Pexider方程是Pexider J V在文献[1]中提出的一组方程，形式如下：

$ f\left( x \right) + g\left( y \right) = h\left( {x + y} \right) $

$ f\left( x \right)g\left( y \right) = h\left( {x + y} \right) $

$ f\left( x \right) + g\left( y \right) = h\left( {xy} \right) $

$ f\left( x \right)g\left( y \right) = h\left( {xy} \right) $

并给出了方程在实数域 $ {\mathbf{R}} $ 上的连续解。一直以来，很多人致力于这组方程的研究，得到了许多有意义的成果，文献[2-7]研究了Pexider方程在不同区域上解的情况，文献[8-9]讨论了可加Pexider方程在集值函数空间的稳定性，文献[10-14]讨论了几类Pexider方程和时滞方程在不同赋范空间上的解及稳定性。本文研究如下形式的广义Pexider方程：

$ \sum\limits_{k = 1}^n {{\alpha _k}{f_k}\left( {{\beta _k}{x_k}} \right)} = {f_{n + 1}}\left( {\sum\limits_{k = 1}^n {{\gamma _k}{x_k}} } \right) \text{，} \forall {x_k} \in {{\mathbf{R}}} $

(1)

$ \prod\limits_{k = 1}^n {f_k^{{\alpha _k}}\left( {{\beta _k}{x_k}} \right)} = {f_{n + 1}}\left( {\sum\limits_{k = 1}^n {{\gamma _k}{x_k}} } \right) \text{，} \forall {x_k} \in {{\mathbf{R}}} $

(2)

$ \sum\limits_{k = 1}^n {{\alpha _k}{f_k}\left( {x_k^{{\beta _k}}} \right)} = {f_{n + 1}}\left( {\prod\limits_{k = 1}^n {x_k^{{\gamma _k}}} } \right) \text{，} \forall {x_k} \in {{\mathbf{R}}^ + } $

(3)

$ \prod\limits_{k = 1}^n {f_k^{{\alpha _k}}\left( {x_k^{{\beta _k}}} \right)} = {f_{n + 1}}\left( {\prod\limits_{k = 1}^n {x_k^{{\gamma _k}}} } \right) \text{，} \forall {x_k} \in {{\mathbf{R}}^ + } $

(4)

其中 $ {\alpha _k},{\beta _k},{\gamma _k} $ 为常数，且 $ \prod\limits_{k = 1}^n {{\alpha _k}{\beta _k}{\gamma _k} \ne 0,\left( {n \geqslant 2} \right)} $ 。通过赋值转化方法，得到了上述方程的通解。

1 引理及基本准备

首先，本文给出下面的引理^[15-16]：

引理1 设 $ f $ 是定义在 ${\mathbf{R}} $ 上的连续函数，如下几个方程

$ f\left( x \right) + f\left( y \right) = f\left( {x + y} \right) \text{，} \forall x,y \in {\mathbf{R}} $

$ f\left( x \right)f\left( y \right) = f\left( {x + y} \right) \text{，} \forall x,y \in {\mathbf{R}} $

$ f\left( x \right) + f\left( y \right) = f\left( {xy} \right) \text{，} \forall x,y \in {{\mathbf{R}}^ + } $

$ f\left( x \right)f\left( y \right) = f\left( {xy} \right) \text{，} \forall x,y \in {{\mathbf{R}}^ + } $

在不考虑平凡解 $ f \equiv 0 $ 的情况下，分别有解为

$ f\left( x \right) = f\left( 1 \right)x \text{，}( x \in {\mathbf{R}} ) $

$ f\left( x \right) = {{\text{e}}^{cx}} = {a^x} \text{，}( c 为任意常数\text{，} x \in {\mathbf{R}} ) $

$ f\left( x \right) = c\ln x \text{，}( c 为任意常数\text{，} x \in {{\mathbf{R}}^ + } ) $

$ f\left( x \right) = {x^c} \text{，}( c 为任意常数\text{，} x \in {{\mathbf{R}}^ + } ) $

下面，将引理1推广为一般的形式，有

引理2 设 $ f $ 是定义在 $ {{{\bf{R}}}} $ 上的连续函数，如下几个方程

$ \sum\limits_{k = 1}^n {f\left( {{x_k}} \right)} = f\left( {\sum\limits_{k = 1}^n {{x_k}} } \right) \text{，} \forall {x_k} \in {\mathbf{R}} $

$ \prod\limits_{k = 1}^n {f\left( {{x_k}} \right)} = f\left( {\sum\limits_{k = 1}^n {{x_k}} } \right) \text{，} \forall {x_k} \in {\mathbf{R}} $

$ \sum\limits_{k = 1}^n {f\left( {{x_k}} \right)} = f\left( {\prod\limits_{k = 1}^n {{x_k}} } \right) \text{，} \forall {x_k} \in {{\mathbf{R}}^ + } $

$ \prod\limits_{k = 1}^n {f\left( {{x_k}} \right)} = f\left( {\prod\limits_{k = 1}^n {{x_k}} } \right) \text{，} \forall {x_k} \in {{\mathbf{R}}^ + } $

在不考虑平凡解 $ f \equiv 0 $ 的情况下，分别有解为

$ f\left( x \right) = f\left( 1 \right)x \text{，}( x \in {\mathbf{R}} ) $

$ f\left( x \right) = {{\text{e}}^{cx}} = {a^x} \text{，}( c 为任意常数\text{，} x \in {\mathbf{R}} ) $

$ f\left( x \right) = c\ln x \text{，}( c 为任意常数\text{，} x \in {\mathbf{R}^ + } ) $

$ f\left( x \right) = {x^c} \text{，}( c 为任意常数\text{，} x \in {\mathbf{R}^ + } ) $

证明 由引理1可得结论成立。

2 主要结论及证明

定理1 设 $ {f_k}\left( {1 \leqslant k \leqslant n + 1,n \geqslant 2} \right) $ 是定义在 ${\mathbf{R}}$ 上的连续函数，广义Pexider可加方程(1)在不考虑平凡解 $ {f_k} \equiv 0\left( {1 \leqslant k \leqslant n + 1,n \geqslant 2} \right) $ 的情况下，有解为

$ {f_k}\left( x \right) = \left[ {{f_k}\left( {\frac{{{\beta _k}}}{{{\gamma _k}}}} \right) - {f_k}\left( 0 \right)} \right]\frac{{{\gamma _k}}}{{{\beta _k}}}x + {f_k}\left( 0 \right) \text{，}( x \in {\mathbf{R}} , 1 \leqslant k \leqslant n ) $

$ {f_{n + 1}}\left( x \right) = \left[ {{f_{n + 1}}\left( 1 \right) - \sum\limits_{k = 1}^n {{\alpha _k}{f_k}\left( 0 \right)} } \right]x + \sum\limits_{k = 1}^n {{\alpha _k}{f_k}\left( 0 \right)} \text{，}( x \in {\mathbf{R}} ) $

证明 在方程(1)中，固定某个 $ {x_k} = \dfrac{1}{{{\gamma _k}}}x $ ,令其他 $ n - 1 $ 个 $ {x_k} = 0 $ ，有

$ \begin{aligned}[b] &{\alpha _1}{f_1}\left( 0 \right) + \cdots + {\alpha _{k - 1}}{f_{k - 1}}\left( 0 \right) + {\alpha _k}{f_k}\left( {\frac{{{\beta _k}}}{{{\gamma _k}}}x} \right) +\\ &{\alpha _{k + 1}}{f_{k + 1}}\left( 0 \right) + \cdots + {\alpha _n}{f_n}\left( 0 \right) = {f_{n + 1}}\left( x \right) \text{，}\\ &1 \leqslant k \leqslant n \end{aligned} $

将上面等式两端同时减去 $ \displaystyle\sum\limits_{k = 1}^n {{\alpha _k}{f_k}\left( 0 \right)} $ ，有

$ \begin{aligned}[b] &{\alpha _k}{f_k}\left( {\frac{{{\beta _k}}}{{{\gamma _k}}}x} \right) - {\alpha _k}{f_k}\left( 0 \right) = \\ &{f_{n + 1}}\left( x \right) - \sum\limits_{k = 1}^n {{\alpha _k}{f_k}\left( 0 \right)} : = \varphi \left( x \right) \text{，} 1 \leqslant k \leqslant n \end{aligned} $

从而有

$ {\alpha _k}{f_k}\left( {\frac{{{\beta _k}}}{{{\gamma _k}}}x} \right) = \varphi \left( x \right) + {\alpha _k}{f_k}\left( 0 \right) \text{，} 1 \leqslant k \leqslant n $

(5)

$ {f_{n + 1}}\left( x \right) = \varphi \left( x \right) + \sum\limits_{k = 1}^n {{\alpha _k}{f_k}\left( 0 \right)} $

(6)

于是，有

$ {\alpha _k}{f_k}\left( {{\beta _k}{x_k}} \right) = \varphi \left( {{\gamma _k}{x_k}} \right) + {\alpha _k}{f_k}\left( 0 \right) \text{，} 1 \leqslant k \leqslant n $

(7)

$ {f_{n + 1}}\left( {\sum\limits_{k = 1}^n {{\gamma _k}{x_k}} } \right) = \varphi \left( {\sum\limits_{k = 1}^n {{\gamma _k}{x_k}} } \right) + \sum\limits_{k = 1}^n {{\alpha _k}{f_k}\left( 0 \right)} $

(8)

将式(7)、(8)代入方程(1)，整理可得

$ \sum\limits_{k = 1}^n {\varphi \left( {{\gamma _k}{x_k}} \right)} = \varphi \left( {\sum\limits_{k = 1}^n {{\gamma _k}{x_k}} } \right) $

由引理2可得

$ \varphi \left( x \right) = \varphi \left( 1 \right)x $

由式(5)、(6)可得

$\begin{aligned}[b] &{f_k}\left( x \right) = \frac{1}{{{\alpha _k}}}\varphi \left( {\frac{{{\gamma _k}}}{{{\beta _k}}}x} \right) + {f_k}\left( 0 \right) = \varphi \left( 1 \right)\frac{{{\gamma _k}}}{{{\alpha _k}{\beta _k}}}x + \\ &{f_k}\left( 0 \right) = \left[ {{f_k}\left( {\frac{{{\beta _k}}}{{{\gamma _k}}}} \right) - {f_k}\left( 0 \right)} \right]\frac{{{\gamma _k}}}{{{\beta _k}}}x + {f_k}\left( 0 \right) \text{，} \\ &1 \leqslant k \leqslant n \end{aligned} $

$ \begin{aligned}[b] &{f_{n + 1}}\left( x \right) = \varphi \left( 1 \right)x + \sum\limits_{k = 1}^n {{\alpha _k}{f_k}\left( 0 \right)} = \\ &\left[ {{f_{n + 1}}\left( 1 \right) - \sum\limits_{k = 1}^n {{\alpha _k}{f_k}\left( 0 \right)} } \right]x + \sum\limits_{k = 1}^n {{\alpha _k}{f_k}\left( 0 \right)} \end{aligned} $

定理2 设 $ {f_k}\left( {1 \leqslant k \leqslant n + 1,n \geqslant 2} \right) $ 是定义在 $ {\mathbf{R}} $ 上的连续函数，广义Pexider指数方程(2)在不考虑平凡解 $ {f_k} \equiv 0\left( {1 \leqslant k \leqslant n + 1,n \geqslant 2} \right) $ 的情况下，有解为

$ {f_k}\left( x \right) = {f_k}\left( 0 \right){a^{\frac{{{\gamma _k}}}{{{\alpha _k}{\beta _k}}}x}} ，( a \gt 0 , x \in {\mathbf{R}} , 1 \leqslant k \leqslant n ) $

$ {f_{k + 1}}\left( x \right) = {a^x}\prod\limits_{k = 1}^n {f_k^{{\alpha _k}}\left( 0 \right)} \text{，} ( a \gt 0 , x \in {\mathbf{R}} ) $

证明 由于不考虑平凡解，在方程(2)中固定某个 $ {x_k} = \dfrac{1}{{{\gamma _k}}}x $ ,令其他 $ n - 1 $ 个 $ {x_k} = 0 $ ，有

$ \begin{aligned}[b] &f_1^{{\alpha _1}}\left( 0 \right) \times \cdots \times f_{_{k - 1}}^{{\alpha _{k - 1}}}\left( 0 \right) \times f_{_k}^{{\alpha _k}}\left( {\frac{{{\beta _k}}}{{{\gamma _k}}}x} \right) \times \\ &f_{_{k + 1}}^{{\alpha _{k + 1}}}\left( 0 \right) \times \cdots \times f_{_n}^{{\alpha _n}}\left( 0 \right) = {f_{n + 1}}\left( x \right) \text{，} 1 \leqslant k \leqslant n \end{aligned} $

将上面 $ n $ 个等式两端同时除 $ \displaystyle\prod\limits_{k = 1}^n {f_{_k}^{{\alpha _k}}\left( 0 \right)} $ ，有

$ \frac{{f_{_k}^{{\alpha _k}}\left( {\dfrac{{{\beta _k}}}{{{\gamma _k}}}x} \right)}}{{f_{_k}^{{\alpha _k}}\left( 0 \right)}} = \frac{{{f_{k + 1}}\left( x \right)}}{{\prod\limits_{k = 1}^n {f_{_k}^{{\alpha _k}}\left( 0 \right)} }} = :\varphi \left( x \right) \text{，} 1 \leqslant k \leqslant n $

从而

$ f_{_k}^{{\alpha _k}}\left( {\frac{{{\beta _k}}}{{{\gamma _k}}}x} \right) = \varphi \left( x \right)f_{_k}^{{\alpha _k}}\left( 0 \right) \text{，} 1 \leqslant k \leqslant n $

(9)

$ {f_{k + 1}}\left( x \right) = \varphi \left( x \right)\prod\limits_{k = 1}^n {f_k^{{\alpha _k}}\left( 0 \right)} $

(10)

于是

$ f_{_k}^{{\alpha _k}}\left( {{\beta _k}{x_k}} \right) = \varphi \left( {{\gamma _k}{x_k}} \right)f_{_k}^{{\alpha _k}}\left( 0 \right) \text{，} 1 \leqslant k \leqslant n $

(11)

$ {f_{k + 1}}\left( {\sum\limits_{k = 1}^n {{\gamma _k}{x_k}} } \right) = \varphi \left( {\sum\limits_{k = 1}^n {{\gamma _k}{x_k}} } \right)\prod\limits_{k = 1}^n {f_{_k}^{{\alpha _k}}\left( 0 \right)} $

(12)

将式(11)、(12)代入方程(2)，整理可得

$ \prod\limits_{k = 1}^n {\varphi \left( {{\gamma _k}{x_k}} \right)} = \varphi \left( {\sum\limits_{k = 1}^n {{\gamma _k}{x_k}} } \right) $

由引理2可得，上式有解

$ \varphi \left( x \right) = {{\text{e}}^{cx}} = {a^x} \text{，} \left( {a \gt 0} \right) $

由式(9)、(10)可得

$ {f_k}\left( x \right) = {f_k}\left( 0 \right){a^{\frac{{{\gamma _k}}}{{{\alpha _k}{\beta _k}}}x}} \text{，} 1 \leqslant k \leqslant n \text{，} \left( {a \gt 0} \right) $

$ {f_{k + 1}}\left( x \right) = {a^x}\prod\limits_{k = 1}^n {f_k^{{\alpha _k}}\left( 0 \right)} \text{，} \left( {a \gt 0} \right) $

定理3 设 $ {f_k}\left( {1 \leqslant k \leqslant n + 1,n \geqslant 2} \right) $ 是定义 ${{\mathbf{R}}^ + }$ 上的连续函数，广义Pexider对数方程(3)在不考虑平凡解 $ {f_k} \equiv 0\left( {1 \leqslant k \leqslant n + 1,n \geqslant 2} \right) $ 的情况下，有解为

$ {f_k}\left( x \right) = \dfrac{{c{\gamma _k}}}{{{\alpha _k}{\beta _k}}}\ln x + {f_k}\left( 1 \right) $ ，( c 为任意常数， $ x \in {{\mathbf{R}}^ + } , 1 \leqslant $ $ k \leqslant n $ )

$ {f_{n + 1}}\left( x \right) = c\ln x + \sum\limits_{k = 1}^n {{\alpha _k}{f_k}\left( 1 \right)} \text{，}( c 为任意常数\text{，} x \in {{\mathbf{R}}^ + } ) $

证明 在方程(3)中固定某个 $ {x_k} = {x^{\frac{1}{{{\gamma _k}}}}} $ ,令其他 $ n - 1 $ 个 $ {x_k} = 1 $ ，有

$ \begin{aligned}[b] &{\alpha _1}{f_1}\left( 1 \right) + \cdots + {\alpha _{k - 1}}{f_{k - 1}}\left( 1 \right) + {\alpha _k}{f_k}\left( {{x^{\frac{{{\beta _k}}}{{{\gamma _k}}}}}} \right) + \\ &{\alpha _{k + 1}}{f_{k + 1}}\left( 1 \right) + \cdots + {\alpha _n}{f_n}\left( 1 \right) = {f_{n + 1}}\left( x \right) \text{，} 1 \leqslant k \leqslant n \end{aligned} $

将上面 $ n $ 个等式两端同时减去 $ \displaystyle\sum\limits_{k = 1}^n {{\alpha _k}{f_k}\left( 1 \right)} $ ，有

$\begin{aligned}[b] &{\alpha _k}{f_k}\left( {{x^{\frac{{{\beta _k}}}{{{\gamma _k}}}}}} \right) - {\alpha _k}{f_k}\left( 1 \right) = \\ &{f_{n + 1}}\left( x \right) - \sum\limits_{k = 1}^n {{\alpha _k}{f_k}\left( 1 \right)} : = \varphi \left( x \right) \text{，} 1 \leqslant k \leqslant n \end{aligned} $

从而有

$ {\alpha _k}{f_k}\left( {{x^{\frac{{{\beta _k}}}{{{\gamma _k}}}}}} \right) = \varphi \left( x \right) + {\alpha _k}{f_k}\left( 1 \right) \text{，} 1 \leqslant k \leqslant n $

(13)

$ {f_{n + 1}}\left( x \right) = \varphi \left( x \right) + \sum\limits_{k = 1}^n {{\alpha _k}{f_k}\left( 1 \right)} $

(14)

于是，有

$ {\alpha _k}{f_k}\left( {x_k^{{\beta _k}}} \right) = \varphi \left( {x_{_k}^{{\gamma _k}}} \right) + {\alpha _k}{f_k}\left( 1 \right) \text{，} 1 \leqslant k \leqslant n $

(15)

$ {f_{n + 1}}\left( {\prod\limits_{k = 1}^n {x_{_k}^{{\beta _k}}} } \right) = \varphi \left( {\prod\limits_{k = 1}^n {x_{_k}^{{\beta _k}}} } \right) + \sum\limits_{k = 1}^n {{\alpha _k}{f_k}\left( 1 \right)} $

(16)

将式(15)、(16)代入方程(3)，整理可得

$ \sum\limits_{k = 1}^n {\varphi \left( {x_{_k}^{{\beta _k}}} \right)} = \varphi \left( {\prod\limits_{k = 1}^n {x_{_k}^{{\beta _k}}} } \right) $

由引理2可得

$ \varphi \left( x \right) = c\ln x $

由式(13)、(14)可得

$ \begin{aligned}[b] &{f_k}\left( x \right) = \frac{1}{{{\alpha _k}}}\varphi \left( {{x^{\frac{{{\gamma _k}}}{{{\beta _k}}}}}} \right) + {f_k}\left( 1 \right) =\\ &\frac{c}{{{\alpha _k}}}\ln {x^{\frac{{{\gamma _k}}}{{{\beta _k}}}}} + {f_k}\left( 1 \right) = \frac{{c{\gamma _k}}}{{{\alpha _k}{\beta _k}}}\ln x + {f_k}\left( 1 \right) \text{，} 1 \leqslant k \leqslant n \end{aligned} $

$ {f_{n + 1}}\left( x \right) = \varphi \left( x \right) + \sum\limits_{k = 1}^n {{\alpha _k}{f_k}\left( 1 \right)} = c\ln x + \sum\limits_{k = 1}^n {{\alpha _k}{f_k}\left( 1 \right)} $

定理4 设 $ {f_k}\left( {1 \leqslant k \leqslant n + 1,n \geqslant 2} \right) $ 是定义在 ${{\mathbf{R}}^ + }$ 上的连续函数，广义Pexider幂函数方程(4)在不考虑平凡解 $ {f_k} \equiv 0\left( {1 \leqslant k \leqslant n + 1,n \geqslant 2} \right) $ 的情况下，有解为

$ {f_k}\left( x \right) = {f_k}\left( 1 \right){x^{\frac{{c{\gamma _k}}}{{{\alpha _k}{\beta _k}}}}} \text{，}( c 为任意常数\text{，} x \in {{\mathbf{R}}^ + } \text{，} 1 \leqslant k \leqslant n ) $

$ {f_{k + 1}}\left( x \right) = {x^c}\prod\limits_{k = 1}^n {f_{_k}^{{\alpha _k}}\left( 1 \right)} \text{，}( c 为任意常数\text{，} x \in {{\mathbf{R}}^ + } ) $

证明 由于不考虑平凡解，在方程(4)中固定某个 $ {x_k} = {x^{\frac{1}{{{\gamma _k}}}}} $ ,令其他 $ n - 1 $ 个 $ {x_k} = 1 $ ，有

$ \begin{aligned}[b] &f_{_1}^{{\alpha _1}}\left( 1 \right) \times \cdots \times f_{_{k - 1}}^{{\alpha _{k - 1}}}\left( 1 \right) \times f_{_k}^{{\alpha _k}}\left( {{x^{\frac{{{\beta _k}}}{{{\gamma _k}}}}}} \right) \times \\ &f_{_{k + 1}}^{{\alpha _{k + 1}}}\left( 1 \right) \times \cdots \times f_{_n}^{{\alpha _n}}\left( 1 \right) = {f_{n + 1}}\left( x \right) \text{，} 1 \leqslant k \leqslant n \end{aligned} $

将上面等式两端同时除 $ \displaystyle\prod\limits_{k = 1}^n {f_k^{{\alpha _k}}\left( 1 \right)} $ ，有

$ \frac{{f_{_k}^{{\alpha _k}}\left( {{x^{\frac{{{\beta _k}}}{{{\gamma _k}}}}}} \right)}}{{f_{_k}^{{\alpha _k}}\left( 1 \right)}} = \frac{{{f_{k + 1}}\left( x \right)}}{{\displaystyle\prod\limits_{k = 1}^n {f_{_k}^{{\alpha _k}}\left( 1 \right)} }} = :\varphi \left( x \right) \text{，} 1 \leqslant k \leqslant n $

从而

$ f_{_k}^{{\alpha _k}}\left( {{x^{\frac{{{\beta _k}}}{{{\gamma _k}}}}}} \right) = \varphi \left( x \right)f_{_k}^{{\alpha _k}}\left( 1 \right) \text{，} 1 \leqslant k \leqslant n $

(17)

$ {f_{k + 1}}\left( x \right) = \varphi \left( x \right)\prod\limits_{k = 1}^n {f_{_k}^{{\alpha _k}}\left( 1 \right)} $

(18)

于是

$ f_{_k}^{{\alpha _k}}\left( {x_{_k}^{{\beta _k}}} \right) = \varphi \left( {x_{_k}^{{\gamma _k}}} \right)f_{_k}^{{\alpha _k}}\left( 1 \right) \text{，} 1 \leqslant k \leqslant n $

(19)

$ {f_{k + 1}}\left( {\prod\limits_{k = 1}^n {x_{_k}^{{\gamma _k}}} } \right) = \varphi \left( {\prod\limits_{k = 1}^n {x_{_k}^{{\gamma _k}}} } \right)\prod\limits_{k = 1}^n {f_{_k}^{{\alpha _k}}\left( 1 \right)} $

(20)

将式(19)、(20)代入方程(4)，整理可得

$ \prod\limits_{k = 1}^n {\varphi \left( {x_{_k}^{{\gamma _k}}} \right)} = \varphi \left( {\prod\limits_{k = 1}^n {x_{_k}^{{\gamma _k}}} } \right) $

由引理2可得，上式有解

$ \varphi \left( x \right) = {x^c} \text{，} \left( {c \in {\mathbf{R}}} \right) $

由式(17)、(18)可得

$ {f_k}\left( x \right) = {\varphi ^{\frac{1}{{{\alpha _k}}}}}\left( {{x^{\frac{{{\gamma _k}}}{{{\beta _k}}}}}} \right){f_k}\left( 1 \right) = {f_k}\left( 1 \right){x^{\frac{{c{\gamma _k}}}{{{\alpha _k}{\beta _k}}}}} \text{，} 1 \leqslant k \leqslant n $

$ {f_{k + 1}}\left( x \right) = \varphi \left( x \right)\prod\limits_{k = 1}^n {f_{_k}^{{\alpha _k}}\left( 1 \right)} = {x^c}\prod\limits_{k = 1}^n {f_{_k}^{{\alpha _k}}\left( 1 \right)} $