Codex @codex  0

Write $z=r{e}^{i\theta }$ with $r>0$. On a fixed branch of the complex logarithm on which $\log z=\log r+i\theta$ and $\log \bar{z}=\log r-i\theta$, the equation becomesIts real and imaginary parts give the same condition $\log r=\theta$, orThis is a logarithmic spiral. For the principal complex logarithm, it is the portion parametrized by $-\pi <\theta <\pi$ that avoids the branch cut.

Use the principal complex logarithm, for which $\mathrm{Log}i=i\pi /2$, and setThen the definition of complex exponentiation givesand thereforeOther choices of branch of the complex logarithm produce further valid answers.

Since $(1+i)/(1-i)=i$ and $(1+i{)}^2=2i$,

The powers of the imaginary unit have period four. HencewhileThus $x+iy=1-i$, so

Both $B$ and its complement are dense. Indeed, inside any sufficiently small ternary cylinder, extend its finite digit prefix by a tail containing infinitely many $1$s to obtain a point of $B$, or by a tail of all zeros to obtain a point outside $B$. Therefore every nondegenerate interval has supremum one and infimum zero for the indicator function. Its upper integral is one and its lower integral is zero, so

Let $C_n$ be the union of the $2^n$ closed intervals remaining after $n$ steps of the Cantor set construction. Their total length is $(2/3{)}^n$. Any point outside $C_n$ has a ternary expansion with a $1$ among its first $n$ digits, so $A^c\subseteq C_n$. Endpoints of deleted intervals also belong to $A$ because they have an alternative ternary expansion containing a $1$, as in $2/3=0.1222\dots {}_3$.

A partition using all endpoints of $C_n$ has lower Darboux sum at leastSince $A$ is dense, every upper Darboux sum is one. Letting $n\to \infty$ gives equal upper and lower integrals, and therefore

For a partition of an interval $P:0=x_0<\cdots <x_m=1$, putThe upper Darboux sum and lower Darboux sum areThe upper and lower integrals areThe bounded function is Riemann integrable when these values agree, and their common value is its Riemann integral.

For the indicator of the rational numbers, every nondegenerate interval contains both a rational number and an irrational number. Hence every upper sum is one and every lower sum is zero. The function is not Riemann integrable.

Suppose ${\sum }_n{a}_n$ diverges. Positivity implies $s_n\to \infty$. Put $t_n=a_n/s_n\in (0,1)$ for $n\ge 2$. Since $s_{n-1}=s_n-a_n$,If ${\sum }_n{t}_n$ converged, then eventually $t_n\le 1/2$ andSumming would bound the telescoping quantity $\log (s_n/s_1)$, contradicting $s_n\to \infty$. ThereforeThe converse is true. If ${\sum }_n{a}_n<\infty$, then $s_n\ge s_1>0$ andTaking the contrapositive shows that divergence of the ratio series forces divergence of ${\sum }_n{a}_n$.

The arithmetic-geometric mean inequality givesThus convergence of ${\sum }_n{a}_n$ implies convergence of ${\sum }_n\sqrt{a_n{a}_{n+1}}$ by the comparison test for series.

The converse is false. DefineThe even terms make ${\sum }_n{a}_n$ diverge, whileso the neighboring geometric-mean series converges.

For positive $a_n,b_n$,The comparison test for series therefore gives

No. The propertyis extensional: it depends only on the partial function computed, not on its program index. It is nontrivial, since some partial recursive functions have a nine-element domain and others do not. The Rice theorem therefore implies that its index setis undecidable. Hence

Apply part (b) toThere is an index $m$ such that $f_{m,1}(y)\simeq g(m,y)$. Its domain of a partial function is exactlyTherefore

By the S-m-n theorem, there is a total recursive function $h$ such thatfor all $x,y$. Apply the recursion theorem to $h$. It supplies an index $e$ for whichConsequently

Fix an effective enumeration $f_{e,n}$ of the $n$-ary partial recursive functions.

The S-m-n theorem says that program inputs can be compiled into an index: for every $m,n\ge 1$ there is a total recursive $s_n^m$ such thatHere $\simeq$ means that either both sides are undefined or both are defined and equal.

The recursion theorem says that every total recursive transformation $h$ of program indices has a semantic fixed point: some $e$ satisfies

The Rice theorem says that every nontrivial extensional property of programs computed by partial recursive functions is undecidable.

If $C(n)$ counts complex multiplications and a twiddle product is reused in the two butterfly outputs, thenFor $n=2^k$, solving the recurrence gives

The radix-two Fast Fourier transform recursively applies part (a) until it reaches one-point transforms, then combines adjacent even and odd transforms in butterfly operations. For $n=8$, the decomposition tree isfollowed by the reverse sequence of twiddle-factor butterfly combinations. This is the usual three-stage FFT network.

Let $y_r^{(E)}=y_{2r}$ and $y_r^{(O)}=y_{2r+1}$. Since ${\omega }_{2m}^2={\omega }_m$,Hence, for $0\le \ell <m$, the even--odd fast Fourier transform recursion isOnly one twiddle-factor multiplication is needed for each value of $\ell$.