Asymptotic sequence of tan(z)

629 Views Asked by At

I have a question about the asymptotic sequence of $\tan(z)$:

$$\tan z \sim ~ z+\frac{1}{3}z^3+\frac{2}{15}z^5 $$ $$\sim~ \sin z+\frac{1}{2}\left(\sin z\right)^3+\frac{3}{8}\left(\sin z\right)^5$$

enter image description here

I have a problem with the second sequence in $(\sin^n(z))_n$. The problem is that I don't see a priori that an asymptotic sequence of $\tan(z)$ in the asymptotic sequence $(\sin^n(z))_n$ will exist. I know how I can calculate the coefficients though (here $1,0,\frac{1}{2} ,0,\frac{3}{8}...)$, given that the sequence will exist.

So the only problem for me to see that the given sequence is indeed an asymptotic sequence for $\tan(z)$ is the question: why is there an asymptotic sequence of $\tan(z)$ in the series $(\sin^n(z))_n$. I hope someone can clarify this for me.

1

There are 1 best solutions below

1
On BEST ANSWER

There are a few ways to verify it, depending on what you're more comfortable with.


The first method is to set $w = \sin z$, so that

$$ \begin{align} \tan z &= \frac{w}{\sqrt{1-w^2}} \\ &= \sum_{k=0}^{\infty} \binom{k-\tfrac{1}{2}}{k} w^{2k+1} \\ &= \sum_{k=0}^{\infty} \binom{k-\tfrac{1}{2}}{k} (\sin z)^{2k+1}, \end{align} $$

the first equality holding for $|z| < \pi/2$. Power series are asymptotic series, so we've got what we want.


A second method takes more advantage of power series. The first thing to note is that since both $\tan$ and $\sin$ are odd, they both only have odd powers in their expansions.

In the base case we know that $\tan z \sim \sin z$ as $z \to 0$. Since they're equal to first order only, their difference will be asymptotic to

$$ \tan z - \sin z \sim c_1 z^3 \sim c_1 (\sin z)^3 $$

for some constant $c_1$.

Now suppose that we have found some sequence of coefficients $c_0, c_1, \ldots, c_n$ with $c_0 = 1$ such that

$$ \tan z - \sum_{k=0}^{n} c_k (\sin z)^{2k+1} \sim C (\sin z)^{2N+1}. $$

for some $C \in \mathbb{R}$ and $N \in \mathbb{N}$ with $N > n$. Then, setting $c_{n+1} = c_{n+2} = \cdots = c_{N-1} = 0$ and $c_N = C$, the new difference,

$$ \tan z - \sum_{k=0}^{N} c_k (\sin z)^{2k+1}, $$

will have a power series expansion beginning with a term of the form $Dz^{2M+1}$ for some $D \in \mathbb{R}$ and some $M \in \mathbb{N}$ with $M > N$. We then have

$$ \tan z - \sum_{k=0}^{N} c_k (\sin z)^{2k+1} \sim Dz^{2M+1} \sim D(\sin z)^{2M+1}. $$

Proceeding by induction we construct the full asymptotic series for $\tan z$ in terms of powers of $\sin z$.


A third way to verify it involves proving that each power of $z$, $z^n$, has an asymptotic series in terms of powers of $\sin z$.

First we wish to show that $z$ has an asymptotic series in terms of powers of $\sin z$. To do this we can use the idea from the first method. Setting $w = \sin z$ we have

$$ \begin{align} z &= \arcsin w \\ &= \sum_{k=0}^{\infty} \frac{\binom{2n}{n}}{4^n (2n+1)} w^{2n+1} \\ &= \sum_{k=0}^{\infty} \frac{\binom{2n}{n}}{4^n (2n+1)} (\sin z)^{2n+1} \end{align} $$

for $|z| < \pi/2$. Taking powers of this we can find coefficients $c_{n,k}$ such that

$$ \begin{align} z^{2n+1} &= \left(\sum_{k=0}^{\infty} \frac{\binom{2k}{k}}{4^k (2k+1)} (\sin z)^{2k+1}\right)^{2n+1} \\ &= \sum_{k=n}^{\infty} c_{n,k} (\sin z)^{2k+1} \end{align} $$

for $|\sin z| < 1$.

From this point we will only use the fact that these series are asymptotic expansions, and we will only assume that the power series for $\tan z$ is asymptotic, not convergent. This is to emphasize that the manipulations from this point on will work the same way for other asymptotic series.

Fix $N \in \mathbb{N}$. We have

$$ \tan z = \sum_{n=0}^{N} b_n z^{2n+1} + O(z^{2N+3}) $$

and for each $n = 0, 1, \ldots, N$ we have

$$ z^{2n+1} = \sum_{k=n}^{N} c_{n,k} (\sin z)^{2k+1} + O\Bigl((\sin z)^{2N+3}\Bigr) $$

as $z \to 0$. Combining these we obtain

$$ \tan z = \sum_{n=0}^{N} d_n (\sin z)^{2n+1} + O\Bigl((\sin z)^{2N+3}\Bigr) + O(z^{2N+3}) $$

for some constants $d_n$. Since $z^{2N+3} = O\Bigl((\sin z)^{2N+3}\Bigr)$ this is

$$ \tan z = \sum_{n=0}^{N} d_n (\sin z)^{2n+1} + O\Bigl((\sin z)^{2N+3}\Bigr) $$

as $z \to 0$. Since $N$ was arbitrary, this is an asymptotic series for $\tan z$.

Using the same idea we could start at this series and find the original asymptotic series in terms of powers of $z$.