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Amplitude amplification increases the probability of obtaining an answer accepted by a specified Boolean function. It requires a coherent preparation $A$, its inverse, and a marked-state phase oracle that negates accepted components. Alternating that sign change with a reflection operator about the prepared state directs interference toward the accepted subspace. Brassard, Høyer, Mosca, and Tapp developed the general framework, extending Grover's algorithm beyond a uniform quantum superposition. ^[ref].

If a normalized quantum state has good amplitude $\sin \theta$, the product of the reflection in the bad axis and the reflection in the state line rotates its good-bad plane by $2\theta$. After $k$ iterations the good probability isso $O(1/\sin \theta )$ iterations raise a small good amplitude close to one. Success subsequently oscillates, making the stopping time part of the algorithm.

Compute the exact Grover search on four entries directly. Order the computational basis as $00,01,10,11$ and mark $11$. The Hadamard transform prepares amplitudes $(1/2,1/2,1/2,1/2)$. The marked-state phase oracle changes these to $(1/2,1/2,1/2,-1/2)$, whose arithmetic mean is $\overline{a}=1/4$. The Grover diffusion operator performs inversion about the mean:The three unmarked amplitudes become $1/2-1/2=0$, while the marked amplitude becomes $1/2-(-1/2)=1$. All three stages have squared norm one. Here $\sin \theta =1/2$, so $\theta =\pi /6$ and one iteration gives ${\sin }^2(3\theta )=1$, agreeing with the explicit calculation.

Grover's original search paper was submitted on 29 May 1996; its algorithm combines selective sign changes with inversion about the mean. The example exposes the beauty of that construction: a change initially invisible to computational-basis measurement becomes complete cancellation of every wrong answer. ^[ref].

The marked-state phase oracle is an assumed operation, whose implementation cost must be counted. Computational-basis measurement yields a candidate label, here $11$; using it to retrieve a record requires the relevant data access. Preparing a quantum superposition does not make all its entries available as classical outputs. The improvement concerns the number of oracle queries needed to obtain an accepted label.

Starting from $0<a_1\le b_1$, replace the lower endpoint by the geometric mean and the upper endpoint by the arithmetic mean. The arithmetic-geometric mean inequality gives $a_n\le a_{n+1}\le b_{n+1}\le b_n$. The bounded monotone sequence theorem gives limits, and the arithmetic recurrence forces them to coincide. The limiting value is the arithmetic-geometric mean of the initial pair.

For a nonempty finite list of real numbers $x_1,\dots ,x_n$, the arithmetic mean isIt is the weighted mean with equal weights, giving each observation equal influence. For $(2,4,9)$, the sum is $15$ and the arithmetic mean is $5$. Replacing every value by $5$ retains the same total. The residuals $-3,-1,4$ sum to zero, so the mean is also the balancing point of these equally weighted observations.

The arithmetic mean minimizes the sum of squared deviations by the weighted mean minimizes squared deviations identity with all weights one. A large input can consequently have a substantial effect: replacing $9$ by $90$ changes the mean to $32$, while the median remains $4$. The choice of summary depends on what aspect of the data matters.

The arithmetic mean of memory access latencies. Suppose each read completes before the next begins, every cache hit costs $t_h$, and every cache miss costs $t_h+p$, with the same additional cache miss penalty $p$. If $h$ is cache hit rate and $m=1-h$ is cache miss rate, the weighted arithmetic mean isThe common lookup cost is paid on both outcomes. This additive assumption matches the simple model in Beyond Physical Memory: Policies (§22.1, equation 22.1, pp. 1–2), where main computer memory acts as a memory cache for memory pages backed by disk; here the same accounting is applied to a toy data memory cache.

Assign synthetic values $t_h=2\mathrm{ns}$ and $p=30\mathrm{ns}$. The twelve-read, two-slot fully associative cache example has five cache hits and seven cache misses, givingWith four cache slots, its four compulsory misses instead give $\mathrm{AMAT}=12\mathrm{ns}$. This comparison holds the timing costs fixed as capacity changes, which is a model assumption rather than a hardware promise.

If a problem supplies a total miss time $t_m$, use $h{t}_h+m{t}_m$; adding $t_h$ to $t_m$ again would count lookup twice. Variable cache miss penalties, multiple consulted levels, or overlapping requests require a richer timing model. Real CPU work can overlap a pending read, so this arithmetic mean cannot by itself predict program speed or stall time. ^[ref]

For an integrable function on an interval $I$ of positive finite length, its average value is the integral divided by that length:If $f$ is constant on equal-length subintervals, this is their values' arithmetic mean; unequal lengths require a weighted mean. For example, values $2$ and $6$ on two equal halves give average value $4$, whereas value $2$ on the first quarter and $6$ on the remaining three quarters give $5$. The integral measures total contribution; the average value of a function divides out the interval length.

For the function $f(x)=2x$ on $[0,1]$, the average value of a function is ${\int }_0^{1}2xdx=1$. Replacing it by the constant $1$ preserves its integral on this interval. This is why the Haar projection uses cell averages to obtain a coarser description while retaining each cell's total contribution.

A categorical distribution describes one trial with one of $k$ labelled outcomes and probabilities $p_1,\dots ,p_k$ summing to one. The labels need not be numbers or have an order. It is the single-trial building block of the multinomial distribution: $n$ independent draws with the same categorical distribution produce multinomial category counts.

For a trial choosing labels $A,B,C$ with probabilities $1/2,3/10,1/5$, the category-indicator random vector is $(1,0,0)$, $(0,1,0)$, or $(0,0,1)$. Its expected value is $(1/2,3/10,1/5)$; its components are dependent because exactly one is one. A numerical encoding of $A,B,C$ is a separate modelling choice: the category labels themselves do not define an arithmetic mean.

The Cesàro mean of the first $n\ge 1$ terms of a sequence is their arithmetic mean. It is a cumulative average: increasing $n$ includes the whole initial segment, rather than a fixed-length sliding window. For $a_k=(-1{)}^k$, the even-length averages are zero and the odd-length averages are $-1/n$, so they tend to zero even though the sequence does not converge.

For a periodic sequence with period $p\ge 1$ and one-cycle arithmetic mean $M$, split an initial segment into full cycles and a remainder of fewer than $p$ terms. The full cycles contribute exactly $M$ per term; the remainder contributes a fixed bounded error divided by $n$. Thus the Cesàro means tend to $M$. The cycle $0,2,1$ in the bounded partial sums need not be Cesàro summable illustration has arithmetic mean one. Treating it as a sequence of partial sums gives a series with first-order Cesàro sum one; averaging a sequence and summing its terms remain different operations.

Cesàro summation assigns a value to a series when the arithmetic means of its partial sums approach a finite limit of a sequence. Here we use the first-order method, often written $(C,1)$. For a real or complex series beginning at index zero, defineIt is Cesàro summable to $L$ when ${\sigma }_N\to L$. The averaging is applied to the partial sums $s_n$, rather than to the original terms $a_n$. This distinction is essential: the terms can average to zero while their accumulated sum keeps growing. ^[ref]

For Grandi's series $1-1+1-1+\cdots$, the partial sums alternate between one and zero. The series is not a convergent series, but the two values occupy asymptotically equal proportions of the averaging window. In the declared index convention,so its first-order Cesàro sum is $1/2$. This names a particular summation method; it does not turn the oscillating partial sums into an ordinarily convergent sequence.

Why is this useful? An oscillating approximation can retain stable information when its cumulative average settles. Because the Cesàro theorem for convergent sequences preserves an existing limit of a sequence, this adds some new cases without changing the sum of any ordinarily convergent series. For Fourier series, averaging the Fourier partial sums gives a Fejér sum; the existing Fejér kernel and approximation articles explain its effect on reconstruction. Averaging does not guarantee success for every oscillating sequence, as the bounded partial sums need not be Cesàro summable example shows.

One historical motivation was the multiplication of series. In 1918, A. F. Andersen recalled Ernesto Cesàro's product theorem and cited his 1890 publication at the start of a study of products involving Cesàro summable series. This connects the theory with preserving useful operations on sums, as well as with assigning a value to an isolated example. His paper treats several orders of summability; the examples here use only $(C,1)$.^[ref]

Let $P$ be the Bernoulli distribution with success probability $0.2$ and $Q$ the Bernoulli distribution with success probability $0.8$. For $0\le s\le 1$,The two summands have product $0.16$. Their arithmetic mean is at least their geometric mean, with equality at $s=1/2$. ThusThe same argument gives $C=-\ln (2\sqrt{p(1-p)})$ for complementary parameters $0<p<1$. At $p=1/2$ the laws agree and $C=0$.

With equal prior probabilities, count the successes $K$ among $n$ observations. The likelihood ratio is $Q^{\otimes n}/P^{\otimes n}=4^{2K-n}$, so choose $Q$ when $K>n/2$ and $P$ when $K<n/2$. If $n$ is even, either decision or a fair tie randomization gives the same average error at $K=n/2$. Under $P$, $K$ has the binomial distribution with parameters $n,0.2$. Symmetry therefore gives the exact optimal average errorThe last term applies only for even $n$; it represents half the tie probability. Direct finite sums give $p_1^{\ast }=0.2$, $p_5^{\ast }=0.05792$ and $p_{20}^{\ast }\simeq 0.00157912$. Twenty observations reduce the optimal average error to about 0.158%. The finite-sample Chernoff error bound at $n=20$ is $\frac{1}{2}(0.8{)}^{20}\simeq 0.00576461$, a guarantee rather than the exact error.

Reducing the exact error from $p_{20}^{\ast }$ to $p_{50}^{\ast }\simeq 1.29363\times 10^{-6}$ illustrates the value of repeated evidence. The example assumes a single fixed coin law throughout the experiment. If repeated sensor readings share an unmodelled common error, increasing their number need not supply this much independent information.

The continuous angular geometric mean of the positive boundary complex modulus isHere the empty product is one. This differs from the angular arithmetic mean $A_f(R)=(2\pi {)}^{-1}{\int }_0^{2\pi }|f(R{e}^{i\theta })|d\theta$. In general $G_f(R)\le A_f(R)$ by the concavity of the logarithm; equality requires constant boundary complex modulus. For a holomorphic function with no zeros of a function in the closed disk, Jensen's formula gives $G_f(R)=|f(0)|$, even when the boundary complex modulus varies with angle. Thus replacing the mean of a logarithm by the logarithm of an arithmetic mean loses the exact zero identity.

Suppose disjoint nonempty groups have sizes $n_j$ and arithmetic means ${\bar{x}}_j$. The arithmetic mean of all observations isEach group contributes the sum $n_j{\bar{x}}_j$; add those totals before dividing by the overall count. This is exactly a weighted mean of group means with weights equal to group sizes. ^[ref]

Take a synthetic example with two observations equal to $90$ and eight equal to $60$. The two group means are $90$ and $60$, and the combined mean is $66$. Giving the two groups equal weights instead produces $(90+60)/2=75$, which summarizes the two group means with equal influence. To summarize all ten individuals with equal influence, use their group counts.

The group means and counts determine the combined arithmetic mean, but do not generally determine the combined variance. Variation within the groups is additional information.