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# 数学代写|实分析代写Real Analysis代考|Math444 Archimedes’ Argument

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## 数学代写|实分析代写Real Analysis代考|Archimedes’ Argument

Let $K$ denote the area bounded by the parabolic arc and the line segment. Archimedes showed that each time we add new triangles, the area of the region inside the parabolic arc that is not covered by our triangles is reduced by more than half (see exercises 2.1.2-2.1.3). It follows that we can make this error as small as we want by taking enough inscribed triangles. If $K$ were larger than $4 / 3$, then we could inscribe triangles until their total area was more than $4 / 3$. This would contradict equation (2.1) which says that the sum of the areas of the inscribed triangles is always strictly less than $4 / 3$. If $K$ were smaller than $4 / 3$, then we could find a $k$ for which $4 / 3-1 /\left(4^k \cdot 3\right)$ is larger than $K$. But then equation (2.1) tells us that the sum of the areas of the corresponding inscribed triangles is strictly larger than $K$. This contradicts the fact that the sum of the areas of inscribed triangles cannot exceed the total area.

This method of calculating areas by summing inscribed triangles is often referred to as the “method of exhaustion.” E. J. Dijksterhuis has pointed out that this is “the worst name that could have been devised.” As Archimedes or Eudoxus of Cnidus ( $c a$. 408-355 в.C.) (the first to employ this method) would have insisted, you never exhaust the area. You only get arbitrarily close to it.

Archimedes argument is important because it points to our modern definition of the infinite series $1+1 / 4+1 / 16+\cdots+1 / 4^n+\cdots$. Just as Archimedes handled his infinite process by producing a value and demonstrating that the answer could be neither greater nor less than this produced value, so Cauchy and others of the early nineteenth century would handle infinite series by producing the desired value and demonstrating that the series could not have a value either greater or less than this.

## 数学代写|实分析代写Real Analysis代考|The Oddity of Infinite Sums

Ordinary sums are very well behaved. They are associative, which means that it does not matter how we group them:
$$(2+3)+5=2+(3+5)$$
and they are commutative, which means that it does not matter how we order them:
$$2+3+5=3+5+2$$
These simple facts do not always hold for infinite sums. If we could group an infinite sum any way we wanted, then we would have that
\begin{aligned} 1 & -1+1-1+1-1+1-1+\cdots \ & =(1-1)+(1-1)+(1-1)+(1-1)+\cdots \ & =0 \end{aligned}

whereas by regrouping we obtain
\begin{aligned} 1 & -1+1-1+1-1+1-1+\cdots \ & =1+(-1+1)+(-1+1)+(-1+1)+(-1+1)+\cdots \ & =1 \end{aligned}

## 数学代写|实分析代写Real Analysis代考|The Oddity of Infinite Sums

$$(2+3)+5=2+(3+5)$$

$$2+3+5=3+5+2$$

$$1-1+1-1+1-1+1-1+\cdots \quad=(1-1)+(1-1)+(1-1)+(1-1)+\cdots=0$$

$$1-1+1-1+1-1+1-1+\cdots \quad=1+(-1+1)+(-1+1)+(-1+1)+(-1+1)+\cdots=1$$

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