From ab2c3626f0e0692a6182ec5052be8faba072610a Mon Sep 17 00:00:00 2001 From: sqrmax <14253201+sqrmax@users.noreply.github.com> Date: Fri, 2 Oct 2020 23:21:20 -0700 Subject: [PATCH 1/2] coefficent -> coefficient --- ads.tex | 4 ++-- sage/ADS_Sets_Combinatorics.sagews | 4 ++-- src/S81.xml | 4 ++-- 3 files changed, 6 insertions(+), 6 deletions(-) diff --git a/ads.tex b/ads.tex index 8e3ccfde2..f5ff13150 100644 --- a/ads.tex +++ b/ads.tex @@ -8386,7 +8386,7 @@ \section*{Goals for Chapter 1} % \end{definition} \par -A word a about definitions: Strictly speaking, when mathematical objects such as binomial coefficents are defined, they should be defined just once. Since we defined binomial coefficients earlier, in \hyperref[binomial-coefficient]{Definition~\ref{binomial-coefficient}}, other statements describing them should be theorems. The theorem, in this case, would be that the ``definition'' above is consistent with the original definition. Our point in this chapter in discussing recursion is to observe alternative definitions that have a recursive nature. In the exercises, you will have the opportunity to prove that the two definitions are indeed equivalent. +A word a about definitions: Strictly speaking, when mathematical objects such as binomial coefficients are defined, they should be defined just once. Since we defined binomial coefficients earlier, in \hyperref[binomial-coefficient]{Definition~\ref{binomial-coefficient}}, other statements describing them should be theorems. The theorem, in this case, would be that the ``definition'' above is consistent with the original definition. Our point in this chapter in discussing recursion is to observe alternative definitions that have a recursive nature. In the exercises, you will have the opportunity to prove that the two definitions are indeed equivalent. % \par Here is how we can apply the recursive definition to compute \(C(5,2)\). @@ -18667,4 +18667,4 @@ \subsection*{A.2.1 Exercises for the Algorithms Appendix} %% The index is here, setup is all in preamble \printindex % -\end{document} \ No newline at end of file +\end{document} diff --git a/sage/ADS_Sets_Combinatorics.sagews b/sage/ADS_Sets_Combinatorics.sagews index d91f19485..70e814495 100644 --- a/sage/ADS_Sets_Combinatorics.sagews +++ b/sage/ADS_Sets_Combinatorics.sagews @@ -210,9 +210,9 @@ binomial(n,2) ︡bfc19f30-4c37-4acb-bff0-a75c1ad87a12︡{"stdout": "1/2*(n - 1)*n"}︡ ︠0cb6f9c7-5463-41c7-94d6-82e0d506f83di︠ %html -

The binomial_coefficents function creates a dictionary of values.

+

The binomial_coefficients function creates a dictionary of values.

-︡2270b35a-79fc-46b1-b199-6e1b06942a69︡{"html": "

The binomial_coefficents function creates a dictionary of values.

"}︡ +︡2270b35a-79fc-46b1-b199-6e1b06942a69︡{"html": "

The binomial_coefficients function creates a dictionary of values.

"}︡ ︠8ae431f5-643d-477c-87c7-6ed1795a654b︠ binomial_coefficients(8) ︡eb850147-51e3-44e9-bb88-58ebf7fae947︡{"stdout": "{(5, 3): 56, (2, 6): 28, (7, 1): 8, (8, 0): 1, (4, 4): 70, (6, 2): 28, (1, 7): 8, (0, 8): 1, (3, 5): 56}"}︡ diff --git a/src/S81.xml b/src/S81.xml index 6aa0c0b92..2033e7970 100644 --- a/src/S81.xml +++ b/src/S81.xml @@ -14,7 +14,7 @@

  • \binom{n}{k} = \binom{n-1}{k}+\binom{n-1}{k-1} if n > k > 0

    • -

    A word about definitions: Strictly speaking, when mathematical objects such as binomial coefficents are defined, they should be defined just once. Since we defined binomial coefficients earlier, in , other statements describing them should be theorems. The theorem, in this case, would be that the definition above is consistent with the original definition. Our point in this chapter in discussing recursion is to observe alternative definitions that have a recursive nature. In the exercises, you will have the opportunity to prove that the two definitions are indeed equivalent. +

    A word about definitions: Strictly speaking, when mathematical objects such as binomial coefficients are defined, they should be defined just once. Since we defined binomial coefficients earlier, in , other statements describing them should be theorems. The theorem, in this case, would be that the definition above is consistent with the original definition. Our point in this chapter in discussing recursion is to observe alternative definitions that have a recursive nature. In the exercises, you will have the opportunity to prove that the two definitions are indeed equivalent.

    Here is how we can apply the recursive definition to compute \binom{5}{2} . @@ -312,4 +312,4 @@ by induction that L_k=7-2^{k+1}.

    \sum_{k=0}^n \binom{n}{k} = 2^n

    - \ No newline at end of file + From cba8d60c408c9d4cd614ad339cf07075ab20ffb0 Mon Sep 17 00:00:00 2001 From: sqrmax <14253201+sqrmax@users.noreply.github.com> Date: Sun, 4 Oct 2020 00:31:53 -0700 Subject: [PATCH 2/2] Fix missing space --- ads.tex | 2 +- src/S83.xml | 4 ++-- 2 files changed, 3 insertions(+), 3 deletions(-) diff --git a/ads.tex b/ads.tex index f5ff13150..1e9010886 100644 --- a/ads.tex +++ b/ads.tex @@ -9451,7 +9451,7 @@ \section*{Goals for Chapter 1} = C^2\), with initial conditions \(B(0) = 1\) and \(B(1) = 1\). If \(C\) is small enough, we might consider approximating the relation by replacing \(1 - C^2\) with 1 and \(C^2\) with 0. Solve the original relation and its approximation. Let \(B_a\) a be the solution of the approximation. Compare closed form expressions for \(B(k)\) and \(B_a(k)\). Their forms are very different because the characteristic roots of the original relation were -close together and the approximation resulted in one double characteristic root.If characteristic roots of a relation +close together and the approximation resulted in one double characteristic root. If characteristic roots of a relation are relatively far apart, this problem will not occur. For example, compare the general solutions of \(\quad\)\(S(k) + 1.001S(k - 1) - 2.004002S(k - 2) = 0.0001\) and \(\quad\)\(S_a(k) + S_a(k - 1) - 2S_a(k - 2) = 0\).% diff --git a/src/S83.xml b/src/S83.xml index 75f03d5cc..b460b8cab 100644 --- a/src/S83.xml +++ b/src/S83.xml @@ -573,7 +573,7 @@ when the value is raised to zero. How much could you borrow if you can afford to = C^2, with initial conditions B(0) = 1 and B(1) = 1. If C is small enough, we might consider approximating the relation by replacing 1 - C^2 with 1 and C^2 with 0. Solve the original relation and its approximation. Let B_a a be the solution of the approximation. Compare closed form expressions for B(k) and B_a(k). Their forms are very different because the characteristic roots of the original relation were -close together and the approximation resulted in one double characteristic root.If characteristic roots of a relation +close together and the approximation resulted in one double characteristic root. If characteristic roots of a relation are relatively far apart, this problem will not occur. For example, compare the general solutions of S(k) + 1.001S(k - 1) - 2.004002 S(k - 2) = 0.0001 and S_a(k) + S_a(k - 1) - 2S_a(k - 2) = 0.

    @@ -600,4 +600,4 @@ are relatively far apart, this problem will not occur. For example, compare the - \ No newline at end of file +