Help finding n-order Maclaurin polynomial

Question

EDIT AND PLEASE NOTE: I DON'T want solutions that are nicer or more elegant but presume knowledge of other infinite series and/or don't come from the nth-derivative because I'm precisely studying how Taylor and Maclaurin series relate to derivatives. My mathematical analysis course hasn't even made it to infinite series yet so there's really no point anyway.

I can't find a pattern in the derivative functions of $(1-x^2)^{-1}$. All I know is that odd terms will probably be 0 and some terms in the derived functions will also be 0 as the derivative of $(1-x^2)^n$ will always have a $2x$ factor, but other than that I'm stumped. Any help will be appreciated.

Answer 1

You could separate in partial fractions, find the derivatives and build the series. But there is another way of doing, much nicer in my opinion: $$f(x) = \frac{1}{1 - x^2} = \frac{1}{(1-x)(1+x)} = \frac{1}{2}\left(\frac{1}{1-x} + \frac{1}{1+x}\right) $$

Then we nicely identify the geometric series where we could use: $$ \sum_{n=0}^\infty x^n = \frac{1}{1-x} \quad\Longrightarrow\quad \sum_{n=0}^\infty (-x)^n = \frac{1}{1+x} $$

Now let's sum up: $$ f(x) = \frac{1}{2}\sum_{n=0}^\infty x^n + \frac{1}{2}\sum_{n=0}^\infty (-1)^n x^n = \frac{1}{2}\sum_{n=0}^\infty \left(1 + (-1)^n\right) x^n $$

And done!

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I see your edit. Once you have separated in partial fractions, the calculation of the $n$th derivatives is quite easy. Notice: $$ g(x) = (1-x)^{-1} \quad\Longrightarrow\quad g^{(n)}(x) = n!(1-x)^{-1-n} $$

You can visualize this by derivating and finding patterns. A more elaborated proof could be done with induction. You can do the same with $h(x) = (1+x)^{-1}$. Once you have it, find Maclaurin polynomial of each fraction, and sum up: $$ f(x) = \frac{1}{2}\left(g(x) + h(x) \right) = \frac{1}{2}\sum_{n=0}^\infty a_n x^n + \frac{1}{2}\sum_{n=0}^\infty b_n x^n = \frac{1}{2}\sum_{n=0}^\infty (a_n + b_n) x^n $$

And done. Now you have your Maclaurin polynomial. Obviously, using $n$th derivatives, you should arrive at the same answer of first method, which is: $a_n = 1$ and $b_n = (-1)^n$.

Answer 2

The higher derivatives can also be found by using the "product rule" for higher derivatives without dealing with series at all. It will be convenient for this function, particularly since you are seeking the values of the derivatives at $ \ x = 0 \ \ . $

We can write the expression $ \ f(x) · (1-x^2) \ = \ 1 \ $ and differentiate it implicitly. The generalization of the product rule of interest here is

$$ \frac{d^n}{dx^n} \ [f(x) · g(x)] \ \ = \ \ \sum_{k \ = \ 0}^{n} \ \binom{n}{k} \ f^{(n-k)}(x) · g^{(k)}(x) \ \ , $$

which resembles a "binomial expansion". What will aid us a good deal in our work is that is that $ \ f(x) \ $ is an even function and is continuous at $ \ x = 0 \ \ , $ so all of the derivatives $ \ f^{(2n-1)}(x) \ $ are odd functions and so $ \ f^{(2n-1)}(0) \ = \ 0 \ \ . $ We will use $ \ g(x) \ = \ 1 - x^2 \ \ , $ for which $ \ g'(x) \ = \ - 2x \ \ , \ \ g''(x) \ = \ - 2 \ \ , $ and $ \ g^{(k)}(x) \ = \ 0 \ \ , \text{for} \ k \ge 3 \ \ ; $ this means that our "binomial expansion" will never use more than three terms.

Our successive differentiations are then

$$ f'(x) · (1-x^2) \ + \ f(x) · (-2x) \ = \ 0 \ \ \Rightarrow \ \ f'(0) · (1-0^2) \ + \ f(0) · (-2·0) \ = \ 0 $$ $$ \Rightarrow \ \ f'(0) \ = \ 0 \ \ , $$

[as expected]

$$ f''(x) · (1-x^2) \ + \ 2·f'(x) · (-2x) \ + \ f(x)· (-2) \ = \ 0 $$ $$ \Rightarrow \ \ f''(0) · (1-0^2) \ + \ f'(0) · (-2·0) \ + \ f(0)· (-2) \ = \ f''(0) \ + \ 0 \ + \ 1 · (-2) \ = \ 0 $$ $$ \Rightarrow \ \ f''(0) \ = \ 2 \ \ , $$ $$ f'''(x) · (1-x^2) \ + \ 3·f''(x) · (-2x) \ + \ 3·f'(x)· (-2) \ + \ f(x) · 0 = \ 0 $$ [since we have now reached a term with $ \ g'''(x) = 0 \ $ ] $$ \Rightarrow \ \ f'''(0) · (1-0^2) \ + \ 3·f''(0) · (-2·0) \ + \ 3·f'(0)· (-2) \ = \ 0 $$ $$ \Rightarrow \ \ f'''(0) \ + \ 3 · 2 · 0 \ + \ 3 · 0 · (-2) \ = \ 0 \ \ \Rightarrow \ \ f'''(0) \ = \ 0 \ \ , $$ [again as it should be] and $$ f^{(4)}(x) · (1-x^2) \ + \ 4·f'''(x) · (-2x) \ + \ 6·f''(x)· (-2) \ = \ 0 $$ $$ \Rightarrow \ \ f^{(4)}(0) · (1-0^2) \ + \ 4·f'''(0) · (-2·0) \ + \ 6·f''(0)· (-2) \ = \ 0 $$ $$ \Rightarrow \ \ f^{(4)}(0) \ + \ 4 · 0 · 0 \ + \ 6 · 2 · (-2) \ = \ 0 \ \ \Rightarrow \ \ f^{(4)}(0) \ = \ 24 \ \ . $$

So we may now suspect that the pattern for our function is $$ f^{(n)}(0) \ \ = \ \ \left\{ \begin{array}{rcl} 0 \ \ , \ \mbox{for} \ n \ \text{odd} \\ n! \ \ , \mbox{for} \ n \ \text{even} \end{array} \right. \ \ . $$ If we examine the non-zero terms in the next non-zero derivative, we have $$ \Rightarrow \ \ \binom{6}{0} · f^{(6)}(0) · (1-0^2) \ + \ \binom{6}{2}·f^{(4)}(0)· (-2) \ = \ 0 $$ $$ \Rightarrow \ \ 1 · f^{(6)}(0) \ = \ \binom{6}{2} · 4! · 2 \ = \ 6! \ \ , $$

and indeed all higher even derivatives are $ \ f^{(2m)}(0) \ = \ \binom{2m}{2} · (2m-2)! · 2 \ = \ (2m)! \ \ . $ Inserting this into the Maclaurin terms for $ \ f(x) \ $ will produce $ \ \frac{f^{(2m)}(0)}{(2m)!} \ x^{2m} \ = \ 1 · x^{2m} \ \ , $ giving us a series containing only even powers of $ \ x \ $ (as we should have for an even function) and for which all of the coefficients equal $ \ 1 \ \ . $