Question

Find the first three terms of the Maclaurin series for f(x) =


`{e}^{ (x)/(2) }`
​

Answer 1

Explanation:

Starting out with the Taylor series,

`\displaystyle f(x) = \sum_(n=0)^(\infty) (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) (f^((n))(0))/(n!)x^n`

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

`f(x) = f(0) + (f'(0))/(1!)x + (f''(0))/(2!)x^2`

Let's find the derivatives of
`e^{(x)/(2)}`:

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

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

We can now write the Maclaurin series for
`e^{(x)/(2)}`as

`e^{(x)/(2)} = 1 + (1)/(2) x + (1)/(8) x^2`