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Confirm the following statements:

i) $$\frac1{1-t^2}=1+t^2+O(t^4)\quad\text{as}\quad t\to0$$

ii) $$\frac1{1-t^2}=-\frac1{t^2}+O\left(\frac1{t^4}\right)\quad\text{as}\quad t\to\infty$$

I'm finding this confusing. The answer I get for i) is $\frac43|t|^4$ but i'm not sure.

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Something being difficult might mean it's complicated, and it might be analysis, but that doesn't mean it's complex analysis. –  Arkamis Mar 8 '13 at 19:07

2 Answers 2

Hint, for i.)

In a neighborhood of $0$, $|t^2| < 1$, so we can use a geometric series for $x=t^2$.

For ii.) let $t = 1/s$, and apply the same argument.

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If $\frac1{1-t^2}-(1+t^2)>ct^4$, then $$ ct^4<\frac1{1-t^2}-\frac{(1+t^2)(1-t^2)}{1-t^2}=\frac{t^4}{1-t^2}$$ i.e. (provided $1-t^2>0$, which holds for all $|t|<1$) $$ c<\frac1{1-t^2}.$$ For all sufficiently small $t$, e.g. $|t|<\frac13$, this implies $c<\frac98$.

By contraposition, if $c\ge\frac98$ (lastly, any $c>1$ would suffice, why?), then $$ \left|\frac1{1-t^2}-(1+t^2)\right|<c\left|t^4\right|$$ and hence $$ \frac1{1-t^2}=1+t^2+O(t^4).$$

Substituting $t\leftarrow \frac1s$ (so that $t\to\infty\iff s\to 0$) we find $$ \frac1{1-t^2}=\frac{1}{1-\frac1{s^2}}=\frac{s^2}{s^2-1}=1-\frac1{1-s^2}=1+s^2+O(s^4)=\frac1{t^2}+O\left(\frac1{t^4}\right).$$

There is another interesting method: As $t\mapsto\frac1{1-t^2}$ is continuous at $t=0$, we have $\frac1{1-t^2}=O(1)$. But then $$\frac1{1-t^2}=\frac{(1-t^2)+t^2}{1-t^2}=1+t^2\cdot\frac{1}{1-t^2}=1+t^2 O(1)=1+O(t^2).$$ By repeating this, $$\frac1{1-t^2}= 1+t^2(1+O(t^2))=1+t^2+O(t^4)$$ and then $$\frac1{1-t^2}= 1+t^2(1+t^2+O(t^4))=1+t^2+t^4+O(t^6)$$ and so on. A similar "trick" can be applied to $t\to \infty$.

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