Expand x+ sqrt(1+x²) in a power series in x, using known
Expand $x+\sqrt{1+x^2}$ in a power series in $x$, using known expansions, and indicate when it is valid.
Explanation of Method
The binomial series $(1+t)^{1/2}=\sum\binom{1/2}nt^n$ applies directly with $t=x^2$; adding the linear term $x$ afterward changes nothing about the region of convergence, since $x$ itself is entire.
- Apply the binomial series to $(1+x^2)^{1/2}$.
- Compute the binomial coefficients $\binom{1/2}n$ explicitly.
- Add $x$ and determine the region of convergence.
Step 1 · Binomial series
With $t=x^2$: $$ \sqrt{1+x^2} = (1+x^2)^{1/2} = \sum_{n=0}^{\infty}\binom{1/2}n(x^2)^n, \qquad |x^2|<1. $$
Step 2 · Compute the coefficients
$$ \binom{1/2}0=1,\quad \binom{1/2}1=\frac12,\quad \binom{1/2}2=-\frac18,\quad \binom{1/2}3=\frac1{16},\quad \binom{1/2}4=-\frac5{128}, \quad\ldots $$ So $$ \sqrt{1+x^2} = 1+\frac{x^2}2-\frac{x^4}8+\frac{x^6}{16}-\frac{5x^8}{128}+\cdots $$
Step 3 · Add $x$ and determine convergence
$$ x+\sqrt{1+x^2} = 1+x+\frac{x^2}2-\frac{x^4}8+\frac{x^6}{16}-\frac{5x^8}{128}+\cdots $$ The binomial series requires $|x^2|<1$, i.e. $|x|<1$; adding the entire function $x$ does not change this region.
Summary Table
| Property | Value |
|---|---|
| Series | $1+x+\sum_{n=1}^{\infty}\binom{1/2}nx^{2n}$ |
| Region of validity | $|x|<1$ |
Additional Notes
- Only even powers of $x$ appear in $\sqrt{1+x^2}$ itself (an even function), so the odd-power content of the full expression $x+\sqrt{1+x^2}$ comes entirely from the added linear term.
- This function is the inverse hyperbolic sine relation in disguise: $\sinh^{-1}x=\ln\bigl(x+\sqrt{1+x^2}\bigr)$, so this series is a direct building block for expanding $\sinh^{-1}x$ by composing with the logarithm series.
- The radius of convergence, $1$, comes from the branch point of $\sqrt{1+x^2}$ at $x=\pm i$ (in the complex plane) — both at distance $1$ from the origin, exactly matching the real-variable radius found here.