coefficent -> coefficient

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 \(C(5,2)\).
@@ -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
 \(\quad\)\(S_a(k) + S_a(k - 1) - 2S_a(k - 2) = 0\).%
@@ -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}
+\end{document}

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
-<p>The <span style="font-family: 'courier new', courier;">binomial_coefficents</span> function creates a dictionary of values.</p>
+<p>The <span style="font-family: 'courier new', courier;">binomial_coefficients</span> function creates a dictionary of values.</p>
 
-︡2270b35a-79fc-46b1-b199-6e1b06942a69︡{"html": "<p>The <span style=\"font-family: 'courier new', courier;\">binomial_coefficents</span> function creates a dictionary of values.</p>"}︡
+︡2270b35a-79fc-46b1-b199-6e1b06942a69︡{"html": "<p>The <span style=\"font-family: 'courier new', courier;\">binomial_coefficients</span> function creates a dictionary of values.</p>"}︡
 ︠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}"}︡

src/S81.xml

@@ -14,7 +14,7 @@
 <li><p><m>\binom{n}{k}  = \binom{n-1}{k}+\binom{n-1}{k-1}</m> if <m>n > k > 0</m></p></li>
 </ul></p>
 </statement></definition>
-<observation><p>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 <xref ref="binomial-coefficient" text="type-global" />, other statements describing them should be theorems.  The theorem, in this case, would be that the <q>definition</q> 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.
+<observation><p>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 <xref ref="binomial-coefficient" text="type-global" />, other statements describing them should be theorems.  The theorem, in this case, would be that the <q>definition</q> 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.
 </p></observation>
 <p>Here is how we can apply the recursive definition to compute <m>\binom{5}{2} </m>.
 <me>
@@ -312,4 +312,4 @@ by induction that <m>L_k=7-2^{k+1}</m>.</p>
 <m>\sum_{k=0}^n \binom{n}{k} = 2^n</m></p></statement>
 </exercise>
 </exercises>
-</section>
+</section>

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</m>, with initial conditions <m>B(0) = 1</m> and <m>B(1) = 1</m>. If <m>C</m> is small enough, we might consider approximating the relation by replacing
 <m>1 - C^2</m> with 1 and <m>C^2</m> with 0. Solve the original relation and its approximation. Let <m>B_a</m> a be the solution of the approximation. Compare
 closed form expressions for <m>B(k)</m> and <m>B_a(k)</m>. 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 
 <m>S(k) + 1.001S(k - 1) - 2.004002 S(k - 2) = 0.0001</m> and
 <m>S_a(k) + S_a(k - 1) - 2S_a(k - 2) = 0</m>.</p>
@@ -600,4 +600,4 @@ are relatively far apart, this problem will not occur. For example, compare the
 
 
 </exercises>
-</section>
+</section>