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I've found two ways to determine the order of the zero $z = 0$ of the function $f(z) = e^{\sin(z)} - e^{\tan(z)}?$

Both ways, described below, are unsatisfactory insofar as I need help either from a calculator like Wolfram Alpha or from a collection of mathematical formulas.

(1) According to Wolfram Alpha the first and second derivatives of f evaluated at z = 0 are zero, while the third derivative is non-zero. Hence the zero has order three.

(2) In a collection of mathematical formulas I've found the first few terms in the Maclaurin expansion of the tangent function. I've used that together with the easy to remember Maclaurin expansions of sin and exp, and have concluded that the Maclaurin expansion of f probably starts with a cubic term, which would mean that the zero has order three, a result which agrees with what we got with method (1).

If I got a problem like this on an exam, I would be in big trouble since I wouldn't have access to any aids like the ones I've mentioned.

It's not possible to calculate all of the first three derivatives of f by hand since repeated applications of the quotient rule would be necessary, which would be just too cumbersome. Also the Maclaurin expansion of tan is not an expansion which you could memorize due to the weird coefficients without a recognizable pattern.

Is there an easy way to find the order of the zero z = 0 of f by hand, without any aids? I haven't been able to find one.

I would like to add that I don't find it possible to calculate by hand the first few terms in the Maclaurin expansion of tan, since also that would require cumbersome, repeated applications of the quotient rule.

Thomas Andrews
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A A
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    You can figure out the first terms of the expansion of $\tan x$ by approximating $\cos x$ as $1-x^2/2$ and thus $\tan x$ as $(x-x^3/6)(1+x^2/2)=x+x^3/3+O(x^5).$ That's enough to work it out. – Thomas Andrews Jul 27 '23 at 23:13
  • Another way would be via the residue theorem, to find the first power $n$ for which $f(z)/z^{n+1}$ has a nonzero contour integral around the origin. But I don't see that actually being so useful in practice here. – Ethan Dlugie Jul 27 '23 at 23:15
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    To get a as many terms as you need of the expansion of $\tan$ just divide the expansions of $\sin$ and $\cos$ Exactly the same way that you divide polynomials (by dividing leading terms by leading terms), but you consider terms larger when they have lower exponent. – NDB Jul 27 '23 at 23:17
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    Ultimate, you'll have $$e^x\left(e^{-x^3/6+O(x^5)}-e^{x^3/3+O(x^5)}\right)$$ The $e^x$ term doesn't affect the zero, and neither do the two $O(x^5)$ values. You get the value is $-x^3/6-x^3/3 +O(x^5)=-x^3/2+O(x^5).$ – Thomas Andrews Jul 27 '23 at 23:19
  • @Thomas Andrews Ok, good, but why did you choose to take two terms in each of the expansions of sin and cos when you made the approximation? Maybe it's enough to use only one term in each expansion, or maybe you need at least three or four? – A A Jul 27 '23 at 23:34
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    In general, if $e^{f(z)}-e^{g(z)}$ will have a non-zero coefficient where the power series of $f$ and $g$ differ. Basically, if $f(z)=\sum_0^\infty a_i z^i$ and $g(z)=\sum_0^\infty b_iz^i,$ with $a_i=b_i$ for $i=0,\dots n-1,$ and $a_n\neq b_n,$ then $$e^{f(z)}-e^{g(z)}=e^{\sum_0^{n-1}a_iz^i}\left(e^{a_nz^n+O(z^{n+1})}-e^{b_nz^n+O(z^{n+1})}\right)\=e^{p(z)}\left(\frac{a_n-b_n}{n!}z^n+O(z^{n+1})\right),$$ where $p(z)$ is a polynomial of degree $n-1.$ So I picked enough of the series to get a difference. – Thomas Andrews Jul 27 '23 at 23:50
  • @Thomas Andrews I see, and how can you be sure that the terms that you get for the expansion of tan are correct if you use only parts of the expansions of sin and cos, and not the full expansions? – A A Jul 27 '23 at 23:56
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    Because the exponents higher than $3$ in $\sin$ and $\cos$ only affect the coefficients of the higher degrees. – Thomas Andrews Jul 28 '23 at 00:11
  • @Thomas Andrews Ok, and how do you know that you are allowed to multiply or divide power series like polynomials in this case? Are you implicitly using the theorem which says that the coefficients in a power series expansion of the product of two absolutely convergent power series are so called Cauchy sums? – A A Jul 28 '23 at 00:20

4 Answers4

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In general, if $f,g$ are analytic in a neighborhood of $0,$ then the degree of the zero at $z=0$ of $e^{f(z)}-e^{g(z)}$ is the smallest degree where the power series of $f$ and $g$ disagree.

If $f(z)=\sum a_iz^i$ and $g(z)=\sum b_iz^i$ and $a_i=b_i$ for $i<n$ and $a_{n}\neq b_n,$ then we get:

$$\begin{align}e^{f(z)}-e^{g(z)}&=e^{\sum_0^{n-1}a_iz^i}\left(e^{a_nz^n+O(z^{n+1})}-e^{b_nz^n+O(z^{n+1})}\right)\\ &=e^{\sum_0^{n-1} a_iz^i}\left(\frac{a_n-b_n}{n!}z^n+O(z^{n+1})\right) \end{align} $$

The exponent doesn't add any zeros.

So you just need to know the first term where the power series for $\sin z$ and $\tan z$ disagree.

There's actually a way to show that $\sin z$ and $\tan z$ disagree first in the $z^3$ term, without knowing the power series of $\tan z.$ (You'll still need to know the power series for $\sin z.$)

First show they agree at $z^1.$ This amounts to knowing $\lim_{z\to 0}\cos z=1.$

Next, $\sin z$ and $\tan z$ are odd functions, so their coefficients of $z^n$ are $0$ for even $n.$

Then you need to know the inequality $\tan x>x$ is true for all $x\in(0,\pi/2).$ But $\sin z=z-z^3/6+O(z ^5)$ and if $\tan z=z-z^3/6+O(z^5),$ we'd have, for small positive $x,$ $\tan x<x.$

Thomas Andrews
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In a half-way approach, consider instead only the series expansion for $e^z$

$$e^{\sin z} - e^{\tan z} = (1 + \sin z + \cdots) - (1 + \tan z + \cdots)$$

$$= \sum_{n=1}^\infty \frac{\sin^nz-\tan^nz}{n!} = \sum_{n=1}^\infty\frac{\tan^nz(\cos^nz-1)}{n!}$$

and by small angle approximation, the lowest order term for each $n$ is

$$\tan^nz(\cos^nz-1) \sim z^n\left(\left(1-\frac{z^2}{2}\right)^n-1\right) \sim z^{3n}$$

therefore a factor of $z^3$ can be pulled from every term in the series for $n\geq 1$

Ninad Munshi
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If $f(z)$ and $g(z)$ are analytic functions with zeros of order $m$ and $n$, respectively, at $z=0$, then:

  • $f(z)g(z)$ has a zero of order $m+n$ at $z=0$ (most people realize this);
  • $f(g(z))$ has a zero of order $mn$ at $z=0$ (this is less well known! but it follows immediately from their series representations).

With this in mind, we can write $e^{\sin z} - e^{\tan z} = e^{\tan z} \bigl( e^{\sin z-\tan z}-1 \bigr) = e^{\tan z} f(g(z))$ where $f(z) = e^z-1$ and $g(z) = \sin z-\tan z$. It's easy to see that ($e^{\tan z}$ doesn't have a zero at $z=0$ and) $f(z)$ has a simple zero at $z=0$; so the order of the zero of $e^{\sin z} - e^{\tan z}$ is the same as the order of the zero of $g(z) = \sin z-\tan z$. (Even here we can avoid taking three derivatives by writing $g(x) = \tan z (\cos z-1)$ and using the first fact above.)

Greg Martin
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$$ \begin{align} e^{\sin(x)}-e^{\tan(x)} &=-e^{\tan(x)}\left(1-e^{\sin(x)-\tan(x)}\right)\tag{1a}\\[6pt] &=-e^{\tan(x)}\left(1-e^{-\frac12x^3+O\left(x^5\right)}\right)\tag{1b}\\ &=-e^{\tan(x)}\left(\frac12x^3+O\!\left(x^5\right)\right)\tag{1c}\\ &=-\frac12x^3+O\!\left(x^4\right)\tag{1d} \end{align} $$ Explanation:
$\text{(1a):}$ factor out $e^{\tan(x)}$
$\text{(1b):}$ $\sin(x)-\tan(x)=-\frac12x^3+O\!\left(x^5\right)$
$\text{(1c):}$ $1-e^{-x}=x+O\!\left(x^2\right)$
$\text{(1d):}$ $e^x=1+x+O\!\left(x^2\right)$

robjohn
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