# A Note on Derivative of Sine Series with Square Root

Authors: Sergiusz Kęska
Journal: Abstract and Applied Analysis
 (2021)

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## Abstract

Chaundy and Jolliffe proved that ifan is a nonnegative, nonincreasing real sequence, then series ∑ansinnx converges uniformly if and only if nan⟶0. The purpose of this paper is to show that if nan is nonincreasing and nan⟶0, then the series fx=∑ansinnx can be differentiated term-by-term on c,d for c,d>0. However, f′0 may not exist.


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## Body

## 1. Introduction

Chaundy and Jolliffe [1] proved the following.Theorem 1.
Ifckk=1∞⊂R+ is decreasing to zero, then ∑k=1∞cksinkx converges uniformly in x if and only if kck⟶0 as k⟶∞.Theorem1 has had numerous generalizations.Leindler [7] verified that in Theorem 1, the monotonicity assumption cn≥cn+1 can be replaced by c∈RBVS, i.e., if the conditions cn⟶0 and ∑k=n∞ck−ck+1≤Kcn hold for all n with constant K=Kc which depends only upon c.The next theorem was indicated in [11].Theorem 2.
Ifck belongs to the class MVBVS, i.e., if there exist constants C and λ≥2, depending only on the sequence ck such that ∑k=n2nck−ck+1≤C/n∑k=λ−1nλnck for all n≥λ, then series ∑k=1∞cksinkx converges uniformly in x if and only if limk⟶∞kck=0.Móricz [8] proves the following theorem.Theorem 3.
Assumef:R+⟶0,∞ with property xfx∈Lloc1R+. If fx is nonincreasing on R+, then the integral ∫0∞fxsintxdx, t∈R+, converges uniformly in t if and only if xfx⟶0 as x⟶∞.A result due to Žak and Šneider [10] holds for double sine series.Theorem 4.
Ifcjkj,k=1∞⊂R+ is monotonically decreasing double sequences, i.e., a sequence of real numbers such that for j,k=1,2,⋯, cjk−cj+1,k≥0, cjk−cj,k+1≥0, and cjk−cj,k+1−cj+1,k+cj+1,k+1≥0, then ∑j=1∞∑k=1∞cjksinjxsinky is uniformly regularly convergent in x,y if and only if jkcjk⟶0 as j+k⟶∞.Theorem4 was generalized by Kórus [6]. He has defined new classes of double sequences (SBVDS1) to obtain those generalizations.Duzinkiewicz and Szal [2] introduce a new class of double sequences called DGMα,β,γ,r, which is a generalization of the class considered by Kórus, and they obtain sufficient and necessary conditions for uniform convergence of double sine series.Dyachenko et al. [3] proved the Chaundy-Jolliffe theorem for GMβ sequences with majorant β having the following form: βn=1/nFna, where Fn is admissible.In the recent paper [5], it was proved that ∑ansinnx converges uniformly on 0,π if and only if nan⟶0.One of the results of paper [9] is that for any α∈0,2, ∑ansinnαx converges uniformly if and only if nan⟶0.


## 2. Main Results

Lemma 5.
Let(1)I∗=cosπ/x0x02m+12/4+j+1x02m+12/4+j+1−4x0πsinπ2x0x02m+124+j+1−x02m+124+j∗cosπ2x0x02m+124+j+1+x02m+124+j,I∗∗=cosπ/x0x02m+12/4−jx02m+12/4−j−4x0πsinπ2x0x02m+124−j−x02m+124−j−1∗cosπ2x0x02m+124−j+x02m+124−j−1,where m∈Odd+=1,3,5,9,⋯.
Leta,b∈R such that 0<a<b. Then, for all x0∈a,b and for all m>max2/x02,2/x0,
(2)∑j=0x02/42m+1−1I∗+I∗∗≤16π2x05m22x04+x02.Proof.
Let(3)I1∗=cosπ/x0x02m+12/4+j+1−cosπ/2x0x02m+12/4+j+1+x02m+12/4+jx02m+12/4+j+1,I2∗=cosπ2x0x02m+124+j+1+x02m+124+j∗1x02m+12/4+j+1−4x0πsinπ2x0x02m+124+j+1−x02m+124+j.
Note that for allm>max2/x02,2/x0,
(4)π2x01x02m+12/4+j+1+x02m+12/4+j−13!π2x01x02m+12/4+j+1+x02m+12/4+j3≤sinπ2x0x02m+124+j+1−x02m+124+j≤π2x01x02m+12/4+j+1+x02m+12/4+j.
In view of (4), the following inequality is satisfied:
(5)I2∗≤π22x053!m3+1x02m+12/4+j+1−2x02m+12/4+j+1+x02m+12/4+j≤π22x053!m3+2m2x04x02m+124+j−x02m+124+j+1≤8π2x05m3.
Let(6)I1∗∗=cosπ/x0x02m+12/4−j−cosπ/2x0x02m+12/4−j+x02m+12/4−j−1x02m+12/4−j,(7)I2∗∗=cosπ2x0x02m+124−j+x02m+124−j−1∗1x02m+12/4−j−4x0πsinπ2x0x02m+124−j−x02m+124−j−1,(8)I1∗+I1∗∗=−2x02/4m+12+j+1sinπ4x0x024m+12+j+1−x024m+12+j∗sinπ4x03x024m+12+j+1+x024m+12+j+−2x02/4m+12−jsinπ4x0x024m+12−j−x024m+12−j−1∗sinπ4x03x024m+12−j+x024m+12−j−1==−2x02/4m+12+j+1sinπ4x0x024m+12+j+1−x024m+12+j∗sinπ4x03x024m+12+j+1+x024m+12+j+sinπ4x03x024m+12−j+x024m+12−j−1+sinπ4x03x024m+12−j+x024m+12−j−1∗−2x02/4m+12−jsinπ4x0x024m+12−j−x024m+12−j−1+2x02/4m+12+j+1sinπ4x0x024m+12+j+1−x024m+12+j=J1+J2.
In analogy with (5), we have
(9)I2∗∗≤8π2x05m3.
IfD∈−3,−2,−1,0, then
(10)J1≤πx03m2sinπ4x02x0m+1+3x024m+12+j+1+x024m+12+j−2x0m+1+sinπ4x02x0m+1+3x024m+12−j+x024m+12−j−1−2x0m+1=πx03m2−1m+1/2sinπ4x03x024m+12+j+1+x024m+12+j−2x0m+1+−1m+1/2sinπ4x03x024m+12−j+x024m+12−j−1−2x0m+1=πx03m2sinπ4x09x02/4m+12+j+1−9/4x02m+123x02/4m+12+j+1+3/2x0m+1+x02/4m+12+j−1/4x02m+12x02/4m+12+j+1/2x0m+1+sinπ4x09x02/4m+12−j−9/4x02m+123x02/4m+12−j+3/2x0m+1+x02/4m+12−j−1−1/4x02m+12x02/4m+12−j−1+1/2x0m+1≤πx03m22sinπ8x03j+3Dx02/4m+12+j+1+1/2x0m+1+j+Dx02/4m+12+j+1/2x0m+1+−3j+3Dx02/4m+12−j+1/2x0m+1+−j+Dx02/4m+12−j−1+1/2x0m+1≤π24x04m23jx02/4m+12−j−x02/4m+12+j+1x02/4m+12−j+1/2x0m+1x02/4m+12+j+1+1/2x0m+1+jx02/4m+12−j−1−x02/4m+12+jx02/4m+12−j−1+1/2x0m+1x02/4m+12+j+1/2x0m+1+2π2x05m3≤16π2x07m5j2+2π2x05m3,(11)J2≤2sinπ/4x0x02/4m+12+j+1−x02/4m+12+jx02/4m+12−j−2sinπ/4x0x02/4m+12−j−x02/4m+12−j−1x02/4m+12−j+2sinπ4x0x024m+12+j+1−x024m+12+j1x02/4m+12+j+1−1x02/4m+12−j≤4x0m2sinπ8x0x024m+12+j+1−x024m+12+j−x024m+12−j+x024m+12−j−1+4πj+1x05m4≤πx02m1x02/4m+12+j+1+x02/4m+12+j−1x02/4m+12−j+x02/4m+12−j−1+4πj+1x05m4≤πx02mx02/4m+12−j+x02/4m+12−j−1−x02/4m+12+j+1−x02/4m+12+jx02m2+4πj+1x05m4≤πx04m32jx0m+2j+2x0m+4πj+1x05m4≤6πj+1x05m4.
In view of (5) and (8)–(11), the following inequality is satisfied:
(12)I∗+I∗∗≤16π2x05m21m+jm2+j2x02m3.
Then, for allx0∈a,b and m>max2/x02,2/x0, we have
(13)∑j=0x02/42m+1−1I∗+I∗∗≤16π2x05m22x04+x02.In the recent paper [4], it was proved, among others, that if cn is a nonincreasing sequence and ∑n=1cncosnx is convergent at any point x≠0, then limn⟶∞ncn=0. We show that if ncn is nonincreasing and limn⟶∞ncn=0, then ∑n=1cncosxn converges uniformly on c,d, where c>0.Theorem 6.
Letc,d∈R such that 0<c<d. If nann=1∞ is nonincreasing and limn⟶∞nan=0, then the series
(14)fx=∑n=1∞ansinxn,can be differentiated term-by-term on c,dor−d,−c. The series
(15)f′x=∑n=1∞anncosxn,converges uniformly on c,dor−d,−c.Proof.
We can finda,b∈R such that 0<a<b, c=π/b, and d=π/a. If ann=1∞ is a nonnegative monotone sequence and limn⟶∞nan=0, then the series (14) converges uniformly [9]. We show that if, in addition, nann=1∞ is nonincreasing, then (for x=π/x0) the series ∑n=1∞anncosπn/x0 converges uniformly on a,b. Let m∈Odd+, bn=cosπn/x0/n, Bk=∑n=1kbn, and Mm=x02/4m2+1,x02/4m2+2,⋯,x02/4m+22. We denote by m0 the minimal odd number for which there holds the following: 4∣m0−1 and m0≥max2/x02,2/x0. Then,
(16)∃H′∀x0∈a,b∑n=1x02m02/4bn<H′.
Let(17)B′′k=∑n=x02/4m02+1kbn.
Note that for allm such that 4m−1 and for all k∈Mm, the following conditions are fulfilled: bk≤0 and
(18)∑n∈⋃r∈m0,m0+2,m0+4,⋯,mMrbn≤B′′k≤∑n∈⋃r∈m0,m0+2,m0+4,⋯,m−2Mrbn.
On the other hand, for allm such that 4∣m−3 and for all k∈Mm, we have bk≥0 and
(19)∑n∈⋃r∈m0,m0+2,m0+4,⋯,m−2Mrbn≤B′′k≤∑n∈⋃r∈m0,m0+2,m0+4,⋯,m−2,mMrbn.
We show that the sequence(20)∑n∈⋃r∈m0,m0+2,m0+4,⋯,m−2,mMrbn,is bounded for 4m−1 and 4m−3.
(21)∑n∈⋃r∈m0,m0+2,m0+4,⋯,m−2,mMrbn=∑r∈Odd+,r=m0,⋯,m∑j=1x02/4r+22−x02/4r2cosπ/x0x02/4r2+jx02/4r2+j=∑r∈Odd+,r=m0,⋯,m∑j=0x02/42r+1−1cosπ/x0x02/4r+12+j+1x02/4r+12+j+1+cosπ/x0x02/4r+12−jx02/4r+12−j+∑j=1x02/4r+12−x02/42r+1−x02/4r2cosπ/x0x02/4r2+jx02/4r2+j+∑j=x02/4r+12+x02/42r+1−x02/4r2x02/4r+22−x02/4r2cosπ/x0x02/4r2+jx02/4r2+j=∑r∈Odd+,r=m0,⋯,mS1+S2+S3.
Note thatx02/2−1≤x02/4r+22−x02/4r+12−x02/42r+1≤x02/2+2. Hence,
(22)S3≤42+x02/2x0rcosπx0x024r+12+x0242r+1≤24+x0x0rcosπx0x02r+2+x024r+12+x0242r+1−x02r+2≤2πx02/2+24+x0x03r2.
Moreover,(23)S2≤2x0r/2cosπx0x024r2+2≤4πx03r2.
Furthermore,(24)S1=∑j=0x02/42r+1−14x0πsinπ2x0x04r+12+j+1−x04r+12+j∗cosπ2x0x04r+12+j+1+x04r+12+j+4x0πsinπ2x0x04r+12−j−x04r+12−j−1∗cosπ2x0x04r+12−j+x04r+12−j−1+I∗+I∗∗=S′1+∑j=0x02/42r+1−1I∗+I∗∗,(25)S′1=∑j=0x02/42r+1−12x0πsinπx0x024r+12+j+1−sinπx0x024r+12+j+2x0πsinπx0x024r+12−j−sinπx0x024r+12−j−1==2x0πsinπx0x024r+12+x0242r+1−sinπx0x024r+12−sinπx0x024r+12−x0242r+1+sinπx0x024r+12==2x0π−1r+1/2cosπx0x024r+12+x0242r+1−r+2π2+−1r+1/2cosπx0x024r+12−x0242r+1−rπ2==2x0π−1r+1/21−1−cosπx0x024r+12+x0242r+1−r+2π2+1−1−cosπx0x024r+12−x0242r+1−rπ2=4x0π−1r+1/2−2x0π−1r+1/21−cos2π/x0x02/4r+12+x02/42r+1−r+2π/21+cosπ/x0x02/4r+12+x02/42r+1−r+2π/2++1−cos2π/x0x02/4r+12−x02/42r+1−rπ/21+cosπ/x0x02/4r+12−x02/42r+1−rπ/2.
Note that(26)∑r∈Odd+r=m0,⋯,m−1r+1/2=−1if4∣m−1,0if4∣m−3.
In view of Lemma5 and (22)–(26), the following inequality is satisfied:
(27)∑r∈Odd+,r=m0,⋯,mS1+S2+S3≤∑r∈Odd+,r=m0,⋯,m4x0π−1r+1/2+∑r∈Odd+,r=m0,⋯,m2x0πsin2πx0x024r+12+x0242r+1−r+2π2+sin2πx0x024r+12−x0242r+1−rπ2+∑j=0x02/42r+1−1I∗+I∗∗+S2+S3≤4x0π+∑r∈Odd+,r=m0,⋯,m16π2r2x052x04+x02+16x0r2ππ24+π2x02+4πr2x03+2πx02/2+24+x0r2x03<H,for all x0∈a,b. Using (17)–(19), (21), and (27), we obtain
(28)∃H′′∀x0∈a,bB′′k<H′′.
In view of (16) and (28), we get
(29)∃G∀x0∈a,bBk<G.
We know thatnan is nonincreasing and limn⟶∞nan=0 and (29) holds. Hence, the series ∑nancosπn/x0/n converges uniformly. This follows from Dirichlet’s test.Remark 7.
The conditions (nan is nonincreasing, and limn⟶∞nan=0) are not sufficient for the differentiability of (14) at the point x=0.Example 1.
Letfx=∑n=4sinxn/nlnn. It converges uniformly on 0,π [5]. We show that f is not differentiable at x=0. Let fΔx=∑n=41/nlnnsinπn/2α for Δx=π/2α and α∈Odd+:4α−1,α≥5. Then, by Lagrange’s theorem,
(30)∑n=4α21nlnnsinπn2α≥1αlnα2∑n=4α21nsinπn2α≥1αlnα2∫3α21nsinπn2αdn=4πlnα2cosπ32α,(31)∑n=α2+12α21nlnnsinπn2α≥12αln2α2∑n=α2+12α21nsinπn2α≥12αln2α2∫α2+12α2+11nsinπn2αdn=−42πln2α2cosπ2α2+12α−cosπα2+12α.
The functionsinx/x has a relative minimum at the point x1∈π,3π/2. Then,
(32)∑n=2α2+1x12α/π21nlnnsinπn2α≥−42πln2α2cosπx12α/π2+12α−cosπ2α2+12α,(33)∑n=x12α/π2+13α21nlnnsinπn2α=∑n=x12α/π2+23α21nlnnsinπn2α+sinπx12α/π2+1/2αx12α/π2+1lnx12α/π2+1≥42πln2α2cosπx12α/π2+12α−1α2lnα2,(34)∑n=3α2+14α21nlnnsinπn2α≥−43πln3α2.
In view of (30)–(34), we obtain
(35)∑n=44α21nlnnsinπn2α≥13πlnα−1α2lnα2.
Ifx′r and xr" are the relative minimum and maximum (respectively) points of sinx/x, such that
(36)xr"∈2π+2rπ,5π2+2rπ,x′r∈3π+2rπ,7π2+2rπ,for r=0,1,2,3,⋯, then
(37)∑n=α2+1∞sinπn/2αnlnn=∑r=0∑n=4r+1α2+1xr"2α/π2sinπn/2αnlnn+∑n=xr"2α/π2+14r+5α2sinπn/2αnlnn+∑n=4r+5α2+14r+6α2sinπn/2αnlnn+∑n=4r+6α2+1xr′2α/π2sinπn/2αnlnn+∑n=xr′2α/π2+14r+7α2sinπn/2αnlnn+∑n=4r+7α2+14r+8α2sinπn/2αnlnn≥∑r=014r+5αln4r+5α2∑n=4r+1α2+1xr"2α/π2sinπn/2αn++∑n=xr"2α/π2+14r+5α2sinπn/2αn+14r+6αln4r+6α2∑n=4r+5α2+14r+6α2sinπn/2αn+∑n=4r+6α2+1xr′2α/π2sinπn/2αn+∑n=xr′2α/π2+14r+7α2sinπn/2αn+14r+7αln4r+7α2∑n=4r+7α2+14r+8α2sinπn/2αn.
In view of (36), we obtain
(38)∑n=4r+1α2+1xr"2α/π2sinπn/2αn+∑n=xr"2α/π2+14r+5α2sinπn/2αn≥∫4r+1α2xr"2α/π2sinπn/2αndn+∫xr"2α/π2+14r+5α2−1sinπn/2αndn=∫4r+1α24r+5α2sinπn/2αndn−∫xr"2α/π2xr"2α/π2+1sinπn/2αndn−∫4r+5α2−14r+5α2sinπn/2αndn≥4απ−2r+1α,(39)∑n=4r+5α2+14r+6α2sinπn/2αn+∑n=4r+6α2+1xr′2α/π2sinπn/2αn+∑n=xr′2α/π2+14r+7α2sinπn/2αn≥∫4r+5α24r+6α2sinπn/2αndn−4r+1α=−4r+1α,(40)∑n=4r+7α2+14r+8α2sinπn/2αn≥∫4r+7α24r+8α2sinπn/2αndn=−4απ.
In view of (37)–(40), we get
(41)∑n=α2+1∞sinπn/2αnlnn≥∑r=04π4r+5ln4r+5α2−4π4r+7ln4r+7α2−6r+12α2≥−π2α2.
Hence,(42)∑n=41nlnnsinπn2α≥13πlnα−π2+1α2,limα⟶∞fΔxΔx=+∞.
This follows from (35) and (41).



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