Julius O
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Julius O >-Fri, 12 Jul 2019 17:35:43 -0400Proof by Induction for the Sum of Squares Formula
http://localhost:4000/2019/07/11/proof-by-induction-for-the-sum-of-squares-formula.html
<h2>Problem</h2>
<p>
Use induction to prove that
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<span class="marginnote">
Sidenotes here and inside the proof will provide commentary, in addition to numbering each step of the proof-building process for easy reference. They are not part of the proof itself, and must be omitted when written.
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</p>
<figure>
$$\sum_{k=0}^{n} k^{2} = \dfrac{n(n+1)(2n+1)}{6}$$
</figure>
<p>
for all \(n \ge 0\).
</p>
<hr>
<p>NOTE: \(k\) is a placeholder variable representing each number in a series starting from \(0\), and up to and including \(n\). Shove the current value of \(n\) into \(k\), evaluate, and do the same for the next value of \(n\), etc. The "E"-shaped figure is an uppercase Sigma from the Greek alphabet. Because Sigma's spelled with an "s", it's used to represent a <strong>s</strong>ummation, or <strong>s</strong>um. The left-hand side (LHS) reads, <em>"the sum from k equals zero to n of k squared."</em></p>
<hr>
<p>
<strong>1. Basis step</strong>
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<span class="marginnote">
Since the formula claims to work for all numbers greater than or equal to (\(\ge\)) \(0\), \(0\) must be tested on both sides. The series on the LHS states to start at \(0\), square \(0\), and stop. The RHS is simply plug and chug.
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</p>
<p>For \(n=0\), the left-hand side (LHS) yields:</p>
<figure>
$$\sum_{k=0}^{0} k^{2} = 0^{2} = 0.$$
</figure>
<p>And the right-hand side (RHS) yields:</p>
<figure>
$$\dfrac{0(0+1)(2 \cdot 0 + 1)}{6}=0.$$
</figure>
<p>So the equation holds on both sides for \(n=0\).</p>
<p>
<strong>2. Assume the result for \(n\)</strong>.
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<span class="marginnote">
With the Basis step verified in Step 1, we assume the result to be true for \(n\), and restate the original problem.
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</p>
<figure>
$$\sum_{k=0}^{n} k^{2} = \dfrac{n(n+1)(2n+1)}{6}.$$
</figure>
<p>
<strong>3. Prove the result for \((n+1)\)</strong>.
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<span class="marginnote">
<em>"If you can show that the next thing exists in the current thing, then you've proved that the current thing works for all subsequent things."</em>
</span>
</p>
<figure>
\begin{align*}
\sum_{k=0}^{n} k^{2} + (n+1)^{2}
&= \dfrac{(n+1)((n+1)+1)(2(n+1)+1)}{6}\\
&= \dfrac{(n+1)(n+2)(2n+2+1)}{6}\\
&= \dfrac{(n+1)(n+2)(2n+3)}{6}.
\end{align*}
</figure>
<hr>
<p>NOTE: Recall that the sum of squares formula on the LHS starts at \(0\), ends after the n-th value is squared and added, and is supposedly equivalent to the algebra on the RHS. For the LHS and RHS to stay equal to each other, any changes made to one side must also be made to the other side. Adding, subtracting, multiplying, or dividing a number on the LHS, must also be done to the opposite side to preserve equality. Observe the last number in the sum of squares formula is \(n\), the "current thing." Since \(n\) can be any number we want that's \(\ge 0\), what would be the NEXT number after \(n\)?</p>
<p>That number would be \(\mathbf{(n+1)}\), or the "next thing" we'll try to coax out from the "current thing." And since we need to square the next number prior to adding it to the series, we'll have to add \(\mathbf{(n+1)^{2}}\) to both sides of the equation. But instead of tacking on \(\mathbf{(n+1)^{2}}\) to the RHS (we'll do this in the next step, the proof itself) as we did to the LHS, we're going to use the RHS of Step 2, and carefully insert and replace all instances of \(n\) with \(n+1\), and simplify. Since the formula claims to work for all \(n \ge 0\), then it should also work for \(n+1\), the next number after the n-th number. </p>
<p>So the proof in the next step must show that the result is equal to</p>
<figure>
\begin{align*}
\dfrac{(n+1)(n+2)(2n+3)}{6},
\end{align*}
</figure>
<p>or the proof fails.</p>
<hr>
<p>
<strong>4. Proof</strong>
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<span class="marginnote">
Each line must be justified, beginning with the Induction Hypothesis, or what we're trying to prove. The rest is algebra and simplification. Notice in the next to last line of the proof, the result must look the same as what was found in Step 3. If pressed for time on a test and can't factor the cubic polynomial quickly, just go back to Step 3, F.O.I.L. and multiply out the terms on scratchpaper to see if the cubic polynomial is the same as what you got in Step 4. If they're equal, then you've got it, and can finish the proof with writing the cubic polynomial in factored form.
</span>
</p>
<figure>
\begin{align*}
\sum_{k=0}^{n} k^{2} + (n+1)^{2}
&= \dfrac{n(n+1)(2n+1)}{6} + (n+1)^{2} && \text{(Induction Hypothesis)}\\
&= \dfrac{n(n+1)(2n+1)+6(n+1)^{2}}{6} && \text{(Algebra)}\\
&= \dfrac{(n^{2} + n)(2n+1)+6(n+1)(n+1)}{6} && \text{(Algebra)}\\
&= \dfrac{2n^{3} + n^{2} + 2n^{2} + n + 6n^{2} + 12n + 6}{6} && \text{(Algebra)}\\
&= \dfrac{2n^{3} + 9n^{2} + 13n + 6}{6} && \text{(Algebra)}\\
&= \dfrac{(n+1)(n+2)(2n+3)}{6}. && \text{(Algebra)}
\end{align*}
</figure>
<p><strong>5. Conclusion</strong></p>
<p>
This proves the result for \((n+1)\), so the result is true for all \(n \ge 0\) by induction. \(\Box\)
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The proof is finished with a concluding statement.
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</p>
<hr>
<p>NOTE: Focusing on the algebra, what induction shows is that inserting the "next thing" into the "current thing" of the formula's RHS representation is equivalent to adding the square of the "next thing" to the "current thing." Amazing!</p>
<figure>
$$\dfrac{(n+1)(n+2)(2n+3)}{6} = \dfrac{n(n+1)(2n+1)}{6} + (n+1)^{2}.$$
</figure>
<hr>
Thu, 11 Jul 2019 15:55:00 -0400
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Induction
http://localhost:4000/2019/07/05/induction.html
<p>Consider the formula for computing the sum of squares:</p>
<figure>
$$\sum_{k=0}^{n} k^{2} = \dfrac{n(n+1)(2n+1)}{6}.$$
</figure>
<p>The formula says there are two ways to find a solution. The conventional way, shown by the left-hand side (LHS) of the equation, says that starting from 0, square 0, then add to it the square of 1, then add to it the square of 2, and continue adding subsequent squares to the series, finally stopping after the n-th number is squared and added. Then just add all the terms in the series together and get an answer.</p>
<p>For example, in calculating the sum of the squares up to and including 10, the LHS method is applied to yield:</p>
<figure>
$$0^{2} + 1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2} + 7^{2} + 8^{2} + 9^{2} + 10^{2} = 385.$$
</figure>
<p>Now the right-hand side (RHS) claims to get the same answer as the LHS simply by substituting the highest term (\(10\) in for \(n\)), and evaluating the expression. With skepticism, substituting and evaluating the RHS yields:</p>
<figure>
$$\dfrac{10(10+1)(2 \cdot 10 + 1)}{6} = \dfrac{10 \cdot 11 \cdot 21}{6} = \dfrac{2310}{6} = 385.$$
</figure>
<p>So while the formula appears to work for a specific example, does it also work with other numbers? Can we be absolutely sure it would work for the first thousand, million, billion, or even trillion square numbers? A reasonable idea would be to keep a continuously expanding book of calculations in the library that we could look up to be sure the formula works for a particular example. Just have a computer continuously calculate the sums of the squares using the LHS and RHS methods, and check to see if both sides equal each other. Surely, performing continuous calculations using larger and larger numbers provides more than ample evidence in proving that the formula is true and universal for other numbers!</p>
<p><span class="marginnote">In 1910, mathematicians Alfred North Whitehead and Bertrand Russell published <a href="https://en.wikipedia.org/wiki/Principia_Mathematica"><em>Principia Mathematica</em></a> to build and prove mathematical foundations from first principles. It took nearly 400 pages of symbolic logic to prove that \(1+1=2\).</span>It turns out that in math, <em>specific examples can't be used as proofs</em>. In writing proofs, only variables, symbols, and previously proven ideas such as axioms, corollaries, lemmas, and theorems can be used as building blocks to show the truth or falsity of a statement. So how do we prove this? The good news is we have a powerful tool at our disposal which will rigorously prove and answer this question.</p>
<p>That tool, is <strong><a href="https://en.wikipedia.org/wiki/Mathematical_induction">INDUCTION</a></strong>. And here's my take on what it means:</p>
<blockquote>
<p><em>"If you can show that the next thing exists in the current thing, then you've proved that the current thing works for all subsequent things."</em></p>
</blockquote>
<p>In the next post, we'll see induction in action, and apply it in proving the sum of squares formula.</p>
Fri, 05 Jul 2019 23:41:00 -0400
http://localhost:4000/2019/07/05/induction.html