# Sharp Refinements for the Inverse Sine Function Related to Shafer-Fink’s Inequality

Authors: Gabriel Bercu
Journal: Mathematical Problems in Engineering
 (2017)

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## Abstract

The aim of our work is to give new sharp refinements of Shafer-Fink’s inequality, using suitable changes of variables.


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## Body

## 1. Introduction and Motivation

The inverse sine function is an elementary function that appears in many fields of engineering. In the communication theory and signal processing, it is used to describe the phase of a complex-valued signal. The inverse sine function also appears in the field of control theory, where a nonlinear network unit is modeled by a nonlinear function.There are many applications in which the inverse sine must be replaced by an approximated function, for example, by a rational function. But we have to mention that finding a replacement of simple form for the inverse sine function is in fact difficult. That is why in our work we focus only on special analytic inequalities which have some interesting properties.The starting point of this paper is Shafer-Fink’s double inequality for the arc sine function:(1)3x2+1-x2≤arcsin⁡x≤πx2+1-x2,0≤x≤1.Furthermore, 3 andπ are best constants in (1). Many refinements and extensions of this inequality have been provided (see, e.g., [1–9] and closely related references therein).


## 2. Results and Discussion

Sincelimx→0arcsin⁡x-3x/(2+1-x2)=0, it results that (1) gives a good results near zero, arcsin⁡x≈3x/(2+1-x2) as x→0.Related to the above approximation, we present the following inequality, which provides improvement of Shafer-Fink’s inequality (1).Theorem 1.
For every real number0≤x≤1, the following two-sided inequality holds: (2)x5180+x7189≤arcsin⁡x-3x2+1-x2≤π-32.Proof.
The function(3)f:0,1⟶R,fx=arcsin⁡x-3x2+1-x2-x5180-x7189 has the derivative (4)f′x=-4x6-3x4108+-x2+2-21-x21-x21-x2+22.
In order to prove thatf′x>0, for every 0≤x≤1, we have to establish that (5)1082-x2-21-x2≥4x6+3x41-x21-x2+22 or equivalently(6)16x8-4x6-12x4-108x2+216≥1-x2-4x8+17x6+15x4+216.
Since both sides of the above inequality are positive for all0≤x≤1, we can rise to the second power and obtain the following true result: (7)x616x10+104x8+177x6-27x4-5373x2+16767≥0.
We thus find thatf′x≥0, which imply that the function f is strictly increasing on [0,1].
For proving the right-hand side inequality from Theorem1, we introduce the function (8)g:0,1⟶R,gx=arcsin⁡x-3x2+1-x2-π-32.
Its derivativeg′x=1-x2-12/1-x21-x2+22 is positive for all 0≤x≤1. Thus the function g is strictly increasing on [0,1] with g1=0, so gx≤0 on [0,1].
The proof is completed.Another way to extend the left-hand side inequality from (1) is to consider the approximation near the origin of the form arcsin⁡x≈(3x+ax)/(1+21-x2), where ax→0 as x→0.The result is stated as Theorem2.Theorem 2.
For every0≤x≤1, one has(9)arcsin⁡x-3x2+1-x2≥ax2+1-x2,where ax=1/60x5+11/840x7.Proof.
The function(10)h:0,1⟶R,fx=arcsin⁡x-3x2+1-x2-1/60x5+11/840x72+1-x2 has the derivative (11)h′x=11-x2-3+1/12x4+11/120x621-x2+1-x2+3x2+1/60x6+11/840x81-x22+1-x22.
For proving thath′x≥0 for all 0≤x≤1, we have to show (12)5-x2+184066x8-21x6-70x4-2520≥21-x21+112x4+11120x6 or equivalently (13)66x8-21x6-70x4-840x2+1680≥121-x211x6+10x4+120.
Since both sides of the above inequality are positive for all0≤x≤1, we can rise to the second power and deduce that (14)x84356x8+20944x6+10605x4-131460x2+759780≥0,which is true for every 0≤x≤1.
The proof of Theorem2 is completed.In the following, we will discuss the right-hand side inequality from (1). More precisely, we state and prove the following results.Theorem 3.
For everyx∈[0,1] in the left-hand side and for every x∈[0,0.871433] in the right-hand side, the following inequalities hold true:(15)1-π3x+16-π18x3≤arcsin⁡x-πx2+1-x2≤1-π3x.Remark 4.
Using MATLAB software, we found that the equation(16)arcsin⁡x-πx2+1-x2-1-π3x=0 has the real roots x1=0, x2,3≈±0.871433.We consider the function(17)p:0,0.871433⟶R,px=arcsin⁡x-πx2+1-x2-1-π3x and its derivative (18)p′x=5-x2-π+4-2π1-x21-x22+1-x22-1-π3.We have to find the real numberα∈[0,0.871433] so that p′x≤0 for all 0≤x≤α or equivalently (19)9-4πx2+π+3≤1-x2π-3x2+π+3,0≤x≤α.Both sides of the above inequality are positive on[0,0.871433], and hence we can rise to the second power and we find (20)x2π-32x4+17π2-66π+54x2+-9π2+81≤0.The polynomial function from the left-hand side has the real rootsx1=x2=0 and x3,4= ±32π2-9/(54-66π+17π2+5π-12π13π-36) ≈ 0.735975.We chooseα=0.735975.Therefore,p′x≤0 for all 0≤x≤α and p′x>0 for all α<x≤0.871433.Sincep0=0 and p0.871433=0, we obtain that px≤0 on [0,0.871433].For proving the left-hand side inequality from Theorem3, we introduce the function (21)s:0,1⟶R,px=arcsin⁡x-πx2+1-x2-1-π3x-16-π18x3.Its derivative is(22)s′x=5-x2-π+4-2π1-x21-x22+1-x22-1-π3-12-π6x2.The inequalitys′x≥0 on [0,1] is equivalent to (23)5-x2-π+4-2π1-x2≥1-x25-x2+41-x2·1-π3+12-π6x2 or (24)12-4πx4+6-4πx2+2π+6≥1-x2π-3x4+9-3πx2+2π+6.Both sides of the above inequality are positive on[0,1]; therefore we rise to the second power and we get the following true inequality on [0,1]: (25)x4π-32x6+9π-3x4+3π-317π-27x2+243+6π-25π2≥0.In Theorem5, we will state the following inequalities which give good results near the number 1 for the approximation arcsin⁡x≈πx/(2+1-x2)+bx.Theorem 5.
For every0≤x≤1, one has(26)π-41-x22≤arcsin⁡x-πx2+1-x2≤π-41-x22+π1-x4.Remark 6.
In order to improve the right-hand side of Shafer-Fink’s inequality (1), we impose that π-41-x/22+π1-x/4≤0, which is true for x∈β,1, where β=(-π2+16π-32)/π2≈0.85068.Proof.
We consider the function(27)r:0,1⟶R,rx=arcsin⁡x-πx2+1-x2-π-41-x22 and its derivative (28)r′x=5-x2-π+4-2π1-x21-x22+1-x22+π-4421-x.
We note thatr1=0. In order to obtain rx≥0 on [0,1], we have to prove that r′x≤0 for all x∈[0,1] or equivalently(29)425-π-x2+4-2π1-x2≤4-π1+x2+1-x22.
The real roots of the function from the left-hand side of (29) are (30)x1,2=±-3+7π-2π2+2-2+π-3+ππ≈±0.8801802.
Case  1. For all 0≤x≤0.8801802, the function from the left-hand side of (29) is nonpositive and the function from the right-hand side of (29) is positive; hence inequality (29) holds true.
Case  2. For 0.8801802<x≤1, both sides of inequality (29) are positive, and thus we can rise to the second power and obtain(31)81-x2π-42x3+-16+8π+π2x2-5π-42x+11π2-72π+80≤π-42x5-16+8π-π2x4-26π-42x3+2208-184π+51π2x2+41π-42x-119π2+504π-656.
The polynomial functionRx=π-42x3+-16+8π+π2x2-5π-42x+11π2-72π+80 has the real roots x1≈-25.9051, x2≈1.3467, or x3≈1.4637; hence Rx<0 for all 0.8801802<x≤1.
The polynomial functionQx=π-42x5-16+8π-π2x4-26π-42x3+2208-184π+51π2x2+41π-42x-119π2+504π-656 has the real roots x1≈2.9092, x2≈-1.0537, x3=1, x4≈2.5536, or x5≈42.8403; hence Qx<0 for all 0.88018026<x≤1.
Therefore, both sides of inequality (31) are nonpositive, so we multiply it by -1 and then we can rise to the second power and obtain (32)1-x·Sx≥0, where (33)Sx=π4-16π3+96π2-256π+256x9+3π4-48π3+224π2-256π-256x8+16π4-256π3+1408π2-3072π+3072x7+296π4-1792π3-128π2+11264π-3072x6+618π4-4000π3+10560π2-13824π+13824x5+-4202π4+38688π3-101312π2+66048π-13824x4+7416π4-34176π3+35712π2-27648π+27648x3+14160π4-84096π3+107904π2+64512π-27648x2+-3699π4+1584π3+42336π2-20736π+20736x+-6417π4+18576π3+34272π2-76032π-20736.
Using MATLAB software, we find that the polynomial functionS has only one real root x1≈-0.9411; hence 1-x·Sx≥0 on [0.88018026,1].
For proving the right-hand side of the inequality from Theorem5 we introduce the function (34)t:0,1⟶R,tx=arcsin⁡x-πx2+1-x2-π-41-x22-π1-x4.
We note thatt1=0; therefore in order to demonstrate the inequality tx≤0 on [0,1], we have to establish that t′x≥0 on [0,1] or equivalently (35)22-421+πx2+162-3π2-π2·x21-x2≥4-π4x+11-x-x2-51+x.
Since both sides of the above inequality are positive on[0,1], we can rise to the second power and get(36)x-181-x2·Ux-Vx≥0,where(37)Ux=2π2+2πx3+3π2-6π+16x2+10π2-50πx+5π2-10π-80,Vx=2π2x5+16-8π+3π2x4-18π2+8πx3+-416+64π-44π2x2+464π-64π2x+656+136π-23π2.
The polynomial functionUx has the real roots x1≈1.43, x2≈-1.10, and x3≈1.5081; hence Ux<0 on [0,1].
The polynomial functionVx has the real roots x1≈-5.53, x2≈-1.13, x3≈-1.0713, x4≈1.1722, and x5≈5.52; hence Vx>0 on [0,1].
Therefore, inequality (36) is true on [0,1].
This completes the proof.Remark 7.
Combining the results of Theorems3 and 5, respectively, we note that we improve the right-hand side of Shafer-Fink’s inequality on [0,1] as follows: (38)arcsin⁡x≤πx2+1-x2+1-π3x∀0≤x≤0.871433,arcsin⁡x≤πx2+1-x2+π-41-x22+π1-x4∀0.85068≤x≤1.


## 3. Conclusion

In the present work we investigated the approximation of the inverse sine function and obtained new bounds. We have deduced lower and upper bounds which are sharp and very accurate and also improve Shafer-Fink’s inequality.



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