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

- anonymous

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

- schrodinger

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- 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

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

It's quite possible. What is the equation?

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## More answers

- 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

yep...

- 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

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

- freckles

the the coefficient of n^5 on bottom is 30

- freckles

so we have 6/30 right? :)

- 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

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

- anonymous

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

- 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

|dw:1336266798515:dw|

- 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

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

- anonymous

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

- freckles

Ok did you see the paragraph i wrote above?

- 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

That one ^

- anonymous

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

- freckles

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

- 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

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

- anonymous

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

- 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

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

- 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

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

- 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

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

- freckles

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

- freckles

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

- 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

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

- anonymous

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

- freckles

6/30

- 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

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

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

- freckles

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

- freckles

definite integrals you only get a number

- 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

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

- anonymous

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

- anonymous

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

- anonymous

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

- 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

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

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

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

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

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

- 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

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

- 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

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

yes, each time....

- anonymous

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

- anonymous

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

- 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

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

that makes sense....

- 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

ok....

- anonymous

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

- anonymous

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

- anonymous

No problem. Glad to be helpful :-) and sorry about the other question, I started answering it but got distracted.

- anonymous

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

- anonymous

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

- anonymous

ok :-)

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