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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. – André Nicolas Sep 17 '11 at 14:44
Why can't I use Taylor? – Brugerfugl Sep 17 '11 at 14:58
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 $1$. – Did Sep 17 '11 at 15:07
@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 $h \to 0$. – Srivatsan Sep 17 '11 at 15:19
@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. – Did Sep 17 '11 at 15:31
up vote 0 down vote accepted

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.

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