Expand f(x)=ln (1+x/1-x) in a Taylor Series about x=0. You must express your answer using summation notation.
---This is what I tried to do---
So I was thinking of taking the derivative of ln(1+x/1-x) and get 2/x^2-1
And then use the identity 1/1-x= sum x^k
So then it will be Sum (-2^k) (x^2k)
To expand the function f(x) = ln(1+x)/(1-x) in a Taylor series about x=0 using summation notation, we can follow these steps:
Step 1: First, let's find the derivatives of f(x) with respect to x.
f(x) = ln(1+x)/(1-x)
Using the quotient rule, we have:
f'(x) = [(1-x)(1/(1+x)) - ln(1+x)(-1)]/(1-x)^2
= [1 - x - ln(1+x)]/(1-x)^2
Step 2: Now, let's find the second derivative.
f''(x) = [(1-x)^2(1 - x - ln(1+x)) - (1 - x - ln(1+x))(2(1-x))]/(1-x)^4
= [(1-x)^2 - 2(1 - x - ln(1+x))]/(1-x)^3
= [1 - 2(1 - x - ln(1+x))]/(1-x)^3
= [1 - 2 + 2x + 2ln(1+x)]/(1-x)^3
= [2x - 1 + 2ln(1+x)]/(1-x)^3
Step 3: Continuing this process, let's find the third derivative.
f'''(x) = [(1-x)^3(2x - 1 + 2ln(1+x)) - (2x - 1 + 2ln(1+x))(3(1-x)^2)]/(1-x)^6
= [(1-x)^3 - 3(2x - 1 + 2ln(1+x))]/(1-x)^4
= [(1-x)^3 - 6x + 3 + 6ln(1+x)]/(1-x)^4
= [1 - 3x + 3x^2 - x^3 - 6x + 3 + 6ln(1+x)]/(1-x)^4
= [-x^3 - 9x + 4 + 6ln(1+x)]/(1-x)^4
Step 4: We can observe a pattern in these derivatives and express the nth derivative as:
f^(n)(x) = [-x^n - (2n-1)x^(n-1) + 6ln(1+x)]/(1-x)^4
Step 5: Now, let's express the Taylor series expansion using the formula:
f(x) = f(0) + f'(0)x + (1/2!)f''(0)x^2 + (1/3!)f'''(0)x^3 + ...
The terms of the series can be written as:
f^(n)(0)x^n/n! = [-x^n - (2n-1)x^(n-1) + 6ln(1+x)]/(1-x)^4 * x^n/n!
Therefore, the Taylor series expansion of f(x) = ln(1+x)/(1-x) about x=0 (in summation notation) is:
f(x) = Σ[-x^n - (2n-1)x^(n-1) + 6ln(1+x)]/(1-x)^4 * x^n/n!,
where the sum is taken from n = 0 to infinity.
you have the right idea.
First, df/dx = -2/(1-x^2), but I'd do it like this:
f(x) = ln(1+x) - ln(1-x)
ln(1+x) = 0+x-x^2/2+x^3/3-x^4/4+x^5/5+...
ln(1-x) = 0-x-x^2/2-x^3/3-x^4/4-x^5/5-...
subtract to get
0 + 2x + 2x^3/3 + 2x^5/5 + ...