How do you find the maclaurin series f(x)=e^5x?

1 Answer
NJ
Apr 29, 2018

#=>e^(5x) = sum_{k=0}^{oo} (5x)^k/(k!)#

Explanation:

MacLaurin series are defined for a function #f(x)# as:

#=>f(x) = sum_{k=0}^{oo} (f^{(k)}(0))/(k!)x^k#

MacLaurin series are a special form of Taylor series, where the derivative of each term is assessed specifically at #x = 0# rather than an arbitrary value.

To find the series of #f(x) = e^(5x)#, there are a couple of approaches.

Approach 1: use a known series

This is a common approach. We choose a series that is close to the one we want and then modify it accordingly. In this case, we could use the MacLaurin series for #f(x) = e^x# and then modify it to account for the difference in the exponent.

The series for #f(x) = e^x# is:

#=> e^x = sum_{k=0}^{oo} x^k/(k!)#

But instead of #x# in the exponent, we have #5x#, so let's substitute that in instead to get the series we want:

#=>e^(5x) = sum_{k=0}^{oo} (5x)^k/(k!)#

Approach 2: from MacLaurin series definition

The more brute-force approach is to start computing terms of the MacLaurin series for our #f(x) = e^(5x)# and simply look for patterns to write the series more concisely. Let's compute some terms:

#k = 0 => (e^(5(0)))/(0!)x^0=1/(0!)x^0=5^0/(0!)x^0#

#k=1 =>(5e^(5(0)))/(1!)x^1=5/(1!)x = 5^1/(1!)x^1#

#k=2 =>(25e^(5(0)))/(2!)x^2=(25)/(2!)x^2=5^2/(2!)x^2#

#k = 3 =>(125e^(5(0)))/(3!)x^3=(125)/(3!)x^3=5^3/(3!)x^3#

Now that we see a pattern, we can write a general term for any #k#:

#" "k"th term" => 5^k/(k!)x^k = (5x)^k/(k!)#

Now we just write this term in a summation:

#=>e^(5x) = sum_{k=0}^{oo} (5x)^k/(k!)#

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