On the quantum differentiation of smooth real-valued functions

TL;DR

This paper introduces a quantum-inspired approach to differentiate smooth real-valued functions by employing $q$-difference operators, Newton's interpolation, and power differences, providing new methods for analyzing derivatives within convergence radii.

Contribution

It presents a novel quantum differentiation method using $q$-difference operators and extends Newton's interpolation to discrete analogs of Taylor series for smooth functions.

Findings

Develops a $q$-difference operator for differentiation.

Introduces $(P,q)$-power difference in Section 5.

Shows interpolation using $q$-difference and power differences.

Abstract

Calculating the value of $C^{k\in\{1,\infty\}}$ class of smoothness real-valued function's derivative in point of $\mathbb{R}^+$ in radius of convergence of its Taylor polynomial (or series), applying an analog of Newton's binomial theorem and $q$-difference operator. $(P,q)$-power difference introduced in section 5. Additionally, by means of Newton's interpolation formula, the discrete analog of Taylor series, interpolation using $q$-difference and $p,q$-power difference is shown.

Equations

$$f(x)=f(a)+\frac{f^{\prime }(a)}{1!}(x-a)+\frac{f^{\prime \prime }(a)}{2!}(x-a{)}^2+\cdots +\frac{f^{(n)}(a)}{n!}(x-a{)}^n+R_{n+1}(x)$$

$$R_{n+1}(x)={\left(\frac{x-a}{x-a}\right)}^p\frac{(x-c{)}^{n+1}}{n!p}{f}^{(n+1)}(c)$$

$$\phi (x,a)=f(a)+\frac{f^{\prime }(a)}{1!}(x-a)+\frac{f^{\prime \prime }(a)}{2!}(x-a{)}^2+\cdots +\frac{f^{(n)}(a)}{n!}(x-a{)}^n$$

$$R_{n+1}(x)=f(x)-\phi (x,a)$$

$$\psi (t)=f(x)-\phi (x,t)-(x-t{)}^{p}Q(x)$$

$$Q(x)=\frac{R_{n+1}(x)}{(x-a{)}^p}$$

$$\psi (t)=f(x)-f(t)-\frac{f^{\prime }(t)}{1!}(x-t)-\frac{f^{\prime \prime }(t)}{2!}(x-t{)}^2-\cdots -\frac{f^{(n)}(t)}{n!}(x-t{)}^n$$

$$-(x-t{)}^{p}Q(x)$$

$$\psi (a)=f(x)-\phi (x,a)-R_{n+1}(x)$$

$${\psi }^{\prime }(c)=0$$

$${\psi }^{\prime }(t)=-f^{\prime }(t)+\frac{f^{\prime }(t)}{1!}-\frac{f^{\prime \prime }(t)}{2!}(x-t)+\frac{f^{\prime \prime }(t)}{2!}2(x-t)-\cdots$$

$$+\frac{f^{(n)}(t)}{n!}n(x-t{)}^{n-1}-\frac{f^{(n+1)}(t)}{n!}(x-t{)}^n+p(x-t{)}^{p-1}Q(x)$$

$${\psi }^{\prime }(t)=-\frac{f^{(n+1)}(t)}{n!}(x-t{)}^n+p\cdot (x-t{)}^{p-1}Q(x)$$

$$Q(x)=\frac{(x-c{)}^{n-p+1}}{n!p}{f}^{(n+1)}(c)$$

$$R_{n+1}(x)=(x-a{)}^{p}Q(x)={\left(\frac{x-a}{x-a}\right)}^p\frac{(x-c{)}^{n+1}}{n!p}{f}^{(n+1)}(c)$$

$$\frac{d}{dx}f(x)=\frac{f^{\prime }(a)}{1!}\frac{d}{dx}(x-a)+\frac{f^{\prime \prime }(a)}{2!}\frac{d}{dx}(x-a{)}^2+\cdots +\frac{f^{(k-1)}(a)}{(k-1)!}\frac{d}{dx}(x-a{)}^k$$

$$+R_{k+1}^{\prime }(x)$$

$$\frac{d}{dx}f(x)=\frac{f^{\prime }(a)}{1!}\frac{d}{dx}(x-a)+\frac{f^{\prime \prime }(a)}{2!}\frac{d}{dx}(x-a{)}^2+\cdots +\frac{f^{(n)}(a)}{n!}\frac{d}{dx}(x-a{)}^n+$$

$$+\cdots$$

$$\frac{d(x^n)}{dx}=\underset{\Delta x\to 0}{\lim }\left\{\sum _{k=1}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)x^{n-k}(\Delta x{)}^{k-1}\right\}=\left(\genfrac{}{}{0ex}{}{n}{1}\right)x^{n-1}$$

$$x^n=x^{n-2}+j\sum _{k\in \mathfrak{C}(x)}k\cdot x^{n-2}-k^2\cdot x^{n-3},x\in \mathbb{N}$$

$$x^n=\sum _{k\in \mathfrak{U}(x)}j\cdot k\cdot x^{n-2}-j\cdot k^2\cdot x^{n-3}+x^{n-3}$$

$$T(x,\mathfrak{U}(x))\equiv T(x,\mathfrak{S}(x)),x\in \mathbb{N}$$

$$U(x,\mathfrak{C}(x))\equiv U(x,\mathfrak{S}(x))\equiv U(x,\mathfrak{U}(x))$$

$$\underset{x^n}{\underbrace{\sum _{t=1}^x\sum _{k=1}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)t^{n-k}}}\equiv \sum _{k=0}^{x-1}j\cdot k\cdot x^{n-2}-j\cdot k^2\cdot x^{n-3}+x^{n-3}$$

$$\Delta (x^n)=\sum _{k=1}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)x^{n-k}\ne j\cdot k\cdot x^{n-2}-j\cdot k^2\cdot x^{n-3}+x^{n-3}$$

$$(D_{q}f)(x)=\frac{f(x)-f(qx)}{(1-q)x},x\ne 0$$

$$\frac{df}{dx}=\underset{q\to 1^{-}}{\lim }(D_{q}f)(x)$$

$$\frac{df(x)}{dx}=\underset{q\to 1^{-}}{\lim }\underset{(D_q{f}^{-})(x)}{\underbrace{\frac{f(x)-f(xq)}{x-xq}}}\equiv \underset{q\to 1^{+}}{\lim }\underset{(D_q{f}^{+})(x)}{\underbrace{\frac{f(xq)-f(x)}{xq-x}}}$$

$$\frac{d(x^n)}{dx}=\underset{q\to 1^{+}}{\lim }(D_q{x}^{n^{+}})(x)=\underset{q\to 1^{+}}{\lim }\frac{x^n(q^n-1)}{x(q-1)}$$

Full text

Keywords. derivative, differential calculus, differentiation, Taylor’s theorem, Taylor’s formula, Taylor’s series, Taylor’s polynomial, power function, Binomial theorem, smooth function, real calculus, Newton’s interpolation formula, finite difference, q-derivative, Jackson derivative, q-calculus, quantum calculus, (p,q)-derivative, (p,q)-Taylor formula

2010 Math. Subject Class. 26A24, 05A30, 41A58

e-mail:  kolosov_94@mail.ua

ORCID:  http://orcid.org/0000-0002-6544-8880

1. 1 Introduction
2. 2 Application of Q-Derivative
3. 3 Application on functions of finite class of smoothness
4. 4 Application on analytic functions
5. 5 Introduction of $(P,\ q)$-power difference
6. 6 Newton’s interpolation formula
7. 7 Conclusion

1. Introduction

Let be Taylor’s theorem (see §7 ”Taylor’s formula”, [1])

Theorem 1.1**.**

Taylor’s theorem. Let be $n\geq 1$ an integer, let function $f(x)$ be $n+1$ times differentiable in neighborhood of $a\in\mathbb{R}$. Let $x$ be an any function’s argument from such neighborhood, $p$ - some positive number. Then, there is exist some $c$ between points $a$ and $x$, such that

[TABLE]

where $R_{n+1}(x)$ - general form of remainder term

[TABLE]

Proof.

Denote $\varphi(x,a)$ polynomial related to $x$ of order $n$, from right part of (1.2), i.e

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Next, denote as $R_{n+1}(x)$ the difference

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Theorem will be proven, if we will find that $R_{n+1}(x)$ is defined by (1.3). Let fix some $x$ in neighborhood, mentioned in theorem 1.1. By definition, let be $x>a$. Denote by $t$ an variable, such that $t\in[a,x]$, and review auxiliary function $\psi(t)$ of the form

[TABLE]

where

[TABLE]

More detailed $\psi(t)$ could be written as

[TABLE]

[TABLE]

Our aim - to express $Q(x)$, going from properties of introduced function $\psi(t)$. Let show that function $\psi(t)$ satisfies to all conditions of Rolle’s theorem [2] on $[a,x]$. From (1.8) and conditions given to function $f(x)$, it’s obvious, that function $\psi(t)$ continuous on $[a,x]$. Given $t=a$ in (1.6) and keeping attention to equality (1.7), we have

[TABLE]

Hence, by means of (1.5) obtain $\psi(a)=0$. Equivalent $\psi(x)=0$ immediately follows from (1.8). So, $\psi(t)$ on segment $[a,x]$ satisfies to all necessary conditions of Rolle’s theorem [2]. By Rolle’s theorem, there is exist some $c\in[a,x]$, such that

[TABLE]

Calculating derivative $\psi^{\prime}(t)$, differentiating equality (1.8), we have

[TABLE]

[TABLE]

It’s seen that all terms in right part of (1.11), except last two items, self-destructs. Hereby,

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Given $t=c$ in (1.12) and applying (1.10), obtain

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By means of (1.13) and (1.7), finally, we have

[TABLE]

Case $x<a$ is reviewed absolutely similarly. (see for reference [1], pp 246-247)

This proves the theorem. ∎

Let function $f(x)\in C^{k}$ class of smoothness and satisfies to theorem 1.1, then its derivative by means of its Taylor’s polynomial centered at $a\in\mathbb{R}$ in radius of convergence with $f(x)$ and linear nature of derivative, $(gf(x)+um(x))^{{}^{\prime}}=gf^{{}^{\prime}}(x)+um^{{}^{\prime}}(x)$, is

[TABLE]

[TABLE]

Otherwise, if $f\in C^{\infty}$ we have derivative of Taylor series [5] of $f$ given the same conditions as (1.15)

[TABLE]

[TABLE]

Hence, derivative of function $f:1\leq C(f)\leq\infty$ 111For example, let $f$ be a $k$-smooth function, then $C(f)=k$. could be reached by differentiating of its Taylor’s polynomial or series in radius of convergence, and consequently summation of power’s derivatives being multiplied by coefficient, according theorem 1.1, over $k$ from $1$ to $t\leq\infty$, depending on class of smoothness. Hereby, the properties of power function’s differentiation holds, in particular, the derivative of power close related to Newton’s binomial theorem [4].

Lemma 1.2**.**

Derivative of power function equals to limit of Binomial expansion of $(x+\Delta x)^{n}$, iterated from $1$ to $n$, divided by $\Delta x:\Delta x\to 0$.

Proof.

[TABLE]

∎

According to lemma (1.17), Binomial expansion is used to reach derivative of power, otherwise, let be introduced expansion, based on forward finite differences, discussed in [3]

[TABLE]

where $j=3!\ \mathrm{and}\ \mathfrak{C}(x):=\{0,\ 1,\ \ldots,\ x\}\subseteq\mathbb{N}$. Particularize222Transferring $x^{n-2}$ under sigma operator, decreasing the power by 1 and taking summation over $k\in\mathfrak{U}(x)$ (1.19), one has

[TABLE]

where $\mathfrak{U}(x):=\{0,\ 1,\ \ldots,\ x-1\}\subseteq\mathbb{N}$.

Property 1.3**.**

Let $\mathfrak{S}(x)$ be a set $\mathfrak{S}(x):=\{1,\ 2,\ \ldots,\ x\}\subseteq\mathbb{N}$, let be (1.20) written as $T(x,\ \mathfrak{U}(x))$, then we have equality

[TABLE]

Let (1.19) be denoted as $U(x,\ \mathfrak{C}(x))$, then

[TABLE]

Proof.

Let be a plot of $jkx^{n-2}-jk^{2}x^{n-3}+x^{n-3}$ by $k$ over $\mathbb{R}_{\leq 10}^{+}$, given $x=10$

[math]2$$4$$6$$8$$10[math]50$$100$$150$$k\in\mathbb{R}_{\leq 10}$$y=jkx^{n-2}-jk^{2}x^{n-3}+x^{n-3}

Figure 1. Plot of $jkx^{n-2}-jk^{2}x^{n-3}+x^{n-3}$ by $k$ over $\mathbb{R}_{\leq 10}^{+}$, $x=10$

Obviously, being a parabolic function, it’s symmetrical over $\tfrac{x}{2}$, hence equivalent $T(x,\ \mathfrak{U}(x))\equiv T(x,\ \mathfrak{S}(x)),\ x\in\mathbb{N}$ follows. Reviewing (1.19) and denote $u(t)=tx^{n-2}-t^{2}x^{n-3}$, we can make conclusion, that $u(0)\equiv u(x)$, then equality of $U(x,\ \mathfrak{C}(x))\equiv U(x,\ \mathfrak{S}(x))\equiv U(x,\ \mathfrak{U}(x))$ immediately follows.

This completes the proof. ∎

By definition we will use set $\mathfrak{U}(x)\subseteq\mathbb{N}$ in our next expressions.

Since, for each $x=x_{0}\in\mathbb{N}$ we have equivalent

Lemma 1.4**.**

$\forall x=x_{0}\in\mathbb{N}$* holds*

[TABLE]

Proof.

Proof can be done by direct calculations. ∎

By lemma 1.24 we have right to substitute (1.20) into limit (1.18), replacing Binomial expansion, and represent derivative of power by means of expression (1.20). Note that,

[TABLE]

As (1.20) is analog of Binomial expansion of power and works only in space of natural numbers, different in sense, that Binomial expansion, for example, could be denoted as $M(x,\ \mathfrak{C}(n))$, where $n$ - exponent. While (1.20) could be denoted $T(x,\ \mathfrak{U}(x)\equiv\mathfrak{S}(x))$, it shows that in case of Binomial expansion the set over which we take summation depends on exponent $n$ of initial function, when for (1.20) it depends on point $x=x_{0}\in\mathbb{N}$. To provide expressions’ (1.20) usefulness333By classical definition of derivative, we have to use upper summation bound $(x+\Delta x)\in\mathbb{R^{+}}$ on (1.20), which turns false result as (1.20) works in space of $\mathbb{N}$. on taking power’s derivative over $\mathbb{R}^{+}$, derivative in terms of quantum calculus should be applied, as next section dedicated to.

2. Application of Q-Derivative

Derivative of the function $f$ defined as limit of division of function’s grow rate by argument’s grow rate, when grow rate tends to zero, and graphically could be interpreted as follows

x$$y[math]x_{0}$$x_{0}+\Delta xtangent line at $x_{0}$ as \Delta x\to 0^{+}$$f(x_{0})$$f(x)$$f(x_{0}+\Delta x)

Figure 2. Geometrical sense of derivative

In 1908 Jackson [10] reintroduced [11], [12] the Euler-Jackson $q$-difference operator [9]

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The limit as $q$ approaches $1^{-}$ is the derivative

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More generalized form of $q$-derivative

[TABLE]

where $(D_{q}f^{+})(x)\ \mathrm{and}\ (D_{q}f^{-})(x)$ forward and backward $q$-differences, respectively. The follow figure shows the geometrical sense of above equation as $q$ tends to $1^{+}$

x$$y[math]x_{0}$$x_{0}\cdot q$$f(x)tangent line at $x_{0}$ as q\to 1^{+}$$f(x_{0})$$f(x_{0}\cdot q)

Figure 3. Geometrical sense of right part of (2.3)

Review the monomial $x^{n}$, where $n$-positive integer and applying right part of (2.3), then in terms of $q$-calculus we have forward $q$-derivative over $\mathbb{R}$

[TABLE]

[TABLE]

Otherwise, see reference [9], equation (109).

Generalized view of high-order power’s derivative by means of (2.4)

[TABLE]

Since, the main property of power is

Property 2.1**.**

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Let be definition

Definition 2.2**.**

By property (2.6) and (1.20), definition of $c=x\cdot t:t\in\mathbb{R},\ x\in\mathbb{N}\Rightarrow c\in\mathbb{R}$ to power $n\in\mathbb{N}$

[TABLE]

Hereby, applying definition (2.7) and (2.4), derivative of monomial $x^{n}:n\in\mathbb{N}$ by $x$ in point $x_{0}\in\mathbb{N}$ is

[TABLE]

Let us approach to extend the definition space of expression (2.9) from $x_{0}\in\mathbb{N}$ to $x_{0}\in\mathbb{R}^{+}$. Let be $x_{0}=\xi(t_{0},\ p)_{1}\in\mathbb{R}^{+}\not\supseteq\mathbb{N}$ as $p\in\mathbb{R}^{+}\not\supseteq\mathbb{N}$ and $t_{0}\in\mathbb{N}$, then applying $(p,q)$-difference discussed in [13]

[TABLE]

by means of definition (2.7) and (2.10), $(p,q)$-differentiating of monomial $x^{n},\ n\in\mathbb{N}$ gives us

[TABLE]

[TABLE]

Geometrical interpretation is shown below

x$$y[math]x_{0}\cdot q$$x_{0}\cdot ptangent line at $x_{0}$ as p\to q^{+}$$f(x)=x^{n}$$\xi(x_{0},\ q)_{n}$$\xi(x_{0},\ p)_{n}

Figure 4. Geometrical interpretation of (2.11)

3. Application on functions of finite class of smoothness

In this section we will get derivative of function $f\in C^{n}$ in point $x_{0}\in\mathbb{R}^{+}$ by means of its Taylor’s polynomial and (2.9), where $n$ - some positive integer. Let $f(x)$ be an $n$-smooth function, then derivative of its Taylor’s polynomial at radius of convergence with $f$ in $x_{0}:(x_{0}-a)\in\mathbb{N}$ is

[TABLE]

[TABLE]

Otherwise, let $(x_{0}-a)$ satisfies to conditions of (2.11), i.e $(x_{0}-a)\in\mathbb{R}^{+}$, then applying operator $\mathscr{D}$, defined in (2.9) we can reach derivative of $f:f\in C^{n}$ in point $x_{0}:(x_{0}-a)\in\mathbb{R}^{+}$, by differentiation of its Taylor’s polynomial in radius of convergence with $f$, that is

[TABLE]

[TABLE]

4. Application on analytic functions

If $f\in C^{\infty}$ (i.e analytic), then approximation by means of Taylor series holds in neighborhood of its center at $a\in\mathbb{R}$. Suppose that $f$ is real-valued and satisfies to conditions of Taylor’s theorem 1.1, then derivative of $f$ at $x_{0}:x<x_{0}<a$ is

[TABLE]

Let $x_{0}$ satisfies to conditions of (3.1), then, applying definition (2.7), we have derivative of $f$ in point $x_{0}\in\mathbb{R}^{+}$

[TABLE]

Otherwise, if $x_{0}$ satisfies to conditions of (3.2) and $x_{0}$ in radius of convergence with $f$, then derivative of $f\in C^{\infty}$, by means of its Taylor’s series and (2.7), is

[TABLE]

5. Introduction of $(P,\ q)$-power difference

Lemma 5.1**.**

Let be $m\in\mathbb{R}/\mathbb{I}$ and $m$ could be represented as $m=at$, then exists some $c\in\mathbb{R}/\mathbb{I}$, such that

[TABLE]

Reviewing (2.3), we can see, that argument’s differential $\Delta x$ is given by $x\cdot q-x$, according to lemma 5.1 $\exists c\in\mathbb{R}/\mathbb{I},\ x\cdot t-x=x^{c}-x$, then, from (2.4) immediately follows $q$-power difference, (see [14], page 2, equation 3)

[TABLE]

As $q$ tends to $1^{+}$ we have reached derivative

[TABLE]

[TABLE]

where $\lim\limits_{q\to 1^{-}}\mathcal{D}_{q<1}f(x)$ denotes the derivative through backward $q$-power difference By lemma 5.1 from (2.10) immediately follows $(p,q)$-power difference

[TABLE]

Hence, for $v=x^{p},\ p\in\mathbb{R}$

[TABLE]

[TABLE]

where $\mathbf{D}_{p\rightarrow q}[f(x)],\ \mathbf{D}_{p\leftarrow q}[f(x)]$ denote derivative through forward and backward $(p,q)$-power differences. Let us to show geometrical interpretation of (5.4) and (5.6)

x$$y[math]x_{0}$$x_{0}^{q}$$f(x)tangent line at $x_{0}$ as q\to 1^{+}$$f(x_{0})$$f(x_{0}^{q})

Figure 5. Geometrical sense of (5.4)

x$$y[math]x_{0}^{q}$$x_{0}^{p}$$f(x)tangent line at $x_{0}$ as p\to q^{+}$$f(x_{0}^{q})$$f(x_{0}^{p})

Figure 6. Geometrical sense of (5.6)

Applying (5.4) with monomial $x^{m}:m\in\mathbb{N}$, we get

[TABLE]

[TABLE]

Note that $\mathbf{D}_{q<1}[x^{m}],\ \mathbf{D}_{q>1}[x^{m}]$ defined by (5.4). The high order $N\leq m$ derivative, derived from (5.7)

[TABLE]

[TABLE]

Let be analytic function $f$ and let $f$ satisfies to Taylor’s theorem 1.1 on segment of $(a,x),\ a\in\mathbb{R}$, then, applying (5.4), in radius of convergence of its Taylor’s series, we obtain derivative

[TABLE]

Using $\mathbf{D}_{p\rightarrow q}[f(x)],\ \mathbf{D}_{p\leftarrow q}[f(x)]$ defined by (5.8), for each $v=x^{p}$, we receive

[TABLE]

Or, by means of definition (2.7) and (5.9), when $(x_{0}-a)\in\mathbb{N}$ derivative could be taken as follows

[TABLE]

Given $x_{0}$, such that $(x_{0}-a)\in\mathbb{R}^{+}$, then conditions of (3.2) is reached, and, applying definition (2.7), derivative $f^{\prime}$ follows

[TABLE]

Otherwise, let be $f:f\in C^{n}$, where $n$ - positive integer, then under similar conditions as (5.11) and (5.13), derivative could be reached by differentiating of $n$-order Taylor’s polynomial of $f$ in terms of $q$-power difference (5.3) under limit notation over $n$

[TABLE]

Similarly, as (5.13), derivative of $f\in C^{n}$ in point $x=x_{0}$, such that $(x_{0}-a)\in\mathbb{N}$

[TABLE]

Otherwise, going from (5.14), $\forall(x_{0}-a)\in\mathbb{R}^{+}$

[TABLE]

6. Newton’s interpolation formula

Being a discrete analog of Taylor’s series, the Newton’s interpolation formula [6], first published in his Principia Mathematica in 1687, hereby, by author’s opinion, supposed to be discussed

[TABLE]

Given $q=const$ in (2.3) divided $q$-difference $f[xq;\ x]$ is reached. Let be $\Delta f=f[xq;\ x](xq-x)$, then, by means of generalized high order forward finite difference $\Delta^{k}f,\ k\geq 2$, ([7], [8]), revised according to (2.3), Newton’s formula (6.1) takes the form

[TABLE]

Review (5.4) and given $q=const$ divided $q$-power difference follows, by similar way as (6.2) reached, (6.1) could be written as

[TABLE]

7. Conclusion

In this paper was discussed a way of obtaining real-valued smooth function’s derivative in radius of convergence of it’s Taylor’s series or polynomial by means of analog of Newton’s binomial theorem (1.20) in terms of $q$-difference (3.1) and $(p,q)$-power difference operators (5.12). In the last section reviewed a discrete analog of Taylor’s series - Newton’s interpolation formula (6.1), and applying operators of $q$-difference, $(p,q)$-power difference interpolation of initial function is shown (6.2), (6.3).