Difference between revisions of "User:Idk12345678"

Claim: $(x+y)^n \equiv x^n + y^n \pmod{n}$ if $n$ is prime. $\[\]$ Proof: Expanding $(x+y)^n$ out, all the coefficients are of the form $n \choose r$ by the binomial theorem. To prove the original result we must show that if $r \neq 1$ and $r \neq n$ , then ${n \choose r} \equiv 0 \pmod{n}$ . Because ${n \choose r} = \frac{n!}{r!(n-r)!}$ , ${n \choose r} \times r!(n-r)! = n!$ , which is divisible by $n$ , so the original expression must be divisible by $n$ . However if $n$ is prime, $\gcd(n, r!(n-r)!) = 1$ , since $r!$ does not contain $n$ (because $r<n$ ). Therefore, in order for ${n \choose r} \times r!(n-r)!$ to be divisible by $n$ , $n \choose r$ is divisible by $n$ . All the coefficients of the expansion(besides the coefficients of $x^n$ and $y^n$ ) are of the form $n \choose r$ , and ${n \choose r} \equiv 0 \pmod{n}$ , so they cancel out and $(x+y)^n \equiv x^n + y^n \pmod{n}$ if $n$ is prime. $\square$

Volume of Cylinder, Cone, and Sphere

If we have a function $f(x)$ , that can be rotated to make a shape, the area underneath it will turn into the volume. However, since we are revolving it in a circular motion, the area will actually become the radius. Another way of seeing this is splitting it into infinite circles and adding up all of them. Therefore, for a function $f(x)$ , we have the volume of the solid of revolution to be $\pi \int_{a}^{b} (f(x))^2 \,dx$ .

Cylinder: A cylinder can be expressed a solid of revolution by revolving the line $y = r$ around the $x$ -axis. To find the volume, we can find the area under the curve, and then when we revolve it, it becomes the volume. The radius is $r$ and the height, $h$ , is the upper bound of integration. We have $\pi \int_{0}^{h} r^2 \,dx$ . Integrating, we get $\pi (r^2h - r^2(0)) = \pi r^2h$ . This is the formula of a cylinder.

Cone: If you are given the height and radius of the cone, and you have the point $(0,0)$ on your line(since the vertex is 0), then $f(h) = r$ , because the height is the x-coordinate and the radius is the y(for the same reason seen above in the cylinder). Now, since we have $(0,0)$ , we know the y-intercept, and we can only have one slope. If $h=x$ , and $m$ is the slope, then we have $r = mh$ , and therefore $m = \frac{r}{h}$ , so the equation is $f(x) = \frac{rx}{h}$ . For the integral, we get $\pi \int_{0}^{h} \frac{r^2x^2}{h^2} \,dx = \pi \frac{r^2h}{3}$ .

Sphere: The equation of a sphere should be a circle, but that is a relation and not a function. Therefore, we can use the top half of a circle, and the bottom half will get filled in when it rotates. Therefore, we get $f(x) = \sqrt{r^2 - x^2}$ . The diameter is $-r$ to $r$ , so that is where we integrate. $\pi \int_{-r}^{r} r^2 - x^2 \,dx = \frac{4\pi r^3}{3}$ .

@@ Line 38: / Line 38: @@
 ==Some Proofs I wrote==
-===<math>(x+y)^n \equiv x^n + y^n \pmod{n}</math> if <math>n</math> is prime. ===
+===Binomial Expansions mod n when n is prime ===
+Claim: <math>(x+y)^n \equiv x^n + y^n \pmod{n}</math> if <math>n</math> is prime. <cmath></cmath>
 Proof: Expanding <math>(x+y)^n</math> out, all the coefficients are of the form <math>n \choose r</math> by the binomial theorem. To prove the original result we must show that if <math>r \neq 1</math> and <math>r \neq n</math>, then <cmath>{n \choose r} \equiv 0 \pmod{n}</cmath>. Because <cmath>{n \choose r} = \frac{n!}{r!(n-r)!}</cmath>, <cmath>{n \choose r} \times r!(n-r)! = n!</cmath>, which is divisible by <math>n</math>, so the original expression must be divisible by <math>n</math>. However if <math>n</math> is prime, <cmath>\gcd(n, r!(n-r)!) = 1</cmath>, since <math>r!</math> does not contain <math>n</math>(because <math>r<n</math>). Therefore, in order for <cmath>{n \choose r} \times r!(n-r)!</cmath> to be divisible by <math>n</math>, <math>n \choose r</math> is divisible by <math>n</math>. All the coefficients of the expansion(besides the coefficients of <math>x^n</math> and <math>y^n</math>)  are of the form <math>n \choose r</math>, and <cmath>{n \choose r} \equiv 0 \pmod{n}</cmath>, so they cancel out and <cmath>(x+y)^n \equiv x^n + y^n \pmod{n}</cmath> if <math>n</math> is prime. <math>\square</math>

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Difference between revisions of "User:Idk12345678"

Revision as of 15:00, 31 March 2025

Contents

My Solutions

Some Proofs I wrote

Binomial Expansions mod n when n is prime

Volume of Cylinder, Cone, and Sphere