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wu :: forums    riddles    easy (Moderators: SMQ, Grimbal, william wu, towr, Icarus, ThudnBlunder, Eigenray)    polynomials « Previous topic | Next topic » Pages: 1  Reply Notify of replies Send Topic Print    Author  Topic: polynomials  (Read 992 times) fatball Senior Riddler Can anyone help me think outside the box please?     Gender: Posts: 315 polynomials   « on: Dec 14th, 2010, 7:40am » Quote Modify can you tellme how to expand the following or which theorem to use for that pleae?   (a1+a2+...+a12)^4   urgent please!  Thanks! IP Logged SMQ wu::riddles Moderator Uberpuzzler     Gender: Posts: 2084 Re: polynomials   « Reply #1 on: Dec 14th, 2010, 8:31am » Quote Modify (a1 + a2 + ... + an)4  = a14 + a24 + ... + a124     + 4a13a2 + 4a13a3 + ... + 4an-13an     + 6a12a22 + 6a12a32 + ... + 6an-12an2     + 4a1a23 + 4a1a33 + ... + 4an-1an3     + 12a12a2a3 + 12a12a2a4 + ... + 12an-22an-1an     + 12a1a22a3 + 12a1a22a4 + ... + 12an-2an-12an     + 12a1a2a32 + 12a1a2a42 + ... + 12an-2an-1an2     + 24a1a2a3a4 + 24a1a2a3a5 + ... + 24an-3an-2an-1an   Edit: got it right the second time, as usual...   --SMQ « Last Edit: Dec 14th, 2010, 11:15am by SMQ » IP Logged --SMQ fatball Senior Riddler Can anyone help me think outside the box please?     Gender: Posts: 315 Re: polynomials   « Reply #2 on: Dec 14th, 2010, 8:51am » Quote Modify general formula or theorem to use? IP Logged towr wu::riddles Moderator Uberpuzzler Some people are average, some are just mean.     Gender: Posts: 13730 Re: polynomials   « Reply #3 on: Dec 14th, 2010, 9:58am » Quote Modify You can do it recursively with the binomial theorem, if you're so inclined.   (a+b)n = sum k=0..n C(n,k) an-k * bk   So  (a1 + (a2+..+am) ) =   1* a14 +   4* a13*(a2+..+am) +   6* a12*(a2+..+am)2  +   4* a1*(a2+..+am)3 +   1* (a2+..+am)4       You could also simply expand the formula   Sum(ai)n =   Sum(ai * Sum(ai)n-1)     [edit] I think you're missing a few terms, SMQ. You seem to only have terms with at most two different factors, but there can be up to 4. [/edit] « Last Edit: Dec 14th, 2010, 10:11am by towr » IP Logged Wikipedia, Google, Mathworld, Integer sequence DB SMQ wu::riddles Moderator Uberpuzzler     Gender: Posts: 2084 Re: polynomials   « Reply #4 on: Dec 14th, 2010, 11:16am » Quote Modify on Dec 14th, 2010, 9:58am, towr wrote: I think you're missing a few terms, SMQ. You seem to only have terms with at most two different factors, but there can be up to 4. Right you are -- fixed now.  That's what I get for generalizing from the binomial case in my head...   --SMQ IP Logged --SMQ pex Uberpuzzler     Gender: Posts: 880 Re: polynomials   « Reply #5 on: Dec 14th, 2010, 11:18am » Quote Modify on Dec 14th, 2010, 8:51am, fatball wrote: general formula or theorem to use? It's called a multinomial series. IP Logged Pages: 1  Reply Notify of replies Send Topic Print Forum Jump: ----------------------------- riddles -----------------------------=> easy   - medium   - hard   - what am i   - what happened   - microsoft   - cs   - putnam exam (pure math)   - suggestions, help, and FAQ   - general problem-solving / chatting / whatever ----------------------------- general -----------------------------  - guestbook   - truth   - complex analysis   - wanted   - psychology   - chinese « Previous topic | Next topic »

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