Apologies about any formatting issues, I am new.
I have to find the first four terms of the Maclaurin series for $$f(x) = \frac{1}{1-x}$$
So first I plug in: 1st term is 1
Then derive $$(1-x)^{-1} = (1-x)^{-2}$$
and keep going
and then 4th term I get as $$\frac{3(1-x)^{-3}}{3!}$$
The answer says there is no denominator? but I'm getting over 2?
Where did I go wrong?
On
Here's a nifty way to do this problem. Recall that, by the infinite geometric series formula,
$$f(x) = \frac{1}{1-x} = 1 + x + x^2 + x^3 + ...$$
From here, you can figure out all of your derivatives.
$$f'(x) = 1 + 2x + 3x^2 + 4x^3 + ...$$
$$f''(x) = 2 + 6x + 12x^2 + ...$$
$$f'''(x) = 6 + 24x + ...$$
Now, just remember the formula for a Maclaurin series (of four terms).
$$f(0) + f'(0)x + \frac{f''(0)x^2}{2!}+\frac{f'''(0)x^3}{3!}$$
$$f(0) = 1$$
$$f'(0) = 1$$
$$f''(0) = 2$$
$$f'''(0) = 6$$
Now, you can just plug in the values for the various derivatives of $f'$ and you should be good to go. You may notice that this function turns out to be the first four terms of the original function. You would be right.
. The third derivative is $-2(1-x)^{-3}$ nd the 4th is hence $6(1-x)^{-4}$