A question about Fermat's method of calculating areas under curves

[murshid]

I am currently reading the book "e: The Story of a Number" by Eli Maor. And I got stuck at something. In chapter 7 of the book, the author described the method Fermat used to calculate areas under curves of the form [imath]y = x^n[/imath], where n is a positive integer. I am quoting the relevant bit here (sorry, I can't show the figure, but from the description, you can easily receate it):

 

Figure 19 shows a portion of the curve [imath]y = x^n[/imath] between the points [imath]x = 0[/imath] and [imath]x = a[/imath] on the x-axis. We imagine that the interval from [imath]x = 0[/imath] to [imath]x = a[/imath] is divided into an infinite number of subintervals by the points ... K, L, M, N, where ON = a. Then, starting at N and working backward, if these intervals are to form a decreasing geometric progression, we have ON = a, OM = ar, OL = ar², and so on, where r is less than 1. The heights (ordinates) to the curve at these points are then [imath]a^n[/imath], [imath](ar)^n[/imath], [imath](ar^{2})^n[/imath], .... From this it is easy to find the area of each rectangle and then sum up the areas, using the summation formula for an infinite geometric series. The result is the formula,

 

[math]A_{r} = \frac{a^{n+1}(1-r)}{1 - r^{n+1}}[/math]

 

where the subscript r under the A indicates that this area still depends on our choice of r.

 

 

Now I can't get to the final formula. The areas of each rectangle I found are [imath]a^{n+1}, (ar)^{n+1}, (ar^{2})^{n+1},[/imath] and so on. Their sum,

 

[math]A_{r} = a^{n+1} + (ar)^{n+1} + (ar^{2})^{n+1} + \cdots[/math]

 

[math]= a^{n+1}\left(1 + r^{n+1} + r^{2(n+1)} + \cdots \right)[/math]

 

[math]= \frac{a^{n+1}}{1 - r^{n+1}}[/math]

 

 

Where am I getting wrong?

.

Edited November 7, 2011 by murshid

[baric]

I am currently reading the book "e: The Story of a Number" by Eli Maor. And I got stuck at something. In chapter 7 of the book, the author described the method Fermat used to calculate areas under curves of the form [imath]y = x^n[/imath], where n is a positive integer. I am quoting the relevant bit here (sorry, I can't show the figure, but from the description, you can easily receate it):

 

 

 

 

Now I can't get to the final formula. The areas of each rectangle I found are [imath]a^{n+1}, (ar)^{n+1}, (ar^{2})^{n+1},[/imath] and so on. Their sum,

 

[math]A_{r} = a^{n+1} + (ar)^{n+1} + (ar^{2})^{n+1} + \cdots[/math]

 

[math]= a^{n+1}\left(1 + r^{n+1} + r^{2(n+1)} + \cdots \right)[/math]

 

[math]= \frac{a^{n+1}}{1 - r^{n+1}}[/math]

 

 

Where am I getting wrong?

.

 

ok, I get a different answer from the book, but trying to sum based on the geometric series you provided. However, I come up with a different answer from yours as well. We are doing something wrong!

 

if given:

[math]A_{r} = a^{n+1} + (ar)^{n+1} + (ar^{2})^{n+1} + \cdots + (ar^{x-1})^{n+1}[/math]

 

then

[math]A_{r}(r^{n+1}) = (ar)^{n+1} + (ar^{2})^{n+1} + \cdots + (ar^{x})^{n+1}[/math]

 

thus

[math]A_{r} - A_{r}(r^{n+1}) = a^{n+1} - (ar^{x})^{n+1}[/math]

 

[math]A_{r} (1 - r^{n+1}) = a^{n+1} (1 - (r^{x})^{n+1})[/math]

 

 

[math]A_{r}= a^{n+1} \frac{ (1 - (r^{x})^{n+1})}{1 - r^{n+1}}[/math]

Edited November 7, 2011 by baric

[murshid]

ok, I get a different answer from the book, but trying to sum based on the geometric series you provided. However, I come up with a different answer from yours as well. We are doing something wrong!

 

You added a finite geometric series. I added an infinite one.

.

[uncool]

I think you may have forgotten the length of the interval in calculating its area.

=Uncool-

[baric]

You added a finite geometric series. I added an infinite one.

.

 

Yeah. I was looking at the book's result that you quoted and assumed they were using the formula for a finite series. That's what their result looks like, anyway.

Edited November 7, 2011 by baric

[DrRocket]

Has anyone noticed that while uncool's observation is correct -- you do get the expression quoted from the book for the sum of the areas of the rectangles associated with the infinite geometric series if you take into acount the bases of those rectangles -- the result is a lousy approximation to the area under the curve. In fact it is high by a factor of approximately n+1 for small values of r ?

 

You can see this from doing the integral. But if you simply consider the case n=a=1 (a triangle over the unit interval) for which any school boy can see that the area is 1/2 , while Ar is about 1 for r close to 0.

[murshid]

I think you may have forgotten the length of the interval in calculating its area.

Has anyone noticed that while uncool's observation is correct -- you do get the expression quoted from the book for the sum of the areas of the rectangles associated with the infinite geometric series if you take into acount the bases of those rectangles -- the result is a lousy approximation to the area under the curve. In fact it is high by a factor of approximately n+1 for small values of r ?

 

You can see this from doing the integral. But if you simply consider the case n=a=1 (a triangle over the unit interval) for which any school boy can see that the area is 1/2 , while Ar is about 1 for r close to 0.

 

Thanks a lot for your help, but I have come here to tell you guys that I've just figured out my mistake. I got the areas of the rectangles wrong. The sum of the areas would be,

 

[math]A_r = (a - ar)a^n + (ar - ar^2)(ar)^n + (ar^2 - ar^3)(ar^2)^n + \cdots[/math]

 

[math]A_r = a^{n+1}(1 - r) \left(1 + r^{n+1} + r^{2(n+1)} + \cdots \right)[/math]

 

[math]A_r = \frac{a^{n+1}(1 - r)}{1 - r^{n+1}}[/math]

 

 

DrRocket, yes, it is a crude approximation, which is why Fermat reasoned that the width of each rectangle must be made small. And for this, r must be close to 1.

 

[math]A_r = \frac{a^{n+1}(1 - r)}{1 - r^{n+1}}[/math]

 

[math]A_r = \frac{a^{n+1}(1 - r)}{(1 - r)(1 + r + r^2 + \cdots + r^n)}[/math]

 

[math]A_r = \frac{a^{n+1}}{1 + r + r^2 + \cdots + r^n}[/math]

 

 

As we let [imath]r \rightarrow 1[/imath], each term is the denominator tends to 1, resulting in,

 

[math]A = \frac{a^{n+1}}{n + 1}[/math]

 

 

And that is the integration formula [math]\int_{0}^{a} x^n dx = \frac{a^{n+1}}{n + 1}[/math]

.

Edited November 8, 2011 by murshid