I would like to know the leading order of $x$ in the expression $\sqrt{c+x^2-\sqrt{c^2+x^2}}$, where $x>0$ and $c\in \mathbb{R}$. I tried

test = Sqrt[c + x^2 - Sqrt[c^2 + x^2]];
ans1 = Series[test, {x, 0, 2}];
ans2 = Series[test, {x, 0, 2}, Assumptions -> {c > 0}];

The result ans1 is \begin{equation} \sqrt{c-\sqrt{c^2}}+\frac{\left(2 \sqrt{c^2}-1\right) x^2}{4\sqrt{c^2} \sqrt{c-\sqrt{c^2}}}+O\left(x^3\right) \end{equation} where the leading order in $x$ is second order, while ans2 is \begin{equation} \frac{\sqrt{(2 c-1) x^2}}{\sqrt{2} \sqrt{c}}+\frac{x^2 \sqrt{(2c-1) x^2}}{8 \sqrt{2} c^{5/2} (2 c-1)}+O\left(x^3\right) \end{equation} which is first order in $x$ when $c\neq \frac{1}{2}$. Why is it the discrepancy between ans1 and ans2?

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    $\begingroup$ Something ain't right with the last one. The function is even, so you'd expect odd-order terms to be zero. Yet, your last expansion has odd order terms. $\endgroup$ – J. M.'s discontentment Mar 29 '18 at 3:14
  • $\begingroup$ I think something went wrong starting in version 8. Version 5 gives the result $$\sqrt{c-\sqrt{c^2}}+\left(\frac{1}{2\sqrt{c-\sqrt{c^2}}}-\frac{\sqrt{c^2}}{4 c^2\sqrt{c-\sqrt{c^2}}}\right) x^2+O\left(x^3\right)$$ for ans2. $\endgroup$ – J. M.'s discontentment Mar 29 '18 at 4:51
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    $\begingroup$ @J.M. When c>0 the version 5 result has major problems with division by 0. $\endgroup$ – Carl Woll Mar 29 '18 at 6:01


The expansion is around branch point hence the result is not unique! The result for $c>0$ and $x$ approaching $0$ along real axis is at the end.

Note: Version 11.3 is used!


The problem stems from the fact that you are expanding about the branch point of square root for $c>0$. It is easy to see this as

$$\sqrt{c+x^2-\sqrt{c^2+x^2}}\simeq\sqrt{c+x^2-\lvert c\rvert-\lvert c\rvert\frac{x^2}{c^2}+O(x^3)}=\left\{\begin{aligned}\sqrt{0+O(x)}\quad\text{for }c>0\\ \sqrt{2c+O(x)}\quad\text{for }c<0\end{aligned}\right.$$

where we know that $0$ is the branch point of square root function.

It is safer to expand around $a$ and then take the limit to see the behavior. We indeed see that

In[44]:= Series[Series[Sqrt[c+x^2-Sqrt[c^2+x^2]],{x,a,2}],{a,0,1}]

$$\left(\sqrt{c-\sqrt{c^2}}+O\left(a^2\right)\right)+(x-a) \left(\left(\frac{1}{\sqrt{c-\sqrt{c^2}}}-\frac{1}{2 \sqrt{c^2} \sqrt{c-\sqrt{c^2}}}\right) a+O\left(a^2\right)\right)+(x-a)^2 \left(\frac{\left(1-2 \sqrt{c^2}\right) \sqrt{c-\sqrt{c^2}}}{4 \sqrt{c^2} \left(\sqrt{c^2}-c\right)}+O\left(a^2\right)\right)+O\left((x-a)^3\right)$$

One can see that the coefficient of linear term is indeterminate for $c>0$ if we take $a\to 0$. Therefore, the true result depends on how you approach the branch point.

Let us consider the linear term. We will take $a\to 0$ in the complex plane such that $$a\to c^2 e^{i \theta } \left(c-\sqrt{c^2+\text{$\kappa $}}\right)^{3/2}$$ with $$\kappa\to 0$$ for arbitrary phase $\theta$ and real parameter $\kappa$. If we insert this, we see that there remains a $\kappa$ independent term in the series expansion, indicating that we have different linear terms for different paths we take to approach to the branch point:

Refine[Series[a (1/Sqrt[c-Sqrt[c^2]]-1/(2 Sqrt[c^2] Sqrt[c-Sqrt[c^2]])) (-a+x)/.a-> E^(I \[Theta]) c^2 (c-Sqrt[c^2+ \[Kappa]])^(3/2)//FullSimplify,{\[Kappa],0,0}],c<0]

$$-(-2 c-1) c^2 e^{i \theta } \left(x-2 i \sqrt{2} \sqrt{-c} c^3 e^{i \theta }\right)+O\left(\kappa ^1\right)$$

On the other hand, for $c>0$, the linear term dies as $\kappa \to 0$:

Refine[Series[a (1/Sqrt[c-Sqrt[c^2]]-1/(2 Sqrt[c^2] Sqrt[c-Sqrt[c^2]])) (-a+x)/.a-> E^(I \[Theta]) c^2 (c-Sqrt[c^2+ \[Kappa]])^(3/2)//FullSimplify,{\[Kappa],0,0}],c>0]

$$O\left(\kappa ^1\right)$$

The series expansion is still even around 0

The appearance of odd terms may naively prompt one to think that the series expansion is not even. However the situation is not so.

As we saw above, expanding exactly at zero is problematic, so let us expand around $\epsilon$ and consider $\epsilon>0$ and $\epsilon<0$ cases separately.


$$O\left(\epsilon ^1\right)+(x-\epsilon ) \left(\frac{\sqrt{2 c-1}}{\sqrt{2} \sqrt{c}}+O\left(\epsilon ^1\right)\right)+O\left(\epsilon ^1\right) (x-\epsilon )^2+(x-\epsilon )^3 \left(\frac{1}{8 \sqrt{2} c^{5/2} \sqrt{2 c-1}}+O\left(\epsilon ^1\right)\right)+O\left((x-\epsilon )^4\right)$$

As $\epsilon\to 0$ from right, we only get odd terms with those coefficients.

If we approach zero from right:


$$O\left(\epsilon ^1\right)+(x-\epsilon ) \left(-\frac{\sqrt{2 c-1}}{\sqrt{2} \sqrt{c}}+O\left(\epsilon ^1\right)\right)+O\left(\epsilon ^1\right) (x-\epsilon )^2+(x-\epsilon )^3 \left(-\frac{1}{8 \left(\sqrt{2} c^{5/2} \sqrt{2 c-1}\right)}+O\left(\epsilon ^1\right)\right)+O\left((x-\epsilon )^4\right)$$

We see that approaching to zero from left and from right produces same series expansion upto an overall negative sign. Since only the odd powers of x are present in the expansion, we get another relative sign between $x<0$ and $x>0$ cases, so we can conclude

$$ \sqrt{c+x^2-\sqrt{c^2+x^2}}= \frac{\sqrt{2 c-1} \left| x\right| }{\sqrt{2} \sqrt{c}}+\frac{\left| x\right| ^3}{8 \sqrt{2} c^{5/2} \sqrt{2 c-1}}+O\left(\left| x\right| ^5\right)$$

That we are working around a branch point is reflected on the fact that our expansion is discontinuous at $x=0$.

| improve this answer | |
  • $\begingroup$ Can you base your "easy to see this as"? TIA. The "coefficient of linear term" is not indeterminate, the one takes complex values in this case. $\endgroup$ – user64494 Mar 29 '18 at 8:38
  • $\begingroup$ You have $\frac{a}{c-\lvert c \rvert}$ which is $0/0$ hence indeterminate at $a=0$. That is why I avoided expanding at $a=0$ at the end of the post and instead expanded at $\epsilon<0$ or $\epsilon>0$ and took them to zero as limiting cases. $\endgroup$ – Soner Mar 29 '18 at 8:45
  • $\begingroup$ Sorry, you still don't base "easy to see this as" in the beginnig of your answer. $\endgroup$ – user64494 Mar 29 '18 at 8:55
  • $\begingroup$ What I mean there is that we are expanding around $\sqrt{0}$ for $c>0$ and around $\sqrt{2c}$ for $c<0$ where $0$ is the branch point of square root function. I guess I was not clear. $\endgroup$ – Soner Mar 29 '18 at 9:01
  • $\begingroup$ Sorry, your explanations still are unclear. What is expanded? In which variable? At which point? $\endgroup$ – user64494 Mar 29 '18 at 9:33


test = Sqrt[c + x^2 - Sqrt[c^2 + x^2]];
ans3 = Series[test, {x, 0, 2}, Assumptions -> c > 0 && x > 0]

$$ \frac{\sqrt{2 c-1} x}{\sqrt{2} \sqrt{c}}+O\left(x^3\right) $$


ans4 = Series[test, {x, 0, 2}, Assumptions -> c < 0 && x > 0]

$$ i \sqrt{2} \sqrt{-c}+\frac{i \sqrt{-c} (2 c+1) x^2}{4 \sqrt{2} c^2}+O\left(x^3\right) $$

Addition. Let us verify ans3:

Sqrt[c + x^2 - Sqrt[c^2 + x^2]] /. {c -> 1, x -> 0.01}


Normal[Series[test, {x, 0, 2},Assumptions -> c > 0 && x > 0]] /. {c -> 1, x -> 0.01}


and let us verify ans4

Sqrt[c + x^2 - Sqrt[c^2 + x^2]] /. {c -> -1, x -> 0.01}

$0.\, +1.4142 i $

Normal[Series[test, {x, 0, 2},Assumptions -> c < 0 && x > 0]] /. {c -> -1, x -> 0.01}

$0.\, +1.4142 i $

| improve this answer | |
  • $\begingroup$ But shouldn't an even function have zero odd-order terms? $\endgroup$ – J. M.'s discontentment Mar 29 '18 at 4:45
  • $\begingroup$ J. M.: Do you take into account branch cuts? The Maple command FunctionAdvisor(branch_cuts,sqrt(1+x^2-sqrt(1+x^2)),plot=2.); shows the ones through the origin for $c=1$. $\endgroup$ – user64494 Mar 29 '18 at 4:53
  • $\begingroup$ See the Maple result here dropbox.com/s/ibqtm350j34fkki/BC2.rtf?dl=0 . $\endgroup$ – user64494 Mar 29 '18 at 5:03

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