����������� ������ 54 (2002), 171–178

UDK 517.28

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MEAN VALUE THEOREMS IN q-CALCULUS

Predrag M. Rajković, Miomir S. Stanković, Sla - dana D. Marinković

Abstract. In this paper, some properties of continuous functions in q-analysis are investigated. The behavior of q-derivative in a neighborhood of a local extreme point is described. Two theorems are proved which are q-analogons of the fundamental theorems of the differential calculus. Also, two q-integral mean value theorems are proved and applied to estimating remainder term in q-Taylor formula. Finally, the previous results are used in considering some new iterative methods for equation solving.

1. Introduction

At the last quarter of the XX century, q-calculus appeared as a connection between mathematics and physics ([5], [6]). It has a lot of applications in different mathematical areas, such as number theory, combinatorics, orthogonal polynomials, basic hyper-geometric functions and other sciences—quantum theory, mechanics and theory of relativity.

Let q ∈ ℝ^{+} \ {1}. A q-natural number [n]_{q} is defined by

[n]_{q} := 1 + q + ⋯ + q^{n−1}, n ∈ ℕ.

Generally, a q-complex number [a]_{q} is [a]_{q} := \frac{1 − q^{a}}{1 − q} , a ∈ ℂ. The factorial of a

number [n]_{q} is [0]_{q}! := 1, [n]_{q}! := [n]_{q}[n − 1]_{q} ⋯ [1]_{q}, n ∈ ℕ.

Let q-derivative of a function f(z) be

(D_{q}f)(z) := \frac{f(z) − f(qz)}{z − qz}, z ≠ 0, (D_{q}f)(0) := lim_{z→0}(D_{q}f)(z),

and high q-derivatives are

D_{q}^{0}f := f, D_{q}^{n}f := D_{q}(D_{q}^{n−1}f), n = 1, 2, 3,….

Notice, that a continuous function on an interval, which does not include 0, is continuous q-differentiable.

AMS Subject Classification: 26 A 99

Keywords and phrases: q-calculus.

Communicated at the 5th International Symposium on Mathematical Analysis and its Applications, Niška banja, Yugoslavia, October, 2–6, 2002.

Supported by Ministry of Sci. & Techn. Rep. Serbia, the projects No. 1409/1379.

2. Extreme values and q-derivative

We will consider relations between, on one side, extreme value of a continuous function and, on the other side, derivatives and q-derivatives.

Theorem 2.1. Let f(x) be a continuous function on a segment [a, b] and let c ∈ (a, b) bea point of its local maximum.

(i) If 0 < a < b, then there exists q̂ ∈ (0, 1) such that

(D_{q}f)(c)\{ \begin{matrix}≥ 0, & ∀q ∈ (q̂, 1) \\ ≤ 0, & ∀q ∈ (1, q̂^{−1}). \end{matrix}

(ii) If a < b < 0, then there exists q̂ ∈ (0, 1) such

(D_{q}f)(c)\{ \begin{matrix}≤ 0, & ∀q ∈ (q̂, 1) \\ ≥ 0, & ∀q ∈ (1, q̂^{−1}). \end{matrix}

Furthermore, (∀q ∈ (q̂, 1) ∪ (1, q̂^{−1}))(∃ξ ∈ (a, b)) (D_{q}f)(ξ) = 0.

Proof. Since the proofs of (i) and (ii) are very similar, we will expose only the first one. Since c is a point of local maximum of the function f(x), there exists ε > 0, such that f(x) ≤ f(c), for all x ∈ (c − ε, c + ε) ⊂ (a, b). Let q_{0} ∈ (0, 1) such that c − ε < q_{0}c < c. Now, for all q ∈ (q_{0}, 1), it is valid qc < c and f(qc) ≤ f(c), wherefrom (D_{q}f)(c) ≥ 0. In a similar way, there exists q_{1} ∈ (0, 1) such that c < c/q_{1} < c + ε and for all q ∈ (1, q_{1}^{−1}) it holds (D_{q}f)(c) ≤ 0. At last, denote by q̂ = max{q_{0}, q_{1}}.

Let q ∈ (q̂, 1) be an arbitrary real number. Then η = c/q ∈ (c, c+ε), wherefrom f(c) ≥ f(η), i.e., f(qη) ≥ f(η). From qη < η we conclude (D_{q}f)(η) ≤ 0. As f(x) is a continuous function, (D_{q}f)(x) is continuous in (a, b), too. Since (D_{q}f)(c) ≥ 0, (D_{q}f)(η) ≤ 0, where c, η ∈ (a, b), there exists ξ ∈ (c, η) ⊂ (a, b), such that (D_{q}f)(ξ) = 0. Analogously, for an arbitrary q ∈ (1, q̂^{−1}), the number η = c/q ∈ (c − ε, c), wherefrom (D_{q}f)(η) ≥ 0. Since (D_{q}f)(c) ≤ 0, we have proved the existence of a zero of (D_{q}f)(x) for q ∈ (1, q̂^{−1}).

Example 2.1. Let us consider f(x) = (x − 1)(3 − x) + 2. Its maximum is at c = 2, but q-derivative is (D_{q}f)(x) = −[2]x + 4 and it vanishes at the point ξ = 4/(1 + q). So, here is q̂ = 1/3. For q = 3/4, we have (D_{3/4}f)(2) = 1/2, η = 2\frac{2}{3} and ξ = 2\frac{2}{7}.

In a similar way, we can prove the next theorem.

Theorem 2.2. Let f(x) be a continuous function on a segment [a, b] and let c ∈ (a, b) be a point of its local minimum.

(i) If 0 < a < b, then there exists q̂ ∈ (0, 1) such that

(D_{q}f)(c)\{ \begin{matrix}≤ 0, & ∀q ∈ (q̂, 1) \\ ≥ 0, & ∀q ∈ (1, q̂^{−1}). \end{matrix}

Mean value theorems in q-calculus 173

(ii) If a < b < 0, then there exists q̂ ∈ (0, 1) such that

(D_{q}f)(c)\{ \begin{matrix}≥ 0, & ∀q ∈ (q̂, 1) \\ ≤ 0, & ∀q ∈ (1, q̂^{−1}). \end{matrix}

Moreover, (∀q ∈ (q̂, q̂^{−1}))(∃ξ ∈ (a, b)) (D_{q}f)(ξ) = 0.

Remark. If f(x) is differentiable for all x ∈ (a, b), then lim_{q↑1} D_{q}f(x) = f′(x). So, if c ∈ (a, b) is a point of local extreme of f(x), we have f′(c) = D_{1}f(c) = 0.

3. Some q-mean value theorems

By using the previous results, we can establish and prove analogons of wellknown mean value theorems in q-calculus.

Theorem 3.1. (q-Rolle) Let f(x) be a continuous function on [a, b] satisfying f(a) = f(b). Then there exists q̂ ∈ (0, 1) such that

(∀q ∈ (q̂, 1) ∪ (1, q̂^{−1}))(∃ξ ∈ (a, b)) : (D_{q}f)(ξ) = 0.

Proof. If f(x) is not a constant function on [a, b], then it attains its extreme value in some point in (a, b). But, according to Theorems 2.1–2, (D_{q})f(x) vanishes at a point ξ ∈ (a, b).

Theorem 3.2 (q-Lagrange) Let f(x) be a continuous function on [a, b]. Then there exists q̂ ∈ (0, 1) such that

(∀q ∈ (q̂, 1) ∪ (1, q̂^{−1}))(∃ξ ∈ (a, b)) : f(b) − f(a) = (D_{q}f)(ξ)(b − a).

Proof. The statement follows by applying the previous theorem to the function f(x) − x(f(b) − f(a))/(b − a).

4. Mean value theorems for q-integrals

In q-analysis, we define q-integral by

I_{q}(f) = ∫_{0}^{a} f(t) d_{q}(t) := a(1 − q) ∑_{n=0}^{∞} f(aq^{n})q^{n}.

Notice that

I(f) = ∫_{0}^{a} f(t) dt = lim_{q↑1} I_{q}(f).

Theorem 4.1 Let f(x) be a continuous function on a segment [0, a] (a > 0). Then

(∀q ∈ (0, 1))(∃ξ ∈ [0, a]) : I_{q}(f) = ∫_{0}^{a} f(t) d_{q}(t) = a f(ξ).

Proof. Since f(x) is a continuous function on the segment [0, a], it attains its minimum m and maximum M and takes all values between. According to assumption 0 < q < 1, we have 0 < aq^{n} < a and m ≤ f(aq^{n}) ≤ M. Now,

a(1 − q) ∑_{n=0}^{∞} mq^{n} ≤ a(1 − q) ∑_{n=0}^{∞} f(aq^{n})q^{n} ≤ a(1 − q) ∑_{n=0}^{∞} Mq^{n},

wherefrom m ≤ \frac{1}{a} I_{q}(f) ≤ M. So, there exists ξ ∈ [0, a] such that a^{−1} I_{q}(f) = f(ξ).

Moreover, if we define

∫_{a}^{b} f(t) d_{q}(t) := ∫_{0}^{b} f(t) d_{q}(t) − ∫_{0}^{a} f(t) d_{q}(t),

then the next theorem is valid.

Theorem 4.2. Let f(x) be a continuous function on a segment [a, b]. Then there exists q̂ ∈ (0, 1) such that

(∀q ∈ (q̂, 1))(∃ξ ∈ (a, b)) : I_{q}(f) = ∫_{a}^{b} f(t) d_{q}(t) = f(ξ)(b − a).

Proof. It is easy to prove that lim_{q↑1} I_{q}(f) = I(f), i.e.

(∀ε > 0)(∃q_{0} ∈ (0, 1))(∀q ∈ (q_{0}, 1)) : I(f) − ε < I_{q}(f) < I(f) + ε.

According to the well known mean value theorem for integrals, we have

(∃c ∈ (a, b)) : I(f) = f(c)(b − a).

Let ε ≤ (b − a) min{M − f(c), f(c) − m}, where m and M are the minimum and maximum of f(x) on [a, b]. Now,

(∃q̂ ∈ (0, 1))(∀q ∈ (q̂, 1)) : f(c) − \frac{ε}{b − a} < \frac{1}{b − a} I_{q}(f) < f(c) + \frac{ε}{b − a},

hence m < I_{q}(f)/(b − a) < M. Since f(x) is a continuous function on the segment [a, b], it takes all values between m and M, i.e.

(∃ξ ∈ (a, b)) : \frac{1}{b − a}I_{q}(f) = f(ξ),

what we wanted to prove.

Theorem 4.3 Let f(x) and g(x) be some continuous functions on a segment [a, b]. Then there exists q̂ ∈ (0, 1) such that

(∀q ∈ (q̂, 1))(∃ξ ∈ (a, b)) : I_{q}(fg) = g(ξ)I_{q}(f).

Proof. According to the second mean value theorem for integrals, we have

(∃c ∈ (a, b)) : I(fg) = g(c)I(f).

Hence lim_{q↑1} I_{q}(fg) = g(c)I(f) = g(c) lim_{q↑1} I_{q}(f), i.e., lim_{q↑1} \frac{I_{q}(fg)}{I_{q}(f)} = g(c). Now,

(∃q_{0} ∈ (0, 1))(∀q ∈ (q_{0}, 1)) : g(c) − ε < \frac{I_{q}(fg)}{I_{q}(f)} < g(c) + ε.

Since g(x) is a continuous function on the segment [a, b], it attains its minimum m_{g} and maximum M_{g}. Let ε ≤ min{M_{g} − g(c), g(c) − m_{g}}. Hence

(∃q̂ ∈ (0, 1))(∀q ∈ (q̂, 1)) : m_{g} < \frac{I_{q}(fg)}{I_{q}(f)} < M_{g}.

Since f(x) takes all values between m_{g} and M_{g}, we conclude that

(∃ξ ∈ (a, b)) : \frac{I_{q}(fg)}{I_{q}(f)} = g(ξ).

5. Estimation of remainder term in q-Taylor formula

Let f(x) be a continuous function on some interval (a, b) and c ∈ [a, b]. Jackson’s q-Taylor formula (see [3], [4] and [2]) is given by

f(z) = ∑_{k=0}^{∞} \frac{(D_{q}^{k}f)(c)}{[k]_{q}!}(z − c)^{(k)}, z ∈ (a, b),

where

(z − c)^{(0)} = 1, (z − c)^{(k)} = ∏_{i=0}^{k−1}(z − cq^{i}) (k ∈ ℕ).

T. Ernst [2] have found the next q-Taylor formula

f(z) = ∑_{k=0}^{n−1} \frac{(D_{q}^{k}f)(c)}{[k]_{q}!}(z − c)^{(k)} + R_{n}(f, z, c, q),  (5.1)

where R_{n}(f, z, c, q) is the remainder term determined by

R_{n}(f, z, c, q) = ∫_{t=c}^{t=z} \frac{(z − t)^{(n)}}{z − t} \frac{(D_{q}^{n}f)(t)}{[n − 1]_{q}!} d_{q}(t).  (5.2)

Theorem 5.1. Let f(x) be a continuous function on [a, b] and R_{n}(f, z, c, q), z, c ∈ (a, b) be the remainder term in q-Taylor formula. Then there exists q̂ ∈ (0, 1) such that for all q ∈ (q̂, 1), ξ ∈ (a, b) can be found between c and z, which satisfies

R_{n}(f, z, c, q) = \frac{(D_{q}^{n}f)(ξ)}{[n − 1]_{q}!} ∫_{t=c}^{t=z} \frac{(z − t)^{(n)}}{z − t} d_{q}(t).  (5.3)

Proof. Since f(x) is a continuous function on [a, b], it can be expanded by q-Taylor formula (5.1) with the remainder term (5.2). Notice that the functions

\frac{(z − t)^{(n)}}{z − t} = ∏_{i=1}^{n−1}(z − tq^{i})

and (D_{q}^{n}f)(t)/[n − 1]_{q}! are continuous on the segment between c and z which is contained in (a, b). According to Theorem 4.3., there exists q̂ ∈ (0, 1), such that for all q ∈ (q̂, 1) can be found ξ between c and z such that (5.3) is valid.

Theorem 5.2. Let f(x) be a continuous function on [a, b] and z, c ∈ (a, b). Then there exists q̂ ∈ (0, 1) such that for all q ∈ (q̂, 1), ξ ∈ (a, b) can be found between c and z, which satisfies

f(z) = ∑_{k=0}^{n−1} \frac{(D_{q}^{k}f)(c)}{[k]_{q}!}(z − c)^{(k)} + \frac{(D_{q}^{n}f)(ξ)}{[n]_{q}!}(z − c)^{(n)}.

Proof. Applying \frac{(z − t)^{(n)}}{z − t} = −D_{q,t}(\frac{(z − t)^{(n)}}{[n]_{q}} ), to the integral in (5.3) we have

∫_{t=c}^{t=z} \frac{(z − t)^{(n)}}{z − t} d_{q}(t) = − ∫_{t=c}^{t=z} D_{q,t}(\frac{(z − t)^{(n)}}{[n]_{q}} ) d_{q}(t)

= −\frac{(z − t)^{(n)}}{[n]_{q}} |_{t=c}^{t=z} = \frac{(z − c)^{(n)}}{[n]_{q}} .

So, R_{n}(f, z, c, q) = \frac{(D_{q}^{n}f)(ξ)}{[n]_{q}!}(z − c)^{(n)}.

Theorem 5.3 Let f(x) be a continuous function on [a, b] and R_{n}(f, z, c, q), z, c ∈ (a, b) be the remainder term in q-Taylor formula. Then there exists q̂ ∈ (0, 1) such that for all q ∈ (q̂, 1), ξ ∈ (a, b) can be found between c and z, which satisfies

R_{n}(f, z, c, q) = \frac{(D_{q}^{n}f)(ξ)}{[n]_{q}!}(z − c)^{(n)}.

Proof. Since f(x) is a continuous function on [a, b], it can be expanded by q-Taylor formula (5.1) with the remainder term (5.2). Notice that the functions

\frac{(z − t)^{(n)}}{z − t} = ∏_{i=1}^{n−1}(z − tq^{i})

and (D_{q}^{n}f)(t)/[n − 1]_{q}! are continuous on the segment between c and z which is contained in (a, b). According to Theorem 4.3., there exists q̂ ∈ (0, 1), such that for all q ∈ (q̂, 1), ξ between c and z can be found such that

R_{n}(f, z, c, q) = \frac{(D_{q}^{n}f)(ξ)}{[n − 1]_{q}!} ∫_{t=c}^{t=z} \frac{(z − t)^{(n)}}{z − t} d_{q}(t).

Applying

\frac{(z − t)^{(n)}}{z − t} = −D_{q,t}(\frac{(z − t)^{(n)}}{[n]_{q}} ),

to the previous integral we have

∫_{t=c}^{t=z} \frac{(z − t)^{(n)}}{z − t} d_{q}(t) = − ∫_{t=c}^{t=z} D_{q,t}(\frac{(z − t)^{(n)}}{[n]_{q}} ) d_{q}(t)

= −\frac{(z − t)^{(n)}}{[n]_{q}} |_{t=c}^{t=z} = \frac{(z − c)^{(n)}}{[n]_{q}} .

So, R_{n}(f, z, c, q) = \frac{(D_{q}^{n}f)(ξ)}{[n]_{q}!}(z − c)^{(n)}.

6. Application

Here we will apply the previous theorems in analyzing an iterative method for solving equations.

Suppose that an equation f(x) = 0 has a unique isolated solution x = τ. If x_{n} is an approximation for the exact solution τ, using Jackson’s q-Taylor formula, we have

0 = f(τ) ≈ f(x_{n}) + (D_{q}f)(x_{n})(τ − x_{n}),

hence τ ≈ x_{n} − \frac{f(x_{n})}{(D_{q}f)(x_{n})}. So, we can construct q-Newton method

x_{n+1} = x_{n} − \frac{f(x_{n})}{(D_{q}f)(x_{n})}.  (6.1)

More simply, it looks like x_{n+1} = x_{n}\{1 − \frac{1 − q}{1 − \frac{f(qx_{n})}{f(x_{n})}} \}. This method written in the form

x_{n+1} = x_{n} − \frac{x_{n} − qx_{n}}{f(x_{n}) − f(qx_{n})}f(x_{n})

reminds to the method of chords (secants).

Theorem 6.1. Suppose that a function f(x) is continuous on a segment [a, b] and that the equation f(x) = 0 has a unique isolated solution τ ∈ (a, b). Let the conditions

|(D_{q}f)(x)| ≥ M_{1} > 0, |(D_{q}^{2}f)(x)| ≤ M_{2}

are satisfied for all x ∈ (a, b). Then there exists q̂ ∈ (0, 1), such that for all q ∈ (q̂, 1), the iterations obtained by q-Newton method satisfy

|τ − x_{k+1}| ≤ \frac{M_{2}}{(1 + q)M_{1}} |(τ − x_{k})^{(2)}|.

Proof. From the formulation of q-Newton method (6.1), we have

x_{k+1} − τ = x_{k} − τ − \frac{f(x_{k})}{(D_{q}f)(x_{k})},

hence f(x_{k})+(D_{q}f)(x_{k})(τ −x_{k}) = (D_{q}f)(x_{k})(τ −x_{k+1}). By using q-Taylor formula at the point x_{k} of order n = 2 for f(τ) we have

f(τ) = f(x_{k}) + (D_{q}f)(x_{k})(τ − x_{k}) + R_{2}(f, τ, x_{k}, q).

Since f(τ) = 0, we obtain (D_{q}f)(x_{k})(τ − x_{k+1}) = −R_{2}(f, τ, x_{k}, q), i.e.

|τ − x_{k+1}| = \frac{|R_{2}(f, τ, x_{k}, q)|}{|(D_{q}f)(x_{k})|}.

According to Theorem 5.1., there exists q̂ ∈ (0, 1) such that for all q ∈ (q̂, 1), ξ ∈ (a, b) can be found such that

R_{2}(f, τ, x_{k}, q) = \frac{(D_{q}^{2}f)(ξ)}{[2]_{q}}(τ − x_{k})^{(2)}.

Now,

|τ − x_{k+1}| = \frac{|(D_{q}^{2}f)(ξ)|}{|(D_{q}f)(x_{k})|} \frac{|(τ − x_{k})^{(2)}|}{1 + q}.

Using the conditions which function f(x) and its q-derivatives satisfy we obtain the statement of the theorem.

Remark. In our papers [7] and [8] we have discussed q-iterative methods in details.

REFERENCES

[1] N. S. Bakhvalov, Numerical Methods, Moscow State University, Moscow, 1977.

[2] T. Ernst, A new notation for q-calculus and a new q-Taylor formula, Uppsala University Report Depart. Math., (1999), 1–28.

[3] F.H. Jackson, A q-form of Taylor’s theorem, Mess. Math. Math. 38, (1909), 62–64.

[4] S. C. Jing and H. Y. Fan, q-Taylor’s formula with its q-remainder, Comm. Theoret. Phys. 23, 1 (1995), 117–120.

[5] E. Koelink, Eight lectures on quantum groups and q-special functions, Revista Colombiana de Matematicas 30, (1996), 93–180.

[6] T. H. Koornwinder and R. F. Swarttow, On q-analogues of the Fourier and Hankel transforms, Trans. Amer. Math. Soc. 333, (1992), 445–461.

[7] P. M. Rajković, M. S. Stanković and S. D. Marinković, On q-iterative methods for equation solving, Conf. Applied Math. PRIM, Zlatibor, Yugoslavia, 2002.

[8] M.S. Stanković, P.M. Rajković and S.D. Marinković, Solving equations by q-iterative methods and q-Sendov conjecture, Constructive Function Theory (Ed. B. Bojanov), Varna Bulgaria 2002.

(received 23.12.2002)

Predrag M. Rajković, Faculty of Mechanical Engineering, University of Niš, 18000 Niš, Yugoslavia E-mail: pecar@masfak.ni.ac.yu Miomir S. Stanković, Faculty of Occupational Safety, University of Niš, 18000 Niš, Yugoslavia E-mail: mstan@znrfak.znrfak.ni.ac.yu Sla - dana D. Marinković, Faculty of Electrical Engineering, University of Niš, 18000 Niš, Yugoslavia E-mail: sladjana@elfak.ni.ac.yu