PRH | Aux | 4.2 • Fermat via Geometry, Blur, and Fourier

Fermat via Geometry, Blur, and Fourier Unconditional spectral walls, analytic reduction, and afinite certificate for the last band Aleksandar Perišić September 2025 Abstract We develop a Fourier–analytic “blur” framework for the geometric avatar of a putative Fermat identity. The target is the cube ring Ra,c = Qc\Qa with QL = [ −L/ 2 , L/ 2] n and integers n≥ 3 , 0 < a < c . We ask whether Ra,c can be exactly realized by the superposition fs,µ = χBs∗µ of bn congruent axis–aligned cubes Bs = [ −s/ 2 , s/ 2] n (“builders”). Two placement–independent spectral obstructions survive Gaussian blur: (i) the zero–hyperplane wall ( s|a and s|c are necessary), and (ii) the DC wall ( snbn = cn−an is necessary). Under these arithmetic walls we study the spectral ratio g ( ξ ) = \χRa,c ( ξ ) /dχBs ( ξ ). Exact equality would force g to be the Fourier transform of a finite positive measure, hence positive–definite (PD). Along a coordinate axis we compute the 3 × 3Toeplitz Gram determinant with sharp Taylor remainders and prove PD–failure on a large, explicit parameter region; what remains is a compact small–xstrip. The new contribution of this note is an analytics–to–certificate reduction tailored for implementation (e.g. in C#): we rigorously shrink the search to a finite set of steps τ and to finitely many parameter boxes, within which an interval arithmetic check of ∆ g ( t ) < 0 (with explicit Taylor remainders) certifies PD–failure. This leaves a finite, fully explicit computer–verifiable certificate as the only remaining step. We also prove a large– n simplification that makes this finite certificate particularly shallow for all n > 50 (fewer τ values and coarser boxes suffice), though still finite rather than purely analytic. 1 Setup: target, builder, blur, and spectra Fix n≥3and integers 0< a < c. Let QL= [−L/2, L/2]nand Ra,c := Qc\Qa, χRa,c =χQc−χQa. For s > 0set Bs= [−s/2, s/2]n. A configuration of bncongruent cubes is fs,µ := χBs∗µ, µ = bn X m=1 δxm. Why we introduce the builder model (and where Fermat enters). The translate– superposition fs,µ = χBs∗µ is a maximally symmetric “reassembly avatar” of an n -th power identity: we ask whether the centered cube ring Ra,c = Qc\Qa can be synthesized purely by translating one fixed axis-aligned cube profile Bs , with no cancellation (so µ≥ 0, and in our toy model µis a sum of Dirac masses). Lemma 1.1 (Fermat ⇒ exact builder tiling at unit scale).Assume a, b, c ∈N satisfy an + bn = cn with n≥3. Set s= 1 and B1= [−1/2,1/2]n. Let δ∈ {0,1 2}be defined by δ:= (0, c odd, 1 2, c even. 1 If a≡c ( mod 2), then there exist points x1, . . . , xbn∈ ( Z + δ ) n such that, with µ = Pbn m=1 δxm , we have χRa,c =χB1∗µalmost everywhere. In particular, any Fermat counterexample (after the harmless scaling ( a, b, c ) 7→ (2 a, 2 b, 2 c )if needed) would produce a perfect builder configuration of unit cubes for Ra,c. Proof. The family of cubes {u + B1}u∈(Z+δ)n tiles Rn disjointly a.e. Moreover, its face hyperplanes occur at coordinates in Z+1 2if δ= 0,and in Zif δ=1 2. With δ chosen as above, the boundaries of Qc = [ −c/ 2 , c/ 2] n lie on this grid, so Qc is a disjoint a.e. union of exactly cn unit cubes u + B1 with u∈ ( Z + δ ) n . If also a≡c ( mod 2), then the same tiling grid aligns with Qa , so Qa is a disjoint a.e. union of exactly an such cubes, and subtracting gives χRa,c =X u∈(Z+δ)n:u+B1⊂Qc\Qa χu+B1=χB1∗X u∈U δu, a.e., for a finite index set U⊂ ( Z + δ ) n of size |U| = cn−an = bn . Rename the elements of U as x1, . . . , xbn. In short: the builder model is not a vague analogy here. If Fermat were false, the corresponding triple ( a, b, c ; n )would force an exact nonnegative superposition representation at scale s = 1; the rest of the paper shows that such a representation is incompatible with the spectral/positivity constraints for n≥3. Fourier convention: b f(ξ) = ZRn f(x)e−2πix·ξdx, sinc t=sin t t,sinc(0) = 1. Then dχQL(ξ) = Ln n Y j=1 sinc(πL ξj),dχBs(ξ) = sn n Y j=1 sinc(πs ξj), [χRa,c (ξ) = cn n Y j=1 sinc(πc ξj)−an n Y j=1 sinc(πa ξj),d fs,µ(ξ) = dχBs(ξ)bµ(ξ). We blur by Gaussians Φ τ ( x ) = (2 πτ2 ) −n/2e−|x|2/(2τ2) , with c Φτ ( ξ ) = e−2π2τ2|ξ|2∈ (0 , 1] and c Φτ→1pointwise as τ↓0. The blurred L2-error is Eτ(s, µ) := ∥(χRa,c −fs,µ)∗Φτ∥2 2=ZRn[χRa,c −dχBsbµ2e−4π2τ2|ξ|2dξ. 2 Two unconditional spectral walls (survive blur) Lemma 2.1 (Zero–hyperplane wall).Let Zs := Sn j=1{ξ : ξj∈1 sZ\{ 0 }} . Then dχBs vanishes on Zs. Moreover, [χRa,c vanishes on Zsiff s|aand s|c. Proposition 2.2 (DC wall).If fs,µ =χRa,c almost everywhere, then bµ(0) = [χRa,c (0) dχBs(0) =cn−an sn=bn, so in particular snbn=cn−an. Corollary 2.3 (Blurred obstruction).If either s∤a or s∤c , or snbn = cn−an , then ∃τ0, η > 0 such that Eτ(s, µ)≥ηfor all µand all 0< τ ≤τ0. 2 3 Dimensionless parameters under the walls Assume from now on the walls: s|a, c and snbn=cn−an. Write R1=cn+2 −an+2 cn−an, R2=cn+4 −an+4 cn−an, R3=cn+6 −an+6 cn−an, and x:= s2 R1 , y := R2 R2 1 , z := R3 R3 1 . Set ρ:= a/c ∈(0,1). Then R1=c21−ρn+2 1−ρn, R2=c41−ρn+4 1−ρn, R3=c61−ρn+6 1−ρn, y(ρ) = (1 −ρn+4)(1 −ρn) (1 −ρn+2)2∈ymin(n),1. (3.1) with ymin(n) = lim ρ→1− y(ρ) = n(n+ 4) (n+ 2)2= 1 −4 (n+ 2)2. Since s|cwrite c=rs with r∈Z≥2, and a=ps with 1≤p<r. Then x=1−ρn r2(1 −ρn+2)∈(0,1 4],because 1−ρn 1−ρn+2 <1and r≥2.(3.2) 4 The PD ratio on a line and a factored Gram determinant Define the spectral ratio where dχBs= 0: g(ξ) := [χRa,c (ξ) dχBs(ξ). If exact equality holds, then bµ = g and g is the FT of a finite positive measure; in particular the normalized function φ ( t ) := g ( te1 ) /g (0) is a characteristic function on R (even, |φ| ≤ 1, φ(0) = 1). Along the x1 –axis set g ( t ) := g ( te1 ). Using sinc ( πLt ) = 1 −π2L2 6t2 + π4L4 120 t4−π6L6 5040 t6 + O ( t8 ), a direct calculation gives g(t) = M1 + α2t2+α4t4+α6t6+O(t8), M =cn−an sn,(4.1) with α2=π2 6s2−R1=π2 6R1(x−1), α4=π4 3607s4−10s2R1+ 3R2=π4 360R2 17x2−10x+ 3y,(4.2) α6=π6 1512031s6−49s4R1+ 21s2R2−3R3=π6 15120R3 131x3−49x2+ 21xy −3z. Normalize φ ( t ) = g ( t ) /g (0) = 1 + α2t2 + α4t4 + α6t6 + O ( t8 ). For characteristic functions one has φ′′(0) = 2α2≤0, φ(4)(0) = 24α4≥0, φ(6)(0) = 720α6≤0. 3 Lemma 4.1 (Factored 3 × 3Gram).For any real a, b, c , det a b c b a b c b a  = ( a−c ) ( a2 + ac − 2 b2 ). Thus, with Gk=g(kt), ∆g(t) := det  G0G1G2 G1G0G1 G2G1G0 =G0−G2G2 0+G0G2−2G2 1.(4.3) Proposition 4.2 (Exact series with remainder).Let u = t2 and assume the Taylor window |πLt| ≤ 1 2for L∈ {a, c, s}. Then ∆g(t) M3= 8 α2α2 2−6α4u3+ 48 α4α2 2−4α4u4+ 72 α2α2 4u5+ 32 α3 4u6+R7(u),(4.4) with a remainder bound |R7(u)| ≤ C∗u7for an explicit C∗=C∗(a, c, s, n). 5 Large unconditional PD–failure region for x≤1 4 Define F(x, y) := −16x2+ 20x+ 5 −9y. From (4.2) , α2α2 2− 6 α4 = π6 1080R3 1 ( x− 1) F ( x, y ). For fixed y∈ ( ymin ( n ) , 1) the concave quadratic F(·, y)has positive set {x:F(x, y)>0}=x−(y), x+(y), x±(y) = 5±3√5−4y 8, x−(y)∈(0,1 4], x+(y)>1. Under the walls we always have x≤1 4 (Eq. (3.2) ), so x∈ (0 ,1 4 ]. Therefore, for any y∈ ( ymin, 1) and any x∈x−(y),1 4we have α2<0, α2 2−6α4>0, and the leading u3coefficient in (4.4) is negative. Theorem 5.1 (PD fails on the whole upper subband).Assume the walls. If x∈x− ( y ) ,1 4 , then ∃t = 0 in the Taylor window with ∆ g ( t ) < 0. Hence g is not PD and no exact equality fs,µ =χRa,c is possible. Proof. Choose t so small that the u3 term in (4.4) dominates the rest by the explicit bound C∗ (take u≤u0 with u0 from a standard domination lemma). The leading coefficient is negative on the stated band, so ∆g(t)<0. 6 The remaining small– x strip and an analytics–to–certificate reduction The parameters not covered by Theorem 5.1 satisfy 0< x ≤x−(y)≤1 4, y ∈ymin(n),1. On this small– x strip the 3 × 3leading coefficient in (4.4) is nonnegative, so a higher–order control is needed. We now reduce the strip to a finite certificate check. 4 6.1 Sharp moment relation and its consequences Set y(ρ) = (1 −ρn+4)(1 −ρn) (1 −ρn+2)2, z(ρ) = (1 −ρn+6)(1 −ρn)2 (1 −ρn+2)3, ρ =a c∈(0,1). Lemma 6.1 (Sharp lower relation z≥θny).Let θn:= n(n+ 6) (n+ 2)(n+ 4) = 1 −8 (n+ 2)(n+ 4) ∈(0,1). Then for all ρ∈(0,1), z(ρ) y(ρ)=(1 −ρn+6)(1 −ρn) (1 −ρn+2)(1 −ρn+4)≥θn.(6.1) Moreover, the ratio is strictly decreasing in ρand attains the minimum θnat ρ→1−. Proof. Differentiate log R ( ρ )for R = z/y and use the monotonicity/convexity in the index of Tm ( ρ ) = mρm−1/ (1 −ρm )as in standard Chebyshev sum comparisons; details are routine and omitted. 6.2 A 4×4Toeplitz minor: exact leading term Let T4(t) = [g(|i−j|t)]3 i,j=0. Writing u=t2and A=α2, B =α4, C =α6: Lemma 6.2 (Leading term and factorization).In the Taylor window, det T4(t) = g(0)4h−1152 (A2−6B) (5AC −2B2)u6+O(u7)i. Proof. Standard Toeplitz elimination (subtract consecutive rows/columns) kills low orders; a short symbolic expansion yields the factorization asserted. On the small– x strip we have A2− 6 B≤ 0, so the sign of the u6 coefficient is the opposite of sgn(5AC −2B2). With (4.2) one finds 5AC −2B2=π8 (6) ·15120 R4 1P(x, y, z), where P(x, y, z) = 25(x−1)31x3−49x2+ 21xy −3z−77x2−10x+ 3y2. Since Pis decreasing in z, the worst case for negativity is z=θnyby Lemma 6.1. Define Pn(x, y) := Px, y, θny. 6.3 What the 4×4minor gives (and what it does not) Proposition 6.3 (A uniform negative slice inside the strip).For every n≥ 3and every y∈(ymin(n),1), Pn 1 8, y<0. Consequently, for all x∈1 8, x− ( y )  one has det T4 ( t ) < 0for all sufficiently small t in the Taylor window. Sketch. Pn ( x, y )is concave in y and continuous in x ; direct evaluation at x = 1 8 and y∈ {ymin(n),1}yields negativity; hence negativity holds for all intermediate y. 5 Proposition 6.4 (The 4 × 4leading sign is not uniform on the whole strip).For all sufficiently large n there exist y∈ ( ymin ( n ) , 1) and x∈ (0 , x− ( y )) with Pn ( x, y ) > 0. Thus the u6 coefficient alone does not decide the sign of det T4(t)on the entire strip. Sketch. Take, e.g., y near 0 . 9and x near x− ( y ); a direct evaluation shows positivity for n moderately large. (The phenomenon is robust and reflects the cancellation in the one–axis “variance” bracket.) 6.4 Analytics–to–certificate reduction (finite) We now give a clean reduction of the remaining set Sn:= n(x, y, z) : 0 < x ≤min{x−(y),1 8}, y ∈(ymin(n),1), z ≥θnyo to a finite sign–certificate for ∆g(t). Theorem 6.5 (Finite certificate reduction).Fix n≥ 3. There exists a finite list of rational steps {τk}K k=1 ⊂(0,1 2], depending only on n, such that the following are equivalent: (i) For every admissible triple ( a, c, s )in the small– x strip Sn there exists t in the Taylor window with ∆g(t)<0. (ii) For every axis–aligned parameter box B in a finite partition of Sn (constructed below) there exists some k∈ { 1 , . . . , K} such that the interval arithmetic evaluation of ∆ g ( t )at t = τk/c (using the quartic Taylor expansions of sinc and cos with explicit remainder bounds) yields an interval strictly contained in (−∞,0) for all (x, y, z)∈B. Moreover, such a partition and such a set {τk} can be generated deterministically by a simple bisection–style refinement that must terminate in finitely many steps. Proof (outline with precise ingredients). Normalization. Set t = τ/c , zs = πτ/r ∈ (0 ,πτ 2 ], za = πτρ , zc = πτ . For any fixed τ≤1 2 the Taylor window holds uniformly: |πLt| ≤ 1 2 for L∈ {a, c, s} . Certified expansions. Use sinc z= 1 −z2 6+z4 120 ±|z|6 5040,cos z= 1 −z2 2±z4 24,sin z=z−z3 6±z5 120, and sinc (2 z ) = sinc z·cos z . Products inherit explicit remainder bounds, so g ( kt )admits interval enclosures depending monotonically on (x, y, z)and τ. Compactness & continuity. The map ( x, y, z, τ ) 7→ ∆ g ( τ/c )is continuous on the compact set Sn×{τ≤1 2} . If (i) holds, then for each point there exists a neighborhood on which ∆ g< 0 for some τ . A standard finite–subcover argument yields finitely many pairs ( B, τk )with strict negativity. Algorithmic construction. Start with a coarse tiling of Sn and a short menu {τk} . For each box B and each τk compute an interval enclosure [∆ − B,k, ∆ + B,k ]using the certified expansions. If some k gives ∆ + B,k < 0, mark B “good”. Otherwise, bisect B (axis–aligned), and repeat. Termination follows from the continuity and compactness argument above (good boxes accumulate to cover Sn). This produces a finite set of boxes {Bj}and witnesses τk(j). Remark 6.6 (Large– n simplification, n > 50).From (3.1) ,1 −ymin ( n )=4 / ( n + 2) 2≤ 1 / 676 for n≥50. Thus across Sn, 0<1−y≤4 (n+ 2)2≤1 676, θn= 1 −8 (n+ 2)(n+ 4) ≥1−8 54 ·56. These bounds force the low–order positive term in B ( t ) = 1+ φ (2 t ) − 2 φ ( t ) 2 to be uniformly small (since α2 2− 6 α4∝F ( x, y ) ≤ 9(1 −y )), while the signed cubic term is controlled by − 3 z≤ − 3 θny (close to − 3). As a result one can fix a short τ -menu (e.g. eight values) and a coarse initial tiling for all n > 50; the refinement depth needed by Theorem 6.5 is strictly smaller than for small n . This is an analytic speedup, not a replacement: the last step remains a finite certificate. 6 7 FLT as a corollary (conditional on the finite certificate) Theorem 7.1 (Conditional on a finite certificate).Suppose that for each n≥ 3the finite certificate of Theorem 6.5 is validated (e.g. by interval arithmetic). Then no coprime integers a, b, c ≥1and n≥3satisfy an+bn=cn. Proof. If either arithmetic wall fails, Corollary 2.3 gives a uniform blur floor. Under the walls, Theorem 5.1 rules out x∈ ( x− ( y ) ,1 4 ]. On the small– x strip Theorem 6.5 reduces the claim to finitely many certified inequalities ∆ g ( t ) < 0. Hence g is not PD in any admissible case, so no exact equality fs,µ = χRa,c is possible; letting τ↓ 0yields the contradiction to an + bn = cn . 8 Rotations (optional) If a single global orientation R∈SO ( n )is permitted for the builders, the analysis applies in the rotated frame. If multiple orientations are allowed, the zero–hyperplane wall becomes the union of rotated families SRZs,R ; except in trivial cases, [χRa,c cannot vanish on such a union, producing an alignment obstruction prior to PD. 9 The case n= 2 For n = 2 the picture is consistent with Pythagoras. The arithmetic walls pass and Gram determinants remain nonnegative; exact tilings exist (the familiar b×bgrid in an ℓ∞annulus). 12. The Frey Curve: same job, higher–level arithmetic view Assume for contradiction that an + bn = cn with n≥ 3and gcd ( a, b, c ) = 1. Attach the Frey curve Ea,b,c :y2=x(x−an) (x+bn). Conceptually, the Frey route and our blur route are two descriptions of the same hidden symmetry. Each proceeds by: an arithmetic fit, local tests, and a global coherence check. For n = 2 both pipelines pass; for n≥3both pipelines fail because the symmetry is over–constrained. 12.1. Two pipelines, side by side Blur pipeline. 1. Arithmetic fit. Impose the divisibility and volume constraints s|a , s|c , and snbn = cn−an (the zero–hyperplane and DC walls). 2. Local tests. Respect all forced Fourier zeros: dχBs vanishes on the coordinate hyperplane lattice ξj∈1 sZ\{0}; any exact tiling must inherit these. 3. Global coherence. Seek a single positive–definite trigonometric measure µ with bµ = g := [χRa,c dχBs off the zero set. Along a coordinate axis the Toeplitz 3 × 3Gram test, with explicit Taylor remainders, forces a negative minor on the entire upper subband x∈x− ( y ) ,1 4 . The remaining small– x strip 0 < x ≤x− ( y )is reduced to a finite, computer–verifiable certificate (Appendix F); once that finite check is validated, no PD µexists. Frey pipeline. 1. Arithmetic fit. Build Ea,b,c from ( a, b, c ; n ). Its invariants (discriminant, conductor) are governed by the prime factors of abc. 7 2. Local tests. For each p|abc the curve has restricted (semistable) local behavior. These mirror the blur–side hyperplane zero constraints. 3. Global coherence. Modularity demands a weight–2newform whose local data match all these prescriptions. Level–lowering packages the local conditions into a single global object. For n≥ 3such a form cannot exist with those simultaneous constraints (the Wiles/Taylor–Wiles mechanism), so the pipeline breaks. 12.2. The dictionary: what equals what • Same target. The identity an + bn = cn encodes a perfect n –power pattern: blur–side, a Cartesian bn –grid decomposition of the cube ring; Frey–side, a cubic with full rational 2–torsion x(x−an)(x+bn). • Same local restrictions. Hyperplane zeros (blur) vs. prescribed reduction at p|abc (Frey). Both extract many independent local constraints from the same triple (a, b, c;n). • Same global test. One PD measure µ fitting all frequency constraints (blur) vs. one modular form fitting all Galois constraints (Frey). The global object exists only if all locals are mutually compatible. • Same pass/fail pattern. For n = 2 the grid symmetry is real and the PD object exists (and the modular picture aligns—Pythagoras). For n≥ 3the symmetry is overconstrained: the 3 × 3Gram minor is negative on x∈x− ( y ) ,1 4 , and the remaining 0 < x ≤x− ( y )is dispatched by a finite certificate; on the Frey side, level–lowering/modularity rules out a compatible form. 12bis. A plausible Fermat thought experiment: one symmetry, one strip What if Fermat had a proto–Fourier trick? Suppose he fixed a single symmetry (Cartesian n –power grid) and focused on the narrow small– x strip where all difficulty concentrates under the arithmetic walls: Sn:= n(x, y) : 0 < x ≤x−(y)≤1 4, ymin(n)< y < 1o, ymin(n) = n(n+ 4) (n+ 2)2. Outside Sn our one–axis Gram analysis already forbids a PD ratio. On Sn the goal would be to show that for some small rational step t (with |πLt| ≤ 1 2 for L∈ {a, c, s} ) the Toeplitz determinant ∆g(t) = det  g(0) g(t)g(2t) g(t)g(0) g(t) g(2t)g(t)g(0)   is negative, or else a slightly larger Toeplitz minor (e.g. 4 × 4) turns negative via elementary trigonometric bounds and careful remainder bookkeeping. A hand–calculable route (plausible, but a stretch). 1. Arithmetic fit. Enforce s|a, c and snbn = cn−an , rephrase the target as “ rn−pn = bn on the s–grid.” 2. One–axis test with elementary trig. Along one coordinate, use z−z3 6≤sin z≤z−z3 6+z5 120,sinc z=sin z zdecreasing, to bound each sinc(πLt)and expand g(t)with explicit error control. 8 3. Finite inequality on Sn .Show that either the 3 × 3or 4 × 4Toeplitz minor is < 0for some small rational t , with the O ( t8 )(or next–order) remainder dominated by explicit rational bounds. This reduces to a finite family of inequalities on (x, y)in Sn. Verdict (with a wink). It is fun to imagine, but historically unlikely: a lone Fermat grinding through uniform trig bounds and remainder estimates on Sn would face a very long proof. Still, the spirit matches our result: reduce everything to a compact region and a finite list of explicit inequalities—now routine to certify by computer, yet transparent enough to be audited by hand if one insists. Summary Under the arithmetic walls we proved x∈ (0 ,1 4 ]and established PD–failure on the entire upper subband x∈x− ( y ) ,1 4 . The remaining small– x strip is reduced to a finite certificate (Theorem 6.5) using only quartic Taylor expansions with explicit remainders and a finite τ -menu. For n > 50 the certificate is notably shallow (coarser tiling suffices), but the very last step remains a finite computer–verifiable check. Once that check is validated (e.g. by a short C# program with interval arithmetic), the Fourier/blur route to Fermat is complete. While our aim here is to understand what Gaussian blur does to the geometric avatar of Fermat rather than to pursue a purely classical proof, the method consistently behaves in the same way across problems we tested: apply blur, prove that blur can be removed with quantitative control, and you either obtain the solution outright or reduce to a finite lemma. If, instead, blur cannot be removed without contradiction, the argument closes by contradiction. Thus the framework is pragmatic: it drives the analysis to a finite, rigorously checkable residue. 10 Community stance on computer–assisted proofs There is a serious and understandable concern about proofs that rely on large computer checks: they can be harder to audit and to teach, and a small implementation error discovered years later can undermine confidence. We share this concern. The “blur” framework addresses it by isolating afinite and explicit certificate whose verification can be performed with transparent, reproducible interval arithmetic at modest precision. Until such a certificate is supplied, one should regard the final step as witnessed rather than fully written out by hand; in that state the proof is best treated as conditional on a finite computation that any reader can reproduce on demand. At the same time, the methodology encourages theory merging: as analytic structure accumulates, the computational burden shrinks and can eventually disappear. Practically, we recommend that any implementation (in C#, or another language) be open–sourced, use outward–rounded interval arithmetic, record all parameters and seeds, and emit human–readable logs of the negative determinant certificates per box. With these safeguards, the remaining check is narrow, auditable, and repeatable, and once completed it elevates the argument to a fully verified proof via blur. A Taylor control for sinc and products For all real z, sinc z= 1 −z2 6+z4 120 −z6 5040 +R8(z),|R8(z)| ≤ |z|8 362880. On [0,π 2], cos z= 1 −z2 2±z4 24,sin z=z−z3 6±z5 120. 9