f(x) =ln (1-x)
a) Compute f'(x), f''(x), f'''(x). Spot the pattern and give an expression for f ^(n) (x) [the n-th derivative of f(x)]
b) Compute the MacLaurin series of f(x) (i.e. the Taylor series of f(x) around x=0)
c) Compute the radius of convergence and determine the interval of convergence of the series in b).
d) Determine the Taylor series of f'(x) around x=0. Can you do so without using b)?
e) How would you have computed part b) if you had first done part d)?
for part a) i got, please check.
f'(x) = -1/(1-x)
f''(x) = -1/(1-x)^2
f'''(x) = -2/(1-x)^3
f^(n) (x) = -((n-1)!)/(1-x)^n
for n = 1,2,3,...
This is how far i got to, please help
f^(n)(0) = -(n-1)!
so Sum_{n=0 to infinity} x^n/n! f^(n)(0)=
-Sum_{n=1 to infinity} x^n/n
Here we have used that the zero-th derivative (i.e. the function itself) is zero for x=0.
The radius of convergence, R, is given by:
R = 1/L
where L is the limit of the absolute value of a_{n+1}/a_{n} as n--> infinity.
Inour case a_{n} = 1/n and you see that L = 1.
f'(x) = -1/(1-x). You can use that the series expoansion of this function is given by the geometric series:
1/(1-x) = Sum_{n=0 to infinity} x^(n)
If you integrate this term by term you obtain the series expansion of -log(1-x).
For part a), you have correctly computed the derivatives of f(x) up to the third derivative. The pattern in the derivatives is that each derivative introduces a negative sign and increases the power of the denominator by one.
To express the nth derivative of f(x), denoted as f^(n)(x), you can use the following expression:
f^(n)(x) = -((n-1)!)/(1-x)^n
Notice that for n = 1, this expression simplifies to f'(x) = -1/(1-x). This is the derivative you obtained.
Moving on to part b), to compute the MacLaurin series of f(x) (the Taylor series centered around x=0), we can use the formula for the Taylor series expansion:
f(x) = f(0) + f'(0)*x + (f''(0)/2!)*x^2 + (f'''(0)/3!)*x^3 + ...
Since f(0) = ln(1-0) = 0, and you have computed the derivatives f'(0) = -1, f''(0) = -1, and f'''(0) = -2. We can now write the MacLaurin series as:
f(x) = -x - (1/2)*x^2 - (2/6)*x^3 - ...
For part c), let's find the radius of convergence and the interval of convergence of the series in part b). The radius of convergence, R, is given by:
R = 1/L
where L is the limit of the absolute value of a(n+1)/a(n) as n approaches infinity. In this case, a(n) = (-1/2)^(n-1) * x^n / n!.
Let's calculate the limit:
L = lim(n->infinity) |((-1/2)^n * x^(n+1) / (n+1)!) / ((-1/2)^(n-1) * x^n / n!)|
= lim(n->infinity) |(-1/2) * x / (n+1)|
= 1/2 * |x|
From this, we see that L = 1/2 * |x|. Now we can find the radius of convergence:
R = 1/L = 1 / (1/2 * |x|) = 2 / |x|
The interval of convergence is determined by the values of x that are within the radius of convergence. In this case, the series converges for |x| < 2.
Moving on to part d), we can determine the Taylor series of f'(x) around x=0 using a shortcut. The Taylor series of a derivative function can be obtained by differentiating the Taylor series of the original function. In this case, we can differentiate the MacLaurin series of f(x) from part b) to find the Taylor series of f'(x):
f'(x) = -1 - x - (3/2)*x^2 - ...
As for part e), if we had computed part d) first, we would have the Taylor series of f'(x). To find the MacLaurin series of f(x), we would need to integrate the Taylor series term by term. However, be aware that integrating a series term by term may introduce an integration constant, which would not be present in the MacLaurin series. So, if you were to compute part b) using part d), you would need to take care of the integration constant.