Assuming this sequence in alpha-sequential and there is an (x-x) in there, then the product of the binomials is 0.Perhaps this is a good way to teach a student to think before diving into a solution to a problem....
what is the product of the binomials
(x-a)(x-b)(x-c)(x-d)(x-e)(x-f)(x-g)(x-h)(x-i)(x-j)(x-k)......(x-y)(x-z)
??
i must admit i got foxed by this one, since Mr chesart1 wisely made the (x-x) inconspicuous.what is the product of the binomials
(x-a)(x-b)(x-c)(x-d)(x-e)(x-f)(x-g)(x-h)(x-i)(x-j)(x-k)......(x-y)(x-z)
??
Yes very sneaky!i must admit i got foxed by this one, since Mr chesart1 wisely made the (x-x) inconspicuous.
26.Yes very sneaky!
Here is a question: Removing the (x-x) binomial, how many sum of products are there when the brackets are multiplied out?
Dave
Is it 26? The order of the sum of products term will be 26, but will it have 26 product terms. Consider:26.
the answer is quite simple isnt it?..but i wont give away any hint now(will edit later perhaps)
i'll be pretty embarrassed if at all i missed the correct answer.
Whether x is constant, variable, real, complex or other x-x = 0.one reason i got tricked was i assumed all of the terms as constants and the 'x' as variable (and was under the impression any other alphabet not shown represent a constant without thinking about the hidden x)
yes but i thought one x will be variable and the other constant thats where i went wrong.Whether x is constant, variable, real, complex or other x-x = 0.
for a quadratic we have three terms,So I go for 27 - nearly the same!
sure is!Its all fun isn't it!
Ok, I see. Anyone using the same characters for two different variables needs detention!yes but i thought one x will be variable and the other constant thats where i went wrong.
D'oh! How dumb am I - and I set the question!for a quadratic we have three terms,
therefore for polynomial of order 25(26-1,since x-x is absent) we should have 26 terms right?....did i go wrong somewhere?
sure is!
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