2. Results and Discussion
Since limx→0arcsinx-3x/(2+1-x2)=0, it results that (1) gives a good results near zero, arcsinx≈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 number 0≤x≤1, the following two-sided inequality holds: (2)x5180+x7189≤arcsinx-3x2+1-x2≤π-32.
Proof.
The function (3)f:0,1⟶R,fx=arcsinx-3x2+1-x2-x5180-x7189 has the derivative (4)f′x=-4x6-3x4108+-x2+2-21-x21-x21-x2+22.
In order to prove that f′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 all 0≤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 that f′x≥0, which imply that the function f is strictly increasing on [0,1].
For proving the right-hand side inequality from Theorem 1, we introduce the function (8)g:0,1⟶R,gx=arcsinx-3x2+1-x2-π-32.
Its derivative g′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 arcsinx≈(3x+ax)/(1+21-x2), where ax→0 as x→0.
The result is stated as Theorem 2.
Theorem 2.
For every 0≤x≤1, one has(9)arcsinx-3x2+1-x2≥ax2+1-x2,where ax=1/60x5+11/840x7.
Proof.
The function(10)h:0,1⟶R,fx=arcsinx-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 that h′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 all 0≤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 Theorem 2 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 every x∈[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≤arcsinx-πx2+1-x2≤1-π3x.
Remark 4.
Using MATLAB software, we found that the equation(16)arcsinx-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=arcsinx-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 number α∈[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 roots x1=x2=0 and x3,4= ±32π2-9/(54-66π+17π2+5π-12π13π-36) ≈ 0.735975.
We choose α=0.735975.
Therefore, p′x≤0 for all 0≤x≤α and p′x>0 for all α<x≤0.871433.
Since p0=0 and p0.871433=0, we obtain that px≤0 on [0,0.871433].
For proving the left-hand side inequality from Theorem 3, we introduce the function (21)s:0,1⟶R,px=arcsinx-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 inequality s′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 Theorem 5, we will state the following inequalities which give good results near the number 1 for the approximation arcsinx≈πx/(2+1-x2)+bx.
Theorem 5.
For every 0≤x≤1, one has(26)π-41-x22≤arcsinx-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=arcsinx-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 that r1=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 function Rx=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 function Qx=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 function S 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 Theorem 5 we introduce the function (34)t:0,1⟶R,tx=arcsinx-πx2+1-x2-π-41-x22-π1-x4.
We note that t1=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 function Ux has the real roots x1≈1.43, x2≈-1.10, and x3≈1.5081; hence Ux<0 on [0,1].
The polynomial function Vx 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 Theorems 3 and 5, respectively, we note that we improve the right-hand side of Shafer-Fink’s inequality on [0,1] as follows: (38)arcsinx≤πx2+1-x2+1-π3x ∀0≤x≤0.871433,arcsinx≤πx2+1-x2+π-41-x22+π1-x4 ∀0.85068≤x≤1.