Hi, iS THERE Any one out there who can provide an expression for the sum of n terms: i*X* COS i*X where X is in radians, and i varies from 1 to n? Thanks in advance for your help. Cheers
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Dear Almo, you can put out the x front of a bracket, I mean: sum i*x*cos(i*x) = x*sum i*cos(i*x) It can be interesting eventuell the derivation of sin(i*x) is.
Originally Posted by Almo Hi, iS THERE Any one out there who can provide an expression for the sum of n terms: i*X* COS i*X where X is in radians, and i varies from 1 to n? Thanks in advance for your help. Cheers I take it that in this case is your counter, not the imaginary number ? So you're asking what does equal right?
G'day Many thanks for your interest. Yes you are correct that i is NOT an imaginary number i.e. sqrt(-1). It is as you say the sum of the terms whereby i is initialy 1 then increases by 1 to the limit n.
How about this use the following lemma So your sum of and apply the above lemma with
Thanks for your suggestion. I'm afraid I don't understand if your last line is my desired answer. Would you please rephrase your solution in terms of my i and X? Cheers
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