Find the Taylor series expansion of 2^x about x = 0.

Step-by-step solution

1. 1

   Convert the base-2 power into a base-e power. Differentiating $2^x$ repeatedly is awkward, but every exponential can be rewritten on base $e$ using $a = e^{\ln a}$:

   $2^x = \left(e^{\ln 2}\right)^x = e^{x\ln 2}$

   Now the problem reduces to a series that is already known.

2. 2

   Substitute into the exponential series. Since $e^{u} = \sum_{n=0}^{\infty} \frac{u^n}{n!}$ for all real $u$, put $u = x\ln 2$:

   $2^x = \sum_{n=0}^{\infty} \frac{(x\ln 2)^n}{n!} = \sum_{n=0}^{\infty} \frac{(\ln 2)^n}{n!}\,x^n$

3. 3

   Confirm the coefficients match the derivative definition. The Maclaurin coefficient is $\frac{f^{(n)}(0)}{n!}$, and $\frac{d^n}{dx^n}2^x = (\ln 2)^n 2^x$, which at $x = 0$ equals $(\ln 2)^n$. Dividing by $n!$ reproduces the same coefficients — the shortcut and the definition agree.

4. 4

   Write out the first few terms. With $\ln 2 \approx 0.6931472$:

   $2^x = 1 + 0.693147\,x + 0.240227\,x^2 + 0.055504\,x^3 + 0.009618\,x^4 + \cdots$

   Unlike $e^x$, whose coefficients shrink purely by the factorial, these shrink faster because $\ln 2 < 1$ contributes an extra decaying factor each term.

5. 5

   State the radius of convergence. The exponential series converges for every real $u$, and $u = x\ln 2$ is finite whenever $x$ is, so the radius is infinite: the expansion is valid for all $x \in \mathbb{R}$.

6. 6

   Spot-check at x = 1. Summing the first eight terms gives $1 + 0.693147 + 0.240227 + 0.055504 + 0.009618 + 0.001333 + 0.000154 + 0.000015 = 1.999999$, converging to $2^1 = 2$ as expected.