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Taylor Series Construction · Tutorial 843 of 998

Square-Root Taylor Series About Positive Centers

Use the generalized binomial formula to expand \(\sqrt{x}\) about a positive center and identify exactly where the resulting series converges.

Intermediate 9 min read

What You'll Learn

  • Rewrite \(\sqrt{x}\) in a form suited to the generalized binomial series.
  • Find the Taylor coefficients for an arbitrary positive center.
  • Determine the radius of convergence from the nearest singularity.
  • Test convergence at both boundary points of the interval.
  • Explain why the series converges to the correct endpoint values.

Why Expand About a Positive Center?

A Taylor series about \(x=0\) is not always the most useful way to represent a function. For \(\sqrt{x}\), the behavior at zero prevents an ordinary Taylor expansion there, but the function is smooth at every positive number. Choosing a positive center \(a\) gives a Taylor series that represents \(\sqrt{x}\) near \(a\), with a convergence range determined by the distance to zero.

The key step is to express \(x\) as \(a\) times a quantity close to \(1\). This lets us apply the generalized binomial Taylor series, whose coefficients are \(\binom{1/2}{n}\).

Definition: For a positive center \(a>0\), the Taylor series of \(\sqrt{x}\) about \(a\) is the power series in \(x-a\) whose coefficients are the derivatives of \(\sqrt{x}\) at \(a\), divided by the corresponding factorials. We will obtain those coefficients by rewriting the function using the generalized binomial formula.

Constructing the Series

Factor \(a\) out of \(x\):

$$ x=a+(x-a) =a\left(1+\frac{x-a}{a}\right). $$

For real \(x\) in a neighborhood of \(a\), this gives

$$ \sqrt{x} = \sqrt{a}\left(1+\frac{x-a}{a}\right)^{1/2}. $$

The generalized binomial formula states that, for \(|u|<1\),

$$ (1+u)^{1/2} = \sum_{n=0}^{\infty}\binom{1/2}{n}u^n, \qquad \binom{1/2}{0}=1. $$

Here \(\binom{1/2}{n}\) is the generalized binomial coefficient: for \(n\geq1\), it is the product \(\frac{(1/2)(1/2-1)\cdots(1/2-n+1)}{n!}\). Substituting \(u=(x-a)/a\) gives the Taylor series about \(a\).

Theorem: For \(a>0\), the Taylor series of \(\sqrt{x}\) about \(a\) is
$$ \sqrt{x} = \sqrt{a}\sum_{n=0}^{\infty} \binom{1/2}{n}\left(\frac{x-a}{a}\right)^n, \qquad |x-a|<a. $$
Equivalently, its \(n\)th coefficient as a power of \(x-a\) is \(\sqrt{a}\binom{1/2}{n}/a^n\).

The first few generalized binomial coefficients are

$$ \binom{1/2}{0}=1,\qquad \binom{1/2}{1}=\frac12,\qquad \binom{1/2}{2}=-\frac18,\qquad \binom{1/2}{3}=\frac1{16},\qquad \binom{1/2}{4}=-\frac5{128}. $$

Thus the first terms, written in powers of \(x-a\), are

$$ \sqrt{x} = \sqrt{a} +\frac{x-a}{2\sqrt{a}} -\frac{(x-a)^2}{8a^{3/2}} +\frac{(x-a)^3}{16a^{5/2}} -\frac{5(x-a)^4}{128a^{7/2}} +\cdots. $$

The alternating signs after the linear term come from the generalized binomial coefficients. The powers of \(a\) in the denominators appear because each power of \((x-a)/a\) contributes a factor of \(a^{-1}\).

Worked Example: Expanding About \(a=4\)

Worked Example: Expanding About \(a=4\)

Set \(a=4\) in the general formula. Since \(\sqrt{4}=2\), the series is

$$ \sqrt{x} = 2\sum_{n=0}^{\infty} \binom{1/2}{n}\left(\frac{x-4}{4}\right)^n. $$

Compute the first terms one at a time. The constant term is \(2\). The linear term is \(2\cdot\frac12\cdot\frac{x-4}{4}=\frac{x-4}{4}\). The quadratic term is \(2\cdot(-\frac18)\cdot\frac{(x-4)^2}{16}=-\frac{(x-4)^2}{64}\). The cubic term is \(2\cdot\frac1{16}\cdot\frac{(x-4)^3}{64}=\frac{(x-4)^3}{512}\). Therefore,

$$ \sqrt{x} = 2+\frac{x-4}{4} -\frac{(x-4)^2}{64} +\frac{(x-4)^3}{512} -\frac{5(x-4)^4}{16384} +\cdots. $$

Check the initial Taylor data: at \(x=4\), every positive power of \(x-4\) vanishes, so the series gives \(2=\sqrt{4}\). Its linear coefficient is \(1/4\), which equals \(\frac{1}{2\sqrt{4}}\), the derivative at \(4\). Its quadratic coefficient is \(-1/64\); multiplying by \(2!\) gives \(-1/32\), the second derivative at \(4\). The radius is \(4\), so the interval of convergence will be \([0,8]\), as established below.

Why the Radius Is \(a\)

The binomial series applies when \(|u|<1\). In this expansion, \(u=(x-a)/a\), and \(a\) is positive. Therefore, the condition becomes

$$ \left|\frac{x-a}{a}\right|<1 \quad\Longleftrightarrow\quad |x-a|<a. $$

So the radius of convergence is \(a\). On the real line, the open interval supplied by this condition is \((0,2a)\). This radius also has a geometric explanation: the function has its nearest singularity at \(x=0\), a distance \(a\) from the center \(x=a\). The square-root function is not analytic at zero, so a Taylor series centered at \(a\) cannot have a radius larger than that distance.

The ratio test also confirms the radius. For \(n\geq1\), the absolute values of successive generalized binomial coefficients satisfy

$$ \frac{\left|\binom{1/2}{n+1}\right|} {\left|\binom{1/2}{n}\right|} = \frac{n-\frac12}{n+1} \longrightarrow 1. $$

For a fixed \(u\), the ratio of successive term magnitudes in the binomial series therefore tends to \(|u|\). The series converges when \(|u|<1\) and diverges when \(|u|>1\). The ratio test does not decide what happens at \(u=1\) or \(u=-1\); those cases must be checked separately.

Derivation: Both Endpoints Converge

At the two boundary points, \(u=\pm1\), the terms have magnitudes \(\left|\binom{1/2}{n}\right|\). We show these magnitudes have a summable upper bound. Set \(b_n=\left|\binom{1/2}{n}\right|\) for \(n\geq1\). From the product definition,

$$ b_1=\frac12, \qquad \frac{b_{n+1}}{b_n} =\frac{n-\frac12}{n+1}. $$

For \(n\geq1\), both sides below are nonnegative, and squaring verifies the inequality:

$$ \left(\frac{n-\frac12}{n+1}\right)^2 \leq \left(\frac{n}{n+1}\right)^3, $$
$$ \left(n-\frac12\right)^2(n+1) =n^3-\frac34n+\frac14 \leq n^3. $$

Consequently, \(b_{n+1}/b_n\leq(n/(n+1))^{3/2}\). Multiplying these inequalities from \(1\) through \(n-1\) gives

$$ b_n \leq b_1\prod_{k=1}^{n-1} \left(\frac{k}{k+1}\right)^{3/2} = \frac{1}{2n^{3/2}}. $$

The series \(\sum_{n=1}^{\infty}1/n^{3/2}\) converges, so \(\sum_{n=1}^{\infty}b_n\) converges. Thus the binomial series converges absolutely at both \(u=-1\) and \(u=1\). In fact, this bound gives uniform convergence for \(-1\leq u\leq1\), because \(|b_nu^n|\leq b_n\). Its sum is therefore continuous on this closed interval. Taking limits from within \((-1,1)\), where the binomial formula already holds, identifies the endpoint sums as

$$ \sum_{n=0}^{\infty}\binom{1/2}{n}(-1)^n=0, \qquad \sum_{n=0}^{\infty}\binom{1/2}{n}= \sqrt{2}. $$

Returning to \(x\), \(u=-1\) corresponds to \(x=0\), while \(u=1\) corresponds to \(x=2a\). The series therefore converges at both endpoints, with values \(0=\sqrt{0}\) and \(\sqrt{a}\sqrt{2}=\sqrt{2a}\), respectively.

Conclusion: For real \(x\), the Taylor series of \(\sqrt{x}\) about \(a>0\) has radius \(a\) and interval of convergence \([0,2a]\). It agrees with \(\sqrt{x}\) throughout that interval. For \(x>2a\), the series diverges; for \(x<0\), the real-valued function \(\sqrt{x}\) is not defined, and the series also lies outside its radius of convergence.

Worked Example: The Endpoint at Zero

Worked Example: The Endpoint at Zero

Take \(a=1\). Then

$$ \sqrt{x} = 1+\frac12(x-1)-\frac18(x-1)^2 +\frac1{16}(x-1)^3-\frac5{128}(x-1)^4+\cdots. $$

The radius is \(1\), so the boundary points are \(x=0\) and \(x=2\). At \(x=0\), the first terms become

$$ 1-\frac12-\frac18-\frac1{16}-\frac5{128}-\cdots. $$

After the constant term, all terms are negative: the signs of the coefficients alternate, and \((x-1)^n=(-1)^n\) at \(x=0\). The endpoint proof shows the series converges absolutely, and continuity from \(0<x<2\) identifies its sum as \(\sqrt{0}=0\). This is consistent with the displayed terms: the negative terms sum to \(-1\), cancelling the initial \(1\).

At \(x=2\), the same series has \(x-1=1\), and the endpoint argument gives the sum \(\sqrt{2}\). Thus the full interval is \([0,2]\), not merely the open interval \((0,2)\). As a check, the endpoint values agree with the original function at both ends: \(0^2=0\) and \((\sqrt2)^2=2\), with both square roots nonnegative.

Worked Example: Evaluating an Interior Point

Worked Example: Evaluating at \(x=16\) About \(a=9\)

For \(a=9\), the expansion is

$$ \sqrt{x} = 3\left[ 1+\frac12\left(\frac{x-9}{9}\right) -\frac18\left(\frac{x-9}{9}\right)^2 +\frac1{16}\left(\frac{x-9}{9}\right)^3-\cdots \right]. $$

At \(x=16\), the binomial variable is \(u=(16-9)/9=7/9\), which satisfies \(|u|<1\). The series converges. Its first four terms, through the cubic term, give

$$ 3\left(1+\frac{7}{18}-\frac{49}{648} +\frac{343}{11664}\right) = 3\left(\frac{11664+4536-882+343}{11664}\right) = \frac{15661}{3888}. $$

This is a finite partial sum, not the exact value; the remaining terms are still part of the series. The generalized binomial formula evaluates the complete sum directly:

$$ 3(1+7/9)^{1/2} =3(16/9)^{1/2} =3\cdot\frac43 =4. $$

Check against the original function: \(\sqrt{16}=4\), and \(4\) is the positive square root because \(4^2=16\). The partial sum is slightly larger than \(4\), which does not contradict the result; it is only a truncation of the convergent series.

A Common Mistake: Stopping at the Open Interval

A frequent error is to apply the binomial formula’s condition \(|u|<1\), obtain \(0<x<2a\), and declare that this is the entire interval of convergence. That condition establishes convergence inside the radius, but the ratio test is inconclusive at its boundary. It does not rule out convergence at either endpoint.

For this particular series, the endpoint coefficient magnitudes are summable, so both endpoints are included. The correct interval is \([0,2a]\). In contrast, for \(|x-a|>a\), the term ratio tends to a number greater than \(1\), so the terms do not tend to zero and the series diverges. Always test the boundary separately rather than treating the open radius interval as the final answer.

Why the Expansion Is Useful

The center \(a\) can be selected to suit the values of \(x\) being studied. The formula is especially useful when \(x\) is close to a convenient positive center, because then \(|(x-a)/a|\) is small and successive powers shrink rapidly. It also makes the limitation clear: moving to or beyond a distance \(a\) from the center requires an endpoint check or leads outside the convergence interval.

The essential relationship is simple: the center is \(a\), the nearest singularity is at zero, and the distance to that singularity is \(a\). The resulting radius is \(a\); careful endpoint analysis then extends convergence to the full real interval from zero through \(2a\).

Check Your Understanding

Use the general expansion and endpoint reasoning to answer these questions.

  1. Write the first four nonzero terms of the Taylor series of \(\sqrt{x}\) about \(a=1\).
  2. For a positive center \(a\), what substitution turns the expansion into the generalized binomial series?
  3. What are the radius and real interval of convergence of the series about \(a\)?
  4. Why does the ratio test alone not determine convergence at \(x=0\) and \(x=2a\)?
  5. What values does the series sum to at \(x=0\) and \(x=2a\)?

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