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Limits

Petrus

Well-known member
Feb 21, 2013
739
Calculate limit



I have none progress, Is there any special rule for this limit?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,885
Calculate limit



I have none progress, Is there any special rule for this limit?
You can find a lower and upper bound with an integral.
 

Petrus

Well-known member
Feb 21, 2013
739

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,885
Calculate limit



I have none progress, Is there any special rule for this limit?
You can approximate a series with an integral:
$$\sum_{k=a}^b f(k) \approx \int_{a-\frac 1 2}^{b+\frac 1 2}f(x)dx$$
More specifically, if f(x) is a strictly descending function, you have:
$$\int_{a}^{b+1}f(x)dx < \sum_{k=a}^b f(k) < \int_{a-1}^{b}f(x)dx$$
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,885
Here's a picture that is somewhat indicative of how it works.



In this example the area of the rectangles is an upper bound of the integral of the function (starting from 1 in this case).
The area of the rectangles is equal to the series of the function.
 

Petrus

Well-known member
Feb 21, 2013
739
You can approximate a series with an integral:
$$\sum_{k=a}^b f(k) \approx \int_{a-\frac 1 2}^{b+\frac 1 2}f(x)dx$$
More specifically, if f(x) is a strictly descending function, you have:
$$\int_{a}^{b+1}f(x)dx < \sum_{k=a}^b f(k) < \int_{a-1}^{b}f(x)dx$$
So if I got this correct.
I will have
\(\displaystyle \int_{2n-\frac{1}{2}}^{6n+\frac{1}{2}}\frac{n}{k^2+n^2}dk\) or what shall I integrate to respect?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,885
So if I got this correct.
I will have
\(\displaystyle \int_{2n-\frac{1}{2}}^{6n+\frac{1}{2}}\frac{n}{k^2+n^2}dk\) or what shall I integrate to respect?
Yep. That's it.
 

Petrus

Well-known member
Feb 21, 2013
739
So if I just antiderivate that It Will be \(\displaystyle \frac{3n}{k^3}\) Is that correct?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,885
So if I just antiderivate that It Will be \(\displaystyle \frac{3n}{k^3}\) Is that correct?
Nope.
Consider what the derivative of \(\displaystyle \frac{3n}{k^3}\) is with respect to k.
It's $-3 \cdot \dfrac{3n}{k^4} \ne \dfrac{n}{k^2+n^2}$
 

Petrus

Well-known member
Feb 21, 2013
739
Nope.
Consider what the derivative of \(\displaystyle \frac{3n}{k^3}\) is with respect to k.
It's $-3 \cdot \dfrac{3n}{k^4} \ne \dfrac{n}{k^2+n^2}$
hmm... \(\displaystyle \frac{n}{k+n^2}\) that is what I get
 

Klaas van Aarsen

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Staff member
Mar 5, 2012
8,885
hmm... \(\displaystyle \frac{n}{k+n^2}\) that is what I get
Can you show your steps starting from \(\displaystyle \frac{d}{dk}\left(\frac {3n}{k^3}\right)\) then?
 

Petrus

Well-known member
Feb 21, 2013
739
Can you show your steps starting from \(\displaystyle \frac{d}{dk}\left(\frac {3n}{k^3}\right)\) then?
I have problem at imagine the other as a constant and then antiderivate/derivate.. but if you ask \(\displaystyle \frac{3}{k^3} <=>3*\frac{1}{k^3} <=> 3*k^{-3}\) so the derivate is \(\displaystyle -9k^{-4} <=> \frac{-9}{k^4}\)
 
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Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,885
I have problem at imagine the other as a constant and then antiderivate/derivate.. but if you ask \(\displaystyle \frac{3}{k^3} <=>3*\frac{1}{k^3} <=> 3*k^{-3}\) so the derivate is \(\displaystyle -9k^{-4} <=> \frac{-9}{k^4}\)
Correct! :)

In your problem the integral is calculated while keeping n constant.
This means that you can treat n in just the same way as you treated the number 3.
So \(\displaystyle \frac{d}{dk}\left(\frac{3n}{k^3}\right) = \frac{-9n}{k^4} \ne \frac{n}{k^2+n^2}\).


So let's do this first with:
$$\int \frac{3}{x^2+3^2}dx$$
Can you find this anti-derivative?
 
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Petrus

Well-known member
Feb 21, 2013
739
Correct! :)

In your problem the integral is calculated while keeping n constant.
This means that you can treat n in just the same way as you treated the number 3.
So \(\displaystyle \frac{d}{dk}\left(\frac{3}{k^3}\right) = \frac{-9n}{k^4} \ne \frac{n}{k^2+n^2}\).


So let's do this first with:
$$\int \frac{3}{x^2+3^2}dx$$
Can you find this anti-derivative?
Nop. But when I look in my book it looks like it will be \(\displaystyle \arctan\)

- - - Updated - - -

Nop. But when I look in my book it looks like it will be \(\displaystyle \arctan\)
or Wait I think I got it now!
 

Petrus

Well-known member
Feb 21, 2013
739
Nop. But when I look in my book it looks like it will be \(\displaystyle \arctan\)

- - - Updated - - -


or Wait I think I got it now!
\(\displaystyle \frac{3}{x^2+3^2}\) if we factour out 3 from top and bot we get \(\displaystyle \frac{3}{3}\int\frac{1}{\frac{x^2}{3}+1^2}\) so our integrate become \(\displaystyle \frac{3}{3}\arctan{\frac{x}{3}}\) Is this correct?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,885
\(\displaystyle \frac{3}{x^2+3^2}\) if we factour out 3 from top and bot we get \(\displaystyle \frac{3}{3}\int\frac{1}{\frac{x^2}{3}+1^2}\) so our integrate become \(\displaystyle \frac{3}{3}\arctan{\frac{x}{3}}\) Is this correct?
Almost.
But your factoring out isn't quite right.
You seem to have made a mistake with the squaring part.

What do you get if you differentiate \(\displaystyle \frac{3}{3}\arctan{\frac{x}{3}}\)?
 

Petrus

Well-known member
Feb 21, 2013
739
Almost.
But your factoring out isn't quite right.
You seem to have made a mistake with the squaring part.

What do you get if you differentiate \(\displaystyle \frac{3}{3}\arctan{\frac{x}{3}}\)?
im confused now haha. 3/3=1 so we got \(\displaystyle \arctan{\frac{x}{3}}\)
edit: if I got this right It is wrong cause if we use chain rule we get \(\displaystyle \frac{1}{x^2+1}\frac{1}{3}\) and that is not same hmm.. how I integrate that then?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,885
im confused now haha. 3/3=1 so we got \(\displaystyle \arctan{\frac{x}{3}}\)
edit: if I got this right It is wrong cause if we use chain rule we get \(\displaystyle \frac{1}{x^2+1}\frac{1}{3}\) and that is not same hmm.. how I integrate that then?
First fix the factoring out.
Your idea is correct, but you execution is not.

Then substitute u=x/3 (or equivalently x=3u).

(Btw, you did correctly find the derivative of \(\displaystyle \arctan{\frac{x}{3}}\). It should give you a hint what the proper anti-derivative is.)
 

Petrus

Well-known member
Feb 21, 2013
739
First fix the factoring out.
Your idea is correct, but you execution is not.

Then substitute u=x/3 (or equivalently x=3u).
Ok now I got pretty unsure. it should be \(\displaystyle \frac{1/3}{1/3}\) that I factour 1/3 from top and botom
 

Klaas van Aarsen

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Staff member
Mar 5, 2012
8,885
Ok now I got pretty unsure. it should be \(\displaystyle \frac{1/3}{1/3}\) that I factour 1/3 from top and botom
Note that \(\displaystyle \frac{3^2}{3} \ne 1^2\).
And if you don't believe me, try calculating both. ;)
 

Petrus

Well-known member
Feb 21, 2013
739
Note that \(\displaystyle \frac{3^2}{3} \ne 1^2\).
And if you don't believe me, try calculating both. ;)
That make sense.. I believe and can see that.. so I should have \(\displaystyle \frac{3}{3}\int\frac{1}{\frac{x^2}{3}+\frac{1^2}{1^2}}\) and it will give me same result with \(\displaystyle \arctan{\frac{x}{3}}\) is that correct?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,885
That make sense.. I believe and can see that.. so I should have \(\displaystyle \frac{3}{3}\int\frac{1}{\frac{x^2}{3}+\frac{1^2}{1^2}}\) and it will give me same result with \(\displaystyle \arctan{\frac{x}{3}}\) is that correct?
No... since \(\displaystyle \frac{3^2}{3} \ne \frac{1^2}{1^2}\) as well.

- - - Updated - - -

You have \(\displaystyle \frac{3}{x^2+3^2} = \frac{3}{3}\cdot\frac{1}{\frac{x^2}{3}+\frac{3^2}{3}}\)

But what you want is something like \(\displaystyle \frac{1}{(\frac{x}{3})^2+1}\)
 

Petrus

Well-known member
Feb 21, 2013
739
No... since \(\displaystyle \frac{3^2}{3} \ne \frac{1^2}{1^2}\) as well.

- - - Updated - - -

You have \(\displaystyle \frac{3}{x^2+3^2} = \frac{3}{3}\cdot\frac{1}{\frac{x^2}{3}+\frac{3^2}{3}}\)

But what you want is something like \(\displaystyle \frac{1}{(\frac{x}{3})^2+1}\)
Do you mean
\(\displaystyle \frac{1}{(\frac{3x}{3})^2+1}\)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,885
Do you mean
\(\displaystyle \frac{1}{(\frac{3x}{3})^2+1}\)
Huh? No.

I mean something like:
\(\displaystyle \frac{3}{x^2+3^2} = \frac{3}{3^2(\frac{x^2}{3^2}+\frac{3^2}{3^2})} = \frac{3}{3^2} \cdot \frac{1}{\frac{x^2}{3^2}+\frac{3^2}{3^2}}\)
 

Petrus

Well-known member
Feb 21, 2013
739
Huh? No.

I mean something like:
\(\displaystyle \frac{3}{x^2+3^2} = \frac{3}{3^2(\frac{x^2}{3^2}+\frac{3^2}{3^2})} = \frac{3}{3^2} \cdot \frac{1}{\frac{x^2}{3^2}+\frac{3^2}{3^2}}\)
Thanks.
So now we got \(\displaystyle \frac{3}{3^2}\int\frac{1}{\frac{x^2}{3^2}+\frac{3^2}{3^2}}\) and if we subsitute \(\displaystyle u=\frac{x}{3}\) we got \(\displaystyle \frac{3}{3^2}\int\frac{1}{u^2+1}\) so we got \(\displaystyle \frac{3}{3^2}\arctan{u}\)