FamousSeries < MSC

MSC Web>MscCapsules>InfiniteSeriesSynopsis>FamousSeries (2008-11-18, DickFurnas) (raw view)EditAttach
---+ Famous Series
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---++ Geometric Series
<latex>1 + x + x^2 + x^3 + ...</latex>

<latex>\left \{ \begin{array}{l} \mbox{ \small $=\frac{1}{1-x}$ if $|x| < 1$ } \\ \mbox{ \small diverges if $|x| > 1$ } \end{array} \right. </latex>

<latex>\mbox{ \small $=\sum_{i=0}^\infty x^i$ }</latex>
---++ P-Series
<latex>1+\frac{1}{2^p} + \frac{1}{3^p} + \frac{1}{4^p} + \frac{1}{5^p} + ...</latex>

<latex>\mbox{ \small diverges for $p\le 1$ }</latex>

<latex>\mbox{ \small $=\sum \frac{1}{n^p}$ }</latex>
---++ Harmonic Series
<latex>1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ...</latex>
   * diverges
<latex>\mbox{ \small $=\sum_{1}^\infty \frac{1}{n}$ }</latex>
   * this is a special case of the P-Series for P=1
---++ Alternating Harmonic Series
<latex>1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - ...</latex>
   * converges to _ln(1+1) = ln(2)_ using series for _ln(1+x)_ below.
<latex>\mbox{ \small $= \sum_{n=1}^\infty (-1)^n\frac{1}{n}$ }</latex>
---++ Exponential
<latex>1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + ...</latex>

<latex>\mbox{ \small $= e^x$ }</latex>

<latex>\mbox{ \small $= \sum_{n=0}^\infty \frac{x^n}{n!}$ }</latex>

<latex>\mbox{ \small $e^x = \cos{ix} + i\sin{ix}$ } </latex> where <latex>\mbox{ \small $i=\sqrt{-1}$ }</latex>
---++ Sin
<latex>x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + ...</latex>

<latex>\mbox{ \small $=\sin{x}$ }</latex>

<latex>\mbox{ \small $=  \sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}$ }</latex>
---++ Cos
<latex>1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + ...</latex>
   * <latex>\mbox{ \small $= \cos{x}$ }</latex>
<latex>\mbox{ \small $= \sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{(2n)!}$ }</latex>
---++ _ln_ (1+x)
<latex>x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ...</latex>

<latex>\mbox{ \small $= \ln{(1+x)} = \sum_{n=1}^\infty (-1)^{n+1}\frac{x^{n}}{n}$ }</latex>
   * you can arrive at this relation by integrating a _Geometric Series_  in _-t_ term-by-term.
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<latex>\mbox{ \small $ln{(1+x)} = \int{\frac{1}{1+t}}dt$ }</latex>

<latex>\mbox{ \small $= \int{\frac{1}{1-(-t)}}dt$ }</latex>

<latex>\mbox{ \small $= \int{1 + (-t) + (-t)^2 + (-t)^3 + ...} dt$ }</latex>

<latex>\mbox{ \small $= \int{1 -t + t^2 - t^3 + ...} dt$ }</latex>

<latex>\mbox{ \small $= x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ...$ }</latex>

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   * if _x=-1_ , i.e. _ln(1+(-1)) = ln(0)_  , this is the negative of the Harmonic Series which diverges toward -&infin;
---++ _arctan_ (x)
<latex>x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + ...</latex>

<latex>\mbox{ \small $= \arctan{x}$ }</latex>

<latex>\mbox{ \small $= \sum_{n=0}^\infty (-1)^{n}\frac{x^{2n+1}}{2n+1}$ }</latex>
   * you can arrive at this relation by integrating a _Geometric Series_  in _-t<sup>2</sup>_ term-by-term.
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<latex>\mbox{ \small $\arctan{(1+x)} = \int{\frac{1}{1+t^2}}dt$ }</latex>

<latex>\mbox{ \small $= \int{\frac{1}{1-(-t^2)}}dt$ }</latex>

<latex>\mbox{ \small $= \int{1 + (-t^2) + (-t^2)^2 + (-t^2)^3 + ...} dt$ }</latex>

<latex>\mbox{ \small $= \int{1 -t^2 + t^4 - t^6 + ...} dt$ }</latex>

<latex>\mbox{ \small $= x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + ...$ }</latex>

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-- Main.DickFurnas - 16 Nov 2008