Question

Turing Test or Zarkon..another question to my Reimann sum problem....

Answer 1

when I simplify that long formula and take it's limit, I get a -1/4 and that's not possible is it??

Answer 2

And how can you do what zarcon suggested, just look at the exponents for the leading n values and know you'll get their product in the numerator? can you do the same to the constants (+1, +1, and -1 to know you'll get a -1 for the final constant in the numerator?

Answer 3

It's quite possible. What is the equation?

Answer 4

Is this what you are evaluating?
\[\int\limits_0^1x^4dx=\lim_{n\to\infty}\sum_{j=1}^nf(\Delta x_j^*)\Delta x=\lim_{n\to\infty}\sum_{j=1}^n\frac{j^4}{n^4}\cdot\frac1n=\lim_{n\to\infty}\sum_{j=1}^n\frac{j^4}{n^5} \]

Answer 5

yep...

Answer 6

\[\lim_{n\to\infty}\frac1{n^5}\sum_{j=1}^\ni^4=\lim_{n\to\infty}\frac1{n^5}{n(n+1)(2n+1)(3n^2 + 3n -1)\over30} \]

Ok so if we multiply the top out we only care about the coefficient of n^5

If we multiply all the crap out what would be the coefficient of n^5

well

what is 1(1)(2)(3) ?

Answer 7

and then is became 1/n^5 n(n+1)(2n+1)(3n^2+3n-1) all /3

Answer 8

the the coefficient of n^5 on bottom is 30

Answer 9

so we have 6/30 right? :)

Answer 10

yeah, see that and that's what I get when I DO multiply it all out, but is that one of the shortcut things you can just use or KNOW...was there a proof on it somewhere?

Answer 11

no...it didn't come out for me that simply!

Answer 12

OH...I forgot to keep bringng down the 30!!

Answer 13

Do you remember how to find horizontal asypmtotes for polynomials/polynomials?

if the degrees on top and bottom are the same then just take the coeffiicent of the term with the highest exponent on top /over/ the coefficient of the term with the highest exponent on bottom

and that would be the limit