A=B (symbolic summation algorithms) by Wilf, Zeilberger.

By Wilf, Zeilberger.

Show description

Read Online or Download A=B (symbolic summation algorithms) PDF

Best mathematics books

The language of mathematics : making the invisible visible

"The nice booklet of nature," stated Galileo, "can be learn merely by way of those that understand the language within which it used to be written. And this language is arithmetic. " within the Language of arithmetic, award-winning writer Keith Devlin finds the important position arithmetic performs in our everlasting quest to appreciate who we're and the realm we are living in.

Springer-Handbuch der Mathematik IV: Begründet von I.N. Bronstein und K.A. Semendjaew Weitergeführt von G. Grosche, V. Ziegler und D. Ziegler Herausgegeben von E. Zeidler

Als mehrbändiges Nachschlagewerk ist das Springer-Handbuch der Mathematik in erster Linie für wissenschaftliche Bibliotheken, akademische Institutionen und Firmen sowie interessierte Individualkunden in Forschung und Lehre gedacht. Es ergänzt das einbändige themenumfassende Springer-Taschenbuch der Mathematik (ehemaliger Titel Teubner-Taschenbuch der Mathematik), das sich in seiner begrenzten Stoffauswahl  besonders an Studierende richtet.

Extra resources for A=B (symbolic summation algorithms)

Example text

2 Identities 21 identity are equal, let’s prove that their ratio is 1 (that will be a frequent tactic in this book). Not only that, we’ll prove that the ratio is 1 by differentiating it and getting 0 (another common tactic here). So define the function F (x, y) = e(x + y)e(−x)e(−y). By direct differentiation we find that Dx F = Dy F = 0. Thus F is constant. Set x = y = 0 to find that the constant is 1. Thus e(x + y)e(−x)e(−y) = 1 for all x, y. Now let y = 0 to find that e(−x) = 1/e(x). Thus e(x + y) = e(x)e(y) for all x, y, as claimed.

To begin, let’s try to simplify some expressions that contain factorials. /n! then what we get back is (1 + n)! , n! which doesn’t help too much. ] then we also get back (1 + n)! , n! so we must be doing something wrong. Well, it turns out that if you would really like to simplify ratios of factorials then the thing to do is to read in the package RSolve, because in that package there lives a command FactorialSimplify, which does the simplification that you would like to see. Out[2] = 28 Tightening the Target So let’s start over, this time with In[1] :=<< DiscreteMath‘RSolve‘.

3F2 d e a! ( a2 − b)! ( a2 − c)! (a − b − c)! (V) Clausen’s 4 F3 identity. If d is a nonpositive integer and a + b + c − d = 12 , and e = a + b + 12 , and a + f = d + 1 = b + g, then 4F3 (2a)|d| (a + b)|d| (2b)|d| a b c d ;1 = . e f g (2a + 2b)|d| a|d| b|d| (VI) Dougall’s 7 F6 identity. If n + 2a1 + 1 = a2 + a3 + a4 + a5 and a1 a6 = 1 + ; a7 = −n; and bi = 1 + a1 − ai+1 (i = 1, . . , 6), 2 then 7 F6 a1 a2 a3 a4 a5 a6 a7 ;1 b1 b2 b3 b4 b5 b6 (a1 + 1)n (a1 − a2 − a3 + 1)n (a1 − a2 − a4 + 1)n (a1 − a3 − a4 + 1)n = .

Download PDF sample

Rated 4.90 of 5 – based on 44 votes