anonymous
  • anonymous
Turing Test or Zarkon..another question to my Reimann sum problem....
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
when I simplify that long formula and take it's limit, I get a -1/4 and that's not possible is it??
anonymous
  • 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?
anonymous
  • anonymous
It's quite possible. What is the equation?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

freckles
  • freckles
Is this what you are evaluating? \[\int\limits_0^1x^4dx=\lim_{n\to\infty}\sum_{j=1}^nf(\Delta x_j^*)\Delta x=\lim_{n\to\infty}\sum_{j=1}^n\frac{j^4}{n^4}\cdot\frac1n=\lim_{n\to\infty}\sum_{j=1}^n\frac{j^4}{n^5} \]
anonymous
  • anonymous
yep...
freckles
  • 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) ?
anonymous
  • anonymous
and then is became 1/n^5 n(n+1)(2n+1)(3n^2+3n-1) all /3
freckles
  • freckles
the the coefficient of n^5 on bottom is 30
freckles
  • freckles
so we have 6/30 right? :)
anonymous
  • 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?
anonymous
  • anonymous
no...it didn't come out for me that simply!
anonymous
  • anonymous
OH...I forgot to keep bringng down the 30!!
freckles
  • 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
freckles
  • freckles
|dw:1336266798515:dw|
anonymous
  • 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..:-(
freckles
  • freckles
\[\lim_{n \rightarrow \infty}\frac{n(n)(2n)(3n^2)}{30n^5}=\lim_{n \rightarrow \infty}\frac{6n^5}{30n^5}\]
anonymous
  • anonymous
now how did you simplify that so easily??? how can I just drop out the linear terms and constant?
freckles
  • freckles
Ok did you see the paragraph i wrote above?
freckles
  • 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
freckles
  • freckles
That one ^
anonymous
  • anonymous
yes, I remember that now from when I tutor precal....duh....thought about that. :-)
freckles
  • freckles
The degree of the top was 5 The degree of the bottom was 5
freckles
  • 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
anonymous
  • anonymous
But I need to calculate the upper reimann sumof all that mess....so what does it simplify down to??
anonymous
  • anonymous
I have 6n^4 + 15n^3 -10n^2 -1/ 30n^4 and can get no further...what am I doing wriong??
freckles
  • 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
anonymous
  • anonymous
that's what I thought I should get, but see what I got above by multiplying it all out and not enough canceled....
freckles
  • 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
freckles
  • freckles
you don't have to multiply all of that out and nor do i recommend it
anonymous
  • 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
freckles
  • freckles
\[\lim_{n \rightarrow \infty}\frac{\frac{6n^4}{n^4}+\frac{15n^3}{n^4}-\frac{10^2}{n^4}-\frac{1}{n^4}}{\frac{n^4}{n^4}}\]
freckles
  • freckles
\[\lim_{n \rightarrow \infty}\frac{6+0-0-0}{30}\] type-o above
freckles
  • freckles
i left out coefficient of n^4/n^4 on bottom
anonymous
  • 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....
anonymous
  • anonymous
WE only need to evaluate that summation for X^4 the Reimann way....how they did it before Leibnitz came on the scene. :-)
anonymous
  • anonymous
so was my 6n^4 etc technically correct for the evaluation of the sum??
freckles
  • freckles
6/30
anonymous
  • 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?
anonymous
  • 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. :-)
freckles
  • freckles
you don't need to multiply it all out to evaluate the limit
freckles
  • freckles
i don't know what you mean looking for a formula for definite integral
freckles
  • freckles
definite integrals you only get a number
anonymous
  • 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.
anonymous
  • anonymous
ok, so you're saying the Reimann sum LPLUS the limit IS the same as the integral!
anonymous
  • anonymous
duh...I was trying to equate the RS ONLY to the integral....left out the limit step. :)
anonymous
  • anonymous
I am saying that after setting up a Riemann sum, if we evaluate it to infinity, that is the Riemann integral. :-)
anonymous
  • anonymous
vs the leibnitz integral...the short cut. right?
anonymous
  • 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.
anonymous
  • 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
anonymous
  • 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.
anonymous
  • 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...
anonymous
  • 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 :-)
anonymous
  • anonymous
AND if the interval value changes, say to n=2? then calculate that sum by plugging 2 into the formula ?
anonymous
  • 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.
anonymous
  • anonymous
were supposed to verify the formula fo rn=2, 3, 5, 8
anonymous
  • 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.
anonymous
  • 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.
anonymous
  • anonymous
yes, each time....
anonymous
  • anonymous
Then, all that changes is that the interval has to be 2/n, 3/n and so on :-)
anonymous
  • anonymous
so do the reimann sum method separately for each n, just like I did for n=1?
anonymous
  • 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.
anonymous
  • 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)
anonymous
  • anonymous
that makes sense....
anonymous
  • 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: \[\Large\int_{0}^{\text{end of interval}}x^4dx \]
anonymous
  • anonymous
ok....
anonymous
  • anonymous
The limit should be : \[ \frac{(\text{end of interval})^{5}}{5} \]
anonymous
  • anonymous
gotcha...thanks SO much !! Sorry I took you away from other people with questions..
anonymous
  • anonymous
No problem. Glad to be helpful :-) and sorry about the other question, I started answering it but got distracted.
anonymous
  • anonymous
that's ok....is there anyway to priint all of this out???
anonymous
  • anonymous
To print the text, I don't think so. Only printing the page, but that is not that much helpful, I think.
anonymous
  • anonymous
ok :-)

Looking for something else?

Not the answer you are looking for? Search for more explanations.