Sjmarf@sh.itjust.works to Math Memes@lemmy.blahaj.zoneEnglish · 4 days agoSimplifysh.itjust.worksimagemessage-square42fedilinkarrow-up1326arrow-down111
arrow-up1315arrow-down1imageSimplifysh.itjust.worksSjmarf@sh.itjust.works to Math Memes@lemmy.blahaj.zoneEnglish · 4 days agomessage-square42fedilink
minus-squareLostXOR@fedia.iolinkfedilinkarrow-up46·4 days agoFun fact, omitting the (x-x) zero term and expanding the entire polynomial, you’d get something with 2^25 = 33,554,432 terms. May be slightly excessive!
minus-squarethreelonmusketeers@sh.itjust.workslinkfedilinkEnglisharrow-up8·3 days agoCouldn’t you combine a lot of like terms as you went along, though? A polynomial of the order x26 would only have 27 terms.
minus-squareLostXOR@fedia.iolinkfedilinkarrow-up2·3 days agoNo, because each coefficient is its own variable; they’re not constants.
minus-squarethreelonmusketeers@sh.itjust.workslinkfedilinkEnglisharrow-up1·edit-23 days agoHuh, I’m so used to polynomials being in the form ax^2 + bx + c that I never considered that every letter might be a variable.
Fun fact, omitting the (x-x) zero term and expanding the entire polynomial, you’d get something with 2^25 = 33,554,432 terms. May be slightly excessive!
Couldn’t you combine a lot of like terms as you went along, though? A polynomial of the order x26 would only have 27 terms.
No, because each coefficient is its own variable; they’re not constants.
Huh, I’m so used to polynomials being in the form ax^2 + bx + c that I never considered that every letter might be a variable.