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 $\sqrt{x}$?
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$\begingroup$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 inMhenni 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 $g(x) = g(x_0) \cdot g(1 + \frac {x-x_0}{x_0})$
$\endgroup$ 3 $\begingroup$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.
$\endgroup$ 1 $\begingroup$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 $h/a$.
$\endgroup$ 3 $\begingroup$The answer is correct.
Note that you can find Taylor series of $\sqrt{x} $ at a point $a>0$ as
$$ \sqrt{x} = \sum _{n=0}^{\infty }\frac{\sqrt {\pi }}{2}\,{\frac {{a}^{\frac{1}{2}-n} \left( x-a\right)^{n}}{\Gamma\left( \frac{3}{2}-n \right)n! }}.$$
See my answer.
$\endgroup$ 3 $\begingroup$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
$\endgroup$ 0 $\begingroup$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.
$\endgroup$ $\begingroup$Well, I know this is an old answered post, but referring to the question that all web pages and books doesn't show examples for $\sqrt{x}$, on book, Thomas' Calculus Twelfth Edition on exercises of section 10.8, problem 9 the question is: find the Taylor polynomial of order 0,1,2, and 3 generated by $f$ at $a$ and the exercises are the following
And the book also gives the answers, those are the following
I've posted the images of the book in order to show that at least there's one book with the worked exercise.
Internet is a very resourceful place, but good calculus books are a very good option to find lots of exercises and examples that have been worked by experts.
$\endgroup$ 2 $\begingroup$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.
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