Difference between revisions of "User:Temperal/The Problem Solver's Resource3"

Latest revision as of 22:18, 10 January 2009

Introduction \| Other Tips and Tricks \| Methods of Proof \| You are currently viewing page 3.

Summations and Products

Definitions

Summations: $\sum_{i=a}^{b}c_i=c_a+c_{a+1}+c_{a+2}...+c_{b-1}+c_{b}$
Products: $\prod_{i=a}^{b}c_i=c_a\cdot c_{a+1}\cdot c_{a+2}...\cdot c_{b-1}\cdot c_{b}$

Rules of Summation

$\sum_{i=a}^{b}f_1(i)+f_2(i)+\ldots f_n(i)=\sum_{i=a}^{b}f(i)+\sum_{i=a}^{b}f_2(i)+\ldots+\sum_{i=a}^{b}f_n(i)$

$\sum_{i=a}^{b}c\cdot f(i)=c\cdot \sum_{i=a}^{b}f(i)$

$\sum_{i=1}^{n} i= \frac{n(n+1)}{2}$ , and in general $\sum_{i=a}^{b} i= \frac{(b-a+1)(a+b)}{2}$

The above should all be self-evident and provable by the reader within seconds.

$\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$

Derivation: We write $n^2$ as $a_1\binom{n}{1}+a_2\binom{n}{2}$ . Substituting n=1 gives $a_1=1$ while substituting n=2 gives $a_2=2$ . Hence, $n^2=\binom{n}{1}+2\binom{n}{2}$ .

Now, $\sum_{i=1}^{n} i^2=\sum_{i=1}^n (\binom{i}{1}+2\binom{i}{2})=\sum_{i=1}^n \binom{i}{1}+2\sum_{i=1}^n \binom{i}2=\binom{n+1}{2}+2\binom{n+1}{3}$ , where we use the Hockey-Stick Identity. After some algebra, this comes out to $\frac{(n)(n+1)(2n+1)}{6}$ .

This method can be generalized nicely; $i^n=\sum_{k=1}^n a_k\binom{i}{k}$ .

Particularly notable is the case $n=3$ ; we get $\sum_{i=1}^{n} i^3 = \left(\sum_{i=1}^{n} i\right)^2 = \left(\frac{n(n+1)}{2}\right)^2$ . The reader can figure this out themselves.

Rules of Products

$\prod_{i=a}^{b}x=x^{(b-a+1)}$

$\prod_{i=a}^{b}x\cdot y=x^{(b-a+1)}y^{(b-a+1)}$

These should be self-evident, as above.

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-{| style='background:lime;border-width: 5px;border-color: limegreen;border-style: outset;opacity: 0.8;width:840px;height:300px;position:relative;top:10px;'
-|+ <span style="background:aqua; border:1px solid black; opacity: 0.6;font-size:30px;position:relative;bottom:8px;border-width: 5px;border-color:blue;border-style: groove;position:absolute;top:50px;right:155px;width:820px;height:40px;padding:5px;">The Problem Solver's Resource</span>
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-| style="background:lime; border:1px solid black;height:200px;padding:10px;" | {{User:Temperal/testtemplate|page 3}}
 ==<span style="font-size:20px; color: blue;">Summations and Products</span>==
-<!-- will fill in soon, if someone else could in the meantime, it would be appreciated. -->
+===Definitions===
+*Summations: <math>\sum_{i=a}^{b}c_i=c_a+c_{a+1}+c_{a+2}...+c_{b-1}+c_{b}</math>
+*Products: <math>\prod_{i=a}^{b}c_i=c_a\cdot c_{a+1}\cdot c_{a+2}...\cdot c_{b-1}\cdot c_{b}</math>
+===Rules of Summation===
+<math>\sum_{i=a}^{b}f_1(i)+f_2(i)+\ldots f_n(i)=\sum_{i=a}^{b}f(i)+\sum_{i=a}^{b}f_2(i)+\ldots+\sum_{i=a}^{b}f_n(i)</math>
+<math>\sum_{i=a}^{b}c\cdot f(i)=c\cdot \sum_{i=a}^{b}f(i)</math>
+<math>\sum_{i=1}^{n} i= \frac{n(n+1)}{2}</math>, and in general <math>\sum_{i=a}^{b} i= \frac{(b-a+1)(a+b)}{2}</math>
+The above should all be self-evident and provable by the reader within seconds.
+<math>\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}</math>
+Derivation: We write <math>n^2</math> as <math>a_1\binom{n}{1}+a_2\binom{n}{2}</math>. Substituting n=1 gives <math>a_1=1</math> while substituting n=2 gives <math>a_2=2</math>. Hence, <math>n^2=\binom{n}{1}+2\binom{n}{2}</math>.
+Now, <math>\sum_{i=1}^{n} i^2=\sum_{i=1}^n (\binom{i}{1}+2\binom{i}{2})=\sum_{i=1}^n \binom{i}{1}+2\sum_{i=1}^n \binom{i}2=\binom{n+1}{2}+2\binom{n+1}{3}</math>, where we use the [[Hockey-Stick Identity]]. After some algebra, this comes out to <math>\frac{(n)(n+1)(2n+1)}{6}</math>.
+This method can be generalized nicely; <math>i^n=\sum_{k=1}^n a_k\binom{i}{k}</math>.
+Particularly notable is the case <math>n=3</math>; we get <math>\sum_{i=1}^{n} i^3 = \left(\sum_{i=1}^{n} i\right)^2 = \left(\frac{n(n+1)}{2}\right)^2</math>. The reader can figure this out themselves.
+===Rules of Products===
+<math>\prod_{i=a}^{b}x=x^{(b-a+1)}</math>
+<math>\prod_{i=a}^{b}x\cdot y=x^{(b-a+1)}y^{(b-a+1)}</math>
+These should be self-evident, as above.
 [[User:Temperal/The Problem Solver's Resource2|Back to page 2]] | [[User:Temperal/The Problem Solver's Resource4|Continue to page 4]]
-|} <br /><br />

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