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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0when I simplify that long formula and take it's limit, I get a 1/4 and that's not possible is it??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It's quite possible. What is the equation?

freckles
 4 years ago
Best ResponseYou've already chosen the best response.1Is this what you are evaluating? \[\int\limits_0^1x^4dx=\lim_{n\to\infty}\sum_{j=1}^nf(\Delta x_j^*)\Delta x=\lim_{n\to\infty}\sum_{j=1}^n\frac{j^4}{n^4}\cdot\frac1n=\lim_{n\to\infty}\sum_{j=1}^n\frac{j^4}{n^5} \]

freckles
 4 years ago
Best ResponseYou've already chosen the best response.1\[\lim_{n\to\infty}\frac1{n^5}\sum_{j=1}^\ni^4=\lim_{n\to\infty}\frac1{n^5}{n(n+1)(2n+1)(3n^2 + 3n 1)\over30} \] 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) ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and then is became 1/n^5 n(n+1)(2n+1)(3n^2+3n1) all /3

freckles
 4 years ago
Best ResponseYou've already chosen the best response.1the the coefficient of n^5 on bottom is 30

freckles
 4 years ago
Best ResponseYou've already chosen the best response.1so we have 6/30 right? :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah, 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no...it didn't come out for me that simply!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0OH...I forgot to keep bringng down the 30!!

freckles
 4 years ago
Best ResponseYou've already chosen the best response.1dw: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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but 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..:(

freckles
 4 years ago
Best ResponseYou've already chosen the best response.1\[\lim_{n \rightarrow \infty}\frac{n(n)(2n)(3n^2)}{30n^5}=\lim_{n \rightarrow \infty}\frac{6n^5}{30n^5}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now how did you simplify that so easily??? how can I just drop out the linear terms and constant?

freckles
 4 years ago
Best ResponseYou've already chosen the best response.1Ok did you see the paragraph i wrote above?

freckles
 4 years ago
Best ResponseYou've already chosen the best response.1Do 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes, I remember that now from when I tutor precal....duh....thought about that. :)

freckles
 4 years ago
Best ResponseYou've already chosen the best response.1The degree of the top was 5 The degree of the bottom was 5

freckles
 4 years ago
Best ResponseYou've already chosen the best response.1Since 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But I need to calculate the upper reimann sumof all that mess....so what does it simplify down to??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have 6n^4 + 15n^3 10n^2 1/ 30n^4 and can get no further...what am I doing wriong??

freckles
 4 years ago
Best ResponseYou've already chosen the best response.1\[\lim_{n \rightarrow \infty}\frac{n(n)(2n)(3n^2)}{30n^5}=\lim_{n \rightarrow \infty}\frac{6n^5}{30n^5} \] so are you asking me to evaluate this limit for you? It is just 6/30 since the n^5 's cancel

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that's what I thought I should get, but see what I got above by multiplying it all out and not enough canceled....

freckles
 4 years ago
Best ResponseYou've already chosen the best response.1Ok 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

freckles
 4 years ago
Best ResponseYou've already chosen the best response.1you don't have to multiply all of that out and nor do i recommend it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@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

freckles
 4 years ago
Best ResponseYou've already chosen the best response.1\[\lim_{n \rightarrow \infty}\frac{\frac{6n^4}{n^4}+\frac{15n^3}{n^4}\frac{10^2}{n^4}\frac{1}{n^4}}{\frac{n^4}{n^4}}\]

freckles
 4 years ago
Best ResponseYou've already chosen the best response.1\[\lim_{n \rightarrow \infty}\frac{6+000}{30}\] typeo above

freckles
 4 years ago
Best ResponseYou've already chosen the best response.1i left out coefficient of n^4/n^4 on bottom

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0when 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....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0WE only need to evaluate that summation for X^4 the Reimann way....how they did it before Leibnitz came on the scene. :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so was my 6n^4 etc technically correct for the evaluation of the sum??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If 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. :)

freckles
 4 years ago
Best ResponseYou've already chosen the best response.1you don't need to multiply it all out to evaluate the limit

freckles
 4 years ago
Best ResponseYou've already chosen the best response.1i don't know what you mean looking for a formula for definite integral

freckles
 4 years ago
Best ResponseYou've already chosen the best response.1definite integrals you only get a number

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Or 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok, so you're saying the Reimann sum LPLUS the limit IS the same as the integral!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0duh...I was trying to equate the RS ONLY to the integral....left out the limit step. :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I am saying that after setting up a Riemann sum, if we evaluate it to infinity, that is the Riemann integral. :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0vs the leibnitz integral...the short cut. right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0limit = 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Maybe 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0new 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...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And, 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 :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0AND if the interval value changes, say to n=2? then calculate that sum by plugging 2 into the formula ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0were supposed to verify the formula fo rn=2, 3, 5, 8

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ah, 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Remember 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Then, all that changes is that the interval has to be 2/n, 3/n and so on :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so do the reimann sum method separately for each n, just like I did for n=1?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yup. 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok, 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)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No. 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: \[\Large\int_{0}^{\text{end of interval}}x^4dx \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The limit should be : \[ \frac{(\text{end of interval})^{5}}{5} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0gotcha...thanks SO much !! Sorry I took you away from other people with questions..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No problem. Glad to be helpful :) and sorry about the other question, I started answering it but got distracted.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that's ok....is there anyway to priint all of this out???

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0To print the text, I don't think so. Only printing the page, but that is not that much helpful, I think.
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