Riemann sum of a double integral

Question

Problem

Explain that

$$R^k_n=\sum_{i=1}^n\sum^n_{j=1}f\left(-k+\frac{2ki}{n},-k+\frac{2kj}{n}\right)\left(\frac{2k}{n}\right)^2$$

given that $$f(x,y)=\frac{1}{x^4+2x^2y^2+y^4+1}$$

is a Riemann sum for $f$ in the square $[-k,k]\times[-k,k]$ for all positive integers $k$ and $n$

Attempt

$$\iint_Qf(x,y)=\sum^n_{i=1}\sum^n_{j=1}f(x_{ij},y_{ij})\Delta x\Delta y$$

where $Q=[-k,k]$

$$\Delta x=\frac{k-(-k)}{n}=\frac{2k}{n}$$ $$\Delta y=\frac{k-(-k)}{n}=\frac{2k}{n}$$

and for $x_{ij}$ and $y_{ij}$ we have that

$$x_{ij}=-k+(\Delta x)i=-k+\frac{2ki}{n}$$

$$y_{ij}=-k+(\Delta y)j=-k+\frac{2kj}{n}$$

$$\therefore R^k_n=\sum_{i=1}^n\sum^n_{j=1}f\left(-k+\frac{2ki}n,-k+\frac{2kj}n\right)\left(\frac{2k}{n}\right)^2$$

Is this enough, or do I have to explain in further details? If so, how do I do that? Thanks in advance

Answer 1

Great. It's OK in general. Some comments and ideas.

• Important, $\iint_Q f(x,y)$ is not equal to some Riemann sum in general. The integral exists if and only if there exists a 'limit' value for those sums in a sense (when the norm of the partition tends to zero).
• Make clear the definition of Riemann sum you have. I would say that if $Q=[a,b]\times[c,d]$, for each $m,n \in \mathbb N$ and each pair of partitions of $[a,b]$, say $\{a=x_0<x_1<\ldots<x_m=b\}$, and of $[c,d]$, say $\{c=y_0<y_1<\ldots<y_n=d\}$, and for each collection of points $\vec c_{ij} \in [x_{i-1},x_i]\times[y_{j-1},y_j]$ for each $i$ and $j$, the corresponding Riemann sum is defined as $$\sum^m_{i=1}\sum^n_{j=1}f(\vec c_{ij})\Delta x_i\Delta y_j.$$
• So now you can make explicit which are in this case: $m$ and $n$; $[a,b]$ and $[c,d]$; $x_i$, $y_j$ and $\vec c_{ij}$.
• Finally, I don't see why you have the expression for $f(x,y)$: are you sure that's it? Or do you have to use it later in the same exercise?