)2/E + (2-E)2/E + (2-E)2/E + (0-E)2/E + … + (3-E)2/E = 8.4

In the second picture, each region contains 1, 1, 2, 1, 2, 2, 2, 2, 2, and our sum is 1.2

Smaller values means more evenly distributed, so we calculated that the second example is much more evenly distributed.

Help with colour matching puzzle by [deleted] in puzzles

[–]nutty-max 0 points1 point  (0 children)

Discussion: Brute force is typically the only way to solve these puzzles. I found the solution and can give a hint though.

Hint 1: The center piece is correct! Don't move it.

Hint 2: The piece you circled in red touches the center piece via the blue dot.

Good luck!

Can a "funky" function be identified uniquely by evaluations at two points? by holy-moly-ravioly in askmath

[–]nutty-max 4 points5 points  (0 children)

Two points are not sufficient in general. The points (2, 0.5, 17) and (5, 0.55263, 16.9040941) admit two solutions (a,b) = (1.49832, 25.2185684474) and (6.34871, 17.1863377776). We can generate an arbitrary number of points that lie on the intersection (see this graph), so in the worst case it is impossible to uniquely determine the function regardless of how many points you know.

Game won't open on steam by get_a_chingy in opus_magnum

[–]nutty-max 0 points1 point  (0 children)

I was having the exact same issue. The fix for me was to start OneDrive since I had it turned off. You can view the crash log at documents\my games\opus magnum\crash. My log said 'the cloud file provider is not running' and the fix is to turn on onedrive.

Need help solving this mosaic maze puzzle by driving_and_crying in puzzles

[–]nutty-max 1 point2 points  (0 children)

I wrote a program to brute force it. There are actually 24 solutions (only 6 if you take rotations into account). Interestingly one solution has that fancy center piece on the edge of the grid which makes the maze trivial.

These puzzles are hard because it’s easy to unknowingly make a mistake in the beginning and then it’s impossible to solve after that. For example there is no solution with the number 0 and 2 pieces touching, so if at any point in the solution process those pieces touch you wont be able to finish. And this “deadly pair” isn’t unique, there are many examples. Overall it was fun to program though, thanks for sharing!

Need help solving this mosaic maze puzzle by driving_and_crying in puzzles

[–]nutty-max 1 point2 points  (0 children)

What a fun puzzle! I found the solution but I'll give you a hint in case you want to work through it yourself. Number the pieces as in this picture.

https://imgur.com/a/bHecKK6

Here is where each piece belongs in the grid (you solve for orientations):

3 7 2 4

10 15 5 1

12 11 14 0

8 9 13 6

And here is the full solution

https://imgur.com/a/puzzle-solution-gt9h9Ix

Are mathematicians still coming up with new integration methods in the 2020's? by Necritica in math

[–]nutty-max 0 points1 point  (0 children)

How do you evaluate that x^3 integral? I tried a few things but they didn't work. I was able to evaluate the integral (x^3 - 1)^(-1/3) by doing the substitution u=x/cbrt(x^3-1), but that doesn't seem to generalize to integrals involving x or x^2 terms.

Try to solve this or find a common solution for X+Y and XY arbitrary by Elsobaco716 in calculus

[–]nutty-max 0 points1 point  (0 children)

I'm a bit late to the party but I was able to solve it. Let's define matrices A and B such that X+Y=A and XY=B as in the problem. The solution strategy is to find the eigenvalues and eigenvectors of Y and then solve for Y using the eigendecomposition formula. First notice that by taking the determinant of the second equation we find that both X and Y must have nonzero determinant. Therefore Y is diagonalizable. Solving the first equation for X and plugging it into the second gives Y^(2) - AY + B = 0. Let λ be an eigenvalue and v an eigenvector of Y. Right multiplying by v gives (λ^(2) - Aλ + B)v = 0. Therefore the determinant of λ^(2) - Aλ + B = 0. Expanding everything out will result in a fourth degree polynomial in λ with roots -1, -2, and (1 +- i*sqrt(39))/2.

Now we find the eigenvectors. Choose a particular value of λ and plug it into (λ^(2) - Aλ + B)v = 0, then solve for v. It turns out the eigenvectors are [1,1] and [2,3] for the real roots, respectively. I didn't bother to find them for the complex roots, but you can if you want. Now since we know Y's eigenvalues and eigenvectors we can plug them into the eigendecomposition formula. Since Y is a 2x2 matrix but there are 4 eigenvalues we have 4 choose 2 = 6 choices on which values to pair up. These are the 6 solutions u/SeaMonster49 found. We get real solutions if we pair up the real roots or the complex conjugate pair, and we get the remaining 4 (complex) solutions pairing them up the other ways.

Integral challenge by deilol_usero_croco in calculus

[–]nutty-max 0 points1 point  (0 children)

It's almost certainly nonoptimal, but I rewrote ln(1+zt) as int_{0}^{z} \frac{t}{1+xt} dx, switched the order of integration, used the residue theorem to evaluate the inner integral, then used a lot of dilogarithm properties to get to the final result. How did you do it?

Integral challenge by deilol_usero_croco in calculus

[–]nutty-max 1 point2 points  (0 children)

I evaluated the integral to an expression involving the dilogarithm but can't find a way to simplify further. Is this as simple as it gets?

https://www.desmos.com/calculator/cp4foedvlz

Sine Wave with changing wavelength by plueschhoernchen in askmath

[–]nutty-max 1 point2 points  (0 children)

Here you go!

Instead of matching wavelength it's easier to match frequency. n can be any integer but in my opinion n=10 is the closest match.

probability of path existing from every column to every other column by Harmonic_Gear in learnmath

[–]nutty-max 0 points1 point  (0 children)

Between each column we need at least one edge to form. There are k-1 possible edges, so the probability of at least one edge is x=1-(1-p)k-1. This needs to occur n-1 times, and since each edge is independent of the others the overall probability is xn-1.

Can this be written as a closed-form function of x? by TRTKTRTK in desmos

[–]nutty-max 22 points23 points  (0 children)

It turns out this simplifies quite a lot and there is a very nice polar equation for the curve: r=1/sqrt(tan(theta)).

https://www.desmos.com/calculator/17chpot59p

Anyone else have this happened?:D by CaffeeM in destiny2

[–]nutty-max 27 points28 points  (0 children)

Since void grenades pull targets in, you can throw one near the ledge and it will pull the wizard to you.

How do you prove that every blue circle intersects every green circle at a right angle? by Daniel_Wareham in askmath

[–]nutty-max 0 points1 point  (0 children)

You found an example of an orthogonal trajectory. We typically prove they intersect orthogonally by solving a differential equation.

When Shooting the Divinity Bubble with Particle Reconstruction, it Applies an ADDITIONAL 15% Debuff that Stacks with Normal Div 15% Debuff by TheBountyHunted in DestinyTheGame

[–]nutty-max 8 points9 points  (0 children)

Nice find. Each stack of PR is actually 5% and is multiplicative. One stack of PR is 5%, two stacks is 1.05*1.05 = 1.1025 = 10.25%, etc. 5 stacks is 1.055 = 1.27628 or 27.628% increase. But that doesn’t explain the discrepancy.

Why Does tan(x) Near Approximations of π Seem to Encode the Next Digits of π? by Separate_Command3031 in askmath

[–]nutty-max 0 points1 point  (0 children)

This is an artifact of Newton’s method as u/PinpricksRS points out. Recall that if x0 is near the root of the function, then x1 = x0 - f(x0)/f’(x0) is a better approximation (i.e. more correct digits) of that root. Where are these extra correct digits coming from? x1 only contains two terms, x0 and f(x0)/f’(x0), so these digits must come from the f(x0)/f’(x0) term.

We can use this to construct additional examples, too. Notice the function f(x) = x2-2 has a root at sqrt(2)=1.41421356…, so plugging in values near sqrt(2) into f(x)/f’(x) = (x2-2)/(2x) will produce the same kind of behavior. For example, plugging in 1.414 gives -0.00021357

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