Deleting the degree-four and degree-five Chebyshev components of gives the displayed cubic. The deleted terms are , whose supremum norm on is exactly , attained at . This bounds the additional polynomial change. It does not make the total uniform error for finite near its singular endpoint .
At zero the indicial equation is . The two Frobenius series have exponents zero and one half, with recurrenceThe denominators never vanish for these exponents. Unless a series terminates its coefficient ratio tends to , giving convergence within distance two of zero. If is a positive integer, the exponent-zero series terminates at degree . The substitution gives the standard Chebyshev differential equation, and this solution is a multiple of . For half-integer , termination can instead occur in the power-series factor of the exponent-one-half solution.
A field has characteristic three when . The binomial identity then becomes , and . Thus a cyclic shift operator on three coordinates has a repeated-root minimal polynomial in this characteristic and cannot be diagonalizable.
In the conjugacy class order , the five complex irreducible characters of the alternating group areThe class sizes are . Restricting the degree-four and degree-five irreducible characters of gives rows of squared character inner product one. The degree-six restriction has squared norm two and is orthogonal to those rows and the trivial row, so it splits into the two remaining irreducibles. The sum of squared degrees forces their degrees to be three. Orthogonality forces their last two values to have sum one and product minus one, producing the displayed roots. Since a normal subgroup is a union of classes and its order divides sixty, these class sizes also prove the Simplicity of the alternating group A5.
A group of order thirty-six with a normal and a cyclic quotient of order four acting freely on its eight nonzero elements has six conjugacy classes of sizes . Its four linear characters are lifted from the cyclic quotient. The remaining irreducible degrees are four and four, by the sum of squared degrees. Column orthogonality forces the last three values of both degree-four rows to vanish. The square of the order-four action is inversion on , making the first two nontrivial class values real. Orthogonality then gives sum minus one and product minus two, so the rows are and . This illustrates why reality must be justified when reconstructing a character table from orthogonality.
For the nontrivial action of on , the conjugacy class sizes are . Three linear characters factor through the quotient , and the remaining two irreducible characters have degree three. Put and let . With suitable class labels the table isOne can derive the last two rows by inducing nontrivial characters of along the two orbits of the order-three automorphism on their exponents. Their degree-three orbit sums vanish off . On a generator they give the two sums of seventh roots over quadratic residues and nonresidues. Their sum is and their product is : the nine pairwise exponent sums between the two triples contain zero three times and every nonzero residue once. Thus they are the roots of . Alternatively column orthogonality forces the zeros on the centralizers of order three, and orthogonality with the identity column interchanges and in the last row. This finite table illustrates how centralizer sizes and a small number of given values determine missing character values.
For a finite group, the sum of products of irreducible character values is zero when are not conjugate, and equals when are conjugate. To prove this from row orthogonality relations for a character table, choose representatives of its conjugacy classes and form . Row orthogonality says . There are equally many irreducible characters and conjugacy classes, so this is a square unitary matrix and as well. Its column equations give the result, using .
Combine finitely many discrete memoryless channels with disjoint input alphabets and disjoint output alphabets into a block-diagonal channel. The block label is determined by both input and output. If it has law , the mutual information is at most , attained by using each block's capacity-achieving conditional input law. Put and . The expression equals , so Gibbs inequality proves the displayed capacity, attained for block probabilities . Even blocks of capacity zero contribute distinguishable block choices; three disjoint zero-capacity blocks give capacity .
Suppose one row of a finite discrete memoryless channel is a convex combination of other rows: . Deleting input leaves the channel capacity unchanged. Given any input law with mass on , move that mass to the other symbols in proportions . The output law stays the same. By the concavity of information entropy, , so this replacement cannot increase conditional entropy and cannot decrease mutual information. Taking maxima proves the claim; the smaller channel is also trivially a restriction of the original one. If the entropy inequality is strict, every positive use of the redundant row is strictly suboptimal compared with its redistribution.
Pass the output of a discrete memoryless channel of channel capacity through an independent finite-alphabet erasure channel with erasure probability . Its capacity becomes .
To prove it, let be the original input, the original output, the erasure indicator, and the final output. Since the erasure marker is distinguishable, is determined by . It is independent of , so the chain rule for mutual information gives . Maximization over the same input distributions proves the formula. For the maximizing distributions are exactly the original channel's maximizers. For , every input distribution attains the zero capacity. Input-dependent erasure need not satisfy this scaling identity.
Adjustment of cerebral vascular resistance that moderates changes in blood flow as perfusion pressure varies. It has finite physiological limits and is not a guarantee of constant flow.
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