Determine the first three terms of the power series representations for the following functions:
a) 2/(1+x)^3
b) 1/1+x^2
c) ln(1+x)
let's look at the last one:
ln (1 + x) = x − x^2/2 + x^3/3 - x^4/4 + x^5/5 .....
of course x ≥ -1
this should be in your text, in your class notes, or you can easily look it up
For the other two, how about just looking at the general binomial expansion:
2/(1+x)^3
= 2(1+x)^-3
= 2(1 + (-3)x + (-3)(-4)/2! x^2 + (-3)(-4)(-5)/3! x^3 ......)
simplify and state the first three terms
in 1/1+x^2 , I will assume you meant 1/(1+x^2)
= (1 + x^2)^-1
= ......
correction:
for ln (1 + x) = x − x^2/2 + x^3/3 - x^4/4 + x^5/5 .....
x is between -1 and +1
e.g. let x = .46
on my calculator: ln(1.46) = .378436...
using the above formula:
ln(1.46) = .46 - (.46)^2/2 + (.46)^3/3 - (.46)^4/4 ....
= .37545... using 4 terms
To determine the first three terms of the power series representations for the given functions, we can use the Taylor series expansion. The Taylor series expansion represents a function as an infinite sum of terms, where each term is a derivative of the function evaluated at a specific point, multiplied by an appropriate power of the difference between the variable and the point of expansion. We can express the Taylor series expansion of a function f(x) centered at a point c as:
f(x) = f(c) + f'(c)(x-c) + f''(c)(x-c)^2/2! + f'''(c)(x-c)^3/3! + ...
Now we will determine the power series representations for each function:
a) 2/(1+x)^3:
To find the power series representation of this function, we start by finding the derivatives of the function at a convenient point. Let's take c = 0:
f(x) = 2/(1+x)^3
f'(x) = -6/(1+x)^4
f''(x) = 24/(1+x)^5
Now substitute these values into the Taylor series expansion formula:
f(x) = f(0) + f'(0)(x-0) + f''(0)(x-0)^2/2! + ...
Substituting the values, we get:
f(x) = 2 - 6x + 12x^2 + ...
So, the first three terms of the power series representation for 2/(1+x)^3 are 2, -6x, and 12x^2.
b) 1/(1+x^2):
Similarly, let's take c = 0 and find the derivatives:
f(x) = 1/(1+x^2)
f'(x) = -2x/(1+x^2)^2
f''(x) = (6x^2 - 2)/(1+x^2)^3
Now substitute these values into the Taylor series expansion formula:
f(x) = f(0) + f'(0)(x-0) + f''(0)(x-0)^2/2! + ...
Substituting the values, we get:
f(x) = 1 - 2x + (6x^2 - 2)x^2 + ...
So, the first three terms of the power series representation for 1/(1+x^2) are 1, -2x, and (6x^2 - 2)x^2.
c) ln(1+x):
For this function, we can use the Maclaurin series expansion, which is a special case of the Taylor series expansion where the point of expansion is c = 0. The Maclaurin series expansion of ln(1+x) is:
ln(1+x) = x - x^2/2 + x^3/3 - x^4/4 + ...
So, the first three terms of the power series representation for ln(1+x) are x, -x^2/2, and x^3/3.
Note: When using Taylor series expansion, it's important to remember that the expansion is only valid within a certain interval around the point of expansion.