state the power series of an appropriate familiar function and use it to calculate the power series of the given function. give answer in sigma notation. All power series have center at 0.
f(x)=(e^(-x)^2) -1+(x^2)
and
g(x)=x cos (x/square root 3)
f(x)=(e^(-x)^2) -1+(x^2)
= 1 - x^2 + x^4/2! - x^6/3! + ... -1 + x^2
= x^4/2! - x^6/3! + x^8/4! - ...
To calculate the power series of a given function, we need to express it as a sum of terms involving powers of x. Here's how we can find the power series for each given function:
1. For f(x) = (e^(-x)^2) - 1 + (x^2):
Step 1: Find the power series for each term separately.
- Power series for (e^(-x)^2):
Recall that the power series for e^x is given by:
e^x = 1 + x + (x^2)/2! + (x^3)/3! + ...
Substituting -x^2 for x^2, we get:
(e^-x^2) = 1 - x^2 + (x^4)/2! - (x^6)/3! + ...
- Power series for (x^2):
This term is already in power series form.
Step 2: Combine the power series obtained from each term.
The power series for f(x) will be the sum of the respective power series for each term. Let's denote the power series for f(x) as S(x). Therefore:
S(x) = (e^-x^2) + (x^2) - 1
In sigma notation, this is:
S(x) = Σ[n=0 to ∞]((-1)^n * x^(2n))/n! + Σ[n=0 to ∞](x^(2n+2))
2. For g(x) = x cos(x/√3):
Step 1: Find the power series for cos(x).
Recall that the power series for cos(x) is given by:
cos(x) = 1 - (x^2)/2! + (x^4)/4! - (x^6)/6! + ...
Step 2: Substitute x/√3 for x in the power series for cos(x).
cos(x/√3) = 1 - ((x/√3)^2)/2! + ((x/√3)^4)/4! - ((x/√3)^6)/6! + ...
Simplifying, we have:
cos(x/√3) = 1 - (x^2)/3! + (x^4)/9! - (x^6)/27! + ...
Step 3: Multiply the power series for cos(x/√3) by x.
x * cos(x/√3) = x - (x^3)/3! + (x^5)/9! - (x^7)/27! + ...
In sigma notation, this is:
S(x) = Σ[n=0 to ∞]((-1)^n * x^(2n+1))/(2n+1)!
Therefore, the power series for g(x) is:
S(x) = x * cos(x/√3) = Σ[n=0 to ∞]((-1)^n * x^(2n+1))/(2n+1)!
I hope this helps! Let me know if you have any further questions.