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My calculus book states that:

We write $f(x)=\mathcal{O}(u(x))$ as $x\to a$ provided that $|f(x)|\le K|u(x)|$

This means that for example as $x \to 0$:

$$e^x-P_1(x)=\mathcal{O}(x^2)$$

...where $P_1(x)$ is the first order Taylor Polynomial for $e^x$.

What I do not understand is what the use of this statement is when talking about Taylor polynomials.

For example: aren't the following statements also true as $x \to 0$?

$$e^x-P_1(x)=\mathcal{O}(x+1)$$

$$e^x-P_1(x)=\mathcal{O}(1)$$

$$e^x-P_1(x)=\mathcal{O}(x^3+1)$$

For all of these examples we can find some $K$ for which the statement is valid close to $0$, isn't it?

I either don't understand something about Big-Oh notation, or I don't understand the use of Big-Oh notation in Taylor Polynomials. What is so special about $e^x-P_1(x)=\mathcal{O}(x^2)$?

Please help me to shed some light on this issue.

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  • $\begingroup$ Your statements are all true. But the one with $x^2$ is much more useful, since $x^2$ goes to zero, unlike all of yours. We assert that $e^x-P_1(x)$ is very small when $x$ is close to zero. Yours would not show that. $\endgroup$ – GEdgar Feb 28 '19 at 21:25
  • $\begingroup$ @GEdgar How about $e^x-P_1(x)=\mathcal{O}(x)$? That one is also valid and goes to zero... $\endgroup$ – GambitSquared Feb 28 '19 at 21:32
  • $\begingroup$ That one is also true... But $\mathcal{O}(x^2)$ tells you more, since $x^2$ is much smaller than $x$, when $x$ is near zero. And $\mathcal{O}(x^3)$ would e even stronger, but we do not use that because it woul be false. $e^x-P_1(x)$ does not go to zero that fast. $\endgroup$ – GEdgar Feb 28 '19 at 21:34

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