0

Let $α ∈ ℝ$, $D = ℝ^n $\ {$0$} and $f_α : D → ℝ$ be defined by $f_α(x) = ||x||_2^α$, where $||•||_2$ denotes the Euclidean norm.

(1) Calculate the first-order partial derivatives $∂_jf_α$ for $j ∈ {1, ... , n}$.

(2) Calculate the second-order partial derivatives $∂_i∂_jf_α$ and $∂_j∂_if_α$ for $i,j ∈ {1, ... , n}$

Attempt

(1) $f_α(x) = ||x||_2^α = (\displaystyle \sum_{j=1}^n x_j^2)$$\frac{α}{2}$

and

$\frac{∂}{∂x_j} = \frac{α}{2}(\displaystyle \sum_{j=1}^n x_j^2)$$\frac{α}{2}-1$ * $2x_j$ $=$ $α(\displaystyle \sum_{j=1}^n x_j^2)$$\frac{α}{2}-1$ * $x_j$

how exactly i have to do the second part for $∂_i∂_jf_α$ and $∂_j∂_if_α$ for $i,j ∈ {1, ... , n}$. I would be glad if you can help me.^^

Vek
  • 303

1 Answers1

1

So, $$\partial_j f_\alpha(\mathbf{x}) = \alpha x_j \left(\sum\limits_{k=1}^n x_k^2\right)^{\frac{\alpha}{2} - 1}.$$ Then we get

$$\partial_i\partial_j f_\alpha(\mathbf{x}) = \alpha (\partial_i x_j)\left(\sum\limits_{k=1}^n x_k^2\right)^{\frac{\alpha}{2} - 1} + \alpha x_j \partial_i\left(\sum\limits_{k=1}^n x_k^2\right)^{\frac{\alpha}{2} - 1} = 2 \alpha \left(\frac{\alpha}{2} - 1\right) x_i x_j \left(\sum\limits_{k=1}^n x_k^2\right)^{\frac{\alpha}{2} - 2} + \alpha \delta_{ij} \left(\sum\limits_{k=1}^n x_k^2\right)^{\frac{\alpha}{2} - 1}, $$

and after swapping indexes

$$\partial_j\partial_i f_\alpha(\mathbf{x}) = 2 \alpha \left(\frac{\alpha}{2} - 1\right) x_j x_i \left(\sum\limits_{k=1}^n x_k^2\right)^{\frac{\alpha}{2} - 2} + \alpha \delta_{ji} \left(\sum\limits_{k=1}^n x_k^2\right)^{\frac{\alpha}{2} - 1} = \partial_i\partial_j f_\alpha(\mathbf{x}),$$ where $\delta_{ji}$ is the Kronecker delta.

jan
  • 567