# Computing the shifted symmetric polynomial $s^*_{(1^2)}(x_1,x_2)$

By definition,

Let $$\lambda=(\lambda_1,...,\lambda_n)$$ be a partition with $$l(\lambda)\leq n$$. We define the shifted Schur polynomial in $$n$$ variables corresponding to $$\lambda$$ as $$\begin{equation*} s_{\lambda}^*(x_1,...,x_n)=\frac{\det(x_i +n-i \downharpoonright \lambda_j +n-j)}{\det(x_i +n-i \downharpoonright n-j)} \end{equation*}$$ where $$1 \leq i,j\leq n$$.

Then, let $$\lambda=(1^2) \vdash 2$$. Then, for $$1 \leq i,j \leq 2$$, $$\begin{equation*} \det(x_i +n-i \downharpoonright \lambda_j +n-j) = \begin{vmatrix} (x_1+1)x_1 & (x_1+1) \\ x_2(x_2-1) & x_2 \end{vmatrix} = -x_1x_2^2-x_2^2+x_1^2x_2+2x_1x_2+x_2 \end{equation*}$$ and $$\begin{equation*} \det(x_i +n-i \downharpoonright \lambda_j +n-j) =\begin{vmatrix} (x_1+1) & 1 \\ x_2 & 1 \end{vmatrix} =x_1-x_2+1 \end{equation*}$$ Hence, $$\begin{equation*} s^*_{(1^2)}(x_1,x_2)= \frac{-x_1x_2^2-x_2^2+x_1^2x_2+2x_1x_2+x_2}{x_1-x_2+1} = x_1x_2+x_2. \end{equation*}$$

QUESTION

Is this a shifted symmetric polynomial? If I change $$x_1,x_2$$ by $$x_1-1+c$$ and $$x_2-2+c$$, I get

$$\begin{equation*} s^*_{(1^2)}(x_1',x_2')=x_1'x_2'+x_2'=\\s^*_{(1^2)}(x_1,x_2)=x_1x_2-2x_1+cx_1+cx_2-2c+c^2 \end{equation*}$$ which is not symmetric. How do you check this kind of thing?