A DOUBLE INEQUALITY FOR REMAINDER OF POWER SERIES OF TANGENT FUNCTION

CHAO-PING CHEN AND FENG QI

Abstract. By mathematical induction, an identity and a double inequality for remainder of power series of tangent function are established.

1. Introduction

It is well known that Bernoulli numbers B_{i} are defined [11] by

\frac{x}{e^{x} − 1} = 1 − \frac{1}{2}x + ∑_{i=1}^{∞}(−1)^{i} \frac{B_{i}}{(2i)!}x^{2i−1}, |x| < 2π.  (1)

About Bernoulli numbers, some new results can be found in [1, 3, 5].

The tangent and cotangent can be expanded into power series with coefficients involving Bernoulli numbers as follows [11, p. 5]:

tan x = ∑_{i=1}^{∞} \frac{2^{2i}(2^{2i} − 1)B_{i}}{(2i)!} x^{2i−1},  |x| < \frac{π}{2} ;  (2)
cot x = \frac{1}{x} − ∑_{i=1}^{∞} \frac{2^{2i}B_{i}}{(2i)!} x^{2i−1},  |x| < π.  (3)

Introduce two notations S_{n}(x) and r_{n}(x) by

S_{n}(x) = ∑_{i=1}^{∞} \frac{2^{2i}(2^{2i} − 1)B_{i}}{(2i)!} x^{2i−1},  (4)
r_{n}(x) = tan x − S_{n}(x)  (5)

Date: May 16, 2002. 2000 Mathematics Subject Classification. Primary 26D05; Secondary 26D15. Key words and phrases. tangent function, a double inequality, power series, remainder,

Bernoulli number.

The authors were supported in part by NNSF (#10001016) of China, SF for the Prominent Youth of Henan Province, SF of Henan Innovation Talents at Universities, Doctor Fund of Jiaozuo Institute of Technology, China.

This paper was typeset using AMS-L^{A}TEX.

2 CH.-P. CHEN AND F. QI

for 0 < x < \frac{π}{2} . Then tan x = lim_{n→∞} S_{n}(x). We call r_{n}(x) the remainder of power series for tangent function.

For elementary functions sin x, cos x, and e^{x}, there are much literature on estimates of their remainder. For examples, see [6, 7, 9]. The methods used in [6, 7, 9] have been applied to construct inequalities of elliptic integrals. See [8, 10]. Some inequalities involving tan x were researched by the second author and others in [2].

In this article, we will establish a double inequality for remainder r_{n}(x) of power series for tan x. That is Theorem 1. For x ∈ (0, \frac{π}{2} ) and n ∈ ℕ, we have

\frac{2^{2n+1}(2^{2n+1} − 1)B_{n+1}}{(2n + 2)!} x^{2n} tan x < tan x − S_{n}(x) < (\frac{2}{π} )^{2n} x^{2n} tan x. (6)

Remark 1. If taking n = 1 in (6), we have for x ∈ (0, 1)

\frac{π}{2} · \frac{x}{1 − \frac{π^{2}}{12} x^{2}} < tan \frac{πx}{2} < \frac{π}{2} · \frac{x}{1 − x^{2}} .  (7)

For 0 < x < \sqrt{3 − \frac{24}{π^{2}}} , the left inequality in (7) is better than the left inequality in the following Becker-Stark inequality [4, p. 351]:

\frac{4}{π} · \frac{x}{1 − x^{2}} < tan \frac{πx}{2} < \frac{π}{2} · \frac{x}{1 − x^{2}} , x ∈ (0, 1).  (8)

If taking n = 2 in (6), we obtain

x + \frac{1}{3}x^{3} + \frac{2}{15}x^{4} tan x < tan x < x + \frac{1}{3}x^{3} + (\frac{2}{π} )^{4} x^{4} tan x, x ∈ (0, \frac{π}{2} ) . (9)

The constants \frac{2}{15} and (\frac{2}{π} )^{4} in (9) are best possible.

For x ∈ (0, \frac{π}{6} ), the Djokvie inequality states [4, p. 350] that

x + \frac{1}{3}x^{3} < tan x < x + \frac{4}{9}x^{3}.  (10)

Since

\frac{1}{3} + (\frac{2}{π} )^{4} x tan x < \frac{1}{3} + (\frac{2}{π} )^{4} · \frac{π}{6} · \frac{1}{\sqrt{3}} < \frac{4}{9},

thus, the inequality in (9) is better than those in (10).

INEQUALITIES FOR REMAINDER OF POWER SERIES OF TANGENT FUNCTION 3

2. Proof of Theorem

Let

h_{n}(x) = \frac{tan x − S_{n}(x)}{x^{2n} tan x}  (11)

for n ∈ ℕ. Then we have the following lemma. Lemma 1. For x ∈ (0, \frac{π}{2} ) and n ∈ ℕ, we have

h_{n}(x) = ∑_{j=1}^{n} \frac{2^{2(n−j+1)}[2^{2(n−j+1)} − 1]B_{n−j+1}}{[2(n − j + 1)]!} ∑_{k=j}^{∞} \frac{2^{2k}B_{k}}{(2k)!} x^{2(k−j)}.  (12)

Proof. We shall prove this lemma by mathematical induction on n. For n = 1, we have

h_{1}(x) = \frac{tan x − S_{1}(x)}{x^{2} tan x}

= \frac{1}{x^{2}} − \frac{cot x}{x}

= \frac{1}{x^{2}} − \frac{1}{x} (\frac{1}{x} − ∑_{k=1}^{∞} \frac{2^{2k}B_{k}}{(2k)!} x^{2k−1})

= ∑_{k=1}^{∞} \frac{2^{2k}B_{k}}{(2k)!} x^{2(k−1)},

the formula (12) holds for n = 1.

For n = 2, we have

h_{2}(x) = \frac{tan x − S_{2}(x)}{x^{4} tan x}

= \frac{1}{x^{4}} − \frac{cot x}{x^{3}} − \frac{cot x}{3x}

= \frac{1}{x^{4}} − \frac{1}{x^{3}} (\frac{1}{x} − \frac{1}{3}x − ∑_{k=2}^{∞} \frac{2^{2k}B_{k}}{(2k)!} x^{2k−1})

− \frac{1}{3x} (\frac{1}{x} − ∑_{k=1}^{∞} \frac{2^{2k}B_{k}}{(2k)!} x^{2k−1})

= ∑_{k=2}^{∞} \frac{2^{2k}B_{k}}{(2k)!} x^{2(k−2)} + ∑_{k=1}^{∞} \frac{2^{2k}B_{k}}{3 · (2k)!}x^{2(k−1)},

the formula (12) holds for n = 2.

Assume formula (12) holds for n = m. Then for n = m + 1, we have

h_{m+1} = \frac{tan x − S_{m+1}(x)}{x^{2(m+1)} tan x}

= \frac{tan x − S_{m}(x) − \frac{2^{2(m+1)}(2^{2(m+1)}−1)B_{m+1}}{[2(m+1)]!} x^{2m+1}}{x^{2(m+1)} tan x}

4 CH.-P. CHEN AND F. QI

= \frac{1}{x^{2}} · \frac{tan x − S_{m}(x)}{x^{2m} tan x} − \frac{2^{2(m+1)}(2^{2(m+1)} − 1)B_{m+1}}{[2(m + 1)]!} · \frac{cot x}{x}

= \frac{1}{x^{2}} ∑_{j=1}^{m} \frac{2^{2(m−j+1)}[2^{2(m−j+1)} − 1]B_{m−j+1}}{[2(m − j + 1)]!} ∑_{k=j}^{∞} \frac{2^{2k}B_{k}}{(2k)!} x^{2(k−j)}

− \frac{2^{2(m+1)}(2^{2(m+1)} − 1)B_{m+1}}{[2(m + 1)]!} · \frac{1}{x} (\frac{1}{x} − ∑_{k=1}^{∞} \frac{2^{2k}B_{k}}{(2k)!} x^{2k−1})

= \frac{1}{x^{2}} ∑_{j=1}^{m} \frac{2^{2j}(2^{2j} − 1)B_{j}}{(2j)!} · \frac{2^{2(m−j+1)}B_{m−j+1}}{[2(m − j + 1)]!}  (13)

+ ∑_{j=2}^{m+1} \frac{2^{2(m−j+2)}(2^{2(m−j+2)} − 1)B_{m−j+2}}{[2(m − j + 2)]!} ∑_{k=j}^{∞} \frac{2^{2k}B_{k}}{(2k)!} x^{2(k−j)}

− \frac{2^{2(m+1)}[2^{2(m+1)} − 1]B_{m+1}}{[2(m + 1)]!} · \frac{1}{x^{2}}

+ \frac{2^{2(m+1)}[2^{2(m+1)} − 1]B_{m+1}}{[2(m + 1)]!} ∑_{k=1}^{∞} \frac{2^{2k}B_{k}}{(2k)!} x^{2(k−1)}

= ∑_{j=1}^{m+1} \frac{2^{2(m−j+2)}[2^{2(m−j+2)} − 1]B_{m−j+2}}{[2(m − j + 2)]!} ∑_{k=j}^{∞} \frac{2^{2k}B_{k}}{(2k)!} x^{2(k−j)}

+ \frac{1}{x^{2}} ∑_{j=1}^{m} \frac{2^{2j}(2^{2j} − 1)B_{j}}{(2j)!} · \frac{2^{2(m−j+1)}B_{m−j+1}}{[2(m − j + 1)]!}

− \frac{2^{2(m+1)}(2^{2(m+1)} − 1)B_{m+1}}{[2(m + 1)]!} · \frac{1}{x^{2}} .

Since tan x cot x = 1, we have

(∑_{i=1}^{∞} \frac{2^{2i}(2^{2i} − 1)B_{i}}{(2i)!} x^{2i−1}) (\frac{1}{x} − ∑_{i=1}^{∞} \frac{2^{2i}B_{i}}{(2i)!} x^{2i−1}) = 1,

which is equivalent to

∑_{i=2}^{∞} \frac{2^{2i}(2^{2i} − 1)B_{i}}{(2i)!} x^{2i−2} = [∑_{i=1}^{∞} \frac{2^{2i}(2^{2i} − 1)B_{i}}{(2i)!} x^{2i−1}] ∑_{i=1}^{∞} \frac{2^{2i}B_{i}}{(2i)!} x^{2i−1},  (14)

equating coefficients of the term x^{2m} on both sides of (14) yields

\frac{2^{2(m+1)}(2^{2(m+1)} − 1)B_{m+1}}{(2(m + 1))!} = ∑_{j=1}^{m} \frac{2^{2j}(2^{2j} − 1)B_{j}}{(2j)!} · \frac{2^{2(m−j+1)}B_{m−j+1}}{[2(m − j + 1)]!} . (15)

Substituting (15) into (13) and simplifying gives us

h_{m+1}(x) = ∑_{j=1}^{m+1} \frac{2^{2(m−j+2)}(2^{2(m−j+2)} − 1)B_{m−j+2}}{[2(m − j + 2)]!} ∑_{k=j}^{∞} \frac{2^{2k}B_{k}}{(2k)!} x^{2(k−j)}.  (16)

By induction, the proof of Lemma 1 is complete. □

Now we give a proof of Theorem 1.

Proof of Theorem 1. From (12), it is deduced that h′_{n}(x) > 0, and h_{n}(x) is strictly increasing in (0, \frac{π}{2} ). Easy computing yields

h_{n}(0 + 0) = \frac{2^{2n+2}(2^{2n+2} − 1)B_{n+1}}{(2n + 2)!} ,
h (\frac{π}{2} − 0) = (\frac{2}{π} )^{2n} .

Therefore, we have

\frac{2^{2n+2}(2^{2n+2} − 1)B_{n+1}}{(2n + 2)!} < h_{n}(x) < (\frac{2}{π} )^{2n} .  (17)

Inequalities in (17) are equivalent to the double inequality (6).

In [4, p. 421], the following inequalities are given

\frac{2}{π^{2n}4^{n}} < \frac{B_{2n}}{(2n)!} < \frac{2}{π^{2n}(4^{n} − 2)}.  (18)

Then we have

\frac{4^{n+1}(4^{n+1} − 1)B_{2n+2}}{(2n + 2)!} > (2 − \frac{1}{2^{2n+1}} ) (\frac{2}{π} )^{2n+2} .  (19)

The first inequality in (6) follows from (19).

The proof of Theorem 1 is complete. □

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(Chen) Department of Applied Mathematics and Informatics, Jiaozuo Institute of Technology, Jiaozuo City, Henan 454000, China

(Qi) Department of Applied Mathematics and Informatics, Jiaozuo Institute of Technology, Jiaozuo City, Henan 454000, China

E-mail address: qifeng@jzit.edu.cn or qifeng618@jzit.edu.cn
URL: http://rgmia.vu.edu.au/qi.html