How do you find the Maclaurin Series for f(x) = e^(3x)?

1 Answer
Jul 13, 2016

e^(3x) = 1 + 3x + 9/2 x^2 + 9/2 x^3 + 27/8 x^4 + ...= sum_(n=0)^(∞) (3x)^(n)/(n!)

Explanation:

There are about two ways we can go about finding a Maclaurin series. However, in some cases it is much easier and less time-consuming to use some known Maclaurin series that have already been derived.

For this particular problem, I'll try to derive the Maclaurin series from scratch. Then, we'll compare it to a known Maclaurin series.

As you may already know, a Maclaurin series is a special case of a Taylor series centered at x = 0.

We can express a Maclaurin series as an infinite sum:

f(0) + (f^((1))(0))/(1!)x+ (f^((2))(0))/(2!)x^2+ (f^((3))(0))/(3!)x^3 + ... (f^((n))(0))/(n!)x^n

= sum_(n=0)^(∞) (f^((n))(0))/(n!)x^n

As this infinite sum already suggests, we have to compute some derivatives. The more terms your Maclaurin series has, the better the approximation. For this problem, I will only compute four, which is enough to spot a pattern.

f(x) = e^(3x) ... ... ................f(0) = e^(0) ...................... = 1
f^(1)(x) = 3e^(3x) ... ... ...........f^(1)(0) = 3*e^(0) ................ = 3
f^(2)(x) = 9e^(3x) ... ... ...........f^(2)(0) = 9*e^(0) ................ = 9
f^(3)(x) = 27e^(3x) ... ... ..........f^(3)(0) = 27 * e^(0) = ..........= 27
f^(4)(x) = 81e^(3x) ... ... ..........f^(4)(0) = 81 * e^(0) = .......... = 81

We can now substitute these f^(n)(0) terms into our infinite sum, giving us

e^(3x) = 1 + (3)/(1!)x + (9)/(2!)x^2 + (27)/(3!)x^3 + (81)/(4!)x^4 + ...

By simplifying the denominator on each term we get

e^(3x) = 1 + 3x + 9/2 x^2 + 27/6 x^3 + 81/24 x^4 + ...

We can simplify even further if we notice that 27/6 and 81/24 are all divisible by 3, so we write our final answer as

e^(3x) = 1 + 3x + 9/2 x^2 + 9/2 x^3 + 27/8 x^4 + ... = sum_(n=0)^(∞) (3x)^(n)/(n!)

Alternatively, if we know that

e^(x) = sum_(n=0)^(∞) (x)^(n)/(n!)

We can simply replace x in e^(x) with 3x to get the same result.

Also, we can check if this approximation looks alright by graphing both the polynomial with four terms, as well as e^(3x).

Graph of e^(3x):

graph{e^(3x) [-10, 10, -5, 5]}

Graph of 1 + 3x + 9/2 x^2 + 9/2 x^3 + 27/8 x^4:

graph{1 + 3x + 9/2 x^2 + 9/2 x^3 + 27/8 x^4 [-10, 10, -5, 5]}

If we overlap them both on one graph, we can see that this infinite sum does, in fact, look alright to the first four terms.

enter image source here

Red: e^(3x)
Blue: 1 + 3x + 9/2 x^2 + 9/2 x^3 + 27/8 x^4