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<title>An example of proving summation identities with computer - You don't need to prove this</title>
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<h1>An example of proving summation identities with computer</h1>
<div class="post-info">
<div class="w3-opacity w3-margin-right w3-margin-bottom" style="flex-grow: 1;">
<span> Posted on Fri 21 December 2018 in <a href="https://newptcai.github.io/category/math.html" style="font-style: italic">math</a>
</span>
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<p>I am trying to learn how to use computers to prove identities like this one
</p>
<div class="math">$$
\sum_{k=m}^n{k\choose k-m}{2n\choose 2k}=4^{n-m}\frac{n(2n-m-1)!}{(2n-2m)!m!},
$$</div>
<p>
which I came across on <a href="https://math.stackexchange.com/q/3047567/1618">math.stackexchange.com</a>. If
you put it Mathematica and simplify, you will simply get a True without any explanation.
But how dose computer know it? In
the following, I will explain one possible way that machines can prove this for us -- using creative
telescoping algorithms.</p>
<p>I am going to use the Mathematica package <a href="https://www3.risc.jku.at/research/combinat/software/Sigma/index.php">Sigma</a> by <a href="https://www3.risc.jku.at/people/cschneid/">Carsten
Schneider</a>.</p>
<p>First the left-hand-side can be written as
</p>
<div class="math">$$
S(n_0)=\underset{k=0}{\overset{n_0}{\sum }}
f(n_0,k)
$$</div>
<p>
where <span class="math">\(n_0=n-m\)</span> and
</p>
<div class="math">$$
f(n_0,k)=
\left(
\begin{array}{c}
k+m \\
k \\
\end{array}
\right) \left(
\begin{array}{c}
2 \left(m+n_0\right) \\
2 (k+m) \\
\end{array}
\right)
.
$$</div>
<p>
Next, let
</p>
<div class="math">$$
c_0(n)=2 (1 + m + n_0) (m + 2 n_0) (1 + m + 2 n_0),
$$</div>
<p>
and
</p>
<div class="math">$$
c_1(n)=-(1 + n_0) (m + n_0) (1 + 2 n_0)
$$</div>
<p>
and
</p>
<div class="math">$$
g(n_0,k)=-\frac{k (2 k+2 m-1) \left(m+n_0+1\right) \binom{k+m}{k} \left(2 k m+4 k n_0+2 k-4 m n_0-3 m-6 n_0^2-7 n_0-2\right) \binom{2 m+2 n_0}{2 k+2 m}}{\left(2 k-2 n_0-1\right) \left(k-n_0-1\right)}.
$$</div>
<p>
Then you can verify in your favorite CAS system that
</p>
<div class="math">$$
g\left(n_0,k+1\right)-g\left(n_0,k\right)=c_0\left(n_0\right) f\left(n_0,k\right)+c_1\left(n_0\right) f\left(n_0+1,k\right)
.
$$</div>
<p>
I did not magically find this recursion myself. This is done by <a href="https://www3.risc.jku.at/research/combinat/software/Sigma/index.php">Sigma</a>.</p>
<p>Summing over <span class="math">\(k=1,..n_0-1\)</span>, the left-hand-side telescopes (perhaps this is where the name creative-telescoping comes from), and we get
</p>
<div class="math">$$
c_0(n_0) S\left(n_0\right)+c_1(n_0) S\left(n_0+1\right)=0
$$</div>
<p>
This is a simple recursion. Put in <span class="math">\(S(0)=f(0,0)=1\)</span>, we get
</p>
<div class="math">$$
S(n_0)=
\frac{4^{n_0} \left(m+n_0\right) \left(m+2 n_0-1\right)!}{m! \left(2 n_0\right)!}
=
\frac{n 4^{n-m} (-m+2 n-1)!}{m! (2 n-2 m)!}
.
$$</div>
<p>
where we substitute by <span class="math">\(n_0=n-m\)</span>. This is exactly the right-hand-side.</p>
<p>All these have been completely automated. What you need to do is only to trust the computer when it says
something is true. See <a href="https://www3.risc.jku.at/research/combinat/software/Sigma/pub/SLC06.pdf">Schneider's
paper</a> for many more
examples!</p>
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