From the Derivative to the Arcsine Series
The derivative of arcsine has a simple algebraic form, and that form can be expanded using the generalized binomial series. Integrating the resulting powers produces a Maclaurin series for \(\arcsin x\).
The method gives more than a list of coefficients. It also raises an important question: the derivative expansion is valid only for \(|x|<1\), so what happens to the integrated series at \(x=1\) and \(x=-1\)? We will determine both endpoint values carefully.
Expand the Derivative
For \(-1<x<1\), the derivative of the inverse sine function is
The generalized binomial series states that, for \(|u|<1\),
Set \(\alpha=-\tfrac12\) and \(u=-x^2\). The two factors of \((-1)^n\) cancel, leaving positive coefficients:
To see the coefficient formula, the \(n\)th coefficient from the binomial expansion is \[ (-1)^n\binom{-1/2}{n} =\frac{1\cdot 3\cdot 5\cdots(2n-1)}{2^n n!} =\frac{(2n)!}{4^n(n!)^2} =\frac{\binom{2n}{n}}{4^n}. \] For \(n=0\), the product is empty and the coefficient is \(1\).
Integrate the Power Series
On any closed interval \([-r,r]\) with \(0<r<1\), the binomial power series converges uniformly. Consequently, its partial sums can be integrated there and their integrals converge to the integral of the series. This justifies integrating term by term on every such interval. Since \(\arcsin 0=0\), integrating from \(0\) to \(x\) gives
Each coefficient is the corresponding derivative-series coefficient divided by the new power’s exponent \(2n+1\). The resulting series is
Worked Example: Compute the First Four Nonzero Terms
Worked Example: Compute the First Four Nonzero Terms
Find the first four nonzero terms of the Maclaurin series for \(\arcsin x\), and check the result by differentiation.
The derivative-series coefficients for \(1/\sqrt{1-x^2}\), for \(n=0,1,2,3\), are \[ 1,\quad \frac12,\quad \frac38,\quad \frac5{16}. \] Integration changes \(x^{2n}\) to \(x^{2n+1}/(2n+1)\). Therefore the four arcsine coefficients are \[ 1,\quad \frac{1/2}{3}=\frac16,\quad \frac{3/8}{5}=\frac3{40},\quad \frac{5/16}{7}=\frac5{112}. \] Thus the first four nonzero terms are
For a check, differentiate the displayed polynomial: \[ \frac{d}{dx}\left(x+\frac{x^3}{6}+\frac{3x^5}{40}+\frac{5x^7}{112}\right) =1+\frac{x^2}{2}+\frac{3x^4}{8}+\frac{5x^6}{16}. \] These are exactly the first four terms of the binomial expansion of \((1-x^2)^{-1/2}\). Also, the polynomial has value \(0\) at \(x=0\), as required by \(\arcsin 0=0\).
Worked Example: Estimate \(\arcsin(1/2)\)
Worked Example: Estimate \(\arcsin(1/2)\)
Use the first four terms of the series to estimate \(\arcsin(1/2)\), and bound the omitted positive tail.
Substituting \(x=\tfrac12\) into the four-term partial sum gives \[ \frac12+\frac{(1/2)^3}{6}+\frac{3(1/2)^5}{40}+\frac{5(1/2)^7}{112} =\frac12+\frac1{48}+\frac3{1280}+\frac5{14336} \approx 0.523525856. \] All omitted terms are positive. Write \[ a_n=\frac{\binom{2n}{n}}{4^n(2n+1)}. \] The ratio of successive terms in absolute value at \(x=\tfrac12\) is \[ \frac{a_{n+1}(1/2)^{2n+3}}{a_n(1/2)^{2n+1}} =\frac14\frac{(2n+1)^2}{(2n+2)(2n+3)} <\frac14. \] The first omitted term, for \(n=4\), is \[ a_4(1/2)^9 =\frac{35}{128\cdot9}\cdot\frac1{512} =\frac{35}{589824} \approx 0.000059339. \] The remaining terms are bounded above by a geometric series with this first term and ratio \(\tfrac14\). Hence the tail is less than \[ \frac{35}{589824}\left(1+\frac14+\frac1{4^2}+\cdots\right) =\frac{35}{589824}\cdot\frac43 <0.00007912. \] So the four-term partial sum is a lower estimate, and the full value is less than \(0.523525856+0.00007912\). The estimate and bound are consistent with the exact value \(\arcsin(1/2)=\pi/6\).
Why the Series Works: A Complete Endpoint Argument
The power series was derived for \(|x|<1\). To determine its behavior at the endpoints, define \[ a_n=\frac{\binom{2n}{n}}{4^n(2n+1)}. \] Every \(a_n\) is positive. For \(0\le r<1\), the already established series identity is \[ \arcsin r=\sum_{n=0}^{\infty}a_n r^{2n+1}. \] We will show that the endpoint sum is \(\pi/2\), rather than assuming that the identity automatically extends to \(r=1\).
For any \(N\), positivity and \(r^{2n+1}\le 1\) give \(\sum_{n=0}^{N}a_n r^{2n+1}\le\arcsin r\). Letting \(r\) increase to \(1\) in this finite sum gives \(\sum_{n=0}^{N}a_n\le\pi/2\). Thus the increasing sequence of partial sums is bounded above, so \(S=\sum_{n=0}^{\infty}a_n\) exists and \(S\le\pi/2\).
For each \(0\le r<1\), \(r^{2n+1}\le1\), so \(\arcsin r=\sum_{n=0}^{\infty}a_n r^{2n+1}\le S\). Letting \(r\) increase to \(1\), continuity of arcsine gives \(\pi/2\le S\).
The two inequalities imply \(S=\pi/2\). Therefore the series converges at \(x=1\) to \(\pi/2\). At \(x=-1\), its terms have the opposite signs, and their absolute values sum to \(S\); it converges absolutely to \(-\pi/2\).
Where the Series Converges
The endpoint argument establishes convergence at \(x=\pm1\). For \(|x|>1\), the series diverges. Indeed, the ratio of successive absolute terms is \[ \frac{a_{n+1}|x|^{2n+3}}{a_n|x|^{2n+1}} =x^2\frac{(2n+1)^2}{(2n+2)(2n+3)} \longrightarrow x^2. \] When \(|x|>1\), this limit is greater than \(1\). The absolute terms eventually grow by a factor greater than \(1\), so they do not tend to zero. A convergent series must have terms tending to zero; therefore the series cannot converge there. Its interval of convergence is exactly \([-1,1]\).
Worked Example: Decide Convergence at \(x=1\), \(x=-1\), and \(x=2\)
Worked Example: Decide Convergence at \(x=1\), \(x=-1\), and \(x=2\)
At \(x=1\), the series becomes \(\sum_{n=0}^{\infty}a_n\), which converges to \(\pi/2\) by the endpoint argument. At \(x=-1\), it becomes \(-\sum_{n=0}^{\infty}a_n\), which converges absolutely because \(\sum a_n=\pi/2\); its sum is \(-\pi/2\).
At \(x=2\), the ratio of successive absolute terms tends to \(2^2=4\). In particular, from some index onward the terms increase by a factor greater than \(2\). They therefore do not approach zero, so the series diverges at \(x=2\). This agrees with the interval of convergence \([-1,1]\).
The checks use the established sum at each endpoint and the necessary condition that the terms of a convergent series tend to zero outside the interval.
A Common Integration Mistake
A frequent error is to integrate each power but leave its coefficient unchanged. For instance, integrating the derivative term \(\tfrac12x^2\) does not give \(\tfrac12x^3\). The power rule requires division by \(3\):
Consequently, keeping the derivative coefficients unchanged would incorrectly produce \(x+\tfrac12x^3+\cdots\). The correct cubic coefficient is \(\tfrac16\). Differentiating the incorrect expression would return a cubic derivative term with the wrong coefficient, making differentiation a useful check.
A different endpoint mistake is to assume that because the derivative series only holds for \(|x|<1\), the integrated series must fail at \(x=\pm1\). Integration changes the coefficients and can change endpoint convergence. Here the derivative expansion does not give a finite value at the endpoints, but the integrated arcsine series converges there to \(\pm\pi/2\).
Why This Construction Matters
The arcsine series illustrates a general strategy for constructing Taylor series: find a related function with a manageable power series, then integrate to recover the desired function. In this instance, the derivative has a binomial expansion, and the condition \(\arcsin 0=0\) determines the integration constant.
It also demonstrates why convergence must be checked for the series actually being used. The derivative series and the integrated series have different coefficients and different endpoint behavior. The interior identity follows from term-by-term integration on closed intervals strictly inside \((-1,1)\); the endpoint identities require a separate argument.
Check Your Understanding
Use the coefficient formula and endpoint argument to answer the following questions.
- What are the first four nonzero terms in the Maclaurin series for \(\arcsin x\)?
- Why is the coefficient of \(x^{2n+1}\) the derivative-series coefficient divided by \(2n+1\)?
- What value does the arcsine series have at \(x=1\), and why can that value not be assumed from the interior identity alone?
- Why does the series converge absolutely at \(x=-1\)?
- What does the ratio of successive absolute terms show when \(|x|>1\)?
