diff --git a/src/algebra/fft.md b/src/algebra/fft.md
index 28fa86899..31c538b93 100644
--- a/src/algebra/fft.md
+++ b/src/algebra/fft.md
@@ -531,12 +531,13 @@ We have to compute the products of $a$ with every cyclic shift of $b$.
We generate two new arrays of size $2n$:
We reverse $a$ and append $n$ zeros to it.
And we just append $b$ to itself.
-When we multiply these two arrays as polynomials, and look at the coefficient $c[n],~ c[n+1],~ c[2n-1]$ of the product $c$, we get:
+When we multiply these two arrays as polynomials, and look at the coefficient $c[n-1],~ c[n],~ c[2n-2]$ of the product $c$, we get:
$$c[k] = \sum_{i+j=k} a[i] b[j]$$
And since all the elements $a[i] = 0$ for $i \ge n$:
$$c[k] = \sum_{i=0}^{n-1} a[i] b[k-i]$$
-It is easy to see that this sum is just the scalar product of the vector $a$ with the $(k - n - 1)$-th cyclic shift.
+It is easy to see that this sum is just the scalar product of the vector $a$ with the $(k - (n - 1))$-th cyclic left shift.
Thus these coefficients are the answer to the problem, and we were still able to obtain it in $O(n \log n)$ time.
+Note here that $c[2n-1]$ also gives us the $n$th cyclic shift but that is the same as the $0$th cyclic shift so we don't need to consider that seperately into our answer.
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