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Integer arithmetic in which storage grows with the number of numerical digits needed, instead of imposing a fixed word size. It allows exact addition, multiplication, and exponentiation on large integers, subject to available memory; operation costs grow with operand length.

A Bernoulli differential equation is a first-order ordinary differential equation of the displayed form, with $P$ and $Q$ continuous functions on an interval and a fixed real exponent $n$. A nonlinear power can hide a first-order linear differential equation: on a suitable branch, the Bernoulli power substitution $z=y^{1-n}$ removes it exactly. The transformed solution of a differential equation can be completely regular even when recovering $y$ produces finite-time blowup.

For example, the reciprocal Bernoulli equation with finite-time blowup $y^{\prime }+y=e^x{y}^2$ becomes $z^{\prime }-z=-e^x$ under $z=1/y$. The initial conditions $y(0)=1/2$ and $y(0)=-1$ select a positive trajectory blowing up at $x=2$ and a negative trajectory tending to zero as $x\to \infty$, respectively. The integrating factor exposes both behaviors with the same short calculation.

The cases $n=0$ and $n=1$ are already first-order linear differential equations: respectively $y^{\prime }+Py=Q$, with the zeroth-power term interpreted as the constant $1$, and $y^{\prime }+(P-Q)y=0$. There is no need for a nonlinear change of variable; for $n=1$, $y^{1-n}=1$ would discard the unknown entirely. For other exponents, the substitution must respect the domain of exponentiation and may lose the zero solution of a differential equation.^[ref]

For arbitrary real $n\ne 0,1$, work first on an interval where $y>0$. Define exponentiation by $y^r=\exp (r\log y)$, using the real logarithm. The change of variable is differentiable and invertible there, and the chain rule givesMultiplying the Bernoulli differential equation by $(1-n)y^{-n}$ yields the first-order linear differential equationThis is an exact transformation of nonzero solutions of a differential equation on the chosen branch; it is valid far from any reference state and does not discard higher powers as a local linearization would.

Given the initial condition $y(x_0)=y_0>0$, the integrating factor and transformed solution of a differential equation areThe product rule verifies $(\mu z{)}^{\prime }=(1-n)\mu Q$. Recover $y=z^{1/(1-n)}$ only on the connected component of $\{x:z(x)>0\}$ containing $x_0$, inside the original coefficient interval. The first-order linear differential equation can have a longer existence interval than this inverse transformation allows.^[ref]

Negative values need a separately justified real branch. For integer $n$, the required powers are real and differentiable for $y<0$ too, but the inverse must retain the sign selected by the initial condition. For $n=3$, $z=y^{-2}>0$ represents either $y=+1/\sqrt{z}$ or $y=-1/\sqrt{z}$; both square roots are taken of positive $z$. For $n=2$, $z=1/y$ retains the sign directly. For a general noninteger exponent, use the positive branch unless a specific differentiable real negative branch has been established. Never evaluate a real fractional power of a negative value merely because the linear formula has become negative.

Division by $y^n$ excludes $y=0$. When $n>0$ and the original real power is defined at zero, $y\equiv 0$ is an additional solution of a differential equation and must be checked in the original equation. When $n<0$, the equation as written is undefined at zero. On a positive branch the right-hand side is locally Lipschitz in $y$, so the Picard-Lindelöf theorem gives local uniqueness. At zero, the same hypothesis needs a fresh check: it holds for the polynomial $n=2$ example, whereas for $0<n<1$ a nonzero power coefficient prevents local Lipschitz continuity there. If $z\downarrow 0$, the inverse sends $y\to \infty$ for $n>1$, but sends $y\downarrow 0$ for $0<n<1$. The latter endpoint can lead to finite-time extinction in a sublinear Bernoulli equation, where the original equation determines how to join to zero.

This extends real exponentiation.

For nonzero complex $z$ and complex $w$, the multivalued power is

An exponent specifies the power in exponentiation: in $a^b$, $a$ is the base and $b$ is the exponent. For a positive integer exponent, $2^3=2\cdot 2\cdot 2=8$. For a positive real number base, a real exponent is defined by $a^b=\exp (b\log a)$ using the exponential function and natural logarithm. The Lp norm uses its exponent $p$ to determine how strongly large absolute values affect the result.

For a nonnegative integer exponent, exponentiation iterates multiplication: $a^0=1$ and $a^{b+1}=a{a}^b$. This convention defines $0^0=1$ in primitive recursive arithmetic. Extensions to other exponents may use different domains and conventions.

A geometric progression is a sequence obtained by repeatedly multiplying by one fixed common ratio $r$:Here $a$ and $r$ may be real numbers or complex numbers. The recurrence is meaningful even when $a=0$ or $r=0$; expressing $r$ as a ratio of successive terms requires a nonzero denominator. As a polynomial convention, $r^0=1$, including at $r=0$.

For example, quantities $100,80,64,51.2,\dots$ retain four fifths of their preceding value at every step. The decrement changes, but the retained fraction does not. With the same initial value and common ratio $2$, the terms are $100,200,400,800,\dots$. This multiplicative structure is useful whenever the same proportional rule repeats: it turns many updates into the single exponentiation $r^n$. An arithmetic progression instead repeats an additive increment.

For nonzero $a$, the modulus of $a_n$ tends to zero when $|r|<1$, stays fixed when $|r|=1$, and grows without bound when $|r|>1$. For a negative real common ratio, successive nonzero terms also alternate in sign. Thus the terms $1,-1/2,1/4,-1/8,\dots$ decay in size while crossing zero at every step. Adding the terms gives a geometric series; the individual terms and their accumulated partial sums are different sequences.

The hyperoperation hierarchy successively iterates basic arithmetic operations. In the usual indexing, successor is followed by addition, multiplication, exponentiation and their higher iterates. A sequence beginning with addition shifts this index by one. Each fixed rank is a primitive recursive function; this does not imply that the evaluator taking the rank as an additional variable is primitive recursive.

For (1)(a), a primitive recursive function is obtained by a finite construction from the zero function, successor function and projection functions, using function composition in recursion theory and primitive recursion. The recursion rule, with parameters $\mathbf{x}$, isHere $g,h$ have already been constructed. All these functions are total on nonnegative integers: the initial functions are total, composition preserves totality, and ordinary mathematical induction proves totality of each recursion.

For (1)(b), use a shifted hyperoperation sequence, beginning with addition. Defineand, for each fixed $r\ge 0$, defineThe first three members are therefore addition, multiplication and exponentiation:The exponentiation convention includes $a^0=1$, hence $0^0=1$. The next member iterates exponentiation, with $H_3(a,0)=1$; there is no need for an exceptional domain restriction at $a=0$.

For (1)(c), $H_0$ is a primitive recursive function by its displayed recursion. If $H_r$ is a primitive recursive function, thenare primitive recursive functions, using constants, projection functions and function composition in recursion theory. Primitive recursion applied to $g_r,h_r$ gives $H_{r+1}$. Mathematical induction on the fixed rank $r$ proves every member of the sequence is primitive recursive. This does not assert that the joint three-variable evaluator $(r,a,b)\mapsto H_r(a,b)$ is a primitive recursive function.

For (2), as written, the answer is no: the empty set $X=\varnothing$ is semidecidable, whereas a total computable function from $\mathbb{N}$ to $\mathbb{N}$ always has a nonempty range. This is the only obstruction. To prove the stronger useful statement, suppose $X$ is a nonempty computably enumerable set and choose $x_0\in X$. Fix a Turing machine which halts exactly on the inputs in $X$. Its bounded halting predicateis primitive recursive: encode configurations arithmetically, iterate its total single-step operation $t$ times by primitive recursion, and check the halt state. A halted configuration is kept fixed. This is a bounded simulation, not unbounded waiting.

Fix a primitive recursive pairing function $⟨x,t⟩$ with primitive recursive inverse coordinate functions. For example, the Cantor pairing function has inverses obtained by bounded minimization. DefineThe two coordinate functions, the bounded predicate, and the finite case distinction are primitive recursive functions, so $f$ is a primitive recursive function. Its values always lie in $X$. Conversely, if $x\in X$, its computation halts within some finite $t$, and $f(⟨x,t⟩)=x$. ThusThe range need not be a computable set; deciding whether a value occurs anywhere in this total enumeration is an unbounded search.

Take the base-ten logarithm of both positive sides. The exponentiation rule for logarithms gives $x{\log }_{10}2=y$, hence

In this positive integer context, a power of two is $2^k$ for a nonnegative integer $k$, giving $1,2,4,8,\dots$. Allowing negative or real exponents gives a broader meaning of exponentiation, which is not intended in the integer divisibility criteria here.

On an interval where the substitution is real and $x\ne 0$, a factor formed by exponentiation can remove inverse powers from the coefficients of a linear ordinary differential equation. Forthe substitution $y=x^{n}z$ givesIt becomes a constant-coefficient differential equation exactly when $2n+b=0$, $an+d=0$ and $n(n-1)+bn+e=0$. Derive this by the product rule; checking only the coefficient of $z^{\prime }$ is insufficient.

A mathematical expression stored in a spreadsheet cell, with references to other spreadsheet cells. In this example =A1+A2 asks for addition, =2*A1 for multiplication, and =A2^2 for exponentiation. The evaluator in spreadsheet dependency evaluation assumes these spreadsheet formulas have already been represented by their ordered lists of referenced spreadsheet cells and operations on the resulting numbers; it does not execute arbitrary formula text.

Copying a spreadsheet formula produces a new expression by adjusting its spreadsheet cell references. Copying only its displayed value produces a number instead. For example, copying the spreadsheet formula =A2*B2 down one row gives =A3*B3; entering its former value $15/2$ in the next row would give neither the required relationship nor this example's correct subtotal $7$.

Repeated exponentiation with right association: ${}^{1}a=a$ and ${}^{h+1}a=a^{{}^{h}a}$. For example, ${}^{3}a=a^{a^a}$, rather than $(a^a{)}^a$. Computing its remainder modulo an integer often uses a chain of smaller exponent moduli; applying Euler's theorem requires checking its coprimality hypothesis at every level.