An ellipse is a stretched-out circle. It's not just any stretched-out circle though; an ellipse is not an oval, not an egg, not a lemon, not a squircle, nor a capsule. To make an ellipse out of a circle, it must be stretched by a consistent factor. The red curve above is what's known as an ellipse. An ellipse has two quantities that are unique from a circle, namely its focal points, and its major and minor axes. The two focal points of an ellipse have a property such that the sum of the distance from any point of the ellipse to each focal point is equal to the length of its major axis, which brings us to our second quantities. The major and minor axes of an ellipse are its longest and shortest "diameter" respectively. To clear up any confusion, the line connecting both purple points in the picture above is its major axis. Meanwhile, the line connecting both green points is its minor axis. Dividing an axis by two gives us a semi-major axis which is the ellipse's "radius". Unlike the three points needed to create a circle, five points are needed to create an ellipse. This is because there are two new geometric properties introduced in an ellipse, which are its aspect ratio and rotation. The aspect ratio of an ellipse is the ratio of its major axis to its minor axis. The rotation of an ellipse is just how tilted it is from the x-axis or y-axis. This property is only significant in analytical geometry. That aside, the five points needed to create an ellipse can be translated into five variables to modify an ellipse. In the equation I made, the five variables I chose are its tilt (ϕ), its aspect ratio (a), its center's x-coordinate (c), its center's y-coordinate (d), and its "vertical" radius (r). with ρ being sqrt(1−ϕ²). (Note: When I use the words "horizontal" or "vertical", I am referring to the orientation of the ellipse's axes with zero tilt.) Honestly, I was going to explain how I got the equation above, but I ain't got time for allat, so I'll just explain what most of the shit here means. We'll start from the (x-c)'s and the (y-d)'s; these are standard translation operations. For the rho's and the phi's, well, they are modified rotation matrices. They used to be cosine and sine functions of an angle, but I decided against using trigonometric functions to keep it fully algebraic. With that, I transformed the sine function into a variable ϕ (which I bounded between -1 and 1). The a in the leftmost side of the equation is to stretch its "horizontal" diameter of the ellipse; the further away from 1 it gets, the more stretched out it becomes. Finally, the right-hand side of the equation is just to scale up the ellipse. I initially filled it out with a zero but then realized it wouldn't work; the ellipse had to have a size. The black lines are made with these two equations. These are standard line equations, but the gradients are... how did I get these gradients? Trial and error, my friend. For the points, I, uh, trial and error again. Sorta. Why there are four focal points is because I couldn't find out a way to switch their locations from the "horizontal" axis to the "vertical" axis. Basically, I made two points for the case where the "horizontal" axis was longer than its "vertical" axis, and vice versa with the curly brackets acting as an if-then statement. The purple points mark the edges of the "vertical" diameter; they are the center of the ellipse translated <horizontal axis length> units away according to the gradient of the corresponding black lines. The green points are the center translated <vertical axis length> units away according to the gradient of the other black line. This is the Desmos link if you want to try out stretching or rotating ellipses! https://www.desmos.com/calculator/zysqxdrkfz(https://www.desmos.com/calculator/zysqxdrkfz)

The theory of special relativity states two postulates: • the laws of physics are equal for every inertial observer, • the speed of light is constant for all inertial observers. Common sense tells us that the sum of two objects' speeds is done by simply adding them together. However, this naïve assumption would break the second postulate of special relativity—resulting in the speed of light changing whenever an object moves. So, how do we derive the correct formula for adding speeds? This is where linear transformations come into play. In two-dimensional space, eigenvectors are two vectors that do not change directions in a linear transformation. Eigenvectors will be crucial in deriving our formula, as our transformation matrix must obey two requirements: • the observer's speed shall become zero, relative to the observer, • the speed of light shall remain constant. We can create a graph of lines, each representing a different speed. The horizontal axis represents time; Meanwhile, the vertical axis represents position. Therefore, the slopes of these lines represent the object's speed. (The two thick black lines represent the speed of light. No, this is not at all related to light dispersion—I just wanted to make it look pretty.) Before we begin, let's assign variables to some terms. c will represent the velocities of light—traveling forward (1, 1) or backward (1, -1), A will be our 2×2 transformation matrix (c is not to be confused with c), λ will be the eigenvalues of A, v will be a vector that represents the observer's velocity from a reference point O, with v being its speed, and v' will be a vector representing the observer's velocity according to themself, with 0 being the speed and p representing the scale (a practically unimportant variable). Our goal is to find a matrix transformation A such that: • The observer's speed becomes zero. In other words, A transforms (1, v) into (p, 0). • The speed of light stays constant. In other words, (1, 1) and (1, -1) are eigenvectors of A. Let us begin. Our second postulate gives the following equation: We can substitute the values of I, A, and c into this equation, giving us λ₁ - a - b = 0 ...(1) -c + λ₁ - d = 0 ...(2) Additionally, we can substitute c with the negative velocity of light, which gets us λ₂ - a + b = 0 ...(3) -c - λ₂ + d = 0 ...(4) Subtracting (3) from (1) gives us λ₁ - a - b - λ₂ + a - b = 0 λ₁ - λ₂ - 2b = 0 b = ½(λ₁ - λ₂) Adding (2) and (4) gives us -c + λ₁ - d - c - λ₂ + d = 0 -2c + λ₁ - λ₂ = 0 c = ½(λ₁ - λ₂) Hence, b = c Adding (1) and (3) gives us λ₁ + λ₂ - 2a = 0 a = ½(λ₁ + λ₂) Subtracting (4) from (2) gives us λ₁ + λ₂ - 2d = 0     d = ½(λ₁ + λ₂) Hence, a = d Our first postulate gives the following equation: With matrix multiplication, we find that     c + dv = 0     c = -dv We can substitute everything we know into our matrix A, which gets us: Let's define two new variables. u: the speed of a moving object from a reference point O u': the speed of the object according to our observer (Δs and Δt are distance and time changes) Now that we've got the values of Δs' and Δt', we can substitute them into u'. Now you know how to add near-light-speed velocities together! This is just the beginning. We haven't even touched time dilation, length contraction, etc. You could derive formulas for those from the matrix we've got, but who the hell wants to do that LMFAO

This is going to be one of my more unique posts, since instead of talking about topics that are in Indonesia's national curriculum, I'll be talking about a specific math problem. The integral of a function is the area below the function's curve. You could calculate an integral by summing small rectangles, which helps approximate the value of the integral. Although, what happens if we do it the other way? What if we used integrals to approximate the area of said rectangles? Chapter I: Discrete summation and the sigma notation Discrete summation is basically the opposite of continuous summation. An integral is a continuous summation, hence it deals with fractions and really small lengths or areas. Meanwhile, discrete summation deals with finite numbers, ones that we could measure or count. The sigma notation is a way to represent summation. It has four main elements: the first term, the last term, the variable, and the expression. In this case, 1 is the first term, 10 is the last term, n is the variable, and 2n is the expression. This whole expression represents the value of 2 + 4 + 6 + ... + 20. Let's see another example: Here, 1 is the first term, 5 is the last term, i is the variable, and Uᵢ is the expression. Therefore, this expression represents the sum of U₁, U₂, all the way to U₅, whatever these variables may represent. Chapter II: Definite Integrals A definite integral is a bounded area under a curve. (Although if the "bound" is infinity, it technically isn't bounded.) Definite integrals are written similarly to the sigma notation: a lower bound, an upper bound, an integration variable, and an expression a.k.a. function. (Note: the lower bound doesn't necessarily have to be a smaller number than the upper bound, this just negates the result.) Here, -1 represents the lower bound, 3 represents the upper bound, x² represents the function, and dx represents the integration variable. There's also a visualization of what an "area under the curve" looks like, just in case you haven't understood what it is yet. Chapter III: Riemann Sum The Riemann sum is a way to approximate the value of a definite integral by calculating the areas of thin rectangles that have varying heights, depending on the value of a function at a certain point. A notable feature of this method is that the value of the Riemann sum gets closer to the integral's true value as the rectangles become thinner. Chapter IV: The Problem "Instead of using a sum to approximate an integral, is it possible to derive a formula that uses an integral to approximate a sum?" This problem can be represented with a graph, with the black graph representing a function f(x) and the blue graph representing the value of f(⌈x⌉) where ⌈x⌉ is the nearest whole number bigger than x. This transforms the problem into deriving a formula that approximates the area under the blue graph (from x₀-1 to x₁) with the integral of the black curve (from x₀-1 to x₁) plus the area of the shaded blue region. Chapter V: Formula Derivation We will now derive a formula to calculate the area beneath the blue graph from (x₀-1) to x₁. The Area Under the Blue Graph In the picture above, you can see ten individual blue rectangles with equal width (notice the small rectangle to the left x₀, but ignore the blue rectangle past x₁). As is seen above, the area under the blue graph from x₀-1 to x₁ is the total area of the ten rectangles that make up the shaded blue region. The area of one rectangle is its width * height, hence: area under blue graph = w₁h₁ + w₂h₂ + … + w₁₀h₁₀ Since they all have the same width, 1 unit wide, we can factor out the w's. area under blue graph = w(h₁ + h₂ + … + h₁₀) area under blue graph = h₁ + h₂ + … + h₁₀ Notice that the height of the first rectangle is f(x₀), the second rectangle is f(x₀+1), and so on. We can then substitute h with f(x): area under blue graph = f(x₀) + f(x₀+1) + … + f(x₁) This expression can be rewritten with the sigma notation. The Area Under the Black Curve Our next task is to find out the area under the black curve from x₀-1 to x. Fortunately, the area under a curve can be calculated with an integral. So: The Area of the Shaded Blue Region If you look closely, the shaded blue regions look like upside-down right triangles. Although this is not entirely true, it is a really good approximation, especially as x approaches large values. (Note: Δh is the height of one triangle) area of blue region ≈ ½⋅b₁Δh₁ + ½⋅b₂Δh₂ + … + ½⋅b₁₀Δh₁₀ area of blue region ≈ ½(b₁Δh₁ + b₂Δh₂ + … + b₁₀Δh₁₀) Since all of these triangles have the same base width, 1 unit wide, we can factor it out. area of blue region ≈ ½(Δh₁ + Δh₂ + … + Δh₁₀) The height of each individual triangle is the height difference between the rectangle below it and the rectangle to its left. Hence, the first triangle's height (Δh₁) is h₁-h₀, the second triangle's height (Δh₂) is h₂-h₁, and so on. (Note: hᵢ means the height of the i-th rectangle) area of blue region ≈ ½(h₁-h₀ + h₂-h₁ + … + h₁₀-h₉) Matching terms cancel each other out, leaving us with: area of blue region ≈ ½(h₁₀ - h₀) We can substitute h₁₀ with f(x₁) and h₀ with f(x₀-1). Formula Derivation Now we've got everything we need, so we only need to derive the final formula. area under blue graph = area under black curve + area of blue region We can substitute area under blue graph with the sigma notation we've derived, the area under black curve with the integral we've derived, and the area of blue region with the equation above. Chapter VI: Final Thoughts Dang, that was a journey and a half. I've tested this equation several times to check if I've made a mistake in my calculations. And I don't think I have. For example: Plugging in this formula on an inconspicuous-looking quadratic function with 50 as an upper bound returns a pretty good approximated value, with a 0.01% percentage difference. So, is this formula actually useful? I hope it is. I somewhat think that integrating a function is easier to do than adding 50 enormous numbers by hand. Nevertheless, I have not found any real-world usages for this formula, so maybe I'll leave that to you, the reader, to figure out for yourself. ;) Also, for now, try using this formula only on algebraic functions for intervals where the function keeps growing. I'm not sure if it works with trigonometric functions or algebraic function intervals which have negative gradients. Well, I guess that's all! Thanks for reading this long Peridot post. I hope you understand everything I wrote.

A limit is a value of a function as it approaches a certain point. To put it simply, let's say we have a sequence of numbers: 1, 2, 3, 4, x, 6, 7, 8, 9. We'll let x be some undefined number; Nobody knows what it is because, well... it's not defined. But by looking at the sequence, we know it's probably 5. Well, that's kinda what limits are. The image below is the graph of: If you look at it, it's just like any other parabola... until you plug in x = 0. This will cause the denominator to become zero, and we know you can't divide by zero. Hence the hole. By eyeballing it, it's clear that the point looks to be (0,2), but that's not its actual value. That is the 'limit of the function as it approaches x equals zero'. Let's take a look at the following graph: This function does not have a limit as it approaches x equals zero. The limit is simply undefined. Why? For a limit to be deemed valid, it must end up at the same number whether you approach it from the negative or positive sides. Limits aren't only used to calculate the value of a non-continuous function. You can also use them to calculate the value of a function as it approaches infinity. Take a look at this graph: As our function gets closer to infinity, the value gets closer to zero. From this observation, we can conclude that 'the limit of this function as it approaches infinity is zero'. In this case, the x-axis is called an asymptote as the curve keeps getting closer to the x-axis, but they never truly touch. We can also use derivatives to compute limits with L'Hôpital's rule. L'Hôpital's rule state that: I can't think of anything else I should add, so I'll end it here. I hope this post is simple yet clear enough to make you understand limits. If you have any feedback, feel free to post it in the comments section. Good luck with your Calculus classes!

Wouldn't it be much better if you could cross out dy/dx to become y/x? Unfortunately, that's not the case since our good friend Gottfried Leibniz decided that wasn't possible. Though, to be fair it's not his fault. A derivative is an expression representing the rate of change of a function with respect to an independent variable. Basically, it allows us to calculate the gradient—or steepness—of a line that touches a curve at a certain point. What does it mean for a line to "touch" a curve? Take a look at this image. Notice how the green line is parallel with the red curve at point A. From this information, we can conclude that the green line has the same steepness as the curve at point A. This steepness is measured with the derivative of either function at point A. Let's say a variable h exists that has a value of 3. Point B is 3 units to the right of point A. If we draw a line from point A to B, it has a certain gradient. This is not yet the exact steepness of the curve at point A since 3 is still too big. Let's try shrinking h down to 0.1. Now that points A and B are much closer, we're getting closer to our desired steepness. If we continue with this process, shrinking h down and down, we will eventually get the exact gradient of the curve at point A. This is how a derivative is defined. So, a derivative is the gradient of a curve, measured with two points that are close together: Point A whose coordinates are (x, f(x)) and point B whose coordinates are (x+h, f(x+h)). As the value of h approaches zero, our gradient will become more and more accurate until eventually, we get the steepness of the function at point A. We know that the gradient of a function is calculated by (y₂ - y₁) / (x₂ - x₁), so if we substitute the coordinates of points A and B into this function, we get the formal definition of a derivative: I hope you guys enjoyed this brief introduction to derivatives. If you've got any questions, feel free to ask in the comments! I'll try to answer every question. Also, I'm open to feedback!

This level was so insanely hard compared to the other ones lmfao Every single level before this I completed in less than or equal to two attempts but not this one!! This level just decided to be a bitch In total i think I finished it in 6 tries? Im never playing this again Good luck to yall trying to beat this I dont know how moms are so good at the game

Today I went to a volcano. It's an active volcano but it's not in danger of erupting anytime soon. The mountain itself looks beautiful but I think I did a horrible job in taking its picture. The main problem is the fact that the whole thing is covered by fog. I believe the fog is coming for us all. This picture shows how much a simple natural phenomenon can take away from us; How limited our 2-dimensional vision is. It's incredible how such an immense mountain could disappear from our sight because of some fog. At the end of the day, despite all the fog, it was still a fun trip!