# Shape functions of serendipity element as tensor products

For a serendipity element, the shape function of a middle pt on a edge can be formed using the tensor product in each direction, e.g. point 5 on Quad8

$$\phi_5(\xi,\eta)=\frac{1}{2}(1-\xi^2)(1-\eta)$$

but the shape function on corner node (point 1) is $$\phi_1(\xi,\eta)=\frac{1}{4}(1-\xi)(1-\eta)-\frac{1}{4}(1-\xi^2)(1-\eta)-\frac{1}{4}(1-\eta^2)(1-\xi)$$

Why the shape function of the corner point from a serendipity element (Q8) can not be built by the tensor products like this

$$\phi_1(\xi,\eta)=\frac{1}{4}\xi\eta(1-\xi)(1-\eta)$$

Does this mean I just pick up the wrong basis?

One of the several criteria for shape functions is that at any point $\xi$ and $\eta$ in the element, the sum of all shape functions evaluated at that point must equal one.
Consider the center point, $\xi=0, \eta=0$. The Q9 corner point shape functions are all equal zero there. The Q8 edge shape functions (e.g. your first equation) all equal $1/2$ so their sum is two, violating this criterion.