Krdytkyu

0 $begingroup$ I am confused about the result of an odd function being purely imaginary for the DTFT of an odd function. I was under the understanding that the result of a FT was amplitude and phase as a function of frequency, with phase information being imaginary and amplitude being real. The magnitude could then be calculated from this complex number. However, in a purely odd function the DTFT is purely imaginary. Is there then no amplitude information? How is the magnitude found? For instance, if $x[n] = delta(n - pi/2) - delta(n - 3pi/2)$ , the DTFT could be found to be $X(jw) = exp(-jw(pi/2)) - exp(-jw(3pi/2))$ . Even if this is simplified via symmetry I get $2jexp(-jwpi)sin(w*pi/2)$ . This is purely imaginary and I am confused how the results are interpreted. ...

6 $begingroup$ Prove that, for distinct primes $p,q$: $$tag{1} p^{q-1} + q^{p-1} equiv{1} pmod{pq}.$$ This is a pretty straight forward problem, I know, and although the standard proof is simpler, I was wondering if there is any flaw in this slightly different one. Proof: Let $p$ and $q$ be prime numbers, with $plt q$. By Fermat’s Little Theorem, since $p$ and $q$ are distinct: $$begin{align} tag{2} p^{q-1} equiv 1 pmod{q}, \ tag{3} q^{p-1} equiv 1 pmod{p}.end{align}$$ By $(2)$ and $(3)$, there are integers $k_1,k_2$ such that $$begin{align} p^q-p +q^p-q=(k_1 +k_2)pq \ implies tag{4} p^q+q^pequiv p+qpmod{pq}.end{align}$$ Obviously, $pq^{p-1} equiv qp^{q-1}equiv 0pmod{pq},$ and so adding these terms to $(4)$ gives $$tag{5} (p+q)( p^{q-1}+q^{p-1})equiv p+qpmod{pq} .$$ Since $p+qlt 2qleq pq $, it follows ...