Find the first three nonzero terms and the general term of the Maclaurin series generated by e^(-6x).
My answer: 1 + x + (x^2)/2, general term (x^(n-1))/(n-1)!
the expansion is
e^x = 1 + x/1! + x^2/2! + x^3/3! + .... + e^(n-1)x)/(n-1)! + ....
you just stated the first 3 terms and the general term of the general series.
e^(-6x) = 1 + (-6x) + (36x^2)/2 + (-216x^3)/6 + ....
= 1 - 6x + 18x^2 - 36x^3 + ....
general term : e^(n-1)(-6x)/(n-1)!
To find the Maclaurin series of e^(-6x), we can use the formula for the Maclaurin series expansion of the exponential function:
e^(-6x) = 1 - 6x + (36x^2)/2 - (216x^3)/6 + ...
The first three nonzero terms of this series are:
1, -6x, (36x^2)/2
The general term of the Maclaurin series can be determined by examining the pattern:
The coefficient in front of each term is (-6)^n divided by n!. So, the general term of the series is given by:
((-6)^n * x^n) / n!
Therefore, the general term of the Maclaurin series of e^(-6x) is:
((-6)^n * x^n) / n!
To find the Maclaurin series of a function, we can use the formula:
f(x) = f(0) + f'(0)x + (f''(0)x^2)/2! + (f'''(0)x^3)/3! + ...
First, let's find the first three nonzero terms:
Step 1: Find the function at 0
f(0) = e^(-6 * 0) = e^0 = 1
The first term is 1.
Step 2: Find the first derivative and evaluate it at 0
f'(x) = (-6)e^(-6x)
f'(0) = (-6)e^(-6 * 0) = -6
The second term is -6x.
Step 3: Find the second derivative and evaluate it at 0
f''(x) = (-6)^2e^(-6x)
f''(0) = (-6)^2e^(-6 * 0) = 36
The third term is (36x^2)/2 = 18x^2.
So the first three nonzero terms of the Maclaurin series are 1, -6x, and 18x^2/2.
Now, let's find the general term of the Maclaurin series. The general term can be written as:
T(n) = (f^n(0)x^n)/n!
where f^n(0) denotes the nth derivative at 0.
In this case, the general term is:
T(n) = (f^n(0)x^n)/n!
= ((-6)^n e^(-6x) x^n)/n!
So the general term of the Maclaurin series of e^(-6x) is ((-6)^n e^(-6x) x^n)/n!.