# Taylor series for $\sqrt{x}$?

I'm trying to figure Taylor series for $\sqrt{x}$. Unfortunately all web pages and books show examples for $\sqrt{x+1}$. Is there any particular reason no one shows Taylor series for exactly $\sqrt{x}$?

• The Taylor series is only defined for smooth functions. The function $x \mapsto \sqrt{x}$ is not differentiable at $x=0$. (Also, it is not defined for $x<0$.) – copper.hat Jan 16 '15 at 6:17

Short answer: The Taylor series of $\sqrt x$ at $x_0 = 0$ does not exist because $\sqrt x$ is not differentiable at $0$. For any $x_0 > 0$, the Taylor series of $\sqrt x$ at $x_0$ can be computed using the Taylor series of $\sqrt{1 + u}$ at $u_0 = 0$.

Long answer: The Taylor series of a function $f$ that is infinitely differentiable at a point $x_0$ is defined as

$$\sum_{n=0}^\infty \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n = f(x_0) + \frac{f'(x_0)}{1!}(x-x_0) + \frac{f''(x_0)}{2!}(x-x_0)^2 + \ldots \quad .$$ Therefore:

• Asking for "the Taylor series of $f$" makes only sense if you specify the point $x_0$. (Often this point is implicitly assumed as $x_0 = 0$, in this case it is also called the Maclaurin series of $f$.)
• The Taylor series of $f$ at $x_0$ is only defined if $f$ is infinitely differentiable at $x_0$. (But the Taylor series need not be convergent for any $x \ne x_0$, and even if it converges in a neighborhood of $x_0$, the limit can be different from the given function $f$.)

Each Taylor series is a power series $\sum_{n=0}^\infty a_n (x-x_0)^n$ and the connection is roughly the following: If there exists a power series such that $$f(x) = \sum_{n=0}^\infty a_n (x-x_0)^n \text{ in a neighborhood of x_0}$$ then

• $f$ is infinitely differentiable at $x_0$, and
• $a_n = {f^{(n)}(x_0)}/{n!}$ for all $n$, i.e. the power series is exactly the Taylor series.

Now applying that to your question: You are asking for the Taylor series of $f(x) = \sqrt{ x}$. If you meant the Taylor series at $x_0 = 0$: It is not defined because $\sqrt {x}$ is not differentiable at $x_0 = 0$. For the same reason, there is no power series which converges to $f$ in a neighborhood of $0$.

But $f(x) = \sqrt{ x}$ can be developed into a Taylor series at any $x_0 > 0$. The general formula is given in Mhenni Benghorbal's answer. The reason that often only the Taylor series for $\sqrt{1 + x}$ is given in the books is that – for the square-root function – the general case can easily be reduced to the special case: $$\sqrt {\mathstrut x} = \sqrt {\mathstrut x_0 + x - x_0} = \sqrt {\mathstrut x_0}\sqrt {1 + \frac {\mathstrut x-x_0}{x_0}}$$ and now you can use the Taylor series of $\sqrt{1+u}$ at $u_0 = 0$.

The same "trick" would work for functions like $g(x) = x^\alpha$ because $g(x) = g(x_0) \cdot g(1 + \frac {x-x_0}{x_0})$

• I wonder if I can use this "trick" for any function that is not differentiable at $x_0 = 0$? I could always find Taylor series for f(x+1), and then substitute $u = x+1$ to get back to original function f(x). – bodacydo Jan 17 '15 at 7:30
• @bodacydo: No. Here we have used $f(x) = f(x_0) \cdot f(1 + \frac {x-x_0}{x_0})$ and that is not generally true for arbitrary functions. It also does not help to find the Taylor series of $f(x)$ at $x_0 = 0$ because using this would correspond to the Taylor series of $f(1 + u)$ at $u_0 = -1$ which does not exist. – Martin R Jan 17 '15 at 8:26
• @bodacydo: I have expanded the answer and tried to explain it better. – Martin R Jan 17 '15 at 9:38
• @the_candyman: Thanks for fixing my stupid copy/paste error! – Martin R Jan 17 '15 at 10:41

I assume you are talking about the Taylor series at $0$ for $\sqrt{x}$. Let's try to compute the Taylor series at $0$: $$f(x)=f(0)+f'(0)(x-0)+f''(0)\frac{(x-0)^2}2+\dots$$ $f(0)=0$, but $f'(x)=\frac1{2\sqrt{x}}$ blows up at $x=0$. Since $\sqrt{x}$ doesn't have a first derivative at $0$, it doesn't have a Taylor series there.

Note: Strictly speaking, what is proved below is that $\sqrt{x}$ cannot have an asymptotic expansion of the form $a_0 + a_1 x + o(x)$ as $x \to 0$.

There is no Taylor series for it at $0$. If there were, it would be $$\sqrt{x} = a_0 + a_1 x + a_2 x^2 + \dots.$$

Obviously, $a_0$ would have to be $0$, but $\sqrt{x}$ is much larger as $x \to 0$ than any expansion starting with $a_1 x$. For example, we'd have $$\frac{1}{\sqrt{x}} = \frac{\sqrt{x}}{x} = a_1 + a_2 x + \dots \rightarrow a_1,$$ as $x \to 0$, but $\frac{1}{\sqrt{x}}$ doesn't have a finite limit as $x \to 0$.

On the other hand, it's easy to obtain the Taylor expansion for $\sqrt{x}$ at $a > 0$ from the one for $\sqrt{1 + x}$ at $0$. Setting $h = x - a$, you have $$\sqrt{x} = \sqrt{a + h} = \sqrt{a}\sqrt{1 + h/a},$$ and then you expand $\sqrt{1 + h/a}$ in powers of $h/a$.

• Strictly speaking, the definition of a Taylor series does not require that it converges to the given function or converges at all. – Martin R Jan 16 '15 at 8:09
• In all honesty, I'm not really viewing it as a convergent series, but rather as an asymptotic expansion in powers of $x$. I concede that I used what is perhaps misleading notation (as a series), because the question seemed to be asking for intuition on the question. Despite the technical inaccuracy, the answer makes the essential point. Strictly speaking, what I have shown is that $\sqrt{x}$ doesn't have an asymptotic expansion of the form $a_0 + a_1 x + o(x)$ as $x \to 0$. – user208259 Jan 16 '15 at 9:36

The answer is correct.

Note that you can find Taylor series of $\sqrt{x}$ at a point $a>0$ as

$$\sqrt{x} = \sum _{n=0}^{\infty }\frac{\sqrt {\pi }}{2}\,{\frac {{a}^{\frac{1}{2}-n} \left( x-a\right)^{n}}{\Gamma\left( \frac{3}{2}-n \right)n! }}.$$

• So it can have a Taylor series but not a Maclaurin series? – BCLC Jul 12 '15 at 10:55
• @BCLC Yes, that's correct. – Jam Nov 26 '19 at 15:23

if $$\sqrt{x}=a_0+a_1x+a_2x^2+\dots$$ then $$x=a_0^2+(a_0a_1+a_1a_0)x+(a_0a_2+a_1a_1+a_2a_0)x^2+\dots$$ and if you want the identity theorem to hold this is impossible because $a_0=0$ would imply that the coeff of $x$ is zero

As in the other answers, $f:\mathbb{R}^+\bigcup\{0\}\to\mathbb{R}^+\bigcup\{0\};\,f(x)=\sqrt{x}$ has no derivative at $x=0$, so no Taylor expansion around $x=0$.

It's worth noting, however, that the signularity at $x=0$ is a different kind of singularity from the singularity $g:\mathbb{R}\to\mathbb{R}\{0\};\,g(x)=\frac{1}{x}$ that denies us a Taylor expansion for $g$ at $x=0$. This one is simpler to understand and is called a pole.

But your singularity is called a Branch Point and it is where two "branches" of a multi-valued function are joined in an essential way. Recall that $f_\pm(x)=\pm\sqrt{x}$ are both functions which are partial inverses to $x\mapsto x^2$. They "join" at $x=0$. Functions with branch points involving $n^{th}$ roots like yours can have a well-defined value at their branch points (unlike the pole example, which blows up to $\infty$ as one approaches the pole), but some derivative of the function fails to be defined at the branch point. For example, $x\to x^{\frac{3}{2}}$ is well defined at $x=0$, and also has a well defined derivative $x\to \frac{3}{2}x^{\frac{1}{2}}$ at $x=0$. But the second derivative is undefined there.

Let $u = x+1$. Then just substitute into that other Taylor Series. The reason it is found everywhere is simply because it is easy to calculate.

• No, it does not exist. – copper.hat Jan 16 '15 at 6:20
• @copper.hat Maclaurin series dne. Are you saying Taylor series dne for any $x_0 \in \ \mathbb{R}$ ? – BCLC Jul 12 '15 at 11:19
• @bclc: No, it doesn't exist for $x_0=0$, it does exist for $x_0>0$. – copper.hat Jul 14 '15 at 0:03
• @copper.hat You said 'it' dne. What is 'it' ? – BCLC Jul 14 '15 at 16:10
• @bclc: The Maclaurin series... – copper.hat Jul 15 '15 at 17:12