User:Idk12345678

My Solutions

Some Proofs I wrote

$(x+y)^n \equiv x^n + y^n \pmod{n}$ 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 <cmath> \int_{a}^{b} (f(x))^2 \,dx </cmath>.

Cylinder: A cylinder can be expressed a solid of revolution by revolving the line$ (Error compiling LaTeX. Unknown error_msg)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 <cmath>\pi <cmath> \int_{0}^{h} r^2 \,dx </cmath></cmath>. Integrating, we get <cmath>\pi (r^2h - r^2(0)) = \pi r^2h</cmath>. 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$ (Error compiling LaTeX. Unknown error_msg)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 <cmath>\pi [ \int_{0}^{h} \frac{r^2x^2}{h^2} \,dx \] = \pi \frac{r^2h}{3}</cmath>.

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$ (Error compiling LaTeX. Unknown error_msg)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}\] (Error compiling LaTeX. Unknown error_msg)

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User:Idk12345678

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My Solutions

Some Proofs I wrote

$(x+y)^n \equiv x^n + y^n \pmod{n}$ if $n$ is prime.

Volume of Cylinder, Cone, and Sphere

Page

Toolbox

Search

User:Idk12345678

Contents

My Solutions

Some Proofs I wrote

if is prime.

Volume of Cylinder, Cone, and Sphere

$(x+y)^n \equiv x^n + y^n \pmod{n}$ if $n$ is prime.