Apostol Calculus, Method of Exhaustion
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In Apostol's Calculus, he goes through the method of exhaustion to find the area under a parabola from $0 to b$. Using the fact that,
begin{align}
&1^2+2^2+...+(n-1)^2 < frac{n^3}{3} < 1^2+2^2+...+n^2label{1} \
&Rightarrow frac{b^3}{n^3}(1^2+2^2+...+(n-1)^2) < frac{b^3}{3} < frac{b^3}{n^3}(1^2+2^2+...+n^2) nonumber \
&Rightarrow s_{n} < frac{b^3}{3} < S_{n} nonumber
end{align}
where $s_{n}$ and $S_{n}$ are the lower and upper approximation (rectangles), respectively, for the area under the parabola. We then have to show that $A=frac{b^3}{3}$ is the only number that satisfies
begin{equation}
s_{n}<A<S_{n} label{2}
end{equation}
for every $ngeq1$. Using the left-most side of the first inequality, he adds $n^2$ and then multiplies both sides by $frac{b^3}{n^3}$,
begin{align*}
&frac{b^3}{n^3}(1^2+2^2+...+n^2)<frac{b^3}{3}+frac{b^3}{n^2} \
&Rightarrow S_{n}<frac{b^3}{3}+frac{b^3}{n^2}
end{align*}
for the right-most side of the inequality, he subtracts $n^2$ and multiplies by $frac{b^3}{n^3}$,
begin{align*}
&frac{b^3}{3}-frac{b^3}{n^2}<frac{b^3}{n^3}(1^2+2^2+...+(n-1)^2)\
&Rightarrow frac{b^3}{3}-frac{b^3}{n^3}<s_{n}
end{align*}
this implies,
begin{align*}
frac{b^3}{3}-frac{b^3}{n^2}<A<frac{b^3}{3}+frac{b^3}{n^2}
end{align*}
This is where it gets confusing. He says the only possibilities are:
$$begin{array}{ccc}
A>frac{b^3}{3},& A<frac{b^3}{3},& A=frac{b^3}{3}
end{array}$$
and proceeds to show that $A=frac{b^3}{3}$ via contradictions for the first two cases. This is fine, but what about $frac{b^3}{n^2}$ in the inequality? why don't we have to consider the possible relationships between $A$ and $frac{b^3}{n^2}$?
calculus area
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In Apostol's Calculus, he goes through the method of exhaustion to find the area under a parabola from $0 to b$. Using the fact that,
begin{align}
&1^2+2^2+...+(n-1)^2 < frac{n^3}{3} < 1^2+2^2+...+n^2label{1} \
&Rightarrow frac{b^3}{n^3}(1^2+2^2+...+(n-1)^2) < frac{b^3}{3} < frac{b^3}{n^3}(1^2+2^2+...+n^2) nonumber \
&Rightarrow s_{n} < frac{b^3}{3} < S_{n} nonumber
end{align}
where $s_{n}$ and $S_{n}$ are the lower and upper approximation (rectangles), respectively, for the area under the parabola. We then have to show that $A=frac{b^3}{3}$ is the only number that satisfies
begin{equation}
s_{n}<A<S_{n} label{2}
end{equation}
for every $ngeq1$. Using the left-most side of the first inequality, he adds $n^2$ and then multiplies both sides by $frac{b^3}{n^3}$,
begin{align*}
&frac{b^3}{n^3}(1^2+2^2+...+n^2)<frac{b^3}{3}+frac{b^3}{n^2} \
&Rightarrow S_{n}<frac{b^3}{3}+frac{b^3}{n^2}
end{align*}
for the right-most side of the inequality, he subtracts $n^2$ and multiplies by $frac{b^3}{n^3}$,
begin{align*}
&frac{b^3}{3}-frac{b^3}{n^2}<frac{b^3}{n^3}(1^2+2^2+...+(n-1)^2)\
&Rightarrow frac{b^3}{3}-frac{b^3}{n^3}<s_{n}
end{align*}
this implies,
begin{align*}
frac{b^3}{3}-frac{b^3}{n^2}<A<frac{b^3}{3}+frac{b^3}{n^2}
end{align*}
This is where it gets confusing. He says the only possibilities are:
$$begin{array}{ccc}
A>frac{b^3}{3},& A<frac{b^3}{3},& A=frac{b^3}{3}
end{array}$$
and proceeds to show that $A=frac{b^3}{3}$ via contradictions for the first two cases. This is fine, but what about $frac{b^3}{n^2}$ in the inequality? why don't we have to consider the possible relationships between $A$ and $frac{b^3}{n^2}$?
calculus area
You have a confusing typo. The far LHS and far RHS of the inequality above "This is where it get confusing" are identical. Please edit.
– DanielWainfleet
yesterday
I see! I have fixed it, thank you very much.
– Jonathan Duran
yesterday
1
For $bne 0$ we have $|b^{-3}(A-b^3/3|<1/n^2$ for $all$ $nin Bbb Z^+$, which is not possible unless $b^{-3}(A-b^3/3)=0$, which is not possible unless $A=b^3/3$. Note that $A$ and $b$ have no relation to $n$.
– DanielWainfleet
yesterday
What about $A<frac{b^3}{3}$, wouldn't that satisfy the inequality in your response?
– Jonathan Duran
14 hours ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In Apostol's Calculus, he goes through the method of exhaustion to find the area under a parabola from $0 to b$. Using the fact that,
begin{align}
&1^2+2^2+...+(n-1)^2 < frac{n^3}{3} < 1^2+2^2+...+n^2label{1} \
&Rightarrow frac{b^3}{n^3}(1^2+2^2+...+(n-1)^2) < frac{b^3}{3} < frac{b^3}{n^3}(1^2+2^2+...+n^2) nonumber \
&Rightarrow s_{n} < frac{b^3}{3} < S_{n} nonumber
end{align}
where $s_{n}$ and $S_{n}$ are the lower and upper approximation (rectangles), respectively, for the area under the parabola. We then have to show that $A=frac{b^3}{3}$ is the only number that satisfies
begin{equation}
s_{n}<A<S_{n} label{2}
end{equation}
for every $ngeq1$. Using the left-most side of the first inequality, he adds $n^2$ and then multiplies both sides by $frac{b^3}{n^3}$,
begin{align*}
&frac{b^3}{n^3}(1^2+2^2+...+n^2)<frac{b^3}{3}+frac{b^3}{n^2} \
&Rightarrow S_{n}<frac{b^3}{3}+frac{b^3}{n^2}
end{align*}
for the right-most side of the inequality, he subtracts $n^2$ and multiplies by $frac{b^3}{n^3}$,
begin{align*}
&frac{b^3}{3}-frac{b^3}{n^2}<frac{b^3}{n^3}(1^2+2^2+...+(n-1)^2)\
&Rightarrow frac{b^3}{3}-frac{b^3}{n^3}<s_{n}
end{align*}
this implies,
begin{align*}
frac{b^3}{3}-frac{b^3}{n^2}<A<frac{b^3}{3}+frac{b^3}{n^2}
end{align*}
This is where it gets confusing. He says the only possibilities are:
$$begin{array}{ccc}
A>frac{b^3}{3},& A<frac{b^3}{3},& A=frac{b^3}{3}
end{array}$$
and proceeds to show that $A=frac{b^3}{3}$ via contradictions for the first two cases. This is fine, but what about $frac{b^3}{n^2}$ in the inequality? why don't we have to consider the possible relationships between $A$ and $frac{b^3}{n^2}$?
calculus area
In Apostol's Calculus, he goes through the method of exhaustion to find the area under a parabola from $0 to b$. Using the fact that,
begin{align}
&1^2+2^2+...+(n-1)^2 < frac{n^3}{3} < 1^2+2^2+...+n^2label{1} \
&Rightarrow frac{b^3}{n^3}(1^2+2^2+...+(n-1)^2) < frac{b^3}{3} < frac{b^3}{n^3}(1^2+2^2+...+n^2) nonumber \
&Rightarrow s_{n} < frac{b^3}{3} < S_{n} nonumber
end{align}
where $s_{n}$ and $S_{n}$ are the lower and upper approximation (rectangles), respectively, for the area under the parabola. We then have to show that $A=frac{b^3}{3}$ is the only number that satisfies
begin{equation}
s_{n}<A<S_{n} label{2}
end{equation}
for every $ngeq1$. Using the left-most side of the first inequality, he adds $n^2$ and then multiplies both sides by $frac{b^3}{n^3}$,
begin{align*}
&frac{b^3}{n^3}(1^2+2^2+...+n^2)<frac{b^3}{3}+frac{b^3}{n^2} \
&Rightarrow S_{n}<frac{b^3}{3}+frac{b^3}{n^2}
end{align*}
for the right-most side of the inequality, he subtracts $n^2$ and multiplies by $frac{b^3}{n^3}$,
begin{align*}
&frac{b^3}{3}-frac{b^3}{n^2}<frac{b^3}{n^3}(1^2+2^2+...+(n-1)^2)\
&Rightarrow frac{b^3}{3}-frac{b^3}{n^3}<s_{n}
end{align*}
this implies,
begin{align*}
frac{b^3}{3}-frac{b^3}{n^2}<A<frac{b^3}{3}+frac{b^3}{n^2}
end{align*}
This is where it gets confusing. He says the only possibilities are:
$$begin{array}{ccc}
A>frac{b^3}{3},& A<frac{b^3}{3},& A=frac{b^3}{3}
end{array}$$
and proceeds to show that $A=frac{b^3}{3}$ via contradictions for the first two cases. This is fine, but what about $frac{b^3}{n^2}$ in the inequality? why don't we have to consider the possible relationships between $A$ and $frac{b^3}{n^2}$?
calculus area
calculus area
edited yesterday
asked 2 days ago
Jonathan Duran
504
504
You have a confusing typo. The far LHS and far RHS of the inequality above "This is where it get confusing" are identical. Please edit.
– DanielWainfleet
yesterday
I see! I have fixed it, thank you very much.
– Jonathan Duran
yesterday
1
For $bne 0$ we have $|b^{-3}(A-b^3/3|<1/n^2$ for $all$ $nin Bbb Z^+$, which is not possible unless $b^{-3}(A-b^3/3)=0$, which is not possible unless $A=b^3/3$. Note that $A$ and $b$ have no relation to $n$.
– DanielWainfleet
yesterday
What about $A<frac{b^3}{3}$, wouldn't that satisfy the inequality in your response?
– Jonathan Duran
14 hours ago
add a comment |
You have a confusing typo. The far LHS and far RHS of the inequality above "This is where it get confusing" are identical. Please edit.
– DanielWainfleet
yesterday
I see! I have fixed it, thank you very much.
– Jonathan Duran
yesterday
1
For $bne 0$ we have $|b^{-3}(A-b^3/3|<1/n^2$ for $all$ $nin Bbb Z^+$, which is not possible unless $b^{-3}(A-b^3/3)=0$, which is not possible unless $A=b^3/3$. Note that $A$ and $b$ have no relation to $n$.
– DanielWainfleet
yesterday
What about $A<frac{b^3}{3}$, wouldn't that satisfy the inequality in your response?
– Jonathan Duran
14 hours ago
You have a confusing typo. The far LHS and far RHS of the inequality above "This is where it get confusing" are identical. Please edit.
– DanielWainfleet
yesterday
You have a confusing typo. The far LHS and far RHS of the inequality above "This is where it get confusing" are identical. Please edit.
– DanielWainfleet
yesterday
I see! I have fixed it, thank you very much.
– Jonathan Duran
yesterday
I see! I have fixed it, thank you very much.
– Jonathan Duran
yesterday
1
1
For $bne 0$ we have $|b^{-3}(A-b^3/3|<1/n^2$ for $all$ $nin Bbb Z^+$, which is not possible unless $b^{-3}(A-b^3/3)=0$, which is not possible unless $A=b^3/3$. Note that $A$ and $b$ have no relation to $n$.
– DanielWainfleet
yesterday
For $bne 0$ we have $|b^{-3}(A-b^3/3|<1/n^2$ for $all$ $nin Bbb Z^+$, which is not possible unless $b^{-3}(A-b^3/3)=0$, which is not possible unless $A=b^3/3$. Note that $A$ and $b$ have no relation to $n$.
– DanielWainfleet
yesterday
What about $A<frac{b^3}{3}$, wouldn't that satisfy the inequality in your response?
– Jonathan Duran
14 hours ago
What about $A<frac{b^3}{3}$, wouldn't that satisfy the inequality in your response?
– Jonathan Duran
14 hours ago
add a comment |
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You have a confusing typo. The far LHS and far RHS of the inequality above "This is where it get confusing" are identical. Please edit.
– DanielWainfleet
yesterday
I see! I have fixed it, thank you very much.
– Jonathan Duran
yesterday
1
For $bne 0$ we have $|b^{-3}(A-b^3/3|<1/n^2$ for $all$ $nin Bbb Z^+$, which is not possible unless $b^{-3}(A-b^3/3)=0$, which is not possible unless $A=b^3/3$. Note that $A$ and $b$ have no relation to $n$.
– DanielWainfleet
yesterday
What about $A<frac{b^3}{3}$, wouldn't that satisfy the inequality in your response?
– Jonathan Duran
14 hours ago