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Judith McQueen

Answered question

2021-12-12

How to find the Maclaurin series for

e^{x}

Answer & Explanation

boronganfh

Beginner2021-12-13Added 33 answers

The Power Series can be used to obtain the Maclaurin series:
$f (x) = f (0) + f^{'} (0) \frac{x}{0!} + f^{'} (0) \frac{x^{2}}{2!} + f^{'} (0) \frac{x^{3}}{3!} + \dots$
As we have $f (x) = e^{x}$ , then:
$f (x) = e^{x} \Rightarrow f (0) = 1$
$f^{'} (x) = e^{x} \Rightarrow f^{'} (0) = 1$
$f^{″} (x) = e^{x} \Rightarrow f^{″} (0) = 1$
$f^{3} (x) = e^{x} \Rightarrow f^{3} (0) = 1$
$f^{n} (x) = e^{x} \Rightarrow f^{n} (0) = 1$
The Maclaurin series is as follows:
$e^{x} = 1 + 1 \frac{x}{0!} + 1 \frac{x^{2}}{2!} + 1 \frac{x^{3}}{3!} + 1 \frac{x^{4}}{4!} + \dots + 1 \frac{x^{n}}{n!}$

Joseph Fair

Beginner2021-12-14Added 34 answers

Taking the derivatives of the given function and using x=0, we have:

f (x) = e^{x}, f (0) = 1

f^{'} (x) = e^{x}, f^{'} (0) = 1

f^{″} (x) = e^{x}, f^{″} (0) = 1

f^{3} (x) = e^{x}, f^{3} (0) = 1

f^{n} (x) = e^{x}, f^{n} (0) = 1

Now using Maclaurin’s series expansion function, we have:

f (x) = f (0) + f^{'} (0) \frac{x}{0!} + f^{'} (0) \frac{x^{2}}{2!} + f^{'} (0) \frac{x^{3}}{3!} + \dots

Putting the values in the above series, we have:

e^{x} = 1 + 1 \frac{x}{0!} + 1 \frac{x^{2}}{2!} + 1 \frac{x^{3}}{3!} + 1 \frac{x^{4}}{4!} + \dots + 1 \frac{x^{n}}{n!}

e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!} + \dots

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