Reimann Sum Continues.

- anonymous

Reimann Sum Continues.

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- amistre64

ahhh, much better lol

- anonymous

lol

- anonymous

so i wrote down sum (-2+3i/n)(3/n)

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## More answers

- amistre64

good, im reviewing some notes to bring me up to speed

- anonymous

k can you also explain to me the point of what we are doing

- amistre64

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

- amistre64

-3 lim(sum) x (3/n) is the easiest one to do, so lets start with that one :)

- amistre64

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

- anonymous

(-2+3i/n)(3/n)

- amistre64

(sum) (-6/n^2 + 9i/n^2)

- anonymous

so -6/n+9i/n

- anonymous

9i/n^2

- amistre64

your right... I got lost there for a sec :)

- amistre64

it says to split this one up into:
(sum) -6/n+ (sum)9i/n^2

- anonymous

why?

- anonymous

adding rule?

- amistre64

yeah, I am suuming the adding rule :) it gets us to the (sum) i formula

- anonymous

k i split is up as you said

- amistre64

(9/n^2) (sum)i is...
9 n(n+1)
---- ------
n^2 2

- anonymous

ok

- anonymous

do we substitute that?

- amistre64

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 :)

- anonymous

k haha

- anonymous

so we are at (-6+(9(n+1))/2

- amistre64

9(n+1)
------ - 6
2n^2
-6n^2 9n 9
----- + ---- + ---
n^2 n^2 n^2

- anonymous

oh you just lost me

- anonymous

you just distributed?

- amistre64

lost myself too, let me redo that last part :)

- amistre64

9n+9
------ - 6
2n
-12n 9n 9
----- + ---- + ---
2n 2n 2n
might be better

- amistre64

-3 9
--- + --- = -3 as n-> inf
2 2n

- anonymous

yup

- anonymous

hold up what?

- anonymous

lim as n goes to infinity sum =-3?

- amistre64

-3/2 ..... might be a better answer for the right side of our original equation....

- amistre64

I forgot the denominator of 2 :)

- anonymous

oh because 9/2n goes to 0?

- amistre64

yep, when a fraction gets bigg on the bottom what happens?
1
--------------------- becomes infinitly small
100000000000000000 it approaches zero

- amistre64

.0000000000.....00000000000000001

- anonymous

yea i know, sorry stupid question

- amistre64

now if our f(a + i/x\) stuff was correct, then this is our right side part

- anonymous

we are halfway done. Should we take the cube now

- amistre64

ummm...... we need to drag that -3 back over and multiply it I think :)
-9/2

- anonymous

cool 9/2

- anonymous

not negative

- amistre64

-3(-3/2) = 9/2 that s right.... tell me im wrong :) i dont mind

- amistre64

since its just: (sum) x^3 we can take our f^(x_1) and work it thru

- anonymous

so (-2+3i/n)^3?

- amistre64

yep, and I am assuming that that is correct :)
otherwise we are getting good practice, but the wrong results :)

- anonymous

at least we have the answer so we will know

- anonymous

its 3/4 if you forgot

- amistre64

whaaa!!????

- anonymous

the entire answer is 3/4

- amistre64

yeah yeah...maybe..well see :)

- anonymous

want me to distribute the cube?

- amistre64

yep, cube it out so we can multiply our 3/n to it and apply the (sum) i^3 formula to it

- anonymous

k give me a second

- amistre64

gonna swig some coffee..... bad mormon....BAD!!

- anonymous

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)

- anonymous

simplified loll (-8+36i/n-54i^2/n^2+27i^3/n^3)(3/n)

- anonymous

distributing the 3/n : (-24/n+108i/n^2-162i^2/n^3+81i^3/n^4)

- anonymous

made me do all the dirty work ahhaah

- amistre64

thats what I get too :)

- anonymous

sweet

- amistre64

now lets divvy put the "i" factors

- anonymous

now we take each summation?

- amistre64

yupp

- anonymous

what does -24/n become

- anonymous

do i just pull it out?

- amistre64

sum_-24/n
+108/n^2 sum_i
-162/n^3sum_i^2
+81/n^4 sum_i^3

- anonymous

thats what i got

- amistre64

muplitiply each one by the appropriate sum i^x formulas :)

- anonymous

i have to go soon, are we almost done?

- amistre64

not even close :) I can do it quicker on the paper probably....
how long you got?

- anonymous

5 minutes

- anonymous

can you attach the image after?

- amistre64

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 :)

- anonymous

sounds good, that would be great. Sorry that i have to go.

- amistre64

'sok, have fun....:)

- anonymous

for that first one i got (108n^2+108)/2n^2

- anonymous

ill come back later to see if you posted and if you are still online, i may ask you questions

- anonymous

take care, you are a Hero for sure! haha

- anonymous

Hey Im back, are you there? Its kinda late but did you finish?

- amistre64

Here it is

##### 1 Attachment

- anonymous

Wow amazing! You broke everything down

- anonymous

You really helped me. I can't thank you enough

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