1. Problem 45

1)Prove that $\displaystyle \sin 1^{\circ}$ and $\displaystyle \cos 1^{\circ}$ are irrational.

3)Suppose that $\displaystyle \sum_{k=0}^n k^m = P_m(n)$ where $\displaystyle P_m$ is an $\displaystyle m+1$-degree polynomial. Prove that the leading coefficient of $\displaystyle P_m$ is $\displaystyle 1/(m+1)$.

2. Originally Posted by ThePerfectHacker
1)Prove that $\displaystyle \sin 1^{\circ}$ and $\displaystyle \cos 1^{\circ}$ are irrational.
Suppose we let:

$\displaystyle f = cos(x),g = sin(x)$

Then:

$\displaystyle e^{xi} = cos(x) + i*sin(x) = f + ig$

Therefore:

$\displaystyle e^{5xi} = cos(5x) + i*sin(5x) = (f + ig)^5$

$\displaystyle (f + ig)^5 = f^5 + 5f^4gi - 10f^3g^2 - 10f^2g^3i + 5fg^4 + g^5$

Now we take the real parts:

$\displaystyle cos(5x) = f^5 - 10f^3g^2 + 5fg^4$

Using $\displaystyle g^2 = 1 - f^2$ throughout:

$\displaystyle cos(5x) = 16f^5 - 20f^3 + 5f$

We know that:

$\displaystyle cos(45-30) = cos(45)cos(30) + sin(45)sin(30) = \frac{\sqrt{3} + 1}{2\sqrt{2}}$

This is clearly irrational, therefore:

$\displaystyle cos(15) = 16cos(3) - 20cos(3) + 5cos(3) = \frac{\sqrt{3} + 1}{2\sqrt{2}}$

If $\displaystyle cos(3)$ was rational, $\displaystyle 16cos(3) - 20cos(3) + 5cos(3)$ would also be rational, which is a contradiction.

Finally:

Using $\displaystyle cos(3x) = 4cos^3\!(x) - 3cos(x)$

We get:

$\displaystyle cos(3) = 4cos^3\!(1) - 3cos(1)$

If $\displaystyle cos(3)$ was rational, then $\displaystyle 4cos^3\!(1) - 3cos(1)$ would also be rational, and if $\displaystyle cos(1)$ was rational, then the statement above would also be rational, we have thus arrived at a contradiction:

$\displaystyle cos(1)$ is irrational.

$\displaystyle Q.E.D.$

I have to go to class someone else do $\displaystyle sin(1)$ or I'll do it later.

3. For $\displaystyle sin(1) = irrational$

Using the math above and taking the imaginary parts:

$\displaystyle isin(5x) = 5f^4gi - 10f^2g^3i$

Using $\displaystyle f^2 = 1 - g^2$

$\displaystyle sin(5x) = 5g^9 - 10g^7 + 15g^5 - 5g$

We know that:

$\displaystyle sin(45 - 30) = sin(45)cos(30) + cos(45)sin(30) = \frac{\sqrt{3} + 1}{2}$

That is clearly irrational.

$\displaystyle sin(15) = 5sin^9\!(3) - 10sin^7\!(3) + 15sin^5\!(3) - 5sin(3) = \frac{\sqrt{3} + 1}{2}$

sin(3) is irrational since sin(15) is irrational.

Finally:

$\displaystyle sin(3x) = 3sin(x) - 4sin^3\!(x)$

Therefore:

$\displaystyle sin(3) = 3sin(1) - 4sin^3\!(1)$

Therefore, since sin(3) is irrational, sin(1) is also irrational.

$\displaystyle Q.E.D.$

Using this as logic:

$\displaystyle sin(nx) \ is \ irrational \Rightarrow sin(x) \ is \ also \ irrational$

Same with cos(nx)

4. 2)We are given that $\displaystyle \sum_{k=1}^n k^m = P_m(n)$ where $\displaystyle P_m$ is a polynomial. Suppose we wish to compute the integral of $\displaystyle f(x) = x^m$ over the interval $\displaystyle [0,1]$. Then $\displaystyle \int_0^1 x^m ~ dx = \frac{1}{m+1}$. Now there is another way to compute this integral, and that is to use a Riemann sum with $\displaystyle n$ equal subdivision point using the right endpoint. In that case we get $\displaystyle \lim ~ \left( \sum_{k=1}^n \frac{k^m}{n^m} \right) \cdot \frac{1}{n} = \lim ~ \frac{1}{n^{m+1}} \sum_{k=1}^n k^m = \lim ~ \frac{P_m(n)}{n^{m+1}}$. This limit needs to be $\displaystyle 1/(m+1)$ since that is the integral. But this means the polynomial $\displaystyle P_m(n)$ must be degree $\displaystyle m+1$ with leading coefficient of $\displaystyle 1/(m+1)$.

In fact, note the following,
$\displaystyle \sum_{k=1}^n k = \frac{n(n+1)}{2} = \boxed{\frac{1}{2}}n^2 + \frac{1}{2}n$.
$\displaystyle \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6} = \boxed{\frac{1}{3}}n^3 + \frac{1}{2}n^2+\frac{1}{6}n$.
$\displaystyle \sum_{k=1}^n k^3 = \frac{n^2(n+1)^2}{4} = \boxed{\frac{1}{4}}n^4 + \frac{1}{2}n^3+\frac{1}{4}n^2$.
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