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
All hints are welcome.


 * $\checkmark (1):$ Prove Greatest Power of Two not Divisor. Use this fact to prove Harmonic Numbers not Integers.


 * $\checkmark (2):$ Prove that:


 * $\displaystyle K_n = 1 + \frac 1 3 + \frac 1 5 + \ldots + \frac 1 {2n+1}$

is not an integer for $n > 1$.


 * $(3):$ Prove that $\dfrac {\ln 2}{\ln 3}$ is irrational. For integers $p$ and $q$, what condition is essential for $\dfrac {\ln p}{\ln q}$ to be irrational? Justify your claims.

(I'm expected to use external sources for this one.)


 * $\dfrac {\ln 2}{\ln 3} = \log_3 2$ from an elementary result you'll find on somewhere (Change of Base of Logarithms or something).  Then you need Irrationality of Logarithm which Ybab321 posted up a couple of months ago. --prime mover (talk) 21:06, 23 November 2014 (UTC)


 * Well that's a lot more direct than I expected it to be, thanks a lot, you and Ybab. --GFauxPas (talk) 21:14, 23 November 2014 (UTC)


 * $(4):$ Use Maple to execute the command below and carefully explain the output.



returns:

What are the solutions to $10^j \equiv 82 \bmod 543$? Is it doable without a computer? How do I ask Maple what $j$s work? --GFauxPas (talk) 02:15, 24 November 2014 (UTC)


 * $(5):$ Explain the behavior of the Euler Phi Function in finding all positive solutions to:


 * $(a): \phi(n) = 6, (b): \phi(n) = 14, (c):\phi(n) = 24$.

Theorem
Let $p, q$ be constants in $\N_{\ge 1}$, $p \equiv q$.

The diophantine equation:


 * $a^p = b^q$

Has solutions precisely when $b$ is a power of $a$ or $a$ is a power of $b$.

Proof
Let $a, b$ be solutions to $a^p = b^q$.

Consider the [[Definition:Prime Factorization|prime factorizations of $a$ and $b$:

(Note that for sufficiently large primes, all powers of those primes become $0$).

If any prime does not divide $a^p$, then that prime also does not divide $b^q$.

So we consider only the equality of the non-zero powers.

An arbitrary equality is of the form:


 * $B_i p = A_i q$.

By hypothesis $p$ and $q$ are coprime.