How do you find the Maclaurin series for #e^x#?

Question

Evelyn Agard · Accepted Answer

# e^x = 1 + x + x^2/(2!) + x^3/(3!) + x^4/(4!) +... + x^n/(n!) +... #

The Maclaurin series is obtained by the Power Series: # f(x) = f(0) + f'(0)x/(0!) + f''(0)x^2/(2!) + f^((3))(0)x^3/(3!) + ... #

So with #f(x)=e^x# 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 # And in clearly: # f^((n))(x)=e^x => f^((n))(0)=1 #

So the Maclaurin series is: # e^x = 1 + 1x/(0!) + 1x^2/(2!) + 1x^3/(3!) + 1x^4/(4!) +... + 1x^n/(n!) +... # # e^x = 1 + x + x^2/(2!) + x^3/(3!) + x^4/(4!) +... + x^n/(n!) +... #

which is most likely among the most significant power series in mathematics!

Ezra Adams · Answer

undefined To find the Maclaurin series for $ e^x $, you can use the definition of the Maclaurin series, which is:

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

For $ e^x $, the derivatives are:

$$ f(x) = e^x $$
$$ f'(x) = e^x $$
$$ f''(x) = e^x $$
$$ f'''(x) = e^x $$
$$ \text{and so on...} $$

Evaluating these derivatives at $ x = 0 $, we get:

$$ f(0) = 1 $$
$$ f'(0) = 1 $$
$$ f''(0) = 1 $$
$$ f'''(0) = 1 $$
$$ \text{and so on...} $$

Substituting these values into the Maclaurin series definition, we get:

$$ e^x = \sum_{n=0}^{\infty} \frac{1}{n!}x^n $$

So, the Maclaurin series for $ e^x $ is:

$$ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} $$