Signed-off-by: Silvio Fanzon <[email protected]>

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@@ -286,7 +286,7 @@ <h1 class="title">Numbers, Sequences and Series</h1>
     <div>
     <div class="quarto-title-meta-heading">Published</div>
     <div class="quarto-title-meta-contents">
-      <p class="date">4 Oct 2023</p>
+      <p class="date">5 Oct 2023</p>
     </div>
   </div>
 

docs/search.json

@@ -81,7 +81,7 @@
     "href": "sections/chap_2.html#triangle-inequality",
     "title": "2  Preliminaries",
     "section": "2.9 Triangle inequality",
-    "text": "2.9 Triangle inequality\nThe triangle inequality relates the absolute value to the sum operation. It is a very important inequality, which we will use a lot in the future.\n\nTheorem 35: Triangle inequalityFor every \\(x, y \\in \\mathbb{R}\\) we have \\[\n| |x| - |y| | \\leq |x-y| \\leq |x| + |y| \\,.\n\\tag{2.7}\\]\n\n\nBefore proceeding with the proof, let us discuss the geometric meaning of the triangle inequality.\n\nRemark 36: Geometric meaning of triangle inequalityThe notion of absolute value can be extended also to vectors in the plane. Suppose that \\(x\\) and \\(y\\) are two vectors in the plane, as in Figure 2.8 below. Then \\(|x|\\) and \\(|y|\\) can be interpreted as the lengths of these vectors.\nUsing the rule of sum of vectors, we can draw \\(x+y\\), as shown in Figure 2.9 below. From the picture it is evident that \\[\n|x+y| \\leq |x| + |y|\\,,\n\\tag{2.8}\\] that is, the length of each side of a triangle does not exceed the sum of the lengths of the two remaining sides. Note that (2.8) is exactly the second inequality in (2.7). This is why (2.7) is called triangle inequality.\n\n\n\n\n\nFigure 2.8: Vectors \\(x\\) and \\(y\\)\n\n\n\n\n\nFigure 2.9: Summing the vectors \\(x\\) and \\(y\\). The triangle inequality relates the length of \\(x+y\\) to the length of \\(x\\) and \\(y\\)\n\n\n\nProof: Proof of Theorem 35Assume that \\(x,y \\in \\mathbb{R}\\). We prove the two inequalities in (2.7) individually.\nStep 1. Proof of the second inequality in (2.7).\nTrivially we have \\[\n|x| \\leq |x| \\,.\n\\] Therefore we can apply Lemma 33 and infer \\[\n-|x| \\leq x \\leq |x| \\,.\n\\tag{2.9}\\] Similarly we have that \\(|y| \\leq |y|\\), and so Lemma 33 implies \\[\n-|y| \\leq y \\leq |y| \\,.\n\\tag{2.10}\\] Summing (2.9) and (2.10) we get \\[\n-(|x| + |y|) \\leq x + y \\leq |x| + |y| \\,.\n\\] We can now again apply Lemma 33 to get \\[\n|x + y| \\leq |x| + |y| \\,,\n\\tag{2.11}\\] which is the second inequality in (2.7).\nStep 2. Proof of the second inequality in (2.7).\nNote that the trivial identity \\[\nx = x+y - y\n\\] always holds. We then have \\[\\begin{align}\n|x| & = |x+y - y| \\\\\n    & = |(x+y) + (-y)| \\\\\n    & = |a+b|\n\\end{align}\\] with \\(a=x+y\\) and \\(b=-y\\). We can now apply (2.11) to \\(a\\) and \\(b\\) to obtain \\[\\begin{align}\n|x| & = |a+b| \\\\\n    & \\leq |a| + |b| \\\\\n    & = |x+y| + |-y| \\\\\n    & = |x+y| + |y|\n\\end{align}\\] Therefore \\[\n|x| - |y| \\leq |x+y| \\,.\n\\tag{2.12}\\] We can now swap \\(x\\) and \\(y\\) in (2.12) to get \\[\n|y| - |x| \\leq |x+y| \\,.\n\\] By rearranging the above inequality we obtain \\[\n-|x+y| \\leq |x| - |y|  \\,.\n\\tag{2.13}\\] Putting together (2.12) and (2.13) yields \\[\n-|x+y| \\leq |x| - |y| \\leq |x+y| \\,.  \n\\] By Lemma 33 the above is equivalent to \\[\n||x| - |y|| \\leq |x+y| \\,,\n\\] which is the first inequality in (2.7).\n\n\nAn immediate consequence of the triangle inequality are the following inequalities, which are left as an exercise.\n\nRemark 37For any \\(x,y \\in \\mathbb{R}\\) it holds \\[\n||x|-|y|| \\leq |x-y| \\leq |x|+|y|\\,.\n\\] Moreover for any \\(x,y,z \\in \\mathbb{R}\\) it holds \\[\n|x-y| \\leq |x-z| + |z-y| \\,.\n\\]"
+    "text": "2.9 Triangle inequality\nThe triangle inequality relates the absolute value to the sum operation. It is a very important inequality, which we will use a lot in the future.\n\nTheorem 35: Triangle inequalityFor every \\(x, y \\in \\mathbb{R}\\) we have \\[\n| |x| - |y| | \\leq |x+y| \\leq |x| + |y| \\,.\n\\tag{2.7}\\]\n\n\nBefore proceeding with the proof, let us discuss the geometric meaning of the triangle inequality.\n\nRemark 36: Geometric meaning of triangle inequalityThe notion of absolute value can be extended also to vectors in the plane. Suppose that \\(x\\) and \\(y\\) are two vectors in the plane, as in Figure 2.8 below. Then \\(|x|\\) and \\(|y|\\) can be interpreted as the lengths of these vectors.\nUsing the rule of sum of vectors, we can draw \\(x+y\\), as shown in Figure 2.9 below. From the picture it is evident that \\[\n|x+y| \\leq |x| + |y|\\,,\n\\tag{2.8}\\] that is, the length of each side of a triangle does not exceed the sum of the lengths of the two remaining sides. Note that (2.8) is exactly the second inequality in (2.7). This is why (2.7) is called triangle inequality.\n\n\n\n\n\nFigure 2.8: Vectors \\(x\\) and \\(y\\)\n\n\n\n\n\nFigure 2.9: Summing the vectors \\(x\\) and \\(y\\). The triangle inequality relates the length of \\(x+y\\) to the length of \\(x\\) and \\(y\\)\n\n\n\nProof: Proof of Theorem 35Assume that \\(x,y \\in \\mathbb{R}\\). We prove the two inequalities in (2.7) individually.\nStep 1. Proof of the second inequality in (2.7).\nTrivially we have \\[\n|x| \\leq |x| \\,.\n\\] Therefore we can apply Lemma 33 and infer \\[\n-|x| \\leq x \\leq |x| \\,.\n\\tag{2.9}\\] Similarly we have that \\(|y| \\leq |y|\\), and so Lemma 33 implies \\[\n-|y| \\leq y \\leq |y| \\,.\n\\tag{2.10}\\] Summing (2.9) and (2.10) we get \\[\n-(|x| + |y|) \\leq x + y \\leq |x| + |y| \\,.\n\\] We can now again apply Lemma 33 to get \\[\n|x + y| \\leq |x| + |y| \\,,\n\\tag{2.11}\\] which is the second inequality in (2.7).\nStep 2. Proof of the first inequality in (2.7).\nNote that the trivial identity \\[\nx = x+y - y\n\\] always holds. We then have \\[\\begin{align}\n|x| & = |x+y - y| \\\\\n    & = |(x+y) + (-y)| \\\\\n    & = |a+b|\n\\end{align}\\] with \\(a=x+y\\) and \\(b=-y\\). We can now apply (2.11) to \\(a\\) and \\(b\\) to obtain \\[\\begin{align}\n|x| & = |a+b| \\\\\n    & \\leq |a| + |b| \\\\\n    & = |x+y| + |-y| \\\\\n    & = |x+y| + |y|\n\\end{align}\\] Therefore \\[\n|x| - |y| \\leq |x+y| \\,.\n\\tag{2.12}\\] We can now swap \\(x\\) and \\(y\\) in (2.12) to get \\[\n|y| - |x| \\leq |x+y| \\,.\n\\] By rearranging the above inequality we obtain \\[\n-|x+y| \\leq |x| - |y|  \\,.\n\\tag{2.13}\\] Putting together (2.12) and (2.13) yields \\[\n-|x+y| \\leq |x| - |y| \\leq |x+y| \\,.  \n\\] By Lemma 33 the above is equivalent to \\[\n||x| - |y|| \\leq |x+y| \\,,\n\\] which is the first inequality in (2.7).\n\n\nAn immediate consequence of the triangle inequality are the following inequalities, which are left as an exercise.\n\nRemark 37For any \\(x,y \\in \\mathbb{R}\\) it holds \\[\n||x|-|y|| \\leq |x-y| \\leq |x|+|y|\\,.\n\\] Moreover for any \\(x,y,z \\in \\mathbb{R}\\) it holds \\[\n|x-y| \\leq |x-z| + |z-y| \\,.\n\\]"
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@@ -1033,7 +1033,7 @@ <h2 data-number="2.9" class="anchored" data-anchor-id="triangle-inequality"><spa
 <p>The triangle inequality relates the absolute value to the sum operation. It is a very important inequality, which we will use a lot in the future.</p>
 <div id="theorem-triangle-inequality" class="Theorem" title="Triangle inequality">
 <p></p><details class="Theorem fbx-simplebox fbx-default" open=""><summary><strong>Theorem 35: </strong>Triangle inequality</summary><div>For every <span class="math inline">\(x, y \in \mathbb{R}\)</span> we have <span id="eq-triangle-inequality"><span class="math display">\[
-| |x| - |y| | \leq |x-y| \leq |x| + |y| \,.
+| |x| - |y| | \leq |x+y| \leq |x| + |y| \,.
 \tag{2.7}\]</span></span><p></p>
 </div></details>
 </div>
@@ -1071,7 +1071,7 @@ <h2 data-number="2.9" class="anchored" data-anchor-id="triangle-inequality"><spa
 \]</span> We can now again apply Lemma <a href="chap_2.html#lemma-absolute-value">33</a> to get <span id="eq-triangle-inequality-5"><span class="math display">\[
 |x + y| \leq |x| + |y| \,,
 \tag{2.11}\]</span></span> which is the second inequality in (<a href="#eq-triangle-inequality"><span>2.7</span></a>).</p>
-<p><em>Step 2. Proof of the second inequality in (<a href="#eq-triangle-inequality"><span>2.7</span></a>).</em><br>
+<p><em>Step 2. Proof of the first inequality in (<a href="#eq-triangle-inequality"><span>2.7</span></a>).</em><br>
 Note that the trivial identity <span class="math display">\[
 x = x+y - y
 \]</span> always holds. We then have <span class="math display">\[\begin{align}