Question

Inverse sine Given the power series1/√(1 − x²) = 1 + (1/2)x² + (1 ⋅ 3)/(2 ⋅ 4) x⁴ + (1 ⋅ 3 ⋅ 5)/(2 ⋅ 4 ⋅ 6) x⁶ +⋯for −1\u003cx\u003c1, find the power series for f(x) = sin ⁻¹ x centered at 0.

Accepted Answer

Recall that the function $$ f(x) = \sin^{-1} x  is $$the integral of its derivative. Since $$ \frac{d}{dx} \sin^{-1} x = \frac{1}{\sqrt{1 - x^2}} $$, we can use the given power series for $$ \frac{1}{\sqrt{1 - x^2}}  to $$find the series for $$ \sin^{-1} x . $$Write the given power series explicitly as:

$$
\frac{1}{\sqrt{1 - x^2}} = 1 + \frac{1}{2} x^2 + \frac{1 \cdot 3}{2 \cdot 4} x^4 + \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} x^6 + \cdots
$$ Integrate the series term-by-term with respect to $$ x $$ from 0 to $$ x  to $$find $$ \sin^{-1} x $$:

$$
\sin^{-1} x = \int_0^x \frac{1}{\sqrt{1 - t^2}} dt = \int_0^x \left(1 + \frac{1}{2} t^2 + \frac{1 \cdot 3}{2 \cdot 4} t^4 + \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} t^6 + \cdots \right) dt
$$ Integrate each term individually:

- The integral of 1 with respect to $$ t  is$$ $$ t .
- $$The integral of $$ t^2  is$$ $$ \frac{t^3}{3} .
- $$The integral of $$ t^4  is$$ $$ \frac{t^5}{5} $$, and so on.

So the series becomes:

$$
\sin^{-1} x = x + \frac{1}{2} \cdot \frac{x^3}{3} + \frac{1 \cdot 3}{2 \cdot 4} \cdot \frac{x^5}{5} + \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \cdot \frac{x^7}{7} + \cdots
$$ Simplify the coefficients to write the power series for $$ \sin^{-1} x $$ centered at 0. This series converges for $$ -1 \u003c x \u003c 1 $$ and represents the inverse sine function as a power series.

Inverse sine Given the power series
1/√(1 − x²) = 1 + (1/2)x² + (1 ⋅ 3)/(2 ⋅ 4) x⁴ + (1 ⋅ 3 ⋅ 5)/(2 ⋅ 4 ⋅ 6) x⁶ +⋯
for −1<x<1, find the power series for f(x) = sin ⁻¹ x centered at 0.

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Key Concepts

Power Series Representation

Relationship Between a Function and Its Derivative

Term-by-Term Integration of Power Series

Inverse sine Given the power series1/√(1 − x²) = 1 + (1/2)x² + (1 ⋅ 3)/(2 ⋅ 4) x⁴ + (1 ⋅ 3 ⋅ 5)/(2 ⋅ 4 ⋅ 6) x⁶ +⋯for −1<x<1, find the power series for f(x) = sin ⁻¹ x centered at 0.