Find $f^{(n)}(1)$ on $f(x)=(1+\sqrt{x})^{2n+2}$

Question

Find $f^{(n)}(1)$ on $f(x)=(1+\sqrt{x})^{2n+2}$ .

Here is a solution by someone:

\begin{align*} f(x)&=(1+\sqrt{x})^{2n+2}=\sum_{k=0}^{2n+2}\binom{2n+2}{k}x^{\frac{k}{2}}\\ &=\sum_{k=0}^{2n+2}\binom{2n+2}{k}\sum_{j=0}^{\infty}\binom{\frac{k}{2}}{j}(x-1)^j\\ &=\sum_{j=0}^{\infty}\sum_{k=0}^{2n+2}\binom{2n+2}{k}\binom{\frac{k}{2}}{j}(x-1)^j. \end{align*}

Hence \begin{align*} f^{(n)}(1)&=n!\sum_{k=0}^{2n+2}\binom{2n+2}{k}\binom{\frac{k}{2}}{n}=n!\cdot4(n+1)^2. \end{align*}

Is it correct? How to compute $$n!\sum_{k=0}^{2n+2}\binom{2n+2}{k}\binom{\frac{k}{2}}{n}=n!\cdot4(n+1)^2?$$

Answer 1

We first rewrite $f^{(n)}(1)$ as $$\begin{align} n!\sum_{k=0}^{2n+2}\binom{2n+2}k\binom{\frac k2}n&=\sum_{k=0}^{2n+2}\binom{2n+2}k\cdot \frac{k}{2}\cdot \left(\frac{k}{2}-1\right)\cdot \left(\frac{k}{2}-2\right)\dots \left(\frac{k}{2}-n+1\right)\\ &=\binom{2n+2}{2n}n! + (n+1)!+\sum_{r=0}^n \binom{2n+2}{2r+1}\frac{(2r+1)(2r-1)\dots(2r+3-2n)}{2^n}\end{align}$$

The above is obtained by noticing that for every even $k$ except $2n,2n+2$ the value is $0$ (the first and second terms correspond to the values obtained at $2n,2n+2$ while the remaining odd terms are generalised by the $r$ series). The first two terms are easily found to equal $n!\cdot2(n+1)^2$ and I will now focus on the $r$ series.

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A complete answer

Notice that if we define $g(x)=(1-\sqrt x)^{2n+2}$, we have

$$g^{(n)}(1)=n!\sum_{k=0}^{2n+2}(-1)^k\binom{2n+2}k\binom{\frac k2}n$$

and since all odd $k$ cancel out, $$f^{(n)}(1)+g^{(n)}(1)=2\left(\binom{2n+2}{2n}n! + (n+1)!\right)=n!\cdot4(n+1)^2$$ To find $g^{(n)}(1)$, we notice that

$$g'(x)=(2n+2)(1-\sqrt x)^{2n+1}\cdot \frac{-1}{2\sqrt x}$$

Has a $(1-\sqrt x)^{2n+1}$ term. Similarly, $g''(x)$ will have a lowest power term of $(1-\sqrt x)^{2n}$, $g'''(x)$ will have $(1-\sqrt x)^{2n-1}$ and so on. $g^{(n)}(x)$ thus has it's lowest power term being $(1-\sqrt x)^{n+2}$. Thus,

$$g^{(n)}(1)=0\implies \boxed {f^{(n)}(1)=n!\cdot4(n+1)^2}$$

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If the above method seemed a little hand-wavey to you, here is a solution without it:

Evaluating the $r$ series (An almost completed answer)

For every $r$, there are $(r+1)$ +ve terms and $(n-r-1)$ -ve terms. We can separate these and write:

$$\sum_{r=0}^n \binom{2n+2}{2r+1}\frac{(2r+1)(2r-1)\dots(2r+3-2n)}{2^n}=\sum_{r=0}^n \binom{2n+2}{2r+1}(-1)^{n-r-1}\frac{(2r+1)!!(2n-2r-3)!!}{2^n}$$

(I shall no longer write the LHS simply because of how much space this will take)

$$\begin{align}&=\sum_{r=0}^n (-1)^{n-r-1} \frac{(2n+2)!}{(2r+1)!(2n-2r+1)!}\cdot\frac{(2r+1)!!(2n-2r-3)!!}{2^n} \\ &= \sum_{r=0}^n (-1)^{n-r-1} \frac{(2n+2)!}{(2r)!!(2n-2r)!!}\cdot\frac{1}{2^n(2n-2r+1)(2n-2r-1)} \end{align}$$

We can factor out a $2^r$ and $2^{n-r}$ (a $2^n$ in total) from the double factorials in the denominator to get

$$=\frac{(2n+2)!}{4^n} \sum_{r=0}^n \frac{(-1)^{n-r-1}}{(2(n-r)+1)(2(n-r)-1)r!(n-r)!}$$

The $r!(n-r)!$ can be used to generate a $\binom nr$. Also, $(n-r)$ can be exchanged with $r$ here (because of the limits of the sum)

$$\begin{align} &=\frac{(2n+2)!}{4^n\cdot n!} \sum_{r=0}^n (-1)^{r-1}\binom nr \frac{1}{(2r+1)(2r-1)}\\ &=\frac{(2n+2)!}{4^n\cdot n!}\sum_{r=0}^n (-1)^{r}(-1)\binom nr \frac12\left(\frac{1}{2r-1}-\frac{1}{2r+1} \right)\\ &=\frac{(2n+2)!}{2\cdot 4^n\cdot n!}\sum_{r=0}^n (-1)^{r}\binom nr \left(\frac{1}{2r+1}-\frac{1}{2r-1} \right) \end{align}$$

I tried for hours to find a solution to this, but was ultimately unsuccessful. The only patterns I could notice were that:

$$\sum_{r=0}^n (-1)^{r}\binom nr \left(\frac{1}{2r+1} \right)=\frac{(2n)!!}{(2n+1)!!}$$

$$\sum_{r=0}^n (-1)^{r}\binom nr \left(-\frac{1}{2r-1} \right)=\frac{(2n)!!}{(2n-1)!!}$$

This however does allow us to simplify to $$\begin{align}\frac{(2n+2)!}{2\cdot 4^n\cdot n!} \left(\frac{(2n)!!}{(2n+1)!!}+\frac{(2n)!!}{(2n-1)!!}\right)&=\frac{(2n+2)!\cdot(2n)!!}{2\cdot 4^n\cdot n!} \left(\frac{1+(2n+1)}{(2n+1)!!}\right)\\ &=\frac{(2n+2)!!(2n)!!}{2\cdot4^n\cdot n!}(2n+2)\\ &=2(n+1)(n+1)!=n!\cdot2(n+1)^2\end{align}$$

This when paired with the original two terms does give $\boxed{n!\cdot4(n+1)^2}$ as required.

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Added: Since $\sum_{r=0}^n (-1)^{r}\binom nrx^{2r}=(1-x^2)^n$, we can write

$$\sum_{r=0}^n (-1)^{r}\binom nr \left(\frac{1}{2r+1}\right)=\int_0^1(1-x^2)^ndx$$

This is possible to evaluate by using a $x=\sin t $ sub and applying Walli's Formula.

$$\int_0^1(1-x^2)^ndx=\int_0^\frac \pi2(\cos t)^{2n+1} dt=W_{2n+1}=\boxed{\frac{(2n)!!}{(2n+1)!!}}$$

Also,$$\sum_{r=1}^n (-1)^{r}\binom nr \left(-\frac{1}{2r-1}\right)=-\int_0^1\frac{(1-x^2)^n-1}{x^2}dx$$

Integrating by parts, $$-\int_0^1\frac{(1-x^2)^n-1}{x^2}dx=\left[\left((1-x^2)^n-1\right)\int-\frac1{x^2}dx\right]_0^1-\int_0^1n(1-x^2)^{n-1}(-2x)\frac1xdx$$

$$\implies \sum_{r=1}^n (-1)^{r}\binom nr \left(-\frac{1}{2r-1}\right)=-1+2\int_0^1n(1-x^2)^{n-1}dx=-1+(2n)W_{2n-1}$$

Since $\sum_{r=0}^n (-1)^{r}\binom nr \left(-\frac{1}{2r-1}\right)=1+\sum_{r=1}^n (-1)^{r}\binom nr \left(-\frac{1}{2r-1}\right)$,

$$\sum_{r=0}^n (-1)^{r}\binom nr \left(-\frac{1}{2r-1}\right)=(2n)W_{2n-1}=\boxed{\frac{(2n)!!}{(2n-1)!!}}$$

Answer 2

An approach using the Lagrange inversion theorem: let $f(z)$ be analytic around $z=z_0$ with $f'(z_0)\neq0$; then $w=f(z)$ has an inverse $z=g(w)$ analytic around $w=f(z_0)$ with $$g^{(n)}\big(f(z_0)\big)=\lim_{z\to z_0}\frac{d^{n-1}}{dz^{n-1}}\left(\frac{z-z_0}{f(z)-f(z_0)}\right)^n.\qquad(n>0)$$

We apply this to $f(z)=\dfrac{\sqrt{z}-1}{\sqrt{z}+1}$ and $z_0=1$, thus $g(w)=\left(\dfrac{1+w}{1-w}\right)^2$ and $$\lim_{z\to1}\frac{d^n}{dz^n}(1+\sqrt{z})^{2n+2}=g^{(n+1)}(0)=\color{blue}{4(n+1)!(n+1)}$$ since $g(w)=1+\displaystyle\frac{4w}{(1-w)^2}=1+4w\frac{d}{dw}\frac1{1-w}=1+4\sum_{n=1}^\infty nw^n$ for $|w|<1$.

Answer 3

This is just a supplement to the nice answer of @Cathedral. Here we close a gap and show \begin{align*} \color{blue}{\sum_{r=0}^n(-1)^r\binom{n}{r}\frac{1}{2r+1}=\frac{(2r)!!}{(2r+1)!!}}\tag{1} \end{align*}

We obtain \begin{align*} \color{blue}{\sum_{r=0}^n}&\color{blue}{(-1)^r\binom{n}{r}\frac{1}{2r+1}}\\ &=\frac{1}{2}(-1)^nn!\sum_{r=0}^n\frac{1}{r+\frac{1}{2}} \prod_{q=0}^{r-1}\frac{1}{r-q}\prod_{q=r+1}^n\frac{1}{r-q}\tag{2.1}\\ &=\frac{1}{2}(-1)^nn!\sum_{r=0}^n\underbrace{\operatorname{res}_{z=r}\left(\frac{1}{z+\frac{1}{2}} \prod_{q=0}^{n}\frac{1}{z-q}\right)}_{f(z)}\tag{2.2}\\ &=\frac{1}{2}(-1)^nn!\left(-\operatorname{res}_{z=-\frac{1}{2}}f(z) -\operatorname{res}_{z=-\infty}f(z)\right)\tag{2.3}\\ &=\frac{1}{2}(-1)^nn!\left(-\lim_{z=-\frac{1}{2}}\prod_{q=0}^n\frac{1}{z-q} +\underbrace{\operatorname{res}_{z=0}\left(\frac{1}{z^2}f\left(\frac{1}{z}\right)\right)}_{=0}\right)\tag{2.4}\\ &=\frac{1}{2}(-1)^{n+1}n!\prod_{q=0}^n\frac{1}{-\frac{1}{2}-q}\\ &=2^nn!\prod_{q=0}^n\frac{1}{2q+1}\\ &=\frac{2^nn!}{(2n+1)!!}\\ &\,\,\color{blue}{=\frac{(2n)!!}{(2n+1)!!}} \end{align*} and the claim (1) follows. Similarly we can show \begin{align*} \color{blue}{\sum_{r=0}^n(-1)^r\binom{n}{r}\frac{1}{2r-1}=-\frac{(2n)!!}{(2n-1)!!}} \end{align*}

Comment:

• In (2.1) we use \begin{align*} \binom{n}{r}&=\frac{n!}{r!(n-r)!}\\ &=n!\prod_{q=0}^{r-1}\frac{1}{r-q}\prod_{q=0}^{n-r-1}\frac{1}{n-r-q}\\ &=n!\prod_{q=0}^{r-1}\frac{1}{r-q}\prod_{q=0}^{n-r-1}\frac{1}{q+1}\tag{$q\to\ n-r-1-q$}\\ &=n!(-1)^{n-r}\prod_{q=0}^{r-1}\frac{1}{r-q}\prod_{q=r+1}^n\frac{1}{r-q} \end{align*}

• In (2.2) we write the summands as residue of a meromorphic function at the pole $z=q$.

• In (2.3) we use the sum of the residues of a meromorphic function at the poles $z=q, 0\leq q\leq r$ and $z=-\frac{1}{2}$ plus the residue at $\infty$ sum up to zero. This way we get rid of the sum and what is left are just two residues, the one at $z=-\frac{1}{2}$ and the one at $z=\infty$.

• In (2.4) we use the identity \begin{align*} \operatorname{res}_{z=\infty}f(z)=\operatorname{res}_{z=0}\left(-\frac{1}{z^2}f\left(\frac{1}{z}\right)\right) \end{align*} which transforms a residue at infinity to a residue at zero. We then find by inspection the residue of $f(z)$ at $z=\infty$ is zero.

Answer 4

Preliminary Fact $\bf{1}$:

Note that the terms with non-integer exponents are cancelled in $$ \begin{align} \hspace{-24pt}{\color{#090}{\left(1+\sqrt{x}\right)^{2n+2}}+\color{#C00}{\left(1-\sqrt{x}\right)^{2n+2}}} &=\overbrace{\color{#090}{\sum_{k=0}^{n+1}\binom{2n+2}{2k}\,x^{n-k+1}}}^\text{even terms}\color{#090}{+}\overbrace{\color{#090}{\sum_{k=0}^n\binom{2n+2}{2k+1}\,x^{n-k+1/2}}}^\text{odd terms}\\ &{}+{}\color{#C00}{\sum_{k=0}^{n+1}\binom{2n+2}{2k}\,x^{n-k+1}-\sum_{k=0}^n\binom{2n+2}{2k+1}\,x^{n-k+1/2}}\tag{1a}\\ &=2\sum_{k=0}^{n+1}\binom{2n+2}{2k}\,x^{n-k+1}\tag{1b}\\ &=\underbrace{2x^{n+1}+2\binom{2n+2}{2}x^n\vphantom{\sum_{k=2}^{n+1}}}_\text{$n^\text{th}$ derivative is non-vanishing}+\underbrace{2\sum_{k=2}^{n+1}\binom{2n+2}{2k}\,x^{n-k+1}}_\text{$n^\text{th}$ derivative vanishes}\tag{1c} \end{align} $$ Explanation:
$\text{(1a):}$ Binomial Theorem
$\text{(1b):}$ add and cancel
$\text{(1c):}$ the $n^\text{th}$ derivatives of the terms with non-negative
$\phantom{\text{(1c):}}$ integer exponents less than $n$ vanish

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Preliminary Fact $\bf{2}$:

Since $1-\sqrt{x}$ has a $0$ at $x=1$, $\left(1-\sqrt{x}\right)^{2n+2}$ has a zero of degree $2n+2$ at $x=1$. Thus, all derivatives up to order $2n+1$ of $\left(1-\sqrt{x}\right)^{2n+2}$ at $x=1$ are $0$.

That is, for $m\le2n+1$, $$ \begin{align} \left.\frac{\mathrm{d}^m}{\mathrm{d}x^m}\left(1-\sqrt{x}\right)^{2n+2}\right|_{x=1} &=\left.\frac{\mathrm{d}^m}{\mathrm{d}x^m}\left(\left(1-x\right)^{2n+2}\left(\tfrac1{1+\sqrt{x}}\right)^{2n+2}\right)\right|_{x=1}\tag{2a}\\ &=\sum_{k=0}^m\left.\binom{m}{k}\frac{\mathrm{d}^k}{\mathrm{d}x^k}\left(1-x\right)^{2n+2}\frac{\mathrm{d}^{m-k}}{\mathrm{d}x^{m-k}}\left(\tfrac1{1+\sqrt{x}}\right)^{2n+2}\right|_{x=1}\tag{2b}\\[3pt] &=0\tag{2c} \end{align} $$ Explanation:
$\text{(2a):}$ $1-\sqrt{x}=\frac{1-x}{1+\sqrt{x}}$
$\text{(2b):}$ Leibniz' Rule
$\text{(2c):}$ $\frac{\mathrm{d}^k}{\mathrm{d}x^k}\left(1-x\right)^{2n+2}=(-1)^k\frac{(2n+2)!}{(2n+2-k)!}\left(1-x\right)^{2n+2-k}=0$
$\phantom{\text{(2c):}}$ for $x=1$ and all $k\le m\le2n+1$

In particular, the $n^\text{th}$ derivative of $\left(1-\sqrt{x}\right)^{2n+2}$ at $x=1$ is $0$.

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The Computation Is Now Simple:

Taking the $n^\text{th}$ derivative at $x=1$ yields $$ \begin{align} \left.\frac{\mathrm{d}^n}{\mathrm{d}x^n}\left(1+\sqrt{x}\right)^{2n+2}\right|_{x=1} &=\left.\frac{\mathrm{d}^n}{\mathrm{d}x^n}\left(\left(1+\sqrt{x}\right)^{2n+2}+\left(1-\sqrt{x}\right)^{2n+2}\right)\right|_{x=1}\tag{3a}\\ &=\left.\frac{\mathrm{d}^n}{\mathrm{d}x^n}\left(2x^{n+1}+2(n+1)(2n+1)x^n\right)\right|_{x=1}\tag{3b}\\[3pt] &=2(n+1)!+2(n+1)(2n+1)n!\tag{3c}\\[9pt] &=\bbox[5px,border:2px solid #C0A000]{4(n+1)(n+1)!}\tag{3d} \end{align} $$ Explanation:
$\text{(3a):}$ the $n^\text{th}$ derivative of $\left(1-\sqrt{x}\right)^{2n+2}$ at $x=1$ is $0$
$\text{(3b):}$ use the terms from $\text{(1c)}$ whose $n^\text{th}$ derivative is non-vanishing
$\text{(3c):}$ evaluate the $n^\text{th}$ derivative at $x=1$
$\text{(3d):}$ simplify

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Proof of the Formula Used in the Approach in the Question

Notice the parallel between this proof and the proof above.

The order $2n+2$ forward difference of a degree $n$ polynomial vanishes, thus $$ \sum_{k=0}^{2n+2}(-1)^k\binom{2n+2}{k}\binom{k/2}{n}=0\tag4 $$ Therefore, $$ \begin{align} \sum_{k=0}^{2n+2}\binom{2n+2}{k}\binom{k/2}{n} &=\sum_{k=0}^{2n+2}\binom{2n+2}{k}\binom{k/2}{n}\left(1+(-1)^k\right)\tag{5a}\\ &=2\sum_{k=0}^{n+1}\binom{2n+2}{2k}\binom{k}{n}\tag{5b}\\ &=\color{#C00}{2\binom{2n+2}{2n+2}\binom{n+1}{n}}+\color{#090}{2\binom{2n+2}{2n}\binom{n}{n}}\\ &+\color{#00F}{2\sum_{k=0}^{n-1}\binom{2n+2}{2k}\binom{k}{n}}\tag{5c}\\[6pt] &=\color{#C00}{2(n+1)}+\color{#090}{2(n+1)(2n+1)}+\color{#00F}{0}\tag{5d}\\[12pt] &=4(n+1)^2\tag{5e} \end{align} $$ Explanation
$\text{(5a):}$ add $(4)$
$\text{(5b):}$ only the even terms remain, so substitute $k\mapsto2k$
$\text{(5c):}$ $\binom{k}{n}=0$ for $k\lt n$, so isolate those terms
$\text{(5d):}$ evaluate
$\text{(5e):}$ simplify

Answer 5

A partial answer

Let $$F=\sum_{k=0}^{2n+2}\binom{2n+2}{k}\binom{\frac{k}{2}}{n}$$ So we can write $F=G+H$, where $$G=\sum_{j=0}^{n+1}{2n+2\choose 2j}{j \choose n}, \quad H=\sum_{j=1}^{n+1} {2n+2 \choose 2j-1}{j-1/2 \choose n}$$ Only 2 terms in $G$ are nonzero, when $j=n,n+1$. Hence $$G={2n+2 \choose 2n} {n \choose n}+{2n+2 \choose 2n+2}{n+1 \choose n}=2(n+1)^2.$$ Though Mathematica gives $H$ in terms of hypergemetric $_pF_q$, which gives $H=2(n+1)^2$ (numerically) again!

I wish to come back.