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ahhh, much better lol
so i wrote down sum (-2+3i/n)(3/n)
good, im reviewing some notes to bring me up to speed
k can you also explain to me the point of what we are doing
the original equation: x^3 -3x can be split into easier parts x^3 and -3x we use the reimann sum formulas for (sum)x^3 and (sum)x to work these to completion
-3 lim(sum) x (3/n) is the easiest one to do, so lets start with that one :)
we determined that f(x_i) = (-2+3i)/n right? so if the notes are correct, and I dont see any reason to dispute them, we multiply: (-2+3i)/n with 3/n
(sum) (-6/n^2 + 9i/n^2)
your right... I got lost there for a sec :)
it says to split this one up into: (sum) -6/n+ (sum)9i/n^2
yeah, I am suuming the adding rule :) it gets us to the (sum) i formula
k i split is up as you said
(9/n^2) (sum)i is... 9 n(n+1) ---- ------ n^2 2
do we substitute that?
9n(n+1) 9(n+1) ------- = ------ 2n^2 2n now work the other part: -6/n -> 1/n (sum) -6 = (1/n)(-6n) = -6 why? dunno, its just the example I got to go by :)
so we are at (-6+(9(n+1))/2
9(n+1) ------ - 6 2n^2 -6n^2 9n 9 ----- + ---- + --- n^2 n^2 n^2
oh you just lost me
you just distributed?
lost myself too, let me redo that last part :)
9n+9 ------ - 6 2n -12n 9n 9 ----- + ---- + --- 2n 2n 2n might be better
-3 9 --- + --- = -3 as n-> inf 2 2n
hold up what?
lim as n goes to infinity sum =-3?
-3/2 ..... might be a better answer for the right side of our original equation....
I forgot the denominator of 2 :)
oh because 9/2n goes to 0?
yep, when a fraction gets bigg on the bottom what happens? 1 --------------------- becomes infinitly small 100000000000000000 it approaches zero
yea i know, sorry stupid question
now if our f(a + i/x\) stuff was correct, then this is our right side part
we are halfway done. Should we take the cube now
ummm...... we need to drag that -3 back over and multiply it I think :) -9/2
-3(-3/2) = 9/2 that s right.... tell me im wrong :) i dont mind
since its just: (sum) x^3 we can take our f^(x_1) and work it thru
yep, and I am assuming that that is correct :) otherwise we are getting good practice, but the wrong results :)
at least we have the answer so we will know
its 3/4 if you forgot
the entire answer is 3/4
yeah yeah...maybe..well see :)
want me to distribute the cube?
yep, cube it out so we can multiply our 3/n to it and apply the (sum) i^3 formula to it
k give me a second
gonna swig some coffee..... bad mormon....BAD!!
k here it is non simplified: ( -8+12i/n+12i/n-18i^2/n^2+12i/n-18i^2/n^2-18i^2/n^2+27i^3/n^3)(3/n)
simplified loll (-8+36i/n-54i^2/n^2+27i^3/n^3)(3/n)
distributing the 3/n : (-24/n+108i/n^2-162i^2/n^3+81i^3/n^4)
made me do all the dirty work ahhaah
thats what I get too :)
now lets divvy put the "i" factors
now we take each summation?
what does -24/n become
do i just pull it out?
sum_-24/n +108/n^2 sum_i -162/n^3sum_i^2 +81/n^4 sum_i^3
thats what i got
muplitiply each one by the appropriate sum i^x formulas :)
i have to go soon, are we almost done?
not even close :) I can do it quicker on the paper probably.... how long you got?
can you attach the image after?
aint gonna happen in 5 minutes :) I can work it out and post it here if youd like, might even figure out where I go wrong and learn something :)
sounds good, that would be great. Sorry that i have to go.
'sok, have fun....:)
for that first one i got (108n^2+108)/2n^2
ill come back later to see if you posted and if you are still online, i may ask you questions
take care, you are a Hero for sure! haha
Hey Im back, are you there? Its kinda late but did you finish?
Wow amazing! You broke everything down
You really helped me. I can't thank you enough