Question

Find the first three terms of the Maclaurin series for f(x) =
${e}^{ \frac{x}{2} }$
​

Answer 1

Step-by-step explanation:

Starting out with the Taylor series,

$\displaystyle f(x) = \sum_{n=0}^{\infty} \dfrac{f^{(n)}(a)}{n!}(x-a)^n$

where $f^{(n)}$ is the nth derivative of f(x) and if we set a = 0, we get the special case of the Taylor series called the Maclaurin series:

$\displaystyle f(x) = \sum_{n=0}^{\infty} \dfrac{f^{(n)}(0)}{n!}x^n$

Expanding this series up to the 1st 3 terms at a = 0,

$f(x) = f(0) + \dfrac{f'(0)}{1!}x + \dfrac{f''(0)}{2!}x^2$

Let's find the derivatives of $e^{\frac{x}{2}}$:

$f'(x) = \frac{d}{dx} (e^{\frac{x}{2}}) = \frac{1}{2}e^{\frac{x}{2}} \Rightarrow f'(0) = \frac{1}{2}$

$f''(x) = \frac{1}{4}e^{\frac{x}{2}} \Rightarrow f''(0) = \frac{1}{4}$

We can now write the Maclaurin series for $e^{\frac{x}{2}}$as

$e^{\frac{x}{2}} = 1 + \frac{1}{2} x + \frac{1}{8} x^2$