Real zeros and merging poles

Reconstructed and self-checked by GPT-6 Astra (OpenAI), Ultra reasoning effort, October 2026. Earlier edition: GPT-6.1 Sol (OpenAI), Ultra reasoning effort. Original exposition: CC0. Supplied foundations retain their stated licences.

At a simple real zero, approaching a reciprocal from above or below leaves the same principal value and opposite point masses. On the line, a critical zero prevents either limit: one fixed positive test detects the divergence. When two simple poles merge, a different question arises. Subtracting explicit point derivatives restores a limit, whose finite part has a sharp test order.

Pairings are complex linear. The complete individual pole proof is U013, Theorem 4.1 and formula (4.5); it includes both signs, symmetric principal values and their bounds. The smooth inverse theorem and absolute coordinate densities are proved in U037, Lemmas 0.1–0.2 and Theorem 1.1. Distribution continuity is U008, Proposition 1.2. We use the scalar foundation, §§12, 13.1–13.5 and 13.7–13.10, and the integration foundation, §§15.0–15.2 and 16, for compactness, calculus, cutoffs, trigonometry and dominated convergence.

We will need an explicit Taylor identity. For a smooth scalar and integer , repeated FTC gives For this is the FTC along the segment . To pass from to , substitute and integrate on the triangle . The constant term is , and the remaining weight is . This proves the formula, including negative . It also proves smooth division by when those first jets vanish, with compact derivative bounds obtained by differentiating the displayed finite integral.

A simple zero is exactly what the limit requires

Theorem 1.1 (the zero criterion on the line). Let and for nonzero real . Either one-sided distributional limit exists if and only if every zero is simple: . In that case both limits exist, the zeros are locally finite, and defines a distribution. With the limit from positive and from negative , Each compact test meets finitely many terms. If there are no zeros, both limits equal the smooth function .

Proof: simple zeros. U037 Lemma 0.1 gives an inverse interval at each simple zero, isolating it. If infinitely many zeros lay in a compact interval, a subsequence of distinct zeros would converge to . Continuity gives , and their difference quotients at are zero, forcing . This contradiction proves local finiteness.

Fix a compact support . Choose finitely many inverse intervals around the zeros meeting , and smooth cutoffs supported there with sum one near those zeros. The explicit finite bump construction of U037's Theorem 1.1 supplies these. The remaining test piece is supported on a compact set where has positive minimum, so dominated convergence applies to its quotient.

For the inverse on one interval, absolute substitution gives The transformed test extends smoothly by zero. U013 Theorem 4.1 gives . At the chart zero , Symmetric deletion is exactly the deletion , so summing gives (1.1)–(1.2). On this fixed support, the transformed norms are bounded by . U013's pole estimate and the bounded separated density therefore prove distribution continuity. The limits of the same smooth functions are independent of all cutoffs.

Proof: a critical zero. Suppose . Fix one nonnegative compact smooth , equal to one for . For every , differentiability gives such that If , then the interval lies where and . Hence All other contributions have the same nonnegative sign. This is a lower bound for every sufficiently small positive , for each ; thus the fixed pairing diverges. The negative side reverses the imaginary sign and gives the same obstruction. This includes flat zeros and intervals on which vanishes.

Example 1.2 (dimension matters). In , both limits for exist and equal , despite its critical zero. Here is a direct integrability proof. On the integrand is at most , and the region lies in a cube of volume . Its integral is therefore at most , a summable geometric series. The origin has measure zero; away from it the function is continuous. Since almost everywhere, dominated convergence on every compact test proves both limits. The zero criterion is a one-dimensional result.

Exact subtractions when two real poles merge

Order at most means that on each fixed compact test support the pairing is bounded by derivatives through , with a support-dependent constant. The same integer must work for all compacts. Exact order is the least such integer.

Theorem 2.1 (all even powers). For an integer and , let . Then For , define The complete renormalized limit is On any fixed compact test support, the remainder is bounded by a norm. Each , , has support and exact order . The limit has full support, singular support , and exact order .

Proof: boundary values and symmetric deletions. The only real zeros are , with absolute slopes ; Theorem 1.1 gives (2.1) with its canonical deletion. We check that ordinary symmetric deletions at the two poles agree with it. At a simple zero of a smooth real , Taylor division gives with smooth . Thus for a smooth test the quotient equals plus a smooth remainder near . The inverse of satisfies ; the twice-FTC formula gives . Consequently deletion has two radii . Their ratio is , so the singular term's logarithmic imbalance is . The bounded remainder loses an interval of length . The difference tends to zero, proving the deletion claim.

Proof: every coefficient. Put . Substitution before the boundary limit, with imaginary parameter , gives For , means the two local principal values, the point terms, and ordinary integrable tails. It is well defined, since the rational density has decay at least .

Let , , and . These are the distinct roots of : their powers follow from the proved exponential period, and there can be no additional root. Indeed a root of a polynomial gives the factor by the identity ; induction bounds the number of distinct roots by the degree. The same argument proves the partial fractions To check this, multiply by the denominator. Both sides become polynomials of degree at most . At the root , the right-hand value is , because . Thus their difference has roots and is zero.

For , , the real logarithmic primitive and imaginary arctangent primitive give The real integral is , which tends to zero. The imaginary integral is that of ; substitution by and the proved arctangent limits give . For a real root the symmetric principal-value integral on is a logarithmic endpoint ratio tending to zero. Removing small intervals about the other real roots changes the nonsingular terms by quantities tending to zero, so finite summation of (2.5) is valid under these principal values.

The full sum of the is zero by the finite geometric identity. The two real terms cancel since is odd. Hence the lower nonreal sum is the negative of the upper sum. Write ; then , , and The last equality follows by factoring from numerator and denominator and using the sine/cosine formulas. When , and both sides are zero. Equation (2.6) now gives The two masses of contribute , since . Dividing this complete moment by gives .

Proof: the remainder and the finite part. Since and are even, let All derivatives of through vanish at zero: the odd ones vanish by evenness. The proved Taylor formula gives , with smooth near zero. Outside a compact set , of degree at most , so is integrable on the full line.

Linearity of the explicitly defined polynomial moments, together with (2.4), makes the left side of (2.3), paired with , exactly . This notation uses local principal values and those integrable rational tails; no action of arbitrary distributions on arbitrary noncompact functions is being introduced.

Choose a fixed even compact cutoff near zero, supported where the smooth quotient is defined, and put . Then For small , the second term is separated from both poles. There the denominator is uniformly comparable to . Dominated convergence gives More precisely, the difference is the ordinary integral of bounded by on a fixed compact test-support class: it is away from zero, and the tail numerator grows at most as .

Multiplication of (2.1) by gives one; the principal value becomes the ordinary test integral and the point terms vanish. Consequently There is a constant independent of the support of such that To see this, choose fixed compact cutoffs near . In each, the simple-zero change of variables and U013's pole bound use only those two norms, with fixed coefficients. The complementary rational density is absolutely integrable, as it is bounded away from its poles and decays like . Its pairing is bounded by its norm times .

For , the right side is bounded by . The Taylor integral gives . Thus the last term of (2.7) is , with that norm. We have proved (2.3), its quantitative remainder, and the formula The near-zero Taylor bound and the integrable polynomial tails also bound this functional by a norm on every fixed compact support.

For identification with (2.2), integrate (2.8) by parts times on the two intervals excluding zero, with finite outer endpoints. At each stage the inner boundary term tends to zero because . The outer boundary term tends to zero because has degree at most . The resulting integral is Indeed , and the derivative of the even part is the odd part of , which has the same symmetric integral divided by . The definition of distributional derivatives now gives exactly the negative derivative formula in (2.2), with no extra point term.

Proof: sharp orders and supports. Each , because its imaginary part is . The action of bounds its order by and detects its support at zero. For , choose a fixed compact with . The tests have a common compact support and bounded derivatives through , while their -th derivative at zero grows as . This proves exact order; has order zero.

Formula (2.8) already proves the order upper bound . For the lower bound choose even smooth , with compact near zero, and for , for . Such functions follow from the supplied bumps. Set They are smooth because they vanish near zero. On the transition annulus , each derivative of order is bounded by a constant times , by the product rule. Away from that annulus the fixed cutoff gives uniform bounds. All jets at zero vanish, so (2.8) yields The transition contributes a bounded integral by , the region from to a fixed radius gives the displayed logarithm, and the outer cutoff contributes a constant. Hence order is impossible.

Off zero the functional is the nonzero smooth density . Its support is therefore the full line, and all singular support lies at zero. The same lower-order test argument can be placed in any neighborhood of zero, excluding a smooth density there. Thus its singular support is exactly .

Exercises

Exercise 1 (basic). Compute both boundary limits for , their real principal value and their jump.

Exercise 2 (intermediate). Determine the jump for . Prove that it is a distribution on compact tests and that no tempered distribution agrees with it there.

Exercise 3 (intermediate). Set , for . Prove smooth flatness and give a quantitative lower bound, using one fixed compact nonnegative test, excluding both boundary limits.

Exercise 4 (intermediate). On , with real and , compute and both boundary limits for .

Exercise 5 (intermediate). Give every counterterm, coefficient and limiting distribution in the quartic merger , together with their exact orders and supports.

Exercise 6 (advanced). Renormalize , retaining every inverse-coordinate coefficient at , the limit and the exact orders.

Exercise 7 (intermediate). For an integer and real , prove , bound its remainder, and compute its exact imaginary part.

Exercise 8 (advanced). Classify the even distributions homogeneous of degree and satisfying . What changes if evenness is dropped?

Complete solutions

Solution 1. The roots have absolute slopes four. The ordinary identity and the deletion comparison in Theorem 2.1 give The opposite ordinary slopes give the same positive absolute weight in both point masses.

Solution 2. The zeros are , , with . Thus Its locally finite sum defines a distribution: on a fixed compact support only finitely many evaluations occur, bounded by a constant times the test supremum.

For completeness, the Schwartz space here has seminorms A continuous linear functional on that space has for some . Indeed continuity at zero gives a neighborhood specified by finitely many seminorms on which ; their increasing family is controlled by one . Rescaling any nonzero test by proves the bound.

Choose a nonnegative compact smooth , supported in , with , and set . The jump pairing has magnitude , whereas This follows because the derivatives are fixed translates and their supports stay within a fixed distance of . As , the exponential dominates that polynomial: from its positive series for , the ratio tends to infinity. The continuous-functional bound is impossible, so there is no tempered extension.

Solution 3. Each derivative off zero is a polynomial in times , by induction using the product and chain rules. These derivatives, and their quotients by , tend to zero. To verify this last statement, set ; for any fixed power , choose an integer and use . This gives . Extending each derivative by zero at zero, the quotient limit proves recursively that it is the derivative of the preceding extension. Thus is smooth and every derivative at zero vanishes.

Fix , compact smooth and equal to one near zero. For sufficiently small , the interval lies in that neighborhood and has . Therefore Writing makes the final expression , whose divergence follows from the same series bound. The negative side changes the imaginary sign and is excluded by this very same test.

Solution 4. On , and , so zero is the only root and its slope is one. Off zero, The deletion comparison already proved in Theorem 2.1 applies to this simple zero; the logarithmic imbalance tends to zero. Hence on The second summand is smooth and remains part of the real answer.

Solution 5. For , the two angles are , whose cotangents are : sine and cosine agree and are positive at by the complementary-angle identity, and reflection changes the cosine sign at . Formula (2.2) gives Thus The exact orders are zero, two and four. The counterterms have support ; the limit has full support and singular support , by Theorem 2.1.

Solution 6. In the coordinate , use , with angles . Their cotangents are . To verify the constants, the addition formulas give . At , the positive number therefore satisfies , so and . Complement and reflection give the stated cotangents. Thus The transformed test is . Its -th derivative at zero is . Hence, with , the coefficients of the powers are The full limit after those three subtractions is To check its scaling directly, use (2.8) on and substitute . Its even part and each Taylor term gain the common factor , its denominator gains , and , leaving times the formula centered at .