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I recently had a problem. I know how to evaluate power series but I cannot seem to find an expansion for $\sqrt{x+1}$.

I've tried differentiating it, in order to bring it in reciprocal form but that didn't help. Due to the presence of square root, I cannot change it in the form of $1/(x+1)$.

Kindly help.

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    http://en.wikipedia.org/wiki/Binomial_series2012-08-04
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    Please excuse my ignorance. Is binomial series the same as power series?2012-08-04
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    The binomial series is one of the power series that you should be very familiar with.2012-08-04
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    I am, I know how to calculate the binomial series. Just a bit rusty with the concepts. So binomial series is a kind of power series. Thank you for the comment! :)2012-08-04
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    Just to check: you know how to interpret things like $\binom{1/2}{k}$, yes?2012-08-04
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    Yes, this is the combination form. (1/2)!/((k!)(1/2-k)!).2012-08-04
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    No sorry, whats the deal with (1/2)! How to evaluate that?2012-08-04
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    Use the [Gamma function](http://en.wikipedia.org/wiki/Gamma_function) $\Gamma(x+1)=x!$ and generalises for $x\in R$.2012-08-04
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    No need to call on the Gamma function. $${1/2\choose k}={(1/2)(-1/2)\cdots((3/2)-k)\over k!}$$2012-08-04
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    In fact, $$\binom{1/2}{k}=\frac1{1-2k}\binom{2k}{k}\left(-\frac14\right)^k$$2012-08-04
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    Okay thank you guys for enlightening me, but you know what? This got me kinda confused, this is something I haven't studied. I'm willing to delve into this topic if only I had an idea where to start. Thank you once again.2012-08-04

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Here's one way. Start with the expansion you want, using $a_0$, $a_1$, $a_2$, $a_3$, etc. for the unknown coefficients:

$$\sqrt{x+1}\;=\;a_{0}\;+\;a_{1}x\;+\;a_{2}x^2\;+\;a_{3}x^3\;+\;a_{4}x^4\;+\;a_{5}x^5\;+ ...$$

Finding $a_0$: Plugging in $x=0$ on both sides leads to $a_{0}=1$.

Finding $a_1$: Differentiate both sides of the expansion. This gives

$$\frac{1}{2}\left(x+1\right)^{-\frac{1}{2}}\;\;=\;\;a_{1}\;+\;2a_{2}x\;+\;3a_{3}x^2\;+\;4a_{4}x^3\;+\;5a_{5}x^4\;+ ...$$

Pugging in $x=0$ on both sides leads to $a_{1}=\frac{1}{2}$.

Finding $a_2$: Differentiate 2-times both sides of the expansion. This gives

$$\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(x+1\right)^{-\frac{3}{2}}\;\;=\;\;2a_{2}\;+\;(2)(3)a_{3}x\;+\;(3)(4)a_{4}x^2\;+\;(4)(5)a_{5}x^3\;+ ...$$

Pugging in $x=0$ on both sides leads to $a_{2}=-\frac{1}{8}$.

Finding $a_3$: Differentiate 3-times both sides of the expansion. This gives

$$\left(-\frac{3}{2}\right)\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(x+1\right)^{-\frac{5}{2}}\;\;=\;\;(2)(3)a_{3}\;+\;(2)(3)(4)a_{4}x\;+\;(3)(4)(5)a_{5}x^2\;+ ...$$

Pugging in $x=0$ on both sides leads to $a_{3}=\frac{1}{16}$.

Finding $a_4$: Differentiate 4-times both sides of the expansion. This gives

$$\left(-\frac{5}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(x+1\right)^{-\frac{7}{2}}\;\;=\;\;(2)(3)(4)a_{4}\;+\;(2)(3)(4)(5)a_{5}x\;+ ...$$

Pugging in $x=0$ on both sides leads to $a_{4}=-\frac{5}{128}$.

Finding $a_5$: Differentiate 5-times both sides of the expansion. This gives

$$\left(-\frac{7}{2}\right)\left(-\frac{5}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(x+1\right)^{-\frac{9}{2}}\;\;=\;\;(2)(3)(4)(5)a_{5}\;+ ...$$

Pugging in $x=0$ on both sides leads to $a_{5}=\frac{7}{256}$.

Keep going to get as many coefficients as you want. If you keep careful track of the numbers without reducing the fractional expressions for the coefficients, you can easily determine a pattern (a pattern that can be proved by mathematical induction if you're so inclined).

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    That's right, can you tell me the difference between Maclaurin series and Power series? Cause that's Maclaurin series what you just used.2012-08-06
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    @Adnan, a Maclaurin expansion is a special case of the power series expansion; for Maclaurin, we take the expansion point to be $0$.2012-08-06
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    Yes that's what I'm talking about. When you're asked to expand a function by power series, you can't use Maclaurin series unless explicitly asked to.2012-08-06
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    @Adnan Zahid: It might help if you state exactly what question you want answered. I think most anyone would assume that, given $\sqrt{x+1},$ you'd want the expansion about $x=0.$ Indeed, it's very common in physics and engineering to algebraically approximate something like $\sqrt{u+v},$ where $0$\sqrt{v}\sqrt{\frac{u}{v} + 1}$ and expanding $\sqrt{\frac{u}{v} + 1}$ in powers of $\frac{u}{v}.$2012-08-06
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    I clearly wrote "I know how to evaluate power series but I cannot seem to find an expansion ... ". Anyway, I think what you're referring to now is binomial expansion. Thank you by the way.2012-08-06
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    Adman, the binomial expansion will give you the Maclaurin series.2012-08-07
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    @Adnan Zahid: I thought you meant that you knew how to plug numbers and algebraic expressions into power series (like evaluating a function), but didn't know how to obtain the power series expansion to begin with. Incidentally, the reason I gave the method I gave (instead of just quoting the Taylor series formula) is that I didn't know what "I've tried differentiating it, in order to bring it in reciprocal form" meant. I thought maybe "differentiating it" wasn't what you intended, thinking instead you were trying to rewrite the expression so that it represents the sum of a geometric series.2012-08-07
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    **(continued)** Thus, I thought maybe you only knew various algebraic manipulation techniques, not the general Taylor's expansion, and so I wrote my answer with this in mind (assuming only that you could differentiate expressions raised to constant powers).2012-08-07