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

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==<span style="font-size:20px; color: blue;">Summations and Products</span>==
 
==<span style="font-size:20px; color: blue;">Summations and Products</span>==
 
===Definitions===
 
===Definitions===
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===Rules of Summation===
 
===Rules of Summation===
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<math>\sum_{i=a}^{b}f(i)+g(i)=\sum_{i=a}^{b}f(i)+\sum_{i=a}^{b}g(i)</math>
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<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=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>
 
<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>
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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^2 = \frac{n(n+1)(2n+1)}{6}</math>
  
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<math>\sum_{i=1}^{n} i^3 = \left(\sum_{i=1}^{n} i\right)^2 = \left(\frac{n(n+1)}{2}\right)^2</math>
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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>.
  
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<math> \sum_{i\equal{}1}^n i^4\equal{}\frac{n(n\plus{}1)(2n\plus{}1)(3n^2\plus{}3n\minus{}1)}{30}</math>
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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>.
  
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<math> \sum_{i\equal{}1}^n i^5\equal{}\frac{n^2(n\plus{}1)^2(2n^2\plus{}2n\minus{}1)}{12}</math>
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This method can be generalized nicely; <math>i^n=\sum_{k=1}^n a_k\binom{i}{k}</math>.
  
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<!--there are others, I forgot what they were. Could someone please fill them in? -->
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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===
 
===Rules of Products===
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<math>\prod_{i=a}^{b}x\cdot y=x^{(b-a+1)}y^{(b-a+1)}</math>
 
<math>\prod_{i=a}^{b}x\cdot y=x^{(b-a+1)}y^{(b-a+1)}</math>
  
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<!-- same as above, there are others.... but no fancy ones like divergence/convergence, please -->
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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]]
 
[[User:Temperal/The Problem Solver's Resource2|Back to page 2]] | [[User:Temperal/The Problem Solver's Resource4|Continue to page 4]]
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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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