Power series expansion

Question

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.

Please excuse my ignorance. Is binomial series the same as power series? — Adnan Zahid, Aug 04 '12 at 08:02
The binomial series is one of the power series that you should be very familiar with. — J. M. ain't a mathematician, Aug 04 '12 at 08:03
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! :) — Adnan Zahid, Aug 04 '12 at 08:05
Just to check: you know how to interpret things like $\binom{1/2}{k}$, yes? — J. M. ain't a mathematician, Aug 04 '12 at 08:13
Use the Gamma function $\Gamma(x+1)=x!$ and generalises for $x\in R$. — Mark Hurd, Aug 04 '12 at 08:30
No need to call on the Gamma function. $${1/2\choose k}={(1/2)(-1/2)\cdots((3/2)-k)\over k!}$$ — Gerry Myerson, Aug 04 '12 at 10:13
In fact, $$\binom{1/2}{k}=\frac1{1-2k}\binom{2k}{k}\left(-\frac14\right)^k$$ — J. M. ain't a mathematician, Aug 04 '12 at 10:28
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. — Adnan Zahid, Aug 04 '12 at 10:58

score 5 · Answer 1 · answered Aug 06 '12 at 16:47

5

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).

answered Aug 06 '12 at 16:47

Dave L. Renfro

38,688

That's right, can you tell me the difference between Maclaurin series and Power series? Cause that's Maclaurin series what you just used. – Adnan Zahid Aug 06 '12 at 16:52
@Adnan, a Maclaurin expansion is a special case of the power series expansion; for Maclaurin, we take the expansion point to be $0$. – J. M. ain't a mathematician Aug 06 '12 at 16:54
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. – Adnan Zahid Aug 06 '12 at 16:58
@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 $0<u<v,$ by rewriting it as $\sqrt{v}\sqrt{\frac{u}{v} + 1}$ and expanding $\sqrt{\frac{u}{v} + 1}$ in powers of $\frac{u}{v}.$ – Dave L. Renfro Aug 06 '12 at 18:29
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.
– Adnan Zahid Aug 06 '12 at 18:46
Adman, the binomial expansion will give you the Maclaurin series. – Gerry Myerson Aug 07 '12 at 00:27
@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. – Dave L. Renfro Aug 07 '12 at 14:35
(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). – Dave L. Renfro Aug 07 '12 at 14:37

Power series expansion

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