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[Sunyata]

Hi

How exactly do i do b + c of this question??

 

[Gussy Booo]

Hi

How exactly do i do b + c of this question??

Hi there .

Well part (a) should be pretty straight forward.

Its a rectangular hyperbola, in quadrants 1 and 3, due to it being positive. Now, the question asks for the hyperbola to be drawn from 1-->2. Therefore, we are only taking the first quadrant into consideration. Also, we should draw two straight lines, 1 being x=1, and the other being x=2. Between these two lines, is the area in which we will be WORKING with, in proving the other results.

So part (b) asks us to show an inequality considering THREE results. It also mentions rectangles.

Answering this question lies in understanding how area (and thus integration) is performed and found using methods which are more formally logical.

The first result we will find is ln2. This is your standard integration, following the Fundamental Theorem of Calculus. Furthermore:

2[Int]1 (1/x)dx
= (lnx)2->1
= ln |2/1|
= ln 2.

So we have found 1 result. Ie -> ln2.

Now, integrating and thus finding the area under a curve using the Fundamental Theorem of calculus is the most accurate means of finding the area. This should be kept in mind.

The other 2 results comes from finding the area using RECTANGLES.

So lets picture this curve (following the path of a rectangular hyperbola) from 1->2 (you should draw this on a piece of paper btw, make it big). We are going to try and find the area using rectangles. So we can draw lots of rectangles between 1->2, and since we know that the area of a rectangle is LxB, we can somehow add them all together and thus come to a conclusion.

But there are two types of ways we can draw rectangles to find the area under the curve.

1) We make the rectangles SMALLER than the curve, hence producing an area SMALLER than the specific area required. ( This is done in respect to the WIDTH of the curve)

2) We make the rectangles BIGGER than the curve, hence producing rectangles which all in all produce an area LARGER than the specific area required. ( This is done in respect to the HEIGHT of the curve)

So lets work with (1)

Furthermore, lets divide this area into N EQUAL PARTS. Let the part @ x=1 be n/n parts (for n/n =1), and let the n part @ x=2 be 2n/n (for 2n/n) = 2.

Now, we will be dividing it into n N EQUAL PARTS from n/n to 2n/n.

It will look something like this :

n/n --> n+1/n --> n+2/n --> n+3/n . . . ---> 2n-2/n --> 2n-2/n --> 2n/n

As you can see we I have divided the area of the curve we are investigating with n equal parts. Each part being 1/n BIGGER (remember this). The dot dot dot (...) indicates that in between the division, there are many more n equal parts.

But the area of a triangle is given by LxB. What I have is the WIDTH.

So, in order to find my y value, I simply sub it into the original equation.
So lets say we want to find the HEIGHT of a point AT n+1/n.

It will be y = 1/n+1/n
= n/n+1

Now that I have my height, we can start making some real progression in this question.

So EACH EQUAL PART is (1/n) thus the width is 1/n.
Thus, I'll be multiplying the 1/n (which represents the WIDTH), by the height of each corresponding y value of (n/n --> n+1/n --> n+2/n --> n+3/n . . . ---> 2n-2/n --> 2n-2/n --> 2n/n).

Thus I will have:

(1/n)(n/n+1) + (1/n)(n/n+2) . . . + (1/n)(n/2n-2) + (1/n)(n/2n-1) + (1/n)(n/2n)

I can factorise the (1/n) out.

.'. (1/n)[n/n+2 + n/n+2 + . . . + n/n+2 + n/2n-1 + n/2n].

We have hence achieved a result. Now, like I said previously, this is a smaller area than the required. Therefore it is LESS than ln2.

So, we have :

(1/n)[n/n+1 + n/n+2 + . . . + n/n+2 + n/2n-1 + n/2n] < ln2 < ???

To find ???, we will now investigate the (2), where the rectangles are LARGER than the required area.

The process should be considered exactly as part (1). However, the difference comes with the LAST EQUAL PART. That is 2n/n. Why?

The curve is limited from 1-->2, so there is a difference of 1 unit from n/n -> 2n/n. The last rectangle will stretch past the curve to create a SQUARE. Thus its y coordinate at that point is 1. For the width = height for a square.

So we have:

(1/n)[1 + n/n+1 + n/n+2 + . . . + n/n+2 + n/2n-1]

the n/2n has been transformed to 1, for a square is produced at n/2n, thus corresponding it to have an equal height and width. Since the difference from 1->2 is 1, it transforms to one. This might be the hardest concept to realise. I wish I knew how to explain it mathematically.

Furthermore this result depicts LARGER rectangles, and gives an area larger than required. Therefore.

(1/n)[n/n+1 + n/n+2 + . . . + n/n+2 + n/2n-1 + n/2n] < ln2 < (1/n)[1 + n/n+1 + n/n+2 + . . . + n/n+2 + n/2n-1]

(c) [n/n+1 + n/n+2 + . . . + n/n+2 + n/2n-1 + n/2n],

This is weird, usually you'd come up with a conjecture (which is hard), and then your meant to prove it via mathematical induction. After that you do a substitution and you should come up with a nice ln2. I can't think of any real mathematical way to approach this question. Hopefully someone can.
I can however, try and explain it.

So we know the answer is ln2.
Now ln2 = ln|2n/n| = ln 2n - ln n. Furthermore, we can say that our limits are "n" and "2n".
But x = n and x = 2n, so y = 1/n and y = 1/2n respectively.
Following this. with [n/n+1 + n/n+2 + . . . + n/n+2 + n/2n-1 + n/2n], we can factorise an "n" out, hence giving:
n(1/n+1 + 1/n+2 + ... 1/n-2 + 1/n-1 + 1/2n).
Hopefully you realise that y=1/n and y=1/2n fit in nicely ^^^^. Um, and so, if n --> infinity, that means n->2n, for 2n is our limit......
andddddddd im talking shit now.

 

[Trebla]

For c), isn't there be a 1/n factor missing?