0
$\begingroup$

Let $k \in \mathbb N$, $a_{ij} \in \mathbb R$ for $i,j \in \mathbb N$, $i+j \le k$. A function $f:\mathbb R^2 \to \mathbb R$

$$f(x,y) = \sum_{i+j \le k} a_{ij}x^iy^j$$

is called polynomial of degree $k$ in $x,y$.

I have to show that the partial derivatives of polynomials of degree $k \ge 1$ exist, and the partial derivatives $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}$ are polynomials of degree $k-1$.

I start showing that for a polynomial of degree $1$ its partial derivatives $D_1f(x)$ and $D_1f(y)$exists and in the next step that the $D_if(x)$ and $D_if(y)$ derivative exists, but I dont know how to deal with the matrix and the indices of the sum in the formula, how can I take a take a partial derivative of it? Dealing with the matrix and other variable as a constant and use normal derivative rules for the variable in which I want to take the partial derivative?

Any help is appreciated!!!

$\endgroup$
  • $\begingroup$ I believe that $y$ should be raised to the power $j$ in your formula, as opposed to $i$? $\endgroup$ – Foo Barrigno Apr 24 '14 at 17:28
  • $\begingroup$ yes, I edited it, thanks for the suggestion $\endgroup$ – sj134 Apr 24 '14 at 17:37
0
$\begingroup$

There is at least one caveat to your problem that needs to be stated. At least one $a_{ij}$ such that $i+j=k$ must be non-zero. Otherwise your polynomial is of degree less than $k$. And in order for the partial derivative w.r.t. $x$ to be a polynomial of degree $k-1$ one of the $a_{ij}$ must have $i\ge 1$. Similarly for $y$.

Otherwise, show that if $i+j=k$, the partial derivative of $a_{ij}x^iy^j$ for both $x$ and $y$ is of total degree $i+j-1=k-1$.

I don't think you need to use induction here.

To actually perform the partial derivatives, note that: $$\frac{\partial f}{\partial x} \sum_{i+j \le k} a_{ij}x^iy^j = \sum_{i+j \le k} i*a_{ij}x^{i-1}y^j$$ and vice versa for the partial derivative with respect to y. Just note that in cases where there is no $x$ the partial derivative with respect to $x$ is 0.

$\endgroup$
  • $\begingroup$ Okay, I understand that just taking the partial derivatives in the second part would be sufficient, but how would I prove the first part without induction and how can I take the partial derivatives anyways? $\endgroup$ – sj134 Apr 24 '14 at 18:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.