Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The sum $$g_k=\sum_{j,i:i+j=k} (-1)^j\left(\begin{array}{c} m\\j\end{array}\right) \left(\begin{array}{c} n\\i\end{array}\right),$$ for $0\leq j\leq m$ and $0\leq i \leq n$, is involved in the development of the polynomial $(1+x)^n(1-x)^m$. It seems to me that the sum $$\sum_k\frac{g_k}{k+1}=\frac{2^{n+m}}{(n+m+1)\left(\begin{array}{c} n+m\\n\end{array}\right)}.$$ Could anybody provide a formal proof or a closed-form development of $g_k$?

share|improve this question
    
Yes, thanks, I have corrected my question –  Antonio Sciarretta Mar 31 '12 at 16:02
1  
Write $\sum_k\frac{g_k}{k+1}$ as the integral of a product of two polynomials, then use a substitution. You will recognize an equality involving Gamma and Beta functions. –  Davide Giraudo Mar 31 '12 at 16:14
    
The formula is wrong (try $n=1$, $m=2$). –  Did Mar 31 '12 at 16:50
add comment

2 Answers

On the one hand, $$ \int_0^1 \sum_k g_k x^k \,dx = \left.\sum_k \frac{g_k x^{k+1}}{k+1}\right|_0^1 = \sum_k \frac{g_k}{k+1} $$ On the other, by substituting $u=\tfrac12 (1+x)$ and using the symmetry of the integrand [edit: see below] we get $$\int_0^1 \sum_k g_k x^k \,dx = \int_0^1 (1+x)^n (1-x)^m \,dx = 2^{n+m+1} \int_{1/2}^1 u^n (1-u)^m \,du \\ = 2^{n+m} \int_0^1 u^n (1-u)^m \,du = 2^{n+m} B(n+1,m+1) $$ which is what you wanted. (Here $B(\cdot,\cdot)$ is the Beta function.) (The sums over $k$ are formally infinite but actually finite because $(g_k)_{k\in\mathbb Z}$ has finite support; so no issues of convergence arise.)

Edit: Didier quite rightly points out that the integrand isn't symmetric. See Didier's answer for a correct version of this idea.

share|improve this answer
    
I see that Davide Giraudo gave this answer in a comment, more concisely. Also, I forgot to mention: the key word for searching for material on this kind of technique is "generating function". –  Steven Taschuk Mar 31 '12 at 16:25
    
My comments was more concise since I didn't do the computations. –  Davide Giraudo Mar 31 '12 at 16:31
1  
The integrand is not symmetric when $n\ne m$. –  Did Mar 31 '12 at 16:33
    
Thank you, perfect and clear. I still wonder anyway if $g_k$ could be explicitly calculated, i.e., without summation –  Antonio Sciarretta Mar 31 '12 at 16:44
    
Perfect and clear but wrong, though, as @Didier pointed out. (As for a closed form for $g_k$, I don't know of one.) –  Steven Taschuk Mar 31 '12 at 16:46
show 1 more comment

Writing each $\frac1{k+1}$ as $\int\limits_0^1x^k\mathrm dx$, one gets $S=\int\limits_0^1G(x)\mathrm dx$ with $S=\sum\limits_k\frac{g_k}{k+1}$ and $G(x)=\sum\limits_kg_kx^k$, hence $$ G(x)=\sum\limits_{i,j}(-1)^j{m\choose j}{n\choose i}x^{i+j}=\sum\limits_{i}{n\choose i}x^{i}\sum\limits_{j}(-1)^j{m\choose j}x^{j}=(1+x)^n(1-x)^m. $$ Let $x=1-2t$, then $0\leqslant t\leqslant\frac12$, $\mathrm dx=2\mathrm dt$, $1+x=2(1-t)$ and $1-x=2t$, hence $G(x)=2^{n+m}t^m(1-t)^n$ and $$ S=2^{n+m+1}\int_0^{1/2}t^m(1-t)^n\mathrm dt=2^{n+m+1}\,\mathrm B_{1/2}(m+1,n+1), $$ where $x\mapsto\mathrm B_x(m+1,n+1)$ is the incomplete Beta function of parameters $(m+1,n+1)$.

share|improve this answer
    
Thank you, as for the post above, it answers perfectly to the question. –  Antonio Sciarretta Mar 31 '12 at 16:48
    
This cannot be, since the two answers give different results (as gracefully acknowledged by @Steven). –  Did Apr 20 '12 at 16:21
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.