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

  • #1
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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 [itex]y = x^n[/itex], 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 [itex]y = x^n[/itex] between the points [itex]x = 0[/itex] and [itex]x = a[/itex] on the x-axis. We imagine that the interval from [itex]x = 0[/itex] to [itex]x = a[/itex] 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 [itex]a^n[/itex], [itex](ar)^n[/itex], [itex](ar^{2})^n[/itex], .... 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,

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

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 [itex]a^{n+1}, (ar)^{n+1}, (ar^{2})^{n+1},[/itex] and so on. Their sum,

[tex]A_{r} = a^{n+1} + (ar)^{n+1} + (ar^{2})^{n+1} + \cdots[/tex]
[tex]= a^{n+1}\left(1 + r^{n+1} + r^{2(n+1)} + \cdots \right)[/tex]
[tex]= \frac{a^{n+1}}{1 - r^{n+1}}[/tex]


Where am I getting wrong?
.
 
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Answers and Replies

  • #2
mathman
Science Advisor
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You are using the ordinates to get the heights of the rectangles. I think the formula is based on using the averages of the adjacent ordinates to get the rectangle heights,
 
  • #3
409
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You are using the ordinates to get the heights of the rectangles. I think the formula is based on using the averages of the adjacent ordinates to get the rectangle heights,
Thanks, but that was not it. I've just figured out my mistake. I got the areas of the rectangles wrong. The sum of the areas would be,

[tex]A_r = (a - ar)a^n + (ar - ar^2)(ar)^n + (ar^2 - ar^3)(ar^2)^n + \cdots[/tex]
[tex]A_r = a^{n+1}(1 - r) \left(1 + r^{n+1} + r^{2(n+1)} + \cdots \right)[/tex]
[tex]A_r = \frac{a^{n+1}(1 - r)}{1 - r^{n+1}}[/tex]
 
  • #4
Stephen Tashi
Science Advisor
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The areas of each rectangle I found are [itex]a^{n+1}, (ar)^{n+1}, (ar^{2})^{n+1},[/itex]
.
I don't understand how you got those areas. Does the base of the first rectangle have length = a or does it have length = (a - ar)?
 
  • #5
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