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I have a question about the error evaluation between the double integral and its' riemann sums. It seems that this formula is apperantly not zero:

$$\displaystyle\lim_{n\to\infty}n[\int_0^1 dx\int_0^1 f(x,y)dy-\frac{1}{n^2}\sum_{i=1}^{n}\sum_{j=1}^{n}f(\frac{i}{n},\frac{j}{n})]$$

It assumes that f(x,y) is continuous,integrable and derivable in $(0,1)\times(0,1)$. In my calculation, I acquired it by using the first integral theorem,and attain a result that

$$=\displaystyle\lim_{n\to\infty}-\frac{1}{4n}[f(1,1)-f(0,0)]$$

It still reach zero when $\displaystyle\lim_{n\to\infty}$. I am struggling to find its first-order estimation. Can somebody provide an accurate solution to this?

mumujun
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1 Answers1

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Your result, before (and after) you take the limit $n\to\infty$ looks quite wrong.

You can just try some functions and see what happens. Let $f(x,y)=g(x)h(y)$, with $g$, $h$ real valued and continuously differentiable on $[0,1]$. Obviously $\int_0^1\int_0^1 f(x,y)dydx=\int_0^1 g(x)dx \int_0^1 h(y)dy$. If we use the notations $$I(g)=\int_0^1 g(x)dx\;\text{and}\;I_n(g)=\frac{1}{n}\sum_{i=1}^n g\left(\frac{i}{n}\right),$$ then we have $|I(g)-I_n(g)|\leq M_g/n$ for some $M_g$ (see Right Riemann sum Error bound proof for example). It is straightforward to show that a similar error bound applies to your double integral, i.e.: $$|I(g)I(h)-I_n(g)I_n(h)|\leq C/n$$ for some $C>0$. With $f(x,y)=xy$, or $g(x)=x$ and $h(y)=y$, you can do the calculation explicitly, using $I(f)=I(g)=1/2$ and $I_n(g)=I_n(h)=1/n\sum_{i=1}^n i/n=1/2(1+1/n)$.

In this case the limit would be $$ n\left(I(g)I(h)-I_n(g)I_n(h)\right)\to-\frac{1}{2}\;\text{as $n\to\infty$}$$

awsnap
  • 86