## anonymous 4 years ago Turing Test or Zarkon..another question to my Reimann sum problem....

1. anonymous

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

2. anonymous

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?

3. anonymous

It's quite possible. What is the equation?

4. freckles

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

5. anonymous

yep...

6. freckles

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

7. anonymous

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

8. freckles

the the coefficient of n^5 on bottom is 30

9. freckles

so we have 6/30 right? :)

10. anonymous

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?

11. anonymous

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

12. anonymous

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

13. freckles

|dw:1336266690533:dw| 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

14. freckles

|dw:1336266798515:dw|

15. anonymous

but sstilldon't see how the 30 can cancel.....the n^5 does...eventually down to ...I can't get it to factor so IT will cancel out or simplify..:-(

16. freckles

$\lim_{n \rightarrow \infty}\frac{n(n)(2n)(3n^2)}{30n^5}=\lim_{n \rightarrow \infty}\frac{6n^5}{30n^5}$

17. anonymous

now how did you simplify that so easily??? how can I just drop out the linear terms and constant?

18. freckles

Ok did you see the paragraph i wrote above?

19. freckles

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

20. freckles

That one ^

21. anonymous

yes, I remember that now from when I tutor precal....duh....thought about that. :-)

22. freckles

The degree of the top was 5 The degree of the bottom was 5

23. freckles

Since the degrees are the same we only care about the coefficient of the n^5 on top and the coefficient of n^5 on bottom

24. anonymous

But I need to calculate the upper reimann sumof all that mess....so what does it simplify down to??

25. anonymous

I have 6n^4 + 15n^3 -10n^2 -1/ 30n^4 and can get no further...what am I doing wriong??

26. freckles

$\lim_{n \rightarrow \infty}\frac{n(n)(2n)(3n^2)}{30n^5}=\lim_{n \rightarrow \infty}\frac{6n^5}{30n^5}$ so are you asking me to evaluate this limit for you? It is just 6/30 since the n^5 's cancel

27. anonymous

that's what I thought I should get, but see what I got above by multiplying it all out and not enough canceled....

28. freckles

Ok remember what you have on top and bottom the degrees are the same just take the coeifficent of n^4 on top and on bottom

29. freckles

you don't have to multiply all of that out and nor do i recommend it

30. anonymous

@laurie214 Think about it this way. Let 15n^3 - 10n^2 - 1 = m. Then: $\lim_{n \rightarrow \infty} \frac{6n^4 + m}{30n^4} \implies \lim_{n \rightarrow \infty} \frac{6n^4}{30n^4} + \lim_{n \rightarrow \infty} \frac{m}{30n^4}$Now, the highest deegre term for m is n^3. Do you agree that n^4 goes to infinity much quicker than n^4? That implies that the second limit goes to zero and we are left with$\lim_{n \rightarrow \infty} \frac{6n^4}{30n^4}$as wanted

31. freckles

$\lim_{n \rightarrow \infty}\frac{\frac{6n^4}{n^4}+\frac{15n^3}{n^4}-\frac{10^2}{n^4}-\frac{1}{n^4}}{\frac{n^4}{n^4}}$

32. freckles

$\lim_{n \rightarrow \infty}\frac{6+0-0-0}{30}$ type-o above

33. freckles

i left out coefficient of n^4/n^4 on bottom

34. anonymous

when the prof did the problem for x^3, he got down to (n+1)^2/4n^2 for the evaluation of the summation....shouldn't ok...see what you're doing above here....

35. anonymous

WE only need to evaluate that summation for X^4 the Reimann way....how they did it before Leibnitz came on the scene. :-)

36. anonymous

so was my 6n^4 etc technically correct for the evaluation of the sum??

37. freckles

6/30

38. anonymous

And to get the limit, you wouldn't need to multiply it all out....I see that, but to get a formula or an evaluation, you would, right?

39. anonymous

If the amount of "rectangles" that we want is infinity (as the limit is saying), it should equal the defined integral, that is, it should equal 1/5. The evaluation is the limit. :-)

40. freckles

you don't need to multiply it all out to evaluate the limit

41. freckles

i don't know what you mean looking for a formula for definite integral

42. freckles

definite integrals you only get a number

43. anonymous

Or using a better wording, the limit is the evaluation of the sum, because, in your other question, you have already set up the rectangles and f(x*). Now, to evaluate it, we have to take the limit, that should be equal to the definite integral.

44. anonymous

ok, so you're saying the Reimann sum LPLUS the limit IS the same as the integral!

45. anonymous

duh...I was trying to equate the RS ONLY to the integral....left out the limit step. :)

46. anonymous

I am saying that after setting up a Riemann sum, if we evaluate it to infinity, that is the Riemann integral. :-)

47. anonymous

vs the leibnitz integral...the short cut. right?

48. anonymous

If we consider a finite amount of rectangles, we always have an error. Because n -> infinity, the error is minimal, and we consider it equal to the area under the curve, i.e., the Riemann integral in the case of f(x) = x^4.

49. anonymous

limit = definite integral in other words, like you said above...got it! so this integral = 1/5 either way,,,duh, so the sum of the first n squares is 1/3, the first n cubes is 1/4 (we found on last hmwk) and the sum of the first n^4 is 1/5....denominator is always just one more that the exponent

50. anonymous

Maybe this will make you have more intuition: http://en.wikipedia.org/wiki/File:Riemann.gif We are still dealing with a finite amount of rectangles, but you can see that if we grow and grow the number of rectangles, the area approximation has a smaller and smaller error.

51. anonymous

new I was supposed to get that, but the variable expression fo rthe summation threw me off..I HAD to go another step to get it's limit...

52. anonymous

And, yes, if we are dealing with the interval [0,1] for polynomial functions like those you said, x^5 = 1/6, x^7 = 1/8 :-)

53. anonymous

AND if the interval value changes, say to n=2? then calculate that sum by plugging 2 into the formula ?

54. anonymous

I think the $$\Delta x$$ changes in the Riemann sum, or, if you want to integrate, it will be (2^n)/n. It will be, for instance, (2^4)/4, and so on. I will double check the Riemann sum part.

55. anonymous

were supposed to verify the formula fo rn=2, 3, 5, 8

56. anonymous

Ah, yes. That is correct. For instance, if you have [0,1], you will want to cut the interval in n pieces: 1/n. If it's from [0,2], you will want to cut the interval in n pieces again: 2/n and so on.

57. anonymous

Remember the formula TuringTest showed you? Pick the size of the interval and do it like that: 2/n, 3/n, etc. Albeit, I think this n means the limit. So we aren't really taking it to infinity.

58. anonymous

yes, each time....

59. anonymous

Then, all that changes is that the interval has to be 2/n, 3/n and so on :-)

60. anonymous

so do the reimann sum method separately for each n, just like I did for n=1?

61. anonymous

Yup. It's a bit of a torture, I agree. But always remember that, if we are evaluating the sum to infinity, the result should be the same as the integral.

62. anonymous

ok, that's what it means to evaluate these....but the limit will be 1/5 each time, no matter what the interval is (or width of the rectangle, in other words)

63. anonymous

that makes sense....

64. anonymous

No. The limit does change. Now, instead of the definite integral being : $\int_{0}^{1} x^4 dx$we are dealing now with: $\int_{0}^{2} x^4dx$Or more generally: $\Large\int_{0}^{\text{end of interval}}x^4dx$

65. anonymous

ok....

66. anonymous

The limit should be : $\frac{(\text{end of interval})^{5}}{5}$

67. anonymous

gotcha...thanks SO much !! Sorry I took you away from other people with questions..

68. anonymous

69. anonymous

that's ok....is there anyway to priint all of this out???

70. anonymous

To print the text, I don't think so. Only printing the page, but that is not that much helpful, I think.

71. anonymous

ok :-)