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The expressions $e^h, (1-h^4)^{-1}, \cos(h), 1+\sin(h^3)$ all have he same limits as $h\to 0$. Express each in the following form with the best integer values of $\alpha$ and $\beta$.

$$f(h) = c + O(h^{\alpha}) = c + o(h^{\beta}) .$$

Well I have found the first one $e^h$, but I have problems with $(1-h^4)^{-1}$. I Taylor expand $(1-h^4)^{-1}$ with only one term, so I have the following equation:

$$ (1-h^{4})^{-1}= 1 + E_{0}(h) ,$$ where $E_{0} = \dfrac{4\xi^3h}{(\xi^4-1)^2}$.

Then I let $c$ be $1$ and I therefore have to find the best $\alpha$ in this equation:

$$\dfrac{4\xi ^{3}h}{(\xi ^{4}-1)^{2}} = O(h^{\alpha}) \Rightarrow \exists C,\forall h\neq 0: \left|\dfrac{4\xi ^{3}h}{(\xi ^{4}-1)^{2}}h^{-\alpha}\right|\leq C.$$

And therefore I choose $\alpha$ to be $1$, because I want $\alpha$ be as big as possible and it solves the above equation. The problem is something is wrong. $\alpha$ has to be 4, but I can't see what I have done wrong?

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    Geometric Series: $1/(1-h^4)=1+h^4+h^8+\cdots$ (if $|h|<1$). We don't want to use Taylor's Theorem on *everything*.2011-09-17
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    Why can't I use Taylor?2011-09-17
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    What you got wrong is (mainly) to forget that $\xi$ is constrained by the condition $0\le\xi\le h$. Hence in the last displayed formula of your post, $|\xi^3h|\le h^4$ and $\alpha$ may be much larger than $1$.2011-09-17
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    @Didier's suggestion should solve the issue. But keep in mind that you still must *prove* that $4$ is the *largest* value of $\alpha$. Note that whatever expression for error you have got till now does not tell you that the error is at least a constant times $|h|^4$ as $h \to 0$.2011-09-17
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    @Srivatsan is right, of course. Let me suggest that you write an explicit formula for $E_0(h)=(1-h^4)^{-1}-1$ as a rational function of $h$ only (no $\xi$ involved) and that you prove things from there.2011-09-17
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    @Brugerfugl: You can use Taylor's Theorem. It is just more work.2011-09-17
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    Okay. It's sounds very complicated. Isn't there someone who can write a clear answer? With all the details.2011-09-17
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    @Bru I have written *an* answer but it is not that clean (I mean only the calculation part). Read it and tell me if something is not clear or if you want me to show the calculations. :-)2011-09-17
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    @Brugerfugl, Sorry but my last comment is a clear indication of an easy road to the answer. Just have to R-E-A-D it, slowly and with a pen in your hand...2011-09-17
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    @Didier I am not sure if you consider Taylor's theorem idea a difficult road (:)). In any case, that is a little specialized to rational functions, no? For e.g., the OP wants to solve the problem for $f(x) = 1+\sin (h^3)$ as well. (Added: But yours is a nice suggestion, and I think you should post it as an answer...)2011-09-17
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    @Sri, you are right that Taylor is not difficult either... And yes, the indication I gave is specialized to one function in a list of three. But it works for this function (as you know...) and I am convinced the OP would greatly benefit from writing it in full. Will this happen remains to be seen.2011-09-17

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As André says in his comment, we already know that the correct value of $\alpha$ for $f(h) = (1-h^4)^{-1}$ is $4$, thanks to the geometric series. But the OP wants to justify this using Taylor's theorem.

Note that since we are after the best value of $\alpha$ that works, we really need to show $\alpha$ "works" (i.e., $f(h) = 1+O(h^\alpha)$) and $\alpha' > \alpha$ does not work. Unfortunately, the second part does not follow from the Lagrange form of the remainder $$ E_0(h) = \frac{4\xi^3 h}{1 - h^4}. $$ that the OP has written down in the question. This is because though we can upper bound $E_0(h)$ using $0 \leq |\xi| \leq |h|$, we do not get any lower bound better than $0$.

One fix to this is to prove a stronger statement: $$ f(h) = 1 + h^4 + o(h^4). \tag{1} $$ The advantage of this method is that it shows both that $f(h) = 1 + O(h^4)$ and that $f(h)$ is not $1 + O(h^{4+\rho})$ for any $\rho > 0$. On the other hand, the disadvantage is that we need to know that the correct constant in front of $h^4$ is $1$. But this is usually not a big issue usually.

Ok, now how do we prove $(1)$?


Method 1: Apply higher order Taylor's theorem. Here we want to apply Taylor's theorem for order $k=4$. The Lagrange form of the remainder will actually show that $f(h) = 1 + h^4 + O(h^8)$, whereas the basic form of the theorem will let us conclude $f(h) = 1+h^4+o(h^4)$. Fortunately, the basic form is then sufficient for us.

This method is, of course, systematic. However in order to apply this, one needs to compute four derivatives of the function (at $0$), which is admittedly a daunting task. The only consolation here is that all of those derivatives (except the fourth) should come out to be $0$; so in case we make a mistake, we'll likely realize it soon. :-) I will skip the details of this method since I do not have much to add other than the calculations.


Method 2: Substitution. Make the substitution $x = h^4$, so that the function becomes $(1-x)^{-1}$. Now, our task is to show that $$ \frac{1}{1-x} = 1 + x + o(x). \tag{2} $$ This is quite easy because this follows from the basic form of Taylor's theorem with $k=1$. In particular, we need to calculate just one derivative. The derivative of course if $(1-x)^{-2}$ which evaluates to $1$ at $0$. The statement $(2)$ then follows.

Now, substituting back $x = h^4$, we get $f(h) = 1 + h^4 + o(h^4)$, which is what we wanted to show.