Find the Maclaurin series for f(x) using the definition of a Maclaurin serles. [Assume that f has a power serles expansion. Do not show that R,(x) → 0.]f(x) = e-4xΣf(x) =n = 0Find the associated radius of convergence R.R =

Answered: Find the Maclaurin series for f(x)…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Related questions

Question

Find the Maclaurin series for f(x) using the definition of a Maclaurin serles. [Assume that f has a power serles expansion. Do not show that R,(x) → 0.]
f(x) = e-4x
Σ
f(x) =
n = 0
Find the associated radius of convergence R.
R =

Expert Solution

Step 1

The given function is $f (x) = e^{- 4 x}$ .

The Maclaurin series for a function is given by $f (x) = f (0) + \frac{f' (0)}{1!} x + \frac{f'' (0)}{2!} x^{2} + \frac{f''' (0)}{3!} x^{3} + \frac{f^{(4)} (0)}{4!} x^{4} + \dots$ .

Here, $f (x) = e^{- 4 x}$ and hence

$f (x) = e^{- 4 x} \Rightarrow f' (x) = \frac{d}{d x} (e^{- 4 x}) = - 4 e^{- 4 x} \Rightarrow f'' (x) = \frac{d}{d x} (- 4 e^{- 4 x}) = 16 e^{- 4 x} \Rightarrow f''' (x) = \frac{d}{d x} (16 e^{- 4 x}) = - 64 e^{- 4 x} \Rightarrow f^{(4)} (x) = \frac{d}{d x} (- 64 e^{- 4 x}) = 256 e^{- 4 x}$

Step 2

Therefore,

$f (0) = e^{- 4 \times 0} = e^{0} = 1 f' (0) = - 4 e^{- 4 \times 0} = - 4 e^{0} = - 4 f'' (0) = 16 e^{- 4 \times 0} = 16 e^{0} = 16 f''' (0) = - 64 e^{- 4 \times 0} = - 64 e^{0} = - 64 f^{(4)} (0) = 256 e^{- 4 \times 0} = 256 e^{0} = 256$

Substituting the above values in $f (x) = f (0) + \frac{f' (0)}{1!} x + \frac{f'' (0)}{2!} x^{2} + \frac{f''' (0)}{3!} x^{3} + \frac{f^{(4)} (0)}{4!} x^{4} + \dots$ with $f (x) = e^{- 4 x}$ , we get

$\begin{array}{rcl} f (x) & = & 1 + \frac{- 4}{1!} x + \frac{16}{2!} x^{2} + \frac{- 64}{3!} x^{3} + \frac{256}{4!} x^{4} + \dots \\ = & \frac{{(- 4)}^{0} x^{0}}{0!} + \frac{(- 4) x}{1!} + \frac{{(- 4)}^{2} x^{2}}{2!} + \frac{{(- 4)}^{3} x^{3}}{3!} + \frac{{(- 4)}^{4} x^{3}}{4!} + \dots \\ = & \sum_{n = 0}^{\infty} \frac{{(- 4)}^{n} x^{n}}{n!} \\ = & \sum_{n = 0}^{\infty} (\frac{{(- 4 x)}^{n}}{n!}) \end{array}$

Hence, the Maclaurin series for $f (x) = e^{- 4 x}$ is $\begin{array}{rcl} f (x) & = & \sum_{n = 0}^{\infty} (\frac{{(- 4 x)}^{n}}{n!}) \end{array}$ .

Trending now

This is a popular solution!

Step by step

Solved in 3 steps

Knowledge Booster

$Power Series$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.

Similar questions