How do you differentiate the maclaurin series for 1/(1-x) twice to find the maclaurin series of 1/(1-x)^3.
To find the Maclaurin series of a function, we need to differentiate its Maclaurin series term by term.
Let's start by finding the Maclaurin series for the function f(x) = 1/(1-x). The Maclaurin series for this function is given by:
f(x) = 1 + x + x^2 + x^3 + ...
To differentiate f(x) twice, we apply the differentiation rules to each term of the series.
The first derivative of f(x) is:
f'(x) = 0 + 1 + 2x + 3x^2 + ...
The second derivative of f(x) is:
f''(x) = 0 + 0 + 2 + 6x + ...
Now, we want to find the Maclaurin series for the function g(x) = 1/(1-x)^3. We can obtain this by differentiating the Maclaurin series of f(x) = 1/(1-x) twice.
To differentiate the series for f(x) twice, we differentiate each term of the series term by term again.
The first derivative of f(x) gives us:
f'(x) = 1 + 2x + 3x^2 + ...
Now, differentiate the first derivative of f(x):
f''(x) = 2 + 6x + ...
We have obtained the Maclaurin series for g(x) = 1/(1-x)^3, which is:
g(x) = 1 + 3x + 6x^2 + ...
Thus, the Maclaurin series for 1/(1-x)^3 is 1 + 3x + 6x^2 + ...
To differentiate the Maclaurin series for \(\frac{1}{1-x}\) twice, we can follow the steps below:
Step 1: Find the Maclaurin series for \(\frac{1}{1-x}\).
The Maclaurin series for \(\frac{1}{1-x}\) is given by:
\[\frac{1}{1-x} = 1 + x + x^2 + x^3 + \dots\]
Step 2: Differentiate the series once.
To differentiate the series once, we differentiate each term:
\[\frac{d}{dx}(\frac{1}{1-x}) = \frac{d}{dx}(1 + x + x^2 + x^3 + \dots)\]
Differentiating term by term, we get:
\[0 + 1 + 2x + 3x^2 + \dots\]
Step 3: Differentiate the series again.
To differentiate the series once more, we differentiate each term again:
\[\frac{d^2}{dx^2}(\frac{1}{1-x}) = \frac{d}{dx}(0 + 1 + 2x + 3x^2 + \dots)\]
Differentiating term by term, we get:
\[0 + 0 + 2 + 6x + \dots\]
Step 4: Simplify the series.
Simplifying the series, we have:
\[\frac{1}{(1-x)^3} = 0 + 0 + 2 + 6x + \dots\]
Thus, the Maclaurin series for \(\frac{1}{(1-x)^3}\) is:
\[1 + 2x + 3x^2 + \dots\]