User:GFauxPas/Sandbox

Welcome to my sandbox, you are free to play here as long as you don't track sand onto the main wiki. --GFauxPas 09:28, 7 November 2011 (CST)

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/existence

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/integrand

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/evaluation

User:GFauxPas/Sandbox/Zeta2/FourierSeries/

User:GFauxPas/Sandbox/Zeta2/Informal Proof

Problem Set 2
Due 12/22

Hint for all of these: Most of these are standard problems; Google creatively.

$(1):$

Prove that $\cos 3^\circ$ is irrational and find its value exactly. You may use any formulas derived in class.

Hint: use Triple Angle Formulas/Cosine, Double Angle Formulas/Cosine

Derived in class:


 * $\cos 36^\circ = \dfrac {1 + \sqrt 5} 4$

Extra credit: Do the same for $\cos 1^\circ$. You may use Maple or Wolfram Alpha to solve the cubic that will appear if you do this.

I gave a wrong answer for the proof of irrationality because I couldn't think of a correct one. Anyone have a hint? Please? It has a lot of nested radicals and I don't know what theorem(s) to invoke. --GFauxPas (talk) 12:01, 22 December 2014 (UTC)

$(2):$


 * $\checkmark (a)$ Prove $\phi(n) \ge \frac {\sqrt n} 2$ and deduce the limit $\lim_{n \to \infty} \phi(n)$

Hint: Write $n$ using its unique prime factorization. Examine each term $p_i^{\alpha_i - 1}\left({p_i - 1}\right)$, and compare it to $p_i^{\alpha_i / 2}$


 * $\checkmark (b)$: Show that if $n$ is a composite positive integer and $\phi(n) \vert n - 1$, then $n$ is a square free product of at least 3 distinct primes.

$\checkmark (3):$ Prove that in any block of consecutive positive integers there is a unique integer divisible by a higher power of $2$ than any of the others. Then use this, or any other method, to show there is no integer among the $2^{n+1}$ numbers:


 * $\displaystyle \pm \frac 1 j \pm \frac 1 {j+1}\pm \frac 1 {j+2}\pm\cdots\pm\pm \frac 1 {j+n}$, where all combinations of $+$ and $-$ signs are allowed, where $n$ and $j$ are any positive integers.

This result is a sweeping generalization of Harmonic Numbers not Integers.

$(4):$


 * $\checkmark (a):$ Assuming that $p$ and $q$ are distinct primes, prove:


 * $p^{q-1} + q^{p-1} \equiv 1 \bmod pq$

Hint: use Fermat's Little Theorem


 * $\checkmark (b):$ If $p$ and $q$ are distinct primes, prove that for any integer $a$;


 * $a^{pq} - a^p - a^q - a \equiv 0 \bmod pq$

$\checkmark(6):$

Carefully outline how you would solve the equation:


 * $ax + by + cz = d$

in integers, where $\gcd(a,b,c)\vert d$.

Use your method to find all solutions to $5x + 7y + 12z = 2014$.

$\checkmark(7):$ Define the binomial coefficient for any $n \in \R, j \in \N_{\ge 0}$ as follows:


 * $\displaystyle \binom n j = \frac {n(n-1)(n-2)\cdots(n-j+1)}{j!}$

Under this definition, find simple formulas for $\displaystyle \binom {-1} j$ and $\displaystyle \binom {-1/2} j$.

$(8):$

Prove that $n = 561$ satisfies $a^{n-1} \equiv 1 \bmod n$ for every $a$ with $\gcd(a,n) = 1$.

Hint: Apply Fermat's little theorem with the prime factors of 561, and then the Chinese Remainder Theorem. You may wish to Google "Carmichael Numbers"

$\checkmark (5):$

Let $a > 1$ be a positive integer and $m, n$ arbitrary positive integers. Prove:


 * $\gcd(a^m-1, a^n - 1) = a^{\gcd(m,n)} - 1$

I figured it out. Should I put it up as a page? --GFauxPas (talk) 16:11, 21 December 2014 (UTC)


 * Please do so. It's an interesting, elegant and non-obvious little result. It reminds me a little of the Law of Quadratic Reciprocity which I vaguely remember, to which it may have a link. --prime mover (talk) 16:16, 21 December 2014 (UTC)


 * My proof needs $v^t - 1 \mathop \backslash v^{tw} - 1$ for $w \in \N$, do we have that or something stronger somewhere? --GFauxPas (talk) 17:24, 21 December 2014 (UTC)


 * The something stronger might be $x^n - y^n \mathop \backslash x^{nt}-y^{nt}$. Is that a theorem? I seem to remember it being a theorem. --GFauxPas (talk) 19:51, 21 December 2014 (UTC)


 * We have Integer Less One divides Power Less One/Corollary which is what you are after. Then we have Difference of Two Powers which is easily adjusted by substituting $a = x^n$ and $b = y^n$. Job done. --prime mover (talk) 20:16, 21 December 2014 (UTC)