Towards Higher-Degree Fuzzy Projection

Abstract

Fuzzy projection is a mathematical operator inspired by the inverse fuzzy transform that is used to approximate functions. The fuzzy projection is designed such that the coefficients of the linear combination of the basis functions (fuzzy sets in a fuzzy partition) are optimized to obtain the best approximation of the functions from a global perspective, as opposed to the fuzzy transform, where the approximation focuses on fitting functions locally. The aim of this paper is to extend the fuzzy projection to a higher degree, similarly to the fuzzy transform, where the coefficients of the linear combination of the basis functions are expressed by polynomials. In this way, we can significantly improve the quality of the approximation by combining the settings of the fuzzy partition and the degree of polynomnials. In this paper, we show that a higher-order fuzzy projection can be computed using matrix calculus, leading to an easy algorithmization of the method. We also give its approximation properties and its applicability to discrete functions. The usefulness of higher-order fuzzy projection is demonstrated on two tasks, namely continuous function approximation and audio signal compression and decompression, where the results are compared with other relevant methods.

Appendices

Appendix A Lemma 6

Proof

For any $\phi _{k,i}\in \mathcal {B}^m_{\mathcal {A}}$, $k\in \mathcal {M}$, $i=0,\ldots ,m$, we have

$$\begin{aligned} \langle f,\phi _{k,i}\rangle _{L^2}&=\int _{a}^{b}f(x)\left( \frac{x-a}{h}-k\alpha \right) ^iA_k\left( x\right) dx\\&=\int _{\text{Supp}A_k}f(x)\left( \frac{x-a}{h}-k\alpha \right) ^iA_k\left( x\right) dx. \end{aligned}$$

If $k<\frac{1}{\alpha }$, then $\text{Supp}A_k=[a,x_k+h]$, where $x_k$ is the k-th node of the fuzzy partition. It follows that

$$\begin{aligned} \langle f,\phi _{k,i}\rangle _{L^2}=\int _{a}^{x_k+h}f(x)\left( \frac{x-a}{h}-k\alpha \right) ^iA_k\left( x\right) dx\\ =\int _{a}^{x_k+h}f(x)\left( \frac{x-a}{h}-k\alpha \right) ^iK\left( \frac{x-a}{h}-k\alpha \right) dx. \end{aligned}$$

Putting $u=\frac{x-a}{h}-k\alpha$, we obtain

$$\begin{aligned} \langle f,\phi _{k,i}\rangle _{L^2}&=h\cdot \int _{-k\alpha }^{1}f(a+k\alpha h+hu)u^iK\left( u\right) du\\&=h\cdot \int _{-k\alpha }^{1}f_k(x)x^iK\left( x\right) dx. \end{aligned}$$

The two remaining cases, i.e., $k>N-\frac{1}{\alpha }$ and $\frac{1}{\alpha }\le k\le N-\frac{1}{\alpha }$, can be proved by similar arguments.□

Appendix B Lemma 8

Proof

Assume that $k\ge \ell$. For any $i,j=0,\ldots ,m$, we have

$$\begin{aligned}&\langle \phi _{k,i},\phi _{\ell ,j}\rangle _{L^2}=\int _{a}^{b}\left( \frac{x-a}{h}-\alpha k\right) ^i\left( \frac{x-a}{h}-\alpha \ell \right) ^j\\&\times A_k\left( x\right) A_\ell \left( x\right) dx\\&=\int _{\text{Supp}A_k\cap \text{Supp}A_\ell }\left( \frac{x-a}{h}-\alpha k\right) ^i\left( \frac{x-a}{h}-\alpha \ell \right) ^j\\&\times A_k\left( x\right) A_\ell \left( x\right) dx. \end{aligned}$$

For $k-\ell \ge \frac{2}{\alpha }$, we simply find that $\text{Supp}A_k\cap \text{Supp}A_\ell =\emptyset$. Therefore, $\langle \phi _{k,i},\phi _{\ell ,j}\rangle _{L^2}=0$. Further, let us consider that $k-\ell <\frac{2}{\alpha }$, i.e., $\text{Supp}A_k\cap \text{Supp}A_\ell \not =\emptyset$. Here, we can distinguish three following cases: $\text{Supp}A_k\cap \text{Supp}A_\ell =[a,x_\ell +h]$, if $k<\frac{1}{\alpha }$, $\text{Supp}A_k\cap \text{Supp}A_\ell =[x_k-h,b]$, if $\ell >N-\frac{1}{\alpha }$, and $\text{Supp}A_k\cap \text{Supp}A_\ell =[x_k-h,x_\ell +h]$, otherwise, where $x_k$ and $x_\ell$ are the k-th and $\ell$-th nodes, respectively. Since all cases can be verified by analogy, we prove here only the last case, i.e., $\text{Supp}A_k\cap \text{Supp}A_\ell =[x_k-h,x_\ell +h]$. For this case, we have

$$\begin{aligned}&\langle \phi _{k,i},\phi _{\ell ,j}\rangle _{L^2}=\int _{x_k-h}^{x_\ell +h}\left( \frac{x-a}{h}-k\alpha \right) ^i\left( \frac{x-a}{h}-\ell \alpha \right) ^j\\&\times A_k(x)A_\ell (x)dx\\&=\int _{x_k-h}^{x_\ell +h}\left( \frac{x-a}{h}-\alpha k\right) ^i\left( \frac{x-a}{h}-\alpha \ell \right) ^j\\&\times K\left( \frac{x-a}{h}-\alpha k\right) K\left( \frac{x-a}{h}-\alpha \ell \right) dx. \end{aligned}$$

Put $u=\frac{x-a}{h}-\ell \alpha$. Then, we obtain that

$$\begin{aligned} \langle \phi _{k,i},\phi _{\ell ,j}\rangle _{L^2}&=h\cdot \int _{(k-\ell )\alpha -1}^{1}\left( u-(k-\ell )\alpha \right) ^iu^j\\&\times K\left( u-(k-\ell )\alpha \right) K\left( u\right) du\\&=h\cdot \int _{d_{k\ell }-1}^{1}\left( x-d_{k\ell } \right) ^ix^j\mathcal {K}_{k\ell }(x)dx, \end{aligned}$$

and the proof is finished.□

Appendix C Theorem 15

Proof

Let $x_k$, $k\in \mathcal {M}$, be the k-th node of the fuzzy partition $\mathcal {A}$. Since $f\in H^{m+1}(a,b)$, we obtain from Theorem 5 in Chapter VI of [21] that f can be extended to a function $f^*$ defined on the interval $I=(x_{-M_\alpha }-h,\ldots ,x_{N+M_\alpha }+h)$ such that $f^*\in H^{m+1}(I)$ and $\Vert f^*\Vert _{H^{m+1}(I)}\le B\cdot \Vert f\Vert _{H^{m+1}(a,b)}$, where B is a positive constant. Let $F^m_k[f^*]$, $k\in \mathcal {M}$, be the k-th component of the direct F$^m$-transform of $f^*$ with respect to $\mathcal {A}$, whose basic functions are extended to I. Since the direct F$^m$-transform preserves polynomials of degree up to m, we obtain from Theorem 3.1.4 in [22] that there exists a constant $C_0>0$ independent of h such that

$$\begin{aligned} \Vert f^*-F^m_k[f^*]\Vert _{L^2(I_k)}\le C_0\cdot h^{m+1}\cdot \Vert \partial ^{m+1}f^*\Vert _{L^2(I_k)}, \end{aligned}$$

where $I_k=(x_k-h,x_k+h)$. Denote $\Omega _k=I_k\cap (a,b)$. Since $\Vert \partial ^{m+1}f^*\Vert _{L^2(I_k)}\le \Vert f^*\Vert _{H^{m+1}(I)}$ and $\Vert f^*-F^m_k[f^*]\Vert _{L^2(\Omega _k)}\le \Vert f^*-F^m_k[f^*]\Vert _{L^2(I_k)}$, we find that

$$\begin{aligned} \Vert f^*-F^m_k[f^*]\Vert _{L^2(\Omega _k)}\le C_0\cdot h^{m+1}\cdot B\cdot \Vert f\Vert _{H^{m+1}(a,b)}. \end{aligned}$$

Since $h=\frac{b-a}{\alpha N}$, there exists a constant $C_1>0$ independent of N such that

$$\begin{aligned} \Vert f^*-F^m_k[f^*]\Vert _{L^2(\Omega _k)}\le C_1\cdot N^{-m-1}, \end{aligned}$$

(C1)

for any $k\in \mathcal {M}$. Let $\xi$ be defined as follows:

$$\begin{aligned} \xi (x)=\sum _{k=-M_\alpha }^{N+M_\alpha }F^m_k[f^*](x)A_k(x),\quad x\in [a,b]. \end{aligned}$$

(C2)

We see that $\xi \in G^m_{\mathcal {A}}$. Moreover, for any $x\in (a,b)$, the sum in (C2) has at most $2(M_\alpha +1)$ non-zero summands. Therefore, we obtain that

$$\begin{aligned}&|(f^*-\xi )(x)|^2=\Big |\sum _{k=-M_\alpha }^{N+M_\alpha }\left( f^*-F^m_k[f^*]\right) (x)\cdot A_k(x)\Big |^2\\&\le 2(M_\alpha +1)\sum _{k=-M_\alpha }^{N+M_\alpha }|\left( f^*-F^m_k[f^*]\right) (x)\cdot A_k(x)|^2, \end{aligned}$$

where the inequality follows from the Bunyakovsky-Cauchy-Schwarz inequality. Hence, we obtain that

$$\begin{aligned} \Vert f^*-\xi \Vert ^2_{L^2(a,b)}\le 2(M_\alpha +1)\sum _{k=-M_\alpha }^{N+M_\alpha }\Vert f^*-F^m_k[f^*]\Vert _{L^2(\Omega _k)}^2. \end{aligned}$$

By substituting (C1) into this inequality, we get

$$\begin{aligned}&\Vert f^*-\xi \Vert ^2_{L^2(a,b)}\le 2(M_\alpha +1)\sum _{k=-M_\alpha }^{N+M_\alpha }{C_1}^2\cdot N^{-2(m+1)}\nonumber \\&=2(M_\alpha +1)(2M_\alpha +N+1){C_1}^2\cdot N^{-2(m+1)}\nonumber \\&=2(M_\alpha +1)\left( \frac{2M_\alpha +1}{N}+1\right) {C_1}^2\cdot N^{-2m-1}\nonumber \\&\le 2(M_\alpha +1)\left( M_\alpha +3/2\right) {C_1}^2\cdot N^{-2m-1}. \end{aligned}$$

(C3)

Therefore, there exists a constant $C>0$ independent of N such that $\Vert f^*-\xi \Vert _{L^2(a,b)}\le C\cdot N^{-m-\frac{1}{2}}.$ Putting $L^2=L^2(a,b)$ and considering that f and $f^*$ coincide on (a, b), we obtain that $\Vert f-\xi \Vert _{L^2}\le C\cdot N^{-m-\frac{1}{2}}.$ It follows from (16) that

$$\begin{aligned} \Vert f-P^m[f]\Vert _{L^2}\le \Vert f-\xi \Vert _{L^2}\le C\cdot N^{-m-\frac{1}{2}}, \end{aligned}$$

and the proof is finished.□