Sum of Computable Real Sequences is Computable

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Theorem

Let $\sequence {x_n}$ and $\sequence {y_n}$ be computable real sequences.

Then $\sequence {x_n + y_n}$ is a computable real sequence.


Corollary

Let $\sequence {x_n}$ be a computable real sequence.

Let $c \in \R$ be a computable real number.

Then, $\sequence {x_n + c}$ is a computable real sequence.


Proof 1

By definition, there exist total recursive $f,g : \N^2 \to \N$ such that:

For every $m,n \in \N$, $\map f {m, n}$ and $\map g {m, n}$ respectively code integers $k$ and $\ell$ such that:
$\dfrac {k - 1} {n + 1} < x_m < \dfrac {k + 1} {n + 1}$
$\dfrac {\ell - 1} {n + 1} < y_m < \dfrac {\ell + 1} {n + 1}$

Define $h : \N^2 \to \N$ as:

$\map h {m, n} = \map {\operatorname{quot}_\Z} {\map f {m, 4 n + 3} +_\Z \map g {m, 4 n + 3} +_\Z 2, 4_\Z}$

which is total recursive by:


Now, let $m, n \in \N$ be arbitrary.

Let $k'$ and $\ell'$ be the integers coded by $\map f {m, 4 n + 3}$ and $\map g {m, 4 n + 3}$, respectively.

Then, $\map h {m, n} = \floor {\dfrac {k' + \ell' + 2} 4}$.

Thus, by Floor is between Number and One Less:

$\dfrac {k' + \ell' - 2} 4 < \map h {m, n} \le \dfrac {k' + \ell' + 2} 4$

Hence:

$\map h {m, n} - 1 \le \dfrac {k' + \ell' - 2} 4$
$\dfrac {k' + \ell' + 2} 4 < \map h {m, n} + 1$

From the inequalities above, we have:

\(\ds \dfrac {k' - 1} {\paren {4 n + 3} + 1} + \dfrac {\ell' - 1} {\paren {4 n + 3} + 1}\) \(<\) \(\, \ds x_m + y_m \, \) \(\, \ds < \, \) \(\ds \dfrac {k' + 1} {\paren {4 n + 3} + 1} + \dfrac {\ell' + 1} {\paren {4 n + 3} + 1}\)
\(\ds \dfrac {k' + \ell' - 2} {4 n + 4}\) \(<\) \(\, \ds x_m + y_m \, \) \(\, \ds < \, \) \(\ds \dfrac {k' + \ell' + 2} {4 n + 4}\)
\(\ds \dfrac {\map h {m, n} - 1} {n + 1}\) \(<\) \(\, \ds x_m + y_m \, \) \(\, \ds < \, \) \(\ds \dfrac {\map h {m, n} + 1} {n + 1}\)

Thus, $\sequence {x_n + y_n}$ is computable by definition.

$\blacksquare$


Proof 2

By Computable Real Sequence iff Limits of Computable Rational Sequences, there exist:

such that:

$\forall m, p \in \N: \forall n \ge \map {\psi_x} {m, p}: \size {a_{\map {\phi_x} {m, n}} - x_m} < \dfrac 1 {p + 1}$
$\forall m, p \in \N: \forall n \ge \map {\psi_y} {m, p}: \size {a_{\map {\phi_y} {m, n}} - y_m} < \dfrac 1 {p + 1}$


By Computable Subsequence of Computable Rational Sequence is Computable/Corollary, there exist:

Computable rational sequences $\sequence {a'_k}, \sequence {b'_k}$

such that, for all $m, n \in \N$:

$a_{\map {\phi_x} {m, n}} = a'_{\map \pi {m, n}}$
$b_{\map {\phi_y} {m, n}} = b'_{\map \pi {m, n}}$

By Sum of Computable Rational Sequences is Computable:

$\sequence {a'_k + b'_k}$

is a computable rational sequence.

Define $\psi : \N^2 \to \N$ as:

$\map \psi {m, p} = \map \max {\map {\psi_x} {m, 2 p + 1}, \map {\psi_y} {m, 2 p + 1}}$

which is total recursive by:

If we can show:

$\forall m, p \in \N: \forall n \ge \map \psi {m, p}: \size {\paren {a'_{\map \pi {m, n}} + b'_{\map \pi {m, n}}} - \paren {x_m - y_m}} < \dfrac 1 {p + 1}$

then the result will follow by Computable Real Sequence iff Limits of Computable Rational Sequences.


Let $m, n, p \in \N$ be arbitrary, and suppose that $n \ge \map \psi {m, p}$.

Then:

$n \ge \map {\psi_x} {m, 2 p + 1}$
$n \ge \map {\psi_y} {m, 2 p + 1}$

Thus, by assumption, we have:

$\size {a_{\map {\phi_x} {m, n}} - x_m} < \dfrac 1 {2 p + 2}$
$\size {b_{\map {\phi_y} {m, n}} - y_m} < \dfrac 1 {2 p + 2}$

Hence:

\(\ds \size {\paren {a'_{\map \pi {m, n} } + b'_{\map \pi {m, n} } } - \paren {x_m - y_m} }\) \(=\) \(\ds \size {\paren {a'_{\map \pi {m, n} } - x_m} + \paren {b'_{\map \pi {m, n} } - y_m} }\)
\(\ds \) \(\le\) \(\ds \size {a'_{\map \pi {m, n} } - x_m} + \size {b'_{\map \pi {m, n} } - y_m}\) Triangle Inequality for Real Numbers
\(\ds \) \(=\) \(\ds \size {a_{\map {\phi_x} {m, n} } - x_m} + \size {b_{\map {\phi_y} {m, n} } - y_m}\) Definitions of $\sequence {a'_k}, \sequence {b'_k}$
\(\ds \) \(<\) \(\ds \frac 1 {2 p + 2} + \frac 1 {2 p + 2}\) Above
\(\ds \) \(=\) \(\ds \frac 1 {p + 1}\)

$\blacksquare$

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