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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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    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

1 Answers 1

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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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    **(co$n$tinued)** 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