Talk:Binomial theorem: Difference between revisions
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imported>Anthony Argyriou (checklist) |
imported>David E. Volk (suggest change in formula for clarity) |
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status is really about 2.5 - 2 for the elementary binomial theorem/formula, and 3 for the Newtonian. [[User:Anthony Argyriou|Anthony Argyriou]] 17:23, 18 July 2007 (CDT) | status is really about 2.5 - 2 for the elementary binomial theorem/formula, and 3 for the Newtonian. [[User:Anthony Argyriou|Anthony Argyriou]] 17:23, 18 July 2007 (CDT) | ||
While the definition is strictly true, it seems written backwards, in that if you actually do | |||
the sum for (x+y)^2 the answer you get as the equation is written is y^2 + 2xy + x^2. Of course you can rearrange | |||
to get x^2 + 2xy + y^2. Another way to write it would be | |||
x^(n-k)y^(k), in which case you directly get the answers as shown in the examples. | |||
[[User:David E. Volk|David E. Volk]] |
Revision as of 14:44, 25 July 2007
Workgroup category or categories | Mathematics Workgroup [Categories OK] |
Article status | Developing article: beyond a stub, but incomplete |
Underlinked article? | Yes |
Basic cleanup done? | Yes |
Checklist last edited by | Anthony Argyriou 17:23, 18 July 2007 (CDT) |
To learn how to fill out this checklist, please see CZ:The Article Checklist.
status is really about 2.5 - 2 for the elementary binomial theorem/formula, and 3 for the Newtonian. Anthony Argyriou 17:23, 18 July 2007 (CDT)
While the definition is strictly true, it seems written backwards, in that if you actually do
the sum for (x+y)^2 the answer you get as the equation is written is y^2 + 2xy + x^2. Of course you can rearrange
to get x^2 + 2xy + y^2. Another way to write it would be
x^(n-k)y^(k), in which case you directly get the answers as shown in the examples.
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