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===<math>(x+y)^n \equiv x^n + y^n \pmod{n}</math> if <math>n</math> is prime. === | ===<math>(x+y)^n \equiv x^n + y^n \pmod{n}</math> if <math>n</math> is prime. === | ||
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> | 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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| + | ===Volume of Cylinder, Cone, and Sphere=== | ||
| + | If we have a function <math>f(x)</math>, 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 <math>f(x)</math>, we have the volume of the solid of revolution to be <math>\pi <cmath> \int_{a}^{b} (f(x))^2 \,dx </cmath>. | ||
| + | |||
| + | Cylinder: A cylinder can be expressed a solid of revolution by revolving the line </math>y = r<math> around the </math>x<math>-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 </math>r<math> and the height, </math>h<math>, 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 </math>f(h) = r<math>, 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 </math>(0,0)<math>, we know the y-intercept, and we can only have one slope. If </math>h=x<math>, and </math>m<math> is the slope, then we have </math>r = mh<math>, and therefore </math>m = \frac{r}{h}<math>, so the equation is </math>f(x) = \frac{rx}{h}<math>. 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 </math>f(x) = \sqrt{r^2 - x^2}<math>. The diameter is </math>-r<math> to </math>r$, so that is where we integrate. <cmath>\pi [ \int_{-r}^{r} r^2 - x^2 \,dx \] = \frac{4\pi r^3}{3}</cmath>. | ||
Revision as of 14:40, 9 June 2024
Contents
My Solutions
Some Proofs I wrote
if 
is prime.
if Proof: Expanding 
out, all the coefficients are of the form 
by the binomial theorem. To prove the original result we must show that if 
and 
, then ![\[{n \choose r} \equiv 0 \pmod{n}\]](http://latex.artofproblemsolving.com/c/6/3/c63848a8ef72c7440c7022bd218dad3ced5ab858.png)
. Because ![\[{n \choose r} = \frac{n!}{r!(n-r)!}\]](http://latex.artofproblemsolving.com/3/c/a/3ca41bab24e0ced5a3bd2665f4e96651799fe7cc.png)
, ![\[{n \choose r} \times r!(n-r)! = n!\]](http://latex.artofproblemsolving.com/7/b/e/7bef983037ba8643bc0efb980a61e51f9e25e8b3.png)
, which is divisible by 
, so the original expression must be divisible by . However if is prime, ![\[\gcd(n, r!(n-r)!) = 1\]](http://latex.artofproblemsolving.com/0/d/0/0d095da7ac98ad36f4c2dd61cc45e45e38906992.png)
, since 
does not contain (because 
). Therefore, in order for ![\[{n \choose r} \times r!(n-r)!\]](http://latex.artofproblemsolving.com/6/a/b/6abc4a965f1ab5a8620d40b4d393876dded1647b.png)
to be divisible by , is divisible by . All the coefficients of the expansion(besides the coefficients of 
and 
) are of the form , and , so they cancel out and ![\[(x+y)^n \equiv x^n + y^n \pmod{n}\]](http://latex.artofproblemsolving.com/3/c/1/3c1d9f3704248e7d2b3af8c5b48ebf8acbcd48b6.png)
if is prime. 
Volume of Cylinder, Cone, and Sphere
If we have a function 
, 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 , 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
x
r
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
(0,0)
h=x
m
r = mh
m = \frac{r}{h}
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}
-r
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).
