My book states that for large n and small a, $$\dfrac{1}{(n+a)^2} - \dfrac{1}{(n)^2} \approx -\dfrac{2a}{n^3}$$

Let $f(x) = \dfrac{1}{x^2}$, $$ df = -\dfrac{2}{n^3}dx$$ with $$dx = a$$ and so the above result follows. But the only restriction with using differentials is that a must be small. How do I say mathematically that n must be large for the approximation to be reasonable?

I have tried calculating with n = 0.01 and a = 0.5, I got an error of $10^6$. Gosh!

  • 2
    $\begingroup$ $df$ has to be small, too, which means $f'$ has to be small for the approximation to be good. $\endgroup$ – Gerry Myerson Mar 10 '13 at 10:05

By the definition of differential we know that $$ \lim_{h\to 0} \frac{f(x+h)-f(x)-hf'(x)}{h} = 0 $$ which we can also write using small-o notation: $$ f(x+h) - f(x) = h f'(x) + o(h). $$ So $$ f(n+a) - f(n) = a f'(n) + o(a) $$ which is $$ \frac{1}{(n+a)^1} - \frac 1 {n^2} = -\frac{2a}{n^3} + o(a). $$ This means that for all $n$ the approximation stated is valid (i.e. the error $o(a)$ goes to zero as $a\to 0$).

However it is also true that the velocity in which $o(a)$ tends to zero depends on $x$. To estimate this we can take another term in the Taylor expansion: $$ f(x+h) = f(x) + h f'(x) + h^2 \frac{f''(x)}{2} + o(h^2) $$ which becomes $$ \frac{1}{(n+a)^2} = \frac 1 {n^2} - \frac{2a}{n^3} + \frac{6 a^2}{n^4} + o(a^2). $$ Here you see that the error term is of order $a^2/n ^4$ and hence to measure the error what you really want to be small is $a/n^2$.


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