Properties of power series

3 Properties of power series

Let $P_{1}$ and $P_{2}$ represent two power series with radii of convergence $R_{1}$ and $R_{2}$ respectively. We can combine $P_{1}$ and $P_{2}$ together by addition and multiplication. We find the following properties:

Key Point 12

If $P_{1}$ and $P_{2}$ are power series with respective radii of convergence $R_{1}$ and $R_{2}$ then the sum $(P_{1} + P_{2})$ and the product $(P_{1} P_{2})$ are each power series with the radius of convergence being the smaller of $R_{1}$ and $R_{2}$ .

Power series can also be differentiated and integrated on a term by term basis:

Key Point 13

If $P_{1}$ is a power series with radius of convergence $R_{1}$ then

$\frac{d}{d x} (P_{1})$ and $\int (P_{1}) d x$

are each power series with radius of convergence $R_{1}$

Example 3

Using the known result that ${(1 + x)}^{p} = 1 + p x + \frac{p (p - 1)}{2!} x^{2} + \dots |x| < 1,$

choose $p = \frac{1}{2}$ and by differentiating obtain the power series expression for ${(1 + x)}^{- \frac{1}{2}}$ .

Solution

${(1 + x)}^{\frac{1}{2}} = 1 + \frac{x}{2} + \frac{\frac{1}{2} (- \frac{1}{2})}{2!} x^{2} + \frac{\frac{1}{2} (- \frac{1}{2}) (- \frac{3}{2})}{3!} x^{3} + \dots$

Differentiating both sides: $\frac{1}{2} {(1 + x)}^{- \frac{1}{2}} = \frac{1}{2} + \frac{1}{2} (- \frac{1}{2}) x + \frac{\frac{1}{2} (- \frac{1}{2}) (- \frac{3}{2})}{2} x^{2} + \dots$

Multiplying through by 2: ${(1 + x)}^{- \frac{1}{2}} = 1 - \frac{1}{2} x + \frac{(- \frac{1}{2}) (- \frac{3}{2})}{2} x^{2} + \dots$

This result can, of course, be obtained directly from the expansion for ${(1 + x)}^{p}$ with $p = - \frac{1}{2} .$

Task!

Using the known result that

$\frac{1}{1 + x} = 1 - x + x^{2} - x^{3} + \dots |x| < 1,$

Find an expression for $ln (1 + x)$
Use the expression to obtain an approximation to $ln (1.1)$

Integrate both sides of $\frac{1}{1 + x} = 1 - x + x^{2} - \dots$ and so deduce an expression for $ln (1 + x)$ :

$\int \frac{d x}{1 + x} = ln (1 + x) + c$ where $c$ is a constant of integration,

$\int (1 - x + x^{2} - \dots) d x = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \dots + k$ where $k$ is a constant of integration.

So we conclude $ln (1 + x) + c = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \dots + k if |x| < 1$

Choosing $x = 0$ shows that $c = k$ so they cancel from this equation.
Now choose $x = 0.1$ to approximate $ln (1 + 0.1)$ using terms up to cubic:

$ln (1.1) ≃ 0.0953$ which is easily checked by calculator.