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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@@ Line 7: / Line 7: @@
 ===Rules of Summation===
-<math>\sum_{i=a}^{b}f(i)+g(i)=\sum_{i=a}^{b}f(i)+\sum_{i=a}^{b}g(i)</math>
+<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>
-<math>\sum_{i=1}^{n} i^3 = \left(\sum_{i=1}^{n} i\right)^2 = \left(\frac{n(n+1)}{2}\right)^2</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>.
-<math> \sum_{i=1}^n i^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}</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>.
-<math> \sum_{i=1}^n i^5=\frac{n^2(n+1)^2(2n^2+2n-1)}{12}</math>
+This method can be generalized nicely; <math>i^n=\sum_{k=1}^n a_k\binom{i}{k}</math>.
-<!--there are others, I forgot what they were. Could someone please fill them in? -->
+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===
@@ Line 29: / Line 31: @@
 <math>\prod_{i=a}^{b}x\cdot y=x^{(b-a+1)}y^{(b-a+1)}</math>
-<!-- same as above, there are others.... but no fancy ones like divergence/convergence, please -->
+These should be self-evident, as above.
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