find the maclaurin series for: f(x) = (1-3x)^-5. What is the radius of convergence?
To find the Maclaurin series for \(f(x) = (1 - 3x)^{-5}\), we can use the following steps:
Step 1: Determine the general form of the Maclaurin series.
The general form of the Maclaurin series is given by:
\[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n \]
Step 2: Calculate the derivatives of \(f(x)\).
To calculate the derivatives, we use the chain rule.
\[ f'(x) = -5(1 - 3x)^{-6}(-3) = 15(1 - 3x)^{-6} \]
\[ f''(x) = 15(-6)(1 - 3x)^{-7}(-3) = 270(1 - 3x)^{-7} \]
\[ f'''(x) = 270(-7)(1 - 3x)^{-8}(-3) = -5670(1 - 3x)^{-8} \]
Step 3: Evaluate the derivatives at x = 0.
Substituting x = 0 into the derivatives, we get:
\[ f(0) = (1 - 3(0))^{-5} = 1 \]
\[ f'(0) = 15(1 - 3(0))^{-6} = 15 \]
\[ f''(0) = 270(1 - 3(0))^{-7} = 270 \]
\[ f'''(0) = -5670(1 - 3(0))^{-8} = -5670 \]
Step 4: Write down the Maclaurin series.
Now we can substitute these values into the general form of the Maclaurin series:
\[ f(x) = 1 + 15x + 270x^2 - 5670x^3 + \ldots \]
The Maclaurin series for \(f(x) = (1 - 3x)^{-5}\) is \(1 + 15x + 270x^2 - 5670x^3 + \ldots \).
Step 5: Find the radius of convergence.
To find the radius of convergence, we can use the ratio test. The formula for the radius of convergence is given by:
\[ R = \lim_{{n \to \infty}} \left| \frac{a_n}{a_{n+1}} \right| \]
where \( a_n \) is the coefficient of \( x^n \) in the Maclaurin series.
In our case, \( a_n \) is given by:
\[ a_n = \frac{f^{(n)}(0)}{n!} \]
The terms of the Maclaurin series are:
\( a_0 = \frac{f^{(0)}(0)}{0!} = 1 \)
\( a_1 = \frac{f^{(1)}(0)}{1!} = 15 \)
\( a_2 = \frac{f^{(2)}(0)}{2!} = 135 \)
\( a_3 = \frac{f^{(3)}(0)}{3!} = -945 \)
Using the ratio test formula, we can calculate the radius of convergence:
\[ R = \lim_{{n \to \infty}} \left| \frac{a_n}{a_{n+1}} \right| \]
\[ R = \lim_{{n \to \infty}} \left| \frac{a_n}{a_{n+1}} \right| = \lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| \]
\[ R = \lim_{{n \to \infty}} \left| \frac{135}{-945} \right| = \frac{135}{945} \]
\[ R = \frac{1}{7} \]
Therefore, the radius of convergence is \( R = \frac{1}{7} \).
To find the Maclaurin series for a function, we need to express the function as a power series centered at x = 0.
In this case, we have f(x) = (1 - 3x)^-5. To expand this expression as a power series, we can use the binomial series.
The general formula for the binomial series is:
(1 + x)^n = C(0,n) + C(1,n)x + C(2,n)x^2 + C(3,n)x^3 + ...
where C(r,n) represents the binomial coefficient, given by C(r,n) = n! / (r! * (n-r)!).
Applying this formula to f(x) = (1 - 3x)^-5, we have:
f(x) = (1 + (-3x))^-5
Now, we can substitute (-3x) for x in the binomial series formula to find the Maclaurin series for f(x).
(1 + (-3x))^-5 = C(0,-5) + C(1,-5)(-3x) + C(2,-5)(-3x)^2 + C(3,-5)(-3x)^3 + ...
Next, let's calculate the binomial coefficients for each term in the expansion.
C(0,-5) = (-5)! / (0! * (-5)!) = 1
C(1,-5) = (-5)! / (1! * (-6)!) = -5 / 1 = -5
C(2,-5) = (-5)! / (2! * (-7)!) = 10 / 2 = 5
C(3,-5) = (-5)! / (3! * (-8)!) = -10 / 6 = -5/3
Substituting these values back into the expansion, we have:
f(x) = 1 - 5(-3x) + 5(-3x)^2 - 5/3(-3x)^3 + ...
Simplifying further, we get:
f(x) = 1 + 15x + 45x^2 + 135x^3 + ...
This is the Maclaurin series for f(x) = (1 - 3x)^-5.
Now, let's determine the radius of convergence. The radius of convergence is a measure of how far away from the center (x = 0) the series will converge.
For the binomial series, the radius of convergence is given by the formula: R = 1 / Lim_{n->∞} |C(n+1, n) / C(n, n)|
In our case, we need to evaluate the above formula using the binomial coefficients we calculated earlier.
Let's calculate the radius of convergence:
R = 1 / Lim_{n->∞} |C(n+1, n) / C(n, n)|
= 1 / Lim_{n->∞} |(n+1)! / ((n+1-n)! * n!) / (n! / (n-n)!|
= 1 / Lim_{n->∞} |(n+1)! / (n!*(n+1))|
= 1 / Lim_{n->∞} |(n+1) / (n+1)|
= 1 / 1
= 1
Therefore, the radius of convergence for the Maclaurin series of f(x) = (1 - 3x)^-5 is 1. The series will converge within a radius of 1 unit from the center x = 0.