From a191cdf7895ae7fc8ba66f262967379f2db7b68a Mon Sep 17 00:00:00 2001 From: PixelatedLagg <81400034+PixelatedLagg@users.noreply.github.com> Date: Mon, 12 Aug 2024 01:10:44 -0400 Subject: [PATCH] Updates --- asset-manifest.json | 8 ++++---- imgs/Proofs/3_1.png | Bin 0 -> 10093 bytes imgs/Proofs/3_2.png | Bin 0 -> 14470 bytes index.html | 2 +- ....bdfd4731.chunk.js => 775.7529fa55.chunk.js} | 2 +- static/js/main.786721c1.js | 3 +++ ...LICENSE.txt => main.786721c1.js.LICENSE.txt} | 0 static/js/main.786721c1.js.map | 1 + static/js/main.c1070ac5.js | 3 --- static/js/main.c1070ac5.js.map | 1 - 10 files changed, 10 insertions(+), 10 deletions(-) create mode 100644 imgs/Proofs/3_1.png create mode 100644 imgs/Proofs/3_2.png rename static/js/{775.bdfd4731.chunk.js => 775.7529fa55.chunk.js} (61%) create mode 100644 static/js/main.786721c1.js rename static/js/{main.c1070ac5.js.LICENSE.txt => main.786721c1.js.LICENSE.txt} (100%) create mode 100644 static/js/main.786721c1.js.map delete mode 100644 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\ No newline at end of file +Home
\ No newline at end of file diff --git a/static/js/775.bdfd4731.chunk.js b/static/js/775.7529fa55.chunk.js similarity index 61% rename from static/js/775.bdfd4731.chunk.js rename to static/js/775.7529fa55.chunk.js index 6256141..df55e03 100644 --- a/static/js/775.bdfd4731.chunk.js +++ b/static/js/775.7529fa55.chunk.js @@ -1 +1 @@ -"use strict";(self.webpackChunkmathinfo=self.webpackChunkmathinfo||[]).push([[775],{775:function(s){s.exports=JSON.parse('{"Summary":{"text":"

Proofs

","image":"
\\"\\"

Proofs graphic.

"},"pythagorean-theorem-and-distance-formula-derivation":{"title":"The Pythagorean Theorem and The Distance Formula Derivation","sections":[{"text":"

The Pythagorean Theorem and The Distance Formula Derivation

Alexander DeCarlo

The Pythagorean Theorem is one of the most fundamental theorems in all of Mathematics, but little are familiar with the proof of such a formula. This theorem is used to find the length of any diagonal line which gives us a formula for how to calculate the distance between two points in space. But how was it proven?

Let\'s throw away any knowledge of the Pythagorean Theorem and simply try to answer this question: What is the length of this line on the coordinate plane?

\'A

Looking at the image, we can see that A = (-6, -3), and B = (7, 6). We can easily calculate distance when the points are on the same dimension, but how do we calculate distance when two dimensions are involved? Looking at A and B, we can see they are 13 units away horizontally and 12 units away vertically. Let\'s represent this as a triangle.

","image":"
\\"\\"

Diagram of right triangle.

"},{"text":"

Here, c is our unknown, b is 13 and a is 12. This creates a right triangle. Right triangles are special for many reasons - a big factor being their properties of similarity.

\'A

Above shows how each component angle of each triangle must be all equivalent when Triangle 3 is bisected to where it internally creates Triangle 1 and Triangle 2. What similarity tells us is that the ratios of the side lengths must be equivalent. This is because while the triangles can have different sizes overall, their interior angles dictate the side length proportions. Therefore, when two angles are shown to be equal, the third must also be equal, creating these proportions. For example, in the previous model: a / b = e / d. Right triangles are the only triangles where you can have this and its two components similar, due to the creation of two equal degree angles when bisecting the hypotenuse. You can mess around with these ratios and find many relationships, so let\'s try to expand and discover something we didn\'t expect.

","image":"
\\"\\"

A triangle representation of the distance between points A and B.

"},{"text":"

Consider two ratios: a / c = e / a and b / c = f / b. These ratios are particularly interesting because they contain a and b side lengths of the main triangle twice in their respective equations. By cross multiplying, a2 = ce and b2 = cf. Since c is in both of these equations, let\'s add them together to simplify things: a2 + b2 = ce + cf. It is clear that we can factor out c: a2 + b2 = c(e + f). Looking at the diagram: e + f = c, meaning we can substitute e + f out for c and get a2 + b2 = c2. This relationship is not obvious at all, but can beautifully be derived from very simple similarity relationships. Some good news here is that since c is isolated, we can solve for the diagonal side length of a triangle given only vertical and horizontal lengths, being c = sqrt(a2 + b2). We can finally get that equation for the distance of any two points on a plane.



let D be the distance
C = sqrt(a2 + b2) (Substitute out hypotenuse length c.)
D = sqrt(a2 + b2) (a is our vertical length and b is our horizontal length.)
a = | y1 - y2 |
b = | x1 - x2 |
D = sqrt((y2 - y1)2 + (x2 - x1)2) (The absolute values are removed because the whole thing is squared, and since negative * negative = positive, the order of variables doesn\'t matter.)



Now we can find the distance between A = (-6, -3) and B = (7, 6). With proper substitution we get:



D = sqrt((-3 - 6)2 + (-6 - 7)2)
D = sqrt(81 + 169)
D = sqrt(250) = \u224815.811","image":""}]},"three-dimensional-pythagorean-theorem":{"title":"Three Dimensional Pythagorean Theorem","sections":[{"text":"

Three Dimensional Pythagorean Theorem

Alexander DeCarlo

Deriving the Pythagorean Theorem in three dimensions is surprisingly simple, as it builds on the 2D Pythagorean Theorem. The Pythagorean Theorem in two dimensions states: a2 + b2 = c2, where a and b are the legs of a right triangle, and c is the hypotenuse.

\'A

From this three dimensional illustration, we want to find what s is from non-angled lines: x, y and z.With help from the Pythagorean Theorem, we can deduce: c2 = x2 + y2. We can also notice that on the rectangular prism, z stays completely vertical. With that information, it is time to use the Pythagorean Theorem on the triangle containing s.



z2 + (sqrt(x2 + y2))2 = s2
x2 + y2 + z2 = s2 (Absolute value signs are not needed when canceling out due to x and y both squared.)



With changing the variable names to match the Pythagorean Theorem, we get: a2 + b2 + c2 = d2. There is a clear pattern when increasing one dimension, and it\'s by simply adding a new variable. To find a distance formula in 3D space, the process is exactly the same:



D = sqrt((x2 - x1)2 + (y2 - y1)2 + (z2 - z1)2)

When going up a dimension, all you need to do is add the subtracted coordinates squared. You can see this in action when going down dimensions. If D = sqrt((x2 - x1)2 + (y2 - y1)2 + (z2 - z1)2), decreasing to a single dimension gives us D = sqrt(x2 - x1)2), which when canceled is: D = | x2 - x1 |, the formula for finding the distance between two points on the number line.

Therefore, for n dimensions, the Pythagorean Theorem theoretically would be: a12 + \u2026 + an2 = c2, where n is the amount of dimensions, each term is the distance in its dimension, and c is the hypotenuse.

","image":"
\\"Diagram

Diagram of right triangle.

"}]}}')}}]); \ No newline at end of file +"use strict";(self.webpackChunkmathinfo=self.webpackChunkmathinfo||[]).push([[775],{775:function(e){e.exports=JSON.parse('{"Summary":{"text":"

Proofs

","image":"
\\"\\"

Proofs graphic.

"},"pythagorean-theorem-and-distance-formula-derivation":{"title":"The Pythagorean Theorem and The Distance Formula Derivation","sections":[{"text":"

The Pythagorean Theorem and The Distance Formula Derivation

Alexander DeCarlo

The Pythagorean Theorem is one of the most fundamental theorems in all of Mathematics, but little are familiar with the proof of such a formula. This theorem is used to find the length of any diagonal line which gives us a formula for how to calculate the distance between two points in space. But how was it proven?

Let\'s throw away any knowledge of the Pythagorean Theorem and simply try to answer this question: What is the length of this line on the coordinate plane?

\'A

Looking at the image, we can see that A = (-6, -3), and B = (7, 6). We can easily calculate distance when the points are on the same dimension, but how do we calculate distance when two dimensions are involved? Looking at A and B, we can see they are 13 units away horizontally and 12 units away vertically. Let\'s represent this as a triangle.

","image":"
\\"\\"

Diagram of right triangle.

"},{"text":"

Here, c is our unknown, b is 13 and a is 12. This creates a right triangle. Right triangles are special for many reasons - a big factor being their properties of similarity.

\'A

Above shows how each component angle of each triangle must be all equivalent when Triangle 3 is bisected to where it internally creates Triangle 1 and Triangle 2. What similarity tells us is that the ratios of the side lengths must be equivalent. This is because while the triangles can have different sizes overall, their interior angles dictate the side length proportions. Therefore, when two angles are shown to be equal, the third must also be equal, creating these proportions. For example, in the previous model: a / b = e / d. Right triangles are the only triangles where you can have this and its two components similar, due to the creation of two equal degree angles when bisecting the hypotenuse. You can mess around with these ratios and find many relationships, so let\'s try to expand and discover something we didn\'t expect.

","image":"
\\"\\"

A triangle representation of the distance between points A and B.

"},{"text":"

Consider two ratios: a / c = e / a and b / c = f / b. These ratios are particularly interesting because they contain a and b side lengths of the main triangle twice in their respective equations. By cross multiplying, a2 = ce and b2 = cf. Since c is in both of these equations, let\'s add them together to simplify things: a2 + b2 = ce + cf. It is clear that we can factor out c: a2 + b2 = c(e + f). Looking at the diagram: e + f = c, meaning we can substitute e + f out for c and get a2 + b2 = c2. This relationship is not obvious at all, but can beautifully be derived from very simple similarity relationships. Some good news here is that since c is isolated, we can solve for the diagonal side length of a triangle given only vertical and horizontal lengths, being c = sqrt(a2 + b2). We can finally get that equation for the distance of any two points on a plane.



let D be the distance
C = sqrt(a2 + b2) (Substitute out hypotenuse length c.)
D = sqrt(a2 + b2) (a is our vertical length and b is our horizontal length.)
a = | y1 - y2 |
b = | x1 - x2 |
D = sqrt((y2 - y1)2 + (x2 - x1)2) (The absolute values are removed because the whole thing is squared, and since negative * negative = positive, the order of variables doesn\'t matter.)



Now we can find the distance between A = (-6, -3) and B = (7, 6). With proper substitution we get:



D = sqrt((-3 - 6)2 + (-6 - 7)2)
D = sqrt(81 + 169)
D = sqrt(250) = \u224815.811","image":""}]},"three-dimensional-pythagorean-theorem":{"title":"Three Dimensional Pythagorean Theorem","sections":[{"text":"

Three Dimensional Pythagorean Theorem

Alexander DeCarlo

Deriving the Pythagorean Theorem in three dimensions is surprisingly simple, as it builds on the 2D Pythagorean Theorem. The Pythagorean Theorem in two dimensions states: a2 + b2 = c2, where a and b are the legs of a right triangle, and c is the hypotenuse.

\'A

From this three dimensional illustration, we want to find what s is from non-angled lines: x, y and z.With help from the Pythagorean Theorem, we can deduce: c2 = x2 + y2. We can also notice that on the rectangular prism, z stays completely vertical. With that information, it is time to use the Pythagorean Theorem on the triangle containing s.



z2 + (sqrt(x2 + y2))2 = s2
x2 + y2 + z2 = s2 (Absolute value signs are not needed when canceling out due to x and y both squared.)



With changing the variable names to match the Pythagorean Theorem, we get: a2 + b2 + c2 = d2. There is a clear pattern when increasing one dimension, and it\'s by simply adding a new variable. To find a distance formula in 3D space, the process is exactly the same:



D = sqrt((x2 - x1)2 + (y2 - y1)2 + (z2 - z1)2)

When going up a dimension, all you need to do is add the subtracted coordinates squared. You can see this in action when going down dimensions. If D = sqrt((x2 - x1)2 + (y2 - y1)2 + (z2 - z1)2), decreasing to a single dimension gives us D = sqrt(x2 - x1)2), which when canceled is: D = | x2 - x1 |, the formula for finding the distance between two points on the number line.

Therefore, for n dimensions, the Pythagorean Theorem theoretically would be: a12 + \u2026 + an2 = c2, where n is the amount of dimensions, each term is the distance in its dimension, and c is the hypotenuse.

","image":"
\\"Diagram

Diagram of right triangle.

"}]},"proof-of-the-fundamental-theorem-of-calculus-part-i":{"title":"Proof of the Fundamental Theorem of Calculus Part I","sections":[{"text":"

Proof of the Fundamental Theorem of Calculus Part I

Alexander DeCarlo

Proving the fundamental theorem of calculus consists of two parts. Interestingly, the \u201cfirst part\u201d was proved secondly, and the \u201csecond part\u201d was proved first. However, it makes much more sense for any calculus student to learn the first part\'s proof, then the second one, despite the order that they were proven. Let\'s jump right into the first part!

The first part of the fundamental theorem of calculus connects the idea of a derivative and an integral. It states that the derivative of an integral returns the original function. In mathematical notation: \\\\[\\\\frac{d}{dx}\\\\left[\\\\int f(x) \\\\, dx\\\\right] = f(x)\\\\]This can also be used to say that the integral of f(x) (area under the curve) is just the \u201cantiderivative\u201d of f(x), or in simpler terms, a function g(x) where its derivative of f(x). As we can see, proving this theorem will give us a gateway into actually solving integrals, one of the hardest tasks in calculus.

But why would we even consider a relationship in the first place? Well, let\'s consider a very general function, f(x).

\'Graph

Now, consider f(x) and A(x), where A(x) is the area function. Basically, we are assuming that the area under any curve can be considered some function of x. As we see with the blue line to the right, changing that also changes the overall area of the function. If f(x) is decreasing, then the added area to A(x) will become less and less large. Here, we can intuitively see a connection to the derivative of f(x) and A(x). It just so happens that the area function IS the antiderivative of f(x), which is less intuitive to see.

Lets begin with the proof:

Consider A(x) and f(x) where A(x) is the area function to f(x). We are to prove that A\'(x) = f(x).

Let\'s begin with the definition of a derivative, just to see if we can work with that somehow: \\\\[ A\'(x) = \\\\lim_{h \\\\to 0} \\\\left[ \\\\frac{A(x+h) - A(x)}{h} \\\\right] \\\\]

Notice A(x + h) and A(x). These appear to be the difference of 2 areas. Let\'s represent this graphically.

\'Graph

As we see, A(x+h) - A(x) must be some definite area, but what? Here, we will use interesting logic in our proof. First thing, we notice that the value of f(x) at point x shown is at a minimum, and f(x+h) is at a maximum for the interval x to x+h. That seems unique, why isn\'t x+h moved a bit more to the right, where it wouldn\'t be a maximum of the interval? Well, the key is that the expression A(x+h) - A(x) is a limit, so \u201ch\u201d is meant to be a slight nudge to x. It seems a lot more like a nudge on the graph to show the idea of the area under a curve. Since \u201ch\u201d approaches 0, it is meant to be a \u201csufficiently small\u201d value. This means that either f(x) or f(x+h) is a max or a min, depending if the function is increasing or not, because the distance between x and x+h will approach 0. Although it doesn\'t really matter as we will see, the function is increasing so f(x+h) will be the max. and f(x) will be the min.

Now with that settled, why does that even matter? Well, we know the area between x and x+h is an area, and that f(x) is the lowest point on the graph, and f(x+h) is the highest. Now, we know that the area in the interval is some area value, A. Let\'s simplify the area for A greatly. Let\'s say that the area for A is just the area of some rectangle with the width of the interval h. Then, A = h*f(c), where f(c) is some height between f(x) and f(x+h). Ok\u2026 but how does that help us? Well, we know that f(x) is a minimum, so f(x) * h MUST be smaller than the actual area for the interval, because all other points on the graph are greater than f(x). So, if you view the interval as the sum of a bunch of rectangles, we just took the smallest rectangle and spread it across the interval, meaning it must be smaller. With similar logic, since f(x+h) is a maximum, it MUST be larger than the actual area. With that, we know that f(c) must be somewhere between f(x) and f(x+h). Since our function is continuous, we know that every value must exist between f(x) and f(x+h), which means every value must exist between f(x)*h and f(x+h)*h. This means that area A must exist in the form f(c) * h, where c \u2208 [x, x+h]. \\\\[ A\'(x) = \\\\lim_{h \\\\to 0} \\\\left[ \\\\frac{A(x+h) - A(x)}{h} \\\\right] \\\\]

Back to the actual math, we essentially just logically proven that A(x+h) - A(x) = f(c) * h. Let\'s substitute that in! \\\\[ A\'(x) = \\\\lim_{h \\\\to 0} \\\\left[ \\\\frac{f(c) * h}{h} \\\\right] \\\\] \\\\[ A\'(x) = \\\\lim_{h \\\\to 0} \\\\left[ f(c) \\\\right] \\\\]

Now, we know that c \u2208 [x, x + h] then, \\\\[ x \\\\leq c \\\\leq x + h \\\\]

But, h\u21920, so: \\\\[ x \\\\leq c \\\\leq x \\\\]

Which means c = x, meaning f(c) = f(x), solving the limit.

A\'(x) = f(x), which is what we meant to prove.

"}]}}')}}]); \ No newline at end of file diff --git a/static/js/main.786721c1.js b/static/js/main.786721c1.js new file mode 100644 index 0000000..7bdc4d8 --- /dev/null +++ b/static/js/main.786721c1.js @@ -0,0 +1,3 @@ +/*! For license information please see main.786721c1.js.LICENSE.txt */ +!function(){var e={124:function(){},463:function(e,t,n){"use strict";var r=n(791),a=n(296);function o(e){for(var t="https://reactjs.org/docs/error-decoder.html?invariant="+e,n=1;n