Define $$f(k) = - \frac{\Gamma(\frac{k-1}{2}) \pi^{3/2}}{\Gamma(\frac{k}{2}) 2^{\frac{k-1}{2}}} \sum_{m=0}^{\lfloor\frac{k-3}{4}\rfloor} \binom{\frac{k-1}{2}}{2m+1} \binom{2m}{m} 4^{-m},$$ And let $$M(2n+1)= \sum_{k=0}^{n-1} (-1)^{n-1-k} f(2k+5) \binom{n-1}{k},$$ $$F(2n+1)= \sum_{k=0}^{n-1} (-1)^{n-1-k} f(2k+7) \binom{n-1}{k}.$$ Prove that $$F(2n+1)= \frac{n^2+3n+3}{(n+1)(2n+3)}M(2n+1).$$ I have verified that for $n=1,2,3,4,5$ the assertion holds, but I don’t know a general strategy to prove it for all $n$. One may use induction, but the steps are not completely clear to me.
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$\begingroup$ Can you share how this identity arises? $\endgroup$Iosif Pinelis– Iosif Pinelis2025-02-23 01:57:24 +00:00Commented Feb 23 at 1:57
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1$\begingroup$ Why do you want a proof specifically by induction? Would other means count? $\endgroup$Max Alekseyev– Max Alekseyev2025-02-23 04:02:05 +00:00Commented Feb 23 at 4:02
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$\begingroup$ @MaxAlekseyev absolutely other methods are welcome. I have a feeling that one may use generating functions to rewrite $f(k)$, but I am not familiar with these techniques. $\endgroup$Peter Liu– Peter Liu2025-02-23 08:21:10 +00:00Commented Feb 23 at 8:21
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1$\begingroup$ The function $f(k)$ arises when I integrate the following function, $\int_0^{s/2} \frac{u^{n-1} + u^{n-2}(s - u) + \cdots + (s - u)^{n-1}}{\sqrt{u(s - u)}} \, du $, and the result is $\frac{\pi s^{n-1}}{2^{n}} \sum_{m=0}^{\left\lfloor \frac{n-1}{2} \right\rfloor} \binom{n}{2m+1} \binom{2m}{m} \frac{1}{4^{m}} $. $\endgroup$Peter Liu– Peter Liu2025-02-23 08:29:26 +00:00Commented Feb 23 at 8:29
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4$\begingroup$ Then why does the title say "Prove by induction..."? $\endgroup$Max Alekseyev– Max Alekseyev2025-02-23 11:53:39 +00:00Commented Feb 23 at 11:53
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2 Answers
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Here's a partial solution, where I use generating functions and reduce the proof to an integral identity, which I could not prove. Since the proof is quite involved, please let me know if anything is unclear.
Let $\overline M(n-1) := M(2n+1), \overline F(n-1) := F(2n+1)$. By inspecting the definitions and using the identity $\binom{n+1}{k}=\binom{n}{k}+\binom{n}{k-1}$, it is not hard to see that $\overline F(n) = \overline M(n) + \overline M(n+1)$. Hence, we need to prove $$ \overline F(n) = \frac {(n+1)^2+3(n+1)+3}{(n+2)(2(n+1)+3)}\overline M(n) = \frac {n^2+5n+7}{2n^2+9n+10}\overline M(n), $$ which, after rearranging and substituting, becomes $$(2n^2+9n+10)\overline M(n+1) \overset{?}{=} -(n^2+4n+3)\overline M(n).$$
Given a sequence $a(n), n\ge 0$, its exponential generating function is the formal power series $$\mathrm{EGF}[a(n)] := \sum_{n\ge 0} a(n) \frac{x^n}{n!}.$$ It can be shown that that $\mathrm{EGF}[a(n)]\cdot\mathrm{EGF}[b(n)] = \mathrm{EGF}[c(n)]$, where $$c(n) := \sum_{k=0}^n \binom{n}{k} a(k) b(n-k),$$ and that $\mathrm{EGF}[na(n)] = x\cdot\mathrm{EGF}[a(n+1)] = x D(\mathrm{EGF}[a(n)])$, where $D$ denotes differentiation. In general, if $p(n)$ is a polynomial, then formally $\mathrm{EGF}[p(n)a(n)] = p(xD)\,\mathrm{EGF}[a(n)]$.
If we consider the exponential generating function $\mu(x) := \mathrm{EGF}[\overline M(n)]$, this is further equivalent to $$ 2 (xD)^2\mu'(x) + 9xD\mu'(x) + 10\mu'(x) = -(xD)^2\mu(x) - 4xD \mu(x) - 3\mu(x), $$ or, after expanding the derivatives, $$ 2x\mu''' + (x^2+11x)\mu'' + (5x+10)\mu' + 3\mu = 0, \tag{$1$} $$ with initial conditions \begin{align} \mu(0) &= \overline M(0) = -2\pi/3, \\ \mu'(0) &= \overline M(1) = \pi/5, \\ \mu''(0) &= \overline M(2) = -8\pi/105. \end{align} Using Mathematica, we find the unique solution to this initial value problem: $$ \mu(x) = -\frac{\pi}{4x}\left[2+\sqrt{\frac{2\pi}{x}}e^{-x/2}(x-1)\operatorname{erfi}\left(\sqrt{x/2}\right)\right],\tag{$2$} $$ where $\operatorname{erfi}$ is the imaginary error function, and the solution can be checked against $(1)$ by using the identity $\frac{d}{dx}\operatorname{erfi}\left(\sqrt{x/2}\right) = e^{x/2}/\sqrt{2\pi x}$.
Now notice that \begin{align} \mu(x) &= \mathrm{EGF}[(-1)^n] \cdot \mathrm{EGF}[f(2n+5)] \\ &= \left(\sum_{k\ge 0} \frac{(-x)^{k}}{k!} \right) \mathrm{EGF}[f(2n+5)] \\ &= e^{-x}\, \mathrm{EGF}[f(2n+5)], \end{align} so we need to prove the equality (in the field of Puiseux series): $$ \mathrm{EGF}[f(2n+5)] \overset{?}{=} -\frac{\pi}{4x}\left[2e^x+\sqrt{\frac{2\pi}{x}}e^{x/2}(x-1)\operatorname{erfi}\left(\sqrt{x/2}\right)\right].\tag{$3$} $$ A rather long, but mundane, computation shows that the right-hand side is the $\mathrm{EGF}$ of the sequence $$ g(n) := -\pi 2^{-n}\sum_{m=0}^n\binom{n}{m} \frac{2m+2}{(2m+1)(2m+3)} \overset{?}{=} f(2n+5). $$ Let us compute its ordinary generating function: \begin{align} \mathrm{GF}[g(n)] :\!\!&= -\pi \sum_{n\ge 0} x^n 2^{-n}\sum_{m\ge 0}\binom{n}{m} \frac{2m+2}{(2m+1)(2m+3)} \\ &= -\frac{\pi}{2} \sum_{m\ge 0} \left[\frac{1}{2m+1}+\frac{1}{2m+3}\right] \sum_{n\ge 0}\binom{n}{m} (x/2)^n \\ &= -\frac{\pi}{2-x} \sum_{m\ge 0} \left[\frac{1}{2m+1}+\frac{1}{2m+3}\right] \left(\frac{x}{2-x}\right)^m \\ &= -\frac{\pi}{x}\left[\frac{1}{\sqrt{x(2-x)}} \log\left(\frac{\sqrt{2-x}+\sqrt{x}}{\sqrt{2-x}-\sqrt{x}}\right)\right]. \end{align}
Now we can write, using the integral provided in Peter's comment (with $s=1$) and doing some manipulations with the gamma functions, $$ f(2n+1) = -\frac{4^n}{n\binom{2n}{n}}\int_0^{1/2}\frac{(1-u)^n-u^n}{(1-2u)\sqrt{u(1-u)}}\,du,\qquad n \ge 1. $$ We can define $\mathrm{GF}[f(2n+1)]$ by starting the sum at $n = 1$. If we take advantage of the integral's symmetry ($u\leftrightarrow 1-u$) and apply Theorem 2.2 in this paper, we have \begin{align} \mathrm{GF}[f(2n+1)] &= \int_0^1 \frac{1}{(1-2u)\sqrt{u(1-u)}}\tag{$4$} \\ &\cdot \left[\sqrt{\frac{xu}{1-xu}} \arctan \sqrt{\frac{xu}{1-xu}} -\sqrt{\frac{x(1-u)}{1-x(1-u)}} \arctan \sqrt{\frac{x(1-u)}{1-x(1-u)}} \right]\, du. \end{align} Using the fact that $f(3) = -\pi$, it suffices to prove that this equals $$ \mathrm{GF}[g(n)]\,x^2+f(3)\,x = \pi\sqrt{\frac{x}{2-x}} \log \left(\frac{\sqrt{2-x}-\sqrt{x}}{\sqrt{2-x}+\sqrt{x}}\right). $$ Numerically, the identity seems to hold. I suspect that it can be proved by contour integration, but the integrand has several branch cuts, which prevent a simple application of the residue theorem.
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$\begingroup$ Thanks! would you mind explaining a bit more the lines at (1) and above? How does the function $\phi$ arise? $\endgroup$Peter Liu– Peter Liu2025-02-24 16:48:25 +00:00Commented Feb 24 at 16:48
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$\begingroup$ @PeterLiu Sorry, that was a typo. I fixed it. $\endgroup$Jacopo G. Chen– Jacopo G. Chen2025-02-24 16:52:22 +00:00Commented Feb 24 at 16:52
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$\begingroup$ And how does the third derivative $\mu'''$ arise? Thanks! $\endgroup$Peter Liu– Peter Liu2025-02-24 17:04:49 +00:00Commented Feb 24 at 17:04
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$\begingroup$ @PeterLiu Another typo: since we have $\overline M(n+1)$ on the left hand side, the equation before (1) should have $\mu'$ on the left. $\endgroup$Jacopo G. Chen– Jacopo G. Chen2025-02-24 17:17:04 +00:00Commented Feb 24 at 17:17
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To complete the final step in Jacopo's answer, let \begin{align*}F(x):=&\int_0^1 \frac{1}{(1-2u)\sqrt{u(1-u)}}\\ & \cdot \left[\sqrt{\frac{xu}{1-xu}} \arctan \sqrt{\frac{xu}{1-xu}} - \sqrt{\frac{x(1-u)}{1-x(1-u)}} \arctan \sqrt{\frac{x(1-u)}{1-x(1-u)}}\right] du, \end{align*} and let \begin{align*} R(x) := \pi \sqrt{\frac{x}{2-x}} \log \left(\frac{\sqrt{2-x} - \sqrt{x}}{\sqrt{2-x} + \sqrt{x}}\right), \quad \left\lvert x \right\rvert <1 \end{align*} We will show that $F(0) = R(0) = 0$ and $F'(x) = R'(x)$ for $\left\lvert x \right\rvert <1 $. Therefore, $F(x) - R(x) \equiv 0$ on $\left\lvert x \right\rvert <1$. Indeed, set \begin{align*} A_x(u) := \sqrt{\frac{xu}{1-xu}}\arctan \sqrt{\frac{xu}{1-xu}},\quad B_x(u) := \sqrt{\frac{x(1-u)}{1-x(1-u)}} \arctan \sqrt{\frac{x(1-u)}{1-x(1-u)}}. \end{align*} Note that $B_x(u) = A_x(1-u)$, so the bracket in the integrand of $F(x)$ is $A_x(u) - B_x(u) = A_x(u) - A_x(1-u)$. For small $x$, $A_x(u) - B_x(u) \sim x(2u-1)$. Hence, \begin{align*} F(x) \sim - \frac{x}{\sqrt{u(1-u)}}. \end{align*} So near $x = 0$, \begin{align*} F(x) \sim -x \int_0^1 \frac{dx}{\sqrt{u(1-u)}} = -\pi x, \end{align*} which means that $F(0) = \lim_{x \to 0}F(x) = 0$. Next, we use the expansion \begin{align*} \log \frac{1-y}{1+y} = -2y + O(y^3), \quad y \to 0, \end{align*} with $y = \sqrt{x/(2-x)}$, we know that $R(x) \sim \pi x$ as $x \to 0$, so $R(0) = 0$ as well. Furthermore, \begin{align*} \frac{\partial A_x(u)}{\partial x} = \frac{u}{2(1-xu)} + \frac{\sqrt{u}}{2\sqrt{x}(1-xu)^{3/2}} \arctan \sqrt{\frac{xu}{1-xu}}, \end{align*} and \begin{align*} \frac{\partial B_x(u)}{\partial x} = \frac{\partial A_x(1-u)}{\partial x}, \end{align*} so that \begin{align*} F'(x) = \int_0^1 \frac{1}{(1-2u)\sqrt{u(1-u)}} \left(\frac{\partial A_x(u)}{\partial x} - \frac{\partial B_x(u)}{\partial x}\right) du. \end{align*} We split the above into the ''algebraic'' part and ''arctan'' part, $F'(x) = I_1(x)+ I_2(x)$ where \begin{align*} I_1(x) &= \frac{1}{2} \int_0^1 \frac{1}{(1-2u)\sqrt{u(1-u)}} \left(\frac{u}{1-xu} - \frac{1-u}{1-x(1-u)}\right) du,\\ I_2(x) &= \frac{1}{2\sqrt{x}} \int_0^1 \frac{1}{(1-2u)\sqrt{u(1-u)}} \\ &\cdot \left( \frac{\sqrt{u}}{(1-xu)^{3/2}} \arctan \sqrt{\frac{xu}{1-xu}} - \frac{\sqrt{1-u}}{(1-x(1-u))^{3/2}}\arctan \sqrt{\frac{x(1-u)}{1-x(1-u)}} \right) du. \end{align*} Note that $I_2(x)$ vanishes due to symmetry. Hence, $F'(x) = I_1(x)$. Evaluating the integral we get \begin{align*} F'(x) &= - \frac{1}{2}\int_0^1 \frac{du}{\sqrt{u(1-u)}(1-xu)(1-x(1-u))}\\ &= - \int_0^{\pi/2} \frac{d\theta}{1-x+ \frac{x^2}{4}\sin^2(2\theta)}. \end{align*} Now, using the identity \begin{align*} \int_0^{\pi/2} \frac{d\theta}{a^2 + b^2 \sin^2 \theta} = \frac{\pi}{2ab}, \quad a,b>0 \end{align*} we have \begin{align*} F'(x) = - \int_0^{\pi/2} \frac{d\theta}{A^2 + B^2 \sin^2 \theta} = \frac{\pi}{4AB} \end{align*} where $A = \sqrt{1-x}$ and $B = x/2$. This means that \begin{align*} F'(x) = -\frac{\pi}{2} \cdot \frac{1}{x\sqrt{1-x}}. \end{align*} On the other hand, for $R(x)$, we rewrite the log using $\log \frac{1-r}{1+r} = -2 \text{arctanh} r$, $\left\lvert x \right\rvert <1 $. With $r = \sqrt{\frac{x}{2-x}}$, we get \begin{align*} R(x) = - 2\pi \sqrt{\frac{x}{2-x}} \text{arctanh} \sqrt{\frac{x}{2-x}}, \end{align*} then a straightforward differentiation gives $F'(x) = R'(x)$ for $\left\lvert x \right\rvert <1$. The proof is complete.