The limit of R_n as n-->inf is...

• Sep 21st 2013, 06:26 PM
dokrbb
The limit of R_n as n-->inf is...
It seems that I have some issues with the sums...

I have the function $\displaystyle f(x) = \sqrt{4 - x^2}$

I need to evaluate it $\displaystyle \displaystyle \int_0^2 (\sqrt{4 - x^2})dx$ by using the definition of Riemann sum for $\displaystyle R_n$ - right end-points,

$\displaystyle \sum_{n=1}^{n} (\sqrt{4 - x^2})$

$\displaystyle \delta x = \frac{2-0}{n} = \frac{2}{n}$

then,

$\displaystyle a) R_n = \sum_{n=1}^{n} \sqrt{4 - (\frac{2i}{n})^2}) *(\frac{2}{n}) =$
$\displaystyle b) R_n = (\frac{2}{n}) \sum_{n=1}^{n} \sqrt{4 - (\frac{4i^2}{n^2})} =$
$\displaystyle c) R_n = (\frac{2}{n}) \sum_{n=1}^{n} \sqrt{4n^2[\frac{1}{n^2} - i^2]} =$
$\displaystyle d) R_n = (\frac{2}{n})(2n) \sum_{n=1}^{n} \sqrt{\frac{1}{n^2} - [\frac{n(n+1)(2n+1)}{6}]} =$...

I'm not sure I did right so far, and how should I proceed further...

thanks for hints,
dokrbb
• Sep 21st 2013, 06:31 PM
HallsofIvy
Re: The limit of R_n as n-->inf is...
Quote:

Originally Posted by dokrbb
It seems that I have some issues with the sums...

I have the function $\displaystyle f(x) = \sqrt{4 - x^2}$

I need to evaluate it $\displaystyle \displaystyle \int_0^2 (\sqrt{4 - x^2})dx$ by using the definition of Riemann sum for $\displaystyle R_n$ - right end-points,

$\displaystyle \sum_{n=1}^{n} (\sqrt{4 - x^2})$

$\displaystyle \delta x = \frac{2-0}{n} = \frac{2}{n}$

then,

$\displaystyle a) R_n = \sum_{n=1}^{n} \sqrt{4 - (\frac{2i}{n})^2}) *(\frac{2}{n}) =$

The sum is over i not n.

Quote:

$\displaystyle b) R_n = (\frac{2}{n}) \sum_{i=1}^{n} \sqrt{4 - (\frac{4i^2}{n^2})} =$
$\displaystyle c) R_n = (\frac{2}{n}) \sum_{i=1}^{n} \sqrt{4n^2[\frac{1}{n^2} - i^2]} =$
$\displaystyle d) R_n = (\frac{2}{n})(2n) \sum_{n=1}^{n} \sqrt{\frac{1}{n^2} - [\frac{n(n+1)(2n+1)}{6}]} =$...
At this point you have already done the sum. You should not have summation sign.

Quote:

I'm not sure I did right so far, and how should I proceed further...

thanks for hints,
dokrbb
• Sep 21st 2013, 06:48 PM
dokrbb
Re: The limit of R_n as n-->inf is...
Quote:

Originally Posted by HallsofIvy
The sum is over i not n.

At this point you have already done the sum. You should not have summation sign.

ok,

$\displaystyle a) R_n = \sum_{i=1}^{n} \sqrt{4 - (\frac{2i}{n})^2}) *(\frac{2}{n}) =$
$\displaystyle b) R_n = (\frac{2}{n}) \sum_{n=1}^{n} \sqrt{4 - (\frac{4i^2}{n^2})} =$
$\displaystyle c) R_n = (\frac{2}{n}) \sum_{n=1}^{n} \sqrt{4n^2[\frac{1}{n^2} - i^2]} =$

$\displaystyle (\frac{2}{n})(2n) \lim_{n \to \infty} \sqrt{\frac{1}{n^2} - [\frac{n(n+1)(2n+1)}{6}]} =$

is that right... $\displaystyle d) (4) \lim_{n \to \infty} \sqrt{\frac{1}{n^2} - [\frac{n(n+1)(2n+1)}{6}]} =$ what can I do with all this under the radical? I have no idea
• Sep 22nd 2013, 03:34 PM
dokrbb
Re: The limit of R_n as n-->inf is...
Quote:

Originally Posted by dokrbb
ok,

$\displaystyle a) R_n = \sum_{i=1}^{n} \sqrt{4 - (\frac{2i}{n})^2}) *(\frac{2}{n}) =$
$\displaystyle b) R_n = (\frac{2}{n}) \sum_{n=1}^{n} \sqrt{4 - (\frac{4i^2}{n^2})} =$
$\displaystyle c) R_n = (\frac{2}{n}) \sum_{n=1}^{n} \sqrt{4n^2[\frac{1}{n^2} - i^2]} =$

$\displaystyle (\frac{2}{n})(2n) \lim_{n \to \infty} \sqrt{\frac{1}{n^2} - [\frac{n(n+1)(2n+1)}{6}]} =$

is that right... $\displaystyle d) (4) \lim_{n \to \infty} \sqrt{\frac{1}{n^2} - [\frac{n(n+1)(2n+1)}{6}]} =$ what can I do with all this under the radical? I have no idea

someone???
• Sep 22nd 2013, 04:28 PM
FelixFelicis28
Re: The limit of R_n as n-->inf is...
Quote:

Originally Posted by dokrbb
someone???

Could you post what the exact question is, as you found it? To compute $\displaystyle \int_0^2 \sqrt{4 - x^2} \ dx$ from first principles would involve deriving a closed form for this partial sum:

$\displaystyle \sum_{i=1}^n \sqrt{n^2 - i^2}$

which seems to be very difficult, I've got nothing...
• Sep 22nd 2013, 05:41 PM
dokrbb
Re: The limit of R_n as n-->inf is...
Quote:

Originally Posted by FelixFelicis28
Could you post what the exact question is, as you found it? To compute $\displaystyle \int_0^2 \sqrt{4 - x^2} \ dx$ from first principles would involve deriving a closed form for this partial sum:

$\displaystyle \sum_{i=1}^n \sqrt{n^2 - i^2}$

which seems to be very difficult, I've got nothing...

I'll try, the question is like this:

The following sum $\displaystyle \sqrt{4 - (\frac{2}{n})^2}*(\frac{2}{n}) + \sqrt{4 - (\frac{4}{n})^2}*(\frac{2}{n}) + ...+ \sqrt{4 - (\frac{2n}{n})^2}*(\frac{2}{n})$

is a Right Riemann sum for the definite integral $\displaystyle \displaystyle \int_0^b (f(x))dx$

where b = ...

and f(x) = ...

I checked my answer for $\displaystyle b=2$ and $\displaystyle f(x) = (\sqrt{4 - x^2})$ and they are correct; The next question is: the limit of these Riemann sums as n-> infinity is....

does that change my approach?
• Sep 22nd 2013, 06:17 PM
FelixFelicis28
Re: The limit of R_n as n-->inf is...
Quote:

does that change my approach?
Yes, this changes everything.

Quote:

Originally Posted by dokrbb
I'll try, the question is like this:

The following sum $\displaystyle \sqrt{4 - (\frac{2}{n})^2}*(\frac{2}{n}) + \sqrt{4 - (\frac{4}{n})^2}*(\frac{2}{n}) + ...+ \sqrt{4 - (\frac{2n}{n})^2}*(\frac{2}{n})$

is a Right Riemann sum for the definite integral $\displaystyle \displaystyle \int_0^b (f(x))dx$

where b = ...

and f(x) = ...

We use the definition of the Riemann sum to be $\displaystyle \sum_{i=1}^n f(x_i) \Delta x$

Clearly:

$\displaystyle \Delta x = \frac{b-a}{n} = \frac{b}{n} = \frac{2}{n} \implies b = 2$

as you've done.

And $\displaystyle x_i = 0 + i\Delta x = \frac{2i}{n} \implies f(x_i) = \sqrt{4 - \left(\frac{2i}{n}\right)^2}$, thus $\displaystyle f(x) = \sqrt{4 - x^2}$ which you've also done correctly.

Quote:

The next question is: the limit of these Riemann sums as n-> infinity is....
When they ask for this, they're not asking you to compute this integral $\displaystyle \int_0^2 \sqrt{4 - x^2} \ dx$ by evaluating this limit:

$\displaystyle \lim_{n\to\infty} \sum_{i=1}^n \sqrt{4 - \left(\frac{2i}{n}\right)^2} \cdot \frac{2}{n}$

But rather, what they're asking you to do is evaluate this limit:

$\displaystyle \lim_{n\to\infty} \sum_{i=1}^n \sqrt{4 - \left(\frac{2i}{n}\right)^2} \cdot \frac{2}{n}$

by using this relation:

$\displaystyle \lim_{n\to\infty} \sum_{i=1}^n f(x_i) \Delta x = \int_a^b f(x) \ dx$

where $\displaystyle \Delta x = \frac{b-a}{n}, \ x_i = a + i\Delta x$.

In other words, they're not asking you to compute the integral by using the Riemann sum, they're asking you to compute the Riemann sum by using the integral.
• Sep 23rd 2013, 03:25 PM
dokrbb
Re: The limit of R_n as n-->inf is...
Quote:

Originally Posted by FelixFelicis28
Yes, this changes everything.

We use the definition of the Riemann sum to be $\displaystyle \sum_{i=1}^n f(x_i) \Delta x$

Clearly:

$\displaystyle \Delta x = \frac{b-a}{n} = \frac{b}{n} = \frac{2}{n} \implies b = 2$

as you've done.

And $\displaystyle x_i = 0 + i\Delta x = \frac{2i}{n} \implies f(x_i) = \sqrt{4 - \left(\frac{2i}{n}\right)^2}$, thus $\displaystyle f(x) = \sqrt{4 - x^2}$ which you've also done correctly.

When they ask for this, they're not asking you to compute this integral $\displaystyle \int_0^2 \sqrt{4 - x^2} \ dx$ by evaluating this limit:

$\displaystyle \lim_{n\to\infty} \sum_{i=1}^n \sqrt{4 - \left(\frac{2i}{n}\right)^2} \cdot \frac{2}{n}$

But rather, what they're asking you to do is evaluate this limit:

$\displaystyle \lim_{n\to\infty} \sum_{i=1}^n \sqrt{4 - \left(\frac{2i}{n}\right)^2} \cdot \frac{2}{n}$

by using this relation:

$\displaystyle \lim_{n\to\infty} \sum_{i=1}^n f(x_i) \Delta x = \int_a^b f(x) \ dx$

where $\displaystyle \Delta x = \frac{b-a}{n}, \ x_i = a + i\Delta x$.

In other words, they're not asking you to compute the integral by using the Riemann sum, they're asking you to compute the Riemann sum by using the integral.

I'm not sure I understand really how it should be done, do I really need in this case to evaluate it as

$\displaystyle \lim_{n\to\infty} \sum_{i=1}^n \left[ \sqrt{4 - \left(\frac{bi}{n}\right)^2} \cdot \frac{b}{n} - \sqrt{4 - \left(\frac{ai}{n}\right)^2} \cdot \frac{a}{n} \right] =$
$\displaystyle \left[ \lim_{n\to\infty} \sum_{i=1}^n \sqrt{4 - \left(\frac{bi}{n}\right)^2} \cdot \frac{b}{n} - \lim_{n\to\infty} \sum_{i=1}^n \sqrt{4 - \left(\frac{ai}{n}\right)^2} \cdot \frac{a}{n} \right] =$ and so on, without using the actual values of $\displaystyle b$ and $\displaystyle a$ ?
• Sep 23rd 2013, 04:28 PM
FelixFelicis28
Re: The limit of R_n as n-->inf is...
Quote:

Originally Posted by dokrbb
I'm not sure I understand really how it should be done, do I really need in this case to evaluate it as

$\displaystyle \lim_{n\to\infty} \sum_{i=1}^n \left[ \sqrt{4 - \left(\frac{bi}{n}\right)^2} \cdot \frac{b}{n} - \sqrt{4 - \left(\frac{ai}{n}\right)^2} \cdot \frac{a}{n} \right] =$
$\displaystyle \left[ \lim_{n\to\infty} \sum_{i=1}^n \sqrt{4 - \left(\frac{bi}{n}\right)^2} \cdot \frac{b}{n} - \lim_{n\to\infty} \sum_{i=1}^n \sqrt{4 - \left(\frac{ai}{n}\right)^2} \cdot \frac{a}{n} \right] =$ and so on, without using the actual values of $\displaystyle b$ and $\displaystyle a$ ?

Hmmm? I'm not quite sure what you're doing...

$\displaystyle R_n = \sum_{i=1}^n \sqrt{4 - \left(\frac{2i}{n}\right)^2} \cdot \frac{2}{n}$.
And they're asking you to evaluate $\displaystyle \lim_{n \to \infty} R_n$.
$\displaystyle \lim_{n\to\infty} \sum_{i=1}^n f(x_i)\Delta x = \int_a^b f(x) \ dx$.
$\displaystyle \lim_{n\to\infty} \sum_{i=1}^n \sqrt{4 - \left(\frac{2i}{n}\right)^2} \cdot \frac{2}{n} = \int_0^2 \sqrt{4 - x^2} \ dx$.