# Expressing "formally" $f(x)=\frac {1}{\sqrt {1-2x}}$ as a power series

I have to express $f(x)=\frac {1}{\sqrt {1-2x}}$ as a power series and give its interval of convergence. Knowing the binomial series is as follows this should be fairly easy:

$$(1+x)^{\alpha}=\sum _{n=0}^{\infty}\binom{\alpha}{n}x^n$$

which converges when $-1<x<1$ (it must be verified if $-1,1$ belong to the interval or not).

Then $$f(x)=\frac {1}{\sqrt {1-2x}}=\ (1-2x)^{-1/2}=\sum _{n=0}^{\infty}\binom{-1/2}{n}(-2x)^n=\sum _{n=0}^{\infty}\binom{-1/2}{n}(2)^n (-1)^n x^n$$ which converges when $-1<-2x<1\iff -1/2 < x < 1/2$. I verified that the series diverges when $x=-1/2$ and given that the function is not defined in $x=1/2$, it can't converge. Hence the interval of convergence is $(-1/2,1/2)$ (it would be greatly appreciated if this can be checked, just to be sure).

All of this is thanks to one knowing the binomial series represents functions of the form $(1+x)^{\alpha}$ but I can't prove this. Firstly I tried to find an $M>0$ such that $|f^n(x)|<M^n$. Because $f$ is infinitely differentiable and bounded, then the Taylor series generated by $f(x)$ would represent $f(x)$. I couldn't find such $M$ in the specific case for the function I'm working with nor the general case $(1+x)^\alpha$.

I then tried to show that the Lagrange remainder converges pointwise to the $0$ function. Then the Taylor Series would represent the function. Again I was not able to in my specific case nor the general case. How can I prove the result in the general case of the binomial series or how can I formally prove in the function I'm given that it is represented by it's Taylor Series? Thank you in advance.

• For clarity, I would recommend using different variables for the general binomial series expansion. Commented Dec 15, 2014 at 1:17

How can I prove the result in the general case of the binomial series or how can I formally prove in the function I'm given that it is represented by it's Taylor Series?

There are several nice and easy ways. Which is easiest depends on what you already know.

One way uses complex analysis. The function $f_\alpha(z) = (1+z)^\alpha = \exp\bigl(\alpha \log (1+z)\bigr)$ is holomorphic on the open unit disk $\mathbb{D} = \{z\in \mathbb{C} : \lvert z\rvert < 1\}$, hence its Taylor series about $0$ converges locally uniformly to $f_\alpha$ on $\mathbb{D}$.

Now one can show

$$\Biggl(\frac{d}{dz}\Biggr)^n f_\alpha(z) = n!\cdot \binom{\alpha}{n}\cdot (1+z)^{\alpha-n}\tag{1}$$

by induction, and sees that the coefficients of the Taylor series about $0$ are $\binom{\alpha}{n}$.

Another way uses the theory of differential equations. If we let $f_\alpha(x) = (1+x)^\alpha$ and

$$g_\alpha(x) = \sum_{n=0}^\infty \binom{\alpha}{n}x^n,$$

then it is easily seen that $f_\alpha(0) = g_\alpha(0) = 1$. And if we differentiate and then multiply with $(1+x)$, we have

$$(1+x)\cdot\frac{d}{dx}f_\alpha(x) = (1+x)\cdot\frac{d}{dx}(1+x)^\alpha = (1+x)\cdot \alpha (1+x)^{\alpha-1} = \alpha(1+x)^\alpha,$$

and

\begin{align} (1+x)\cdot \frac{d}{dx} g_\alpha(x) &= (1+x)\sum_{n=1}^\infty n\binom{\alpha}{n}x^{n-1}\\ &= (1+x) \sum_{n=1}^\infty \alpha\binom{\alpha-1}{n-1}x^{n-1}\\ &= \alpha\cdot(1+x)\sum_{m=0}^\infty \binom{\alpha-1}{m}x^m\\ &= \alpha\Biggl(\sum_{m=0}^\infty \binom{\alpha-1}{m}x^m + \sum_{m=0}^\infty \binom{\alpha-1}{m}x^{m+1}\Biggr)\\ &= \alpha \Biggl(1 + \sum_{m=1}^\infty \biggl(\binom{\alpha-1}{m} + \binom{\alpha-1}{m-1}\biggr)x^m\Biggr)\\ &= \alpha \Biggl(1+\sum_{m=1}^\infty \binom{\alpha}{m} x^m\Biggr)\\ &= \alpha g_\alpha(x). \end{align}

So the two functions satisfy the same differential equation

$$(1+x)\cdot y' = \alpha y\tag{\ast}$$

with the initial condition $y(0) = 1$. By the uniqueness of solutions of $(\ast)$(1), we have $f_\alpha \equiv g_\alpha$ on $\{ x : \lvert x\rvert < 1\}$.

(1) The map $F\colon (x,y) \mapsto \alpha\frac{y}{1+x}$ locally satisfies a Lipschitz condition in $y$ on $(\mathbb{R}\setminus \{-1\})\times \mathbb{R}$, so the Picard-Lindelöf theorem asserts the existence and uniqueness of solutions of $(\ast)$ to any given initial condition $y(x_0) = y_0$ in some neighbourhood of $x_0 \neq -1$.

Not quite as nice and short, but not too bad either is showing that the series converges to $f_\alpha$ by showing that the remainder tends to $0$, where we use the integral form of the remainder term. We assume that $\alpha$ is not a non-negative integer, for in that case, the series is actually a finite sum, and the equality is just the binomial formula.

Like in the complex analysis method, we see by induction that the relation $(1)$ holds, where now we restrict $z$ to be real and $> -1$. Thus the series

$$\sum_{n=0}^\infty \binom{\alpha}{n} x^n$$

is the Taylor series of $f_\alpha$ with centre $0$. The remainder term in the integral form is

$$R_n(x) = \frac{1}{n!} \int_0^x (x-t)^n f_\alpha^{(n+1)}(t)\,dt,$$

using $(1)$, that becomes

$$R_n(x) = (n+1)\binom{\alpha}{n+1}\int_0^x (x-t)^n(1+t)^{\alpha-1-n}\,dt.$$

If $0 \leqslant x < 1$, we let $K = \max \{ 1, (1+x)^\alpha\}$, then $0 < (1+t)^{\alpha-1-n}\leqslant \frac{K}{(1+t)^{n+1}}\leqslant K$ for all $n\in\mathbb{N}$ and $t\in [0,x]$, and we have the estimate

$$\lvert R_n(x)\rvert \leqslant K\cdot(n+1)\left\lvert\binom{\alpha}{n+1}\right\rvert \int_0^x (x-t)^n\,dt = K\left\lvert \binom{\alpha}{n+1}\right\rvert\cdot x^{n+1},$$

and since the series converges for $\lvert y\rvert < 1$, in particular for $y = x$, the terms $\binom{\alpha}{n+1} x^{n+1}$ converge to $0$, so $\lvert R_n(x)\rvert \to 0$.

For $-1 < x < 0$, we write

\begin{align} \lvert R_n(x)\rvert &= (n+1)\left\lvert \binom{\alpha}{n+1} \int_0^x (x-t)^n(1+t)^{\alpha-1-n}\,dt \right\rvert\\ &= (n+1)\left\lvert \binom{\alpha}{n+1} \int_0^{\lvert x\rvert} (x+t)^n(1-t)^{\alpha-1-n}\,dt\right\rvert\\ &= (n+1) \left\lvert\binom{\alpha}{n+1}\right\rvert \int_0^{\lvert x\rvert} (\lvert x\rvert-t)^n(1-t)^{\alpha-1-n}\,dt\\ &= (n+1) \left\lvert\binom{\alpha}{n+1}\right\rvert \int_0^{\lvert x\rvert} \biggl(\frac{\lvert x\rvert-t}{1-t}\biggr)^n (1-t)^{\alpha-1}\,dt. \end{align}

Now we note that $t \mapsto \frac{\lvert x\rvert-t}{1-t}$ is monotonically decreasing on $[0,\lvert x\rvert]$, so we obtain the estimate

$$\lvert R_n(x)\rvert \leqslant (n+1)\left\lvert\binom{\alpha}{n+1}\right\rvert\cdot \lvert x\rvert^n \underbrace{\int_0^{\lvert x\rvert} (1-t)^{\alpha-1}\,dt}_C = C\lvert \alpha\rvert\cdot\left\lvert \binom{\alpha-1}{n} x^n\right\rvert.$$

Since also the series $\sum\limits_{n=0}^\infty \binom{\alpha-1}{n}x^n$ converges on the interval $(-1,1)$, it follows that $R_n(x) \to 0$ also for $x\in (-1,0)$.

So we have shown that $R_n(x) \to 0$ for all $x\in (-1,1)$, hence the identity

$$(1+x)^\alpha = \sum_{n=0}^\infty \binom{\alpha}{n} x^n$$

for $x\in (-1,1)$ is proved.

• Nice argument. It is the most elegant one, in my opinion. Anyway, for the benefit of the OP, it might be useful to mention that the uniqueness of a solution to $(*)$ satisfying the initial condition $y(0) = 1$ is a consequence of the uniqueness portion of the Picard-Lindelöf Theorem. Commented Dec 15, 2014 at 2:23
• I greatly appreciate your answer though I don't know complex analysis nor differential equations. I'll comeback to this when I do, thank you very much! Commented Dec 15, 2014 at 2:54
• @AndrewGSM No problem, we can show the convergence to the function using the integral form of the remainder. It's just not as nice and short. I'll add that (will take a while). Commented Dec 15, 2014 at 3:02
• @AndrewGSM Added. Yes, definitely not as short and nice. Commented Dec 15, 2014 at 3:38
• I barely tried with the integral form of the remainder, now I see why I couldn't go too far. Thank you again, really appreciate your clear answers. Commented Dec 15, 2014 at 3:44