Strategies for constructing Maclaurin polynomials?

[ektrules]

I came across a problem in my homework to construct a MacLaurin polynomial of the nth degree for \sqrt{1+x}, and had some major problems. I gave up and looked up the answer on the internet, which was fairly complex: \sum \frac{(-1)^{n}(2n)!x^{n}}{(1-2n)(n!)^{2}(4^{n})}

Well, I know I couldn't have come up with that myself right now (unless I spent much more time trying to figure it out than I want to), so are there any "tricks" or strategies for coming up with MacLaurin polynomials like this that aren't intuitive?

The only way I know of constructing the polynomials is to work my way from the first term of the MacLaurin polynomial, to the second, ..., and then look for an obvious pattern to create a general formula. This pattern was not obvious to me at all; with the (2n)!, (n!)^2, 4^n, and all that. How the hell am I supposed to guess that, lol.

 

[Mu naught]

there is a specific set of steps for generating a series... you should really read up on this in a calculus textbook and look at this link:

http://mathworld.wolfram.com/MaclaurinSeries.html

 

[Cod]

If you have room in your brain, memorize the Maclaurin Series of common functions. In Calc II, my class utilized the common series for 90% of the related questions. After memorizing / learning the common ones, learn the steps of generating a series like Mu naught stated.

List of common functions: http://en.wikipedia.org/wiki/Taylor_series#List_of_Maclaurin_series_of_some_common_functions

 

[adriank]

In this case, you have
(1+x)^\alpha = \sum_{n=0}^\infty \binom{\alpha}{n} x^n,
so the problem is reduced to showing that
\binom{1/2}{n} := \frac{\Gamma(1/2 + n + 1)}{\Gamma(1/2 + 1)\Gamma(n+1)} = \frac{(-1)^n(2n)!}{(1-2n)(n!)^2(4^n)},
where \Gamma is the Gamma function which satisfies \Gamma(k+1) = k \Gamma(k) for all k, and \Gamma(n+1) = n! for nonnegative integers n. You should be able to prove the above formula with induction. (The precise value of the Gamma function at half integers is immaterial for the proof, but \Gamma(1/2) = \sqrt{\pi}.)

 

[awkward]

In this case, you have
(1+x)^\alpha = \sum_{n=0}^\infty \binom{\alpha}{n} x^n,
so the problem is reduced to showing that
\binom{1/2}{n} := \frac{\Gamma(1/2 + n + 1)}{\Gamma(1/2 + 1)\Gamma(n+1)} = \frac{(-1)^n(2n)!}{(1-2n)(n!)^2(4^n)},
where \Gamma is the Gamma function which satisfies \Gamma(k+1) = k \Gamma(k) for all k, and \Gamma(n+1) = n! for nonnegative integers n. You should be able to prove the above formula with induction. (The precise value of the Gamma function at half integers is immaterial for the proof, but \Gamma(1/2) = \sqrt{\pi}.)

There is really no need to involve the Gamma function in the solution. Just observe that

\binom{1/2}{n} = \frac{(1/2)(1/2-1)(1/2-2) \cdots (1/2 - n + 1)}{n!}

which can be reduced, after some simple algebra, to the form shown in the original post.

 

[adriank]

Indeed.