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Adobe Photoshop Elements is a free, yet robust, image-editing program that contains many of the features that Photoshop includes, but offers a less-powerful version of this program that’s more in line with the power of basic photo editing tools. GIMP (GNU Image Manipulation Program) is a free, open source version of Photoshop and is only available for the PC. It’s designed to be an all-around graphics editor and in some ways it is much more powerful than Adobe Photoshop. GIMP has similar editing tools to Photoshop, but also has tools for more-specialized tasks such as drawing with the brush, creating web graphics, and so on. Paint.NET is a popular, open-source replacement for Windows Paint. It supports all of the layers, tools, and modes that Paint.NET supports, but it offers many more features. This free program is available at www.getpaint.net. PhotoPad++ is a free version of Photoshop that is available for both PC and Mac computers. It has some of the same features as Photoshop but is less powerful. It’s a good program for basic image manipulation and learning to use Photoshop.

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(2n-1)/n)$in$D^+_{\lambda}(\zeta_n)$. The proof is analogous to that of Theorem $thm:symGen$. The fact that$M_n$is$K$-invariant means that the restriction of$M_n$to any of the orbits in${{\mathbb{C}}}^k$will be given by a polynomial with *symmetric* coefficients in each variable$z_i$. The fact that$\deg M_n = n$and that each coefficient of$M_n$is divisible by$n$means that the same is true for the restriction of$M_n$to each orbit in${{\mathbb{C}}}^k$. This can be seen as a consequence of the fact that$(1-\lambda_n/n)^k = (1-\lambda_n/n)(1-\lambda_n/n)\cdots(1-\lambda_n/n)$, where there are$k$factors. This implies that for any$K$-orbit in${{\mathbb{C}}}^k$we have $$M_n = \sum_{j=0}^{k} \left(\sum_{j_1 + \cdots + j_k = j} \frac{a_{j_1,\ldots,j_k}}{(n-\lambda_n)^{j_1} j_1! \cdots (n-\lambda_n)^{j_k} j_k!} \right) (\zeta_n-1)^k.$$ The coefficients$a_{j_1,\ldots,j_k}$are symmetric polynomials in the$\lambda_i$. If$k=1$, then$G$is abelian and so$M_n$consists only of the constant term. When$k=2$, we have$\$\begin{aligned} \begin{split} M_n &= \sum_{j=0}^2 \left( \sum_{j_1 + j_2 = j} \frac{a_{j_1,j_2}}{(n-\lambda_n)^{j_1} j_1! \, j_2

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