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For $α ∈ ℝ$ the function $g_α \colon B_1(0) → ℝ, x ↦ (1+x)^α$ is $C^∞$ and $g_α^{(n)}(x) = n! \tbinom{α}{n}(1+x)^{α-n}$, where $\tbinom{α}{n} = \frac{α(α-1)\cdots(α-n+1)}{n!}$ is the generalized binomial coefficient. I want to show that $g_α$ has a Taylor expansion $g_α (x) = \sum_{k=0}^∞ \tbinom{α}{k}x^k$. I can easily show that the series really converges for $|x| < 1$, but can I use this fact to show that the remainder term $$R_n(x) = \int_0^x \frac{(x-t)^n}{n!} (n+1)! \tbinom{α}{n+1}(1+t)^{α-n-1} dt$$ or in Lagrange form $$\quad R_n(x) = \frac{x^{n+1}}{(n+1)!} (n+1)! \binom{α}{n+1}(1+ξ)^{α-n-1} = x^{n+1} \binom{α}{n+1}(1+ξ)^{α-n-1}$$ converges to zero for $|x| < 1$?

How else can I show that the remainder term converges to zero for $|x| < 1$?


Progress: Using the Lagrange form, for $x ∈ (-1,1)$ and $n ∈ ℕ$ there is an $h=h(n,x) ∈ (0,1)$ such that:

$$|R_n(x)| = \Big| x^{n+1} \binom{α}{n+1}(1+hx)^{α-n-1} \Big| = \Big| \binom{α}{n+1}\big(\frac{x}{1+hx}\big)^{n+1}(1+ξ)^{α} \Big| $$ If $x > 0$, then $\big|\frac{x}{1+hx}\big| < {\big|\frac{x}{1+x}\big|}< 1$ and so for $q := \big|\frac{x}{1+x}\big|$, since $\sum_{k=0}^∞ \tbinom{α}{n} q^n$ converges: $$|R_n(x)| ≤ \Big| \binom{α}{n+1} · q^{n+1} ·(1+ξ)^{α} \Big| \overset{n → ∞}{\longrightarrow} 0$$ The problem is still open for $x < 0$. Mhenni suggested to use Stirling approximation, but I want to do this more elementary and I think it is possible.

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  • $\begingroup$ You can follow the technique done in this problem. $\endgroup$ – Mhenni Benghorbal May 26 '13 at 15:39
  • $\begingroup$ Should it be $\big|\frac{x}{1+hx}\big| < {\big|\frac{x}{1}\big|}< 1$ for $x>0$? $\endgroup$ – Nicholas Nov 19 '17 at 17:15
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Again, I think that I got it. One needs to proceed differently:

Let $x ∈ ℝ$ such that $|x| < 1$. For $t ∈ ℝ$ between $x$ and zero, one has $0 ≤ |t| ≤ |x| < 1$ as well as $|x-t| < |1+t|$, because:

  • for $x > 0$ one has $|x-t| = x - t < 1 < 1 + t = |1 + t|$, and
  • for $x < 0$ one has $|x-t| = t - x < t + 1 = |1 + t|$, since $-x < 1$.

Therefore $\big|\frac{x-t}{1+t}\big|$ assumes a maximum $q<1$ for $t$ on the compact interval between zero and $x$. Then: \begin{align*} |R_n(x)| &= \Big| \int_0^x \frac{(x-t)^n}{n!} g_α^{(n+1)}(t) dt \Big|\\ &≤ \Bigg| \int_0^x \Big| \frac{(x-t)^n}{n!} g_α^{(n+1)}(t) \Big| dt \Bigg| \\ &\overset{(1)}{=} \Bigg| \int_0^x \Big| \frac{(x-t)^n}{(1+t)^n} ·(α-n) · \binom{α}{n} · (1+t)^{α-1} \Big| dt \Bigg|\\ &\overset{(2)}{≤} n · \binom{α}{n} ·|q|^n · \Bigg| \int_0^x |α/n -1| ·|1+t|^{α-1} dt\Bigg| \\ &\overset{(3)}{≤} n · \binom{α}{n} · |q|^n · (|α| +1) · C \overset{n → ∞}{\longrightarrow} 0, \end{align*} where the convergence follows from $|(n+1)\tbinom{α}{n+1}\big/n\tbinom{α}{n}|\overset{n → ∞}{\longrightarrow} 1$ and $|q| < 1$, and

  • (1) $g_α^{(n)}(t) = (n+1)! \tbinom{α}{n+1}(1+t)^{α-n-1}$ and $\tbinom{α}{n+1} = \tfrac{α-n}{n+1} \tbinom{α}{n}$,
  • (2) $\big|\frac{x-t}{1+t}\big|^n ≤ |q|^n$, and
  • (3) $|α/n - 1| ≤ |α| + 1$, and $C := \Big| \int_0^x |1+t|^{α-1} dt\Big|$
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  • $\begingroup$ You should separate the cases $0\leq t\leq x<1$ and $-1<x\leq t\leq 0$ becasue the inequality is not true for all $0\leq |t|\leq |x|<1$. It really takes $x,t$ of the same sign. $\endgroup$ – Julien May 27 '13 at 20:11
  • $\begingroup$ Other than that, it looks good. A nice illustration of the advantage of the integral remainder over the Lagrange remainder. $\endgroup$ – Julien May 27 '13 at 20:19
  • $\begingroup$ @julien Yes, right. But in the integral, $t$ has the same sign as $x$, I was silently assuming this. Thanks for taking a look at it! $\endgroup$ – k.stm May 27 '13 at 20:36
  • $\begingroup$ Might be even easier to start from $R_n(x) = \int_0^x \frac{(x-t)^n}{n!} g_α^{(n+1)}(t) dt=\frac{(x-ξ)^n}{n!} g_α^{(n+1)}(ξ)x$ without even bothering integral remainder. $\endgroup$ – Nicholas Nov 20 '17 at 2:26
  • $\begingroup$ @Nicholas Okay, yes. It seems like you don’t need the integral remainder, but I find it more illustrative since the crucial point is that the integrand attains a maximum less than one on the integral interval. This point is way less prominent if you hide the integrand and the compact interval it lives on within some anonymous $ξ$. But yeah, the integral looks a bit more clunky. $\endgroup$ – k.stm Nov 20 '17 at 18:25
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Let $f(z)=(1+z)^\alpha$ and $g(z)=\sum_{n\geq 0}\binom{\alpha}{n}z^n$.

If $\alpha$ is a nonnegative integer, it is clear that both functions coincide on $\mathbb{C}$ by the binomial theorem. So we assume that $\alpha\not\in\mathbb{N}$ from now on, so that $\binom{\alpha}{n}\neq 0$ for every $n\geq 0$.

Using the principal branch of the complex logarithm, we see that $g(z)=\exp(\alpha\log(1+z))$ is holomorphic on $\mathbb{C}\setminus (-\infty,-1]$, whence on the open unit disk $D$ in particular.

Now observe that $$ \frac{\Big|\binom{\alpha}{n+1}\Big||z|^{n+1}}{\Big|\binom{\alpha}{n}\Big||z|^{n}}=\frac{|\alpha-n||z|}{n+1}\longrightarrow |z|. $$ So the radius of convergence of $g$ is $1$ by the ratio test. Hence $g$ converges to a holomorphic function on $D$.

For every $x>0$, the remainder in the approximation of $f$ by its degree $n$ Taylor polynomial at $0$ is, by Taylor-Lagrange, $$ R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}=\binom{\alpha}{n+1}(1+c)^{\alpha-n-1}x^{n+1}=(1+c)^\alpha\binom{\alpha}{n+1}\left(\frac{x}{1+c} \right)^{n+1} $$ for some $0<c<x$. Since $0<\frac{x}{1+c}<1$, the convergence of $g$ at $\frac{x}{1+c}$ implies that the rhs above tends to $0$, since it is a constant times its general term. So the remainder of the Taylor approximation tends to $0$.

It follows that $f$ and $g$ coincide on $(0,1)$.

By the isolated zero principle, $f$ and $g$ coincide on the whole open unit disk $D$.

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  • $\begingroup$ Okay, but from the convergence of the Taylor series of $(1+x)^α$, how can you conclude that it is represented by its Taylor series? $\endgroup$ – k.stm May 27 '13 at 12:58
  • $\begingroup$ @K.Stm. Here is something better than my previous silly post. If you allow complex analysis. $\endgroup$ – Julien May 27 '13 at 17:55
  • $\begingroup$ Thanks! What is the isolated zero princple, though? Also, I found a solution using only elementary real analysis. $\endgroup$ – k.stm May 27 '13 at 18:21
  • $\begingroup$ @K.Stm. If $f$ is holomorphic on an open connected set $\Omega$ and if there is a point $z_0\in\Omega$ which is a limit point of zeros of $f$, then $f$ is equal to $0$ on $\Omega$. $\endgroup$ – Julien May 27 '13 at 20:05
  • $\begingroup$ Ah, it’s what I know as the identity theorem! Could’ve guessed that from the name. $\endgroup$ – k.stm May 27 '13 at 20:39
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Following the technique in the problem. Let $ g(n,\alpha)= \frac{(n+1)!}{n!} \tbinom{α}{n+1}$, for not to have a mess

$$ \Big|R_n(x)\Big| \leq g(n,\alpha)\int_0^x {(x-t)^n} (1+t)^{α-n-1} dt$$

$$ \leq g(n,\alpha)x^n\int_0^x (1+t)^{α-n-1} dt $$

$$ < g(n,\alpha)\int_0^1 (1+t)^{α-n-1} dt,\quad \mathrm{since}\quad |x|<1,\, \int_{0}^{x}\leq \int_{0}^{1}. $$

$$\implies \Big|R_n(x)\Big| < g(n,\alpha)\frac{2^{\alpha-n}-1}{\alpha-n}$$

$$ \Big|R_n(x)\Big| = \frac{\alpha!(2^{\alpha-n}-1)}{(\alpha-n)!n!}. $$

Now, you can see that the remainder goes to $0$ as $n\to \infty$ with some restrictions on $\alpha$. Try to use Stirling approximation $n! \sim \left(\frac{n}{e}\right)^n\sqrt{2 \pi n}$ and see what you get.

Note:

1) we used the inequality $ x-t \leq x $ which follows from

$$ 0\leq t \leq x \implies -x \leq -t \leq 0 \implies 0<x-t < x. $$

2)

$$ \tbinom{α}{n}(2^{α-n} - 1)= \frac{\alpha!(2^{\alpha-n}-1)}{(\alpha-n)!n!}. $$

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  • $\begingroup$ Can you explain why the last term converges to zero? (By the way, I read it as $\tbinom{α}{n}(2^{α-n} - 1)$. Many thanks, so far.) As $2^{α-n} - 1$ converges to $-1$, $\tbinom{α}{n}$ should be converging to zero, right? $\endgroup$ – k.stm May 26 '13 at 18:41
  • $\begingroup$ @K.Stm.: It is just taking the limit as $n\to \infty$. $\endgroup$ – Mhenni Benghorbal May 26 '13 at 18:50
  • $\begingroup$ I’m unsure, does this also work for the general binomial coefficient? I don’t see it. $\endgroup$ – k.stm May 26 '13 at 19:01
  • $\begingroup$ @K.Stm.: Where is the problem? $\endgroup$ – Mhenni Benghorbal May 26 '13 at 19:05
  • $\begingroup$ For $α = -1$, don’t you have $\tbinom{α}{n} = \tfrac{(-1)·(-1-1)·(-1-2) \cdots (-1-n+1)}{n!} = \tfrac{-n!}{n!} = -1$ constantly? $\endgroup$ – k.stm May 26 '13 at 19:09

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