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I stack about the following question

Use the procedure method to find the Maclaurin Series for $ f(x)= \sqrt{1+2x} $

Also I would like to know what the procedure method is because I couldn't find the method in my text book.... So If you know the method, could you explain about it ?

Thanks !!

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    There could be many "procedure" methods. In fact, any method is a procedure.2012-04-19

2 Answers 2

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As far as I know, there are two possible methods:

  1. Use that for $a \in \Bbb R${\left( {1 + x} \right)^a} = \sum\limits_{n = 0}^\infty {a \choose n}{x^n}

This will yield

$\sqrt{1 + 2x}= \sum\limits_{n = 0}^\infty {1/2 \choose n}2^n{x^n} $

Now we need to find a closed form of a fractional binomial coefficient. This can be done rather "empirically", and I hope you also arrive at

${1/2 \choose n}= {\left( { - 1} \right)^{n + 1}}\frac{{(2n-3)\cdots5\cdot3\cdot1}}{{2^n\cdot n!}} $

which gives you each coefficient of the expansion.

  1. Another procedure would be computing each derivative at $x=0$ and trying to find a pattern. If you do things right, you will arrive at

$f^{(n)}(0)= {\left( { - 1} \right)^{n + 1}}\left[ {\left( {2n - 3} \right) \cdots 5 \cdot 3 \cdot 1} \right]$

i.e. $\left\{ f^{(n)}(0) \right\}=\{1,1,-1,3,-3\cdot 5,3\cdot5\cdot7,\dots\}$

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[Edited in accord with the comments] Sorry, I don't know what "the procedure method" is. I do know one method that wroks quite nicely for this problem - use the binomial theorem which says that for all real $n$, $(1+Q)^n=\sum_{k=0}^{\infty}{n\choose k}Q^k$

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    @Peter, didn't mind in the least, but I'm happier with the edit I've made. I think the comments can stay.2012-04-19