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| =Knuth: Volume 1: Fundamental Algorithms=
| | #REDIRECT [[AOCP/Generating Functions]] |
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| ==Chapter 1: Basic Concepts: Generating Functions==
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| You absolutely, positively must read the preceding chapters to follow Knuth here. There are a number of subtle steps that he moves through quite fast, about as quickly as he moves through the other material, except he's now integrating his method of presenting all 8 prior sections, including both explicit and passing references to prior material.
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| As with the binomial coefficients...
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| =Example Problems=
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| ==Knuth AOCP Section 1.2.9 Exercise 1==
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| '''Exercise 1.''' What is the generating function for <math>\langle2^n + 3^n\rangle = \{ 2, 5, 13, 35, \dots \}</math>?
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| For this problem, don't be fooled by the complexity of the preceding section - it's back to the basics. This sequence of numbers looks a lot like, oh, I dunno, two sequences of numbers added together. Specifically,
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| <pre>
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| 2 4 8 16 32
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| 3 9 27 81 243
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| </pre>
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| If we can find the generating function for 2^n, and find the generating function for 3^n, we can use the addition property of generating functions to combine them.
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| To find the generating function for <math>2^n</math>:
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| <math>
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| \begin{align}
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| G_{2^n}(z) &=& 1 + 2z + 4z^2 + \dots \\
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| z G_{2^n}(z) &=& 0 + z + 2z^2 + \dots \\
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| 2z G_{2^n}(z) &=& 0 + 2z + 4z^2 + \dots
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| \end{align}
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| </math>
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| Now subtracting the first and last quantities gives:
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| <math>
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| (1-2z) G_{2^n}(z) = 1 + 0 + 0 + 0 + \dots
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| </math>
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| which yields the generating function for <math>2^n</math>:
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| <math>
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| G_{2^n}(z) = \dfrac{1}{1 - 2z}
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| </math>
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| Repeating the process for <math>3^n</math> gives:
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| <math>
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| G_{3^n}(z) = \dfrac{1}{1 - 3z}
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| </math>
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| Now to get the overall generating function for <math>\langle 2^n + 3^n \rangle</math> we add these two functions together and combine into a single rational expression:
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| <math>
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| G(z) = \dfrac{1}{1-2z} + \dfrac{1}{1 - 3z} = \dfrac{2-5x}{1-5x+6x^2}
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| </math>
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| Now for the cool part. If we do a Taylor Series expansion of <math>G(z)</math>, we can compute <math>2^n + 3^n</math> simply by obtaining the <math>n^{th}</math> coefficient of the Taylor Series polynomial. While this may seem like a pointless mathematical trick for a function like <math>2^n + 3^n</math>, the real utility of the method comes when we apply this approach to much more general functions.
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| [[Image:MathematicaGeneratingFunction1.png|500px]] | |
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| If we evaluate the 500th term of the Taylor Series expansion, which we can denote <math>G_n(z)[500]</math>, we get:
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| <math>
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| \begin{align}
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| G_n(z)[500] = 3636029179586993684238526707954331911802338502600162304034603583258060 \\
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| 0191583895484198511536369996679450049715724100683352008148159059109207 \\
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| 7819142079301219138181759224025267202176687407229640966567179316611617 \\
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| 95984554271383193123905199377
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| \end{align}
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| </math>
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| which, in case you did not already notice, is <math>2^{500} + 3^{500}</math>
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| =Flags=
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| {{AOCPFlag}}
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| [[Category:Generating Functions]]
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| [[Category:Combinatorics]]
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| [[Category:Math]]
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