Math Ext 2 - Got Some Questions (1 Viewer)

Life'sHard · Aug 8, 2021

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This style of question imo is a pretty basic area integration question. I'm just confused with the approach on this specific question. Like what is 2/n is it an x coordinate and where? Normally I would do base times height until you get a series then put it in sigma notation. Any help is appreciated. Thanks

notme123 · Aug 8, 2021

$\frac{2}{n}$ is the width of each rectangle.

Drongoski · Aug 8, 2021

I think the 2 areas are:

$S_1 = \sum_{i=1}^{n-2} \frac{2}{n}\times f(\frac{2i}{n}) = \frac{2}{n} \sum _{i=1} ^{n-2}\sqrt{\frac {2i}{n}(2-\frac {2i}{n})} = \frac {4}{n^2} \sum _{i=1} ^{n-2}\sqrt{(i(n-i))}\\ \\ S_2 = \sum_{n=1}^n \frac{2}{n} \times f(\frac {2i}{n}) = \cdots = \frac{4}{n^2} \sum _{i=1} ^n \sqrt{(i(n-i))}$

tickboom · Aug 8, 2021

These rectangles and curves questions pop up quite a bit in relation to inequality proofs, so it's good to get comfortable with the technique. Here's a few similar questions I've seen on this front:

CM_Tutor · Aug 8, 2021

I agree with @notme123 that the $\cfrac{2}{n}$ values are the widths of each rectangle.

However, on first looking, there appears to me to be a trick in this question. The function is not strictly increasing or decreasing over the domain. As a consequence, the heights of the upper and lower rectangles are not consistently at the start or end of each rectangle, but switch part way through. I don't see that this has been taken into account in @Drongoski's solution. @tickboom's examples are increasing (first example) and decreasing (second example) throughout their domains, so handling that difficulty does not arise. I also suspect that there is an issue with whether $n$ is odd or even because if the stationary point at $(1,\ 1)$ is inside one of the rectangles, rather than at the edge of one, then the height of the upper rectangle will not be at either edge of the central rectangle.

This is definitely a problem with a challenging aspect to it, and I am wondering whether it might not be easier to divide the region into $2n$ strips so that there is definitely a rectangle edge at $x=1$ .

Life'sHard · Aug 8, 2021

notme123 said:
$\frac{2}{n}$ is the width of each rectangle.

Oh that makes a lot more sense now. Thanks

Life'sHard · Aug 8, 2021

CM_Tutor said:
I agree with @notme123 that the $\cfrac{2}{n}$ values are the widths of each rectangle.

However, on first looking, there appears to me to be a trick in this question. The function is not strictly increasing or decreasing over the domain. As a consequence, the heights of the upper and lower rectangles are not consistently at the start or end of each rectangle, but switch part way through. I don't see that this has been taken into account in @Drongoski's solution. @tickboom's examples are increasing (first example) and decreasing (second example) throughout their domains, so handling that difficulty does not arise. I also suspect that there is an issue with whether $n$ is odd or even because if the stationary point at $(1,\ 1)$ is inside one of the rectangles, rather than at the edge of one, then the height of the upper rectangle will not be at either edge of the central rectangle.

This is definitely a problem with a challenging aspect to it, and I am wondering whether it might not be easier to divide the region into $2n$ strips so that there is definitely a rectangle edge at $n=1$ .

Interesting. In the worked solutions I believe they do take into account when x=1. I would have personally missed it out if I wasn't being careful.

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CM_Tutor · Aug 8, 2021

Life'sHard said:
Interesting. In the worked solutions I believe they do take into account when x=1. I would have personally missed it out if I wasn't being careful.
View attachment 31460

I'm not yet convinced that there is not still a problem depending on whether $n$ is odd or even, especially for the upper rectangles.

As a quick check, suppose $n=1$ ... then the lower rectangles should add to zero and the upper to 2... and that formula then has a sum from 1 to 0...

CM_Tutor · Aug 8, 2021

Life'sHard said:
Interesting. In the worked solutions I believe they do take into account when x=1. I would have personally missed it out if I wasn't being careful.
View attachment 31460

Ok, I've tested this by calculating the areas of the upper and lower rectangles for a few cases, where I found:

For $n=1$ , the area of the lower rectangle is 0 square units and the area of the upper rectangle is 2 square units.
For $n=2$ , the area of the lower rectangles is 0 square units and the area of the upper rectangle is 2 square units.
For $n=3$ , the area of the lower rectangles is $\cfrac{4\sqrt{2}}{9}$ square units and the area of the upper rectangle is $\cfrac{6 + 4\sqrt{2}}{9}$ square units.
For $n=4$ , the area of the lower rectangles is $\cfrac{\sqrt{3}}{2}$ square units and the area of the upper rectangle is $\cfrac{2 + \sqrt{3}}{2}$ square units.
For $n=5$ , the area of the lower rectangles is $\cfrac{16 + 4\sqrt{6}}{25}$ square units and the area of the upper rectangle is $\cfrac{26 + 8\sqrt{6}}{25}$ square units.

Please, try calculating these for yourselves and see if I have made any errors (certainly possible).

Now, let's take the formula that is shown in @Life'sHard's post, which is:

$4\sum_{k=1}^{n-1} \cfrac{\sqrt{k(n-k)}}{n^2} - \cfrac{2}{n}$

Trying this with sequential values of $n$ , I get:

$\begin{align*} \text{Put $n = 2$} \qquad 4\sum_{k=1}^{n-1} \cfrac{\sqrt{k(n-k)}}{n^2} - \cfrac{2}{n} &= 4\sum_{k=1}^{2-1} \cfrac{\sqrt{k(2-k)}}{2^2} - \cfrac{2}{2} \\ &= 4\sum_{k=1}^{1} \cfrac{\sqrt{k(2-k)}}{2^2} - 1 \\ &= 4 \cfrac{\sqrt{1(2-1)}}{2^2} - 1 \\ &= \cfrac{4 \times 1}{4} - 1 \\ &= 0 \end{align*}$

$\begin{align*} \text{Put $n = 3$} \qquad 4\sum_{k=1}^{n-1} \cfrac{\sqrt{k(n-k)}}{n^2} - \cfrac{2}{n} &= 4\sum_{k=1}^{2} \cfrac{\sqrt{k(3-k)}}{3^2} - \cfrac{2}{3} \\ &= 4\left(\cfrac{\sqrt{1(3-1)}}{3^2} + \cfrac{\sqrt{2(3-2)}}{3^2}\right) - \cfrac{2}{3} \\ &= 4 \times \cfrac{\sqrt{2} + \sqrt{2}}{3^2} - \cfrac{2}{3} \\ &= \cfrac{4 \times 2\sqrt{2}}{9} - \cfrac{2}{3} \\ &= \cfrac{8\sqrt{2} - 6}{9} \end{align*}$

$\begin{align*} \text{Put $n = 4$} \qquad 4\sum_{k=1}^{n-1} \cfrac{\sqrt{k(n-k)}}{n^2} - \cfrac{2}{n} &= 4\sum_{k=1}^{3} \cfrac{\sqrt{k(4-k)}}{4^2} - \cfrac{2}{4} \\ &= 4\left(\cfrac{\sqrt{1(4-1)}}{4^2} + \cfrac{\sqrt{2(4-2)}}{4^2} + \cfrac{\sqrt{3(4-3)}}{4^2}\right) - \cfrac{1}{2} \\ &= 4 \times \cfrac{\sqrt{3} + 2 + \sqrt{3}}{4^2} - \cfrac{1}{2} \\ &= \cfrac{2 + 2\sqrt{3}}{4} - \cfrac{1}{2} \\ &= \cfrac{2 + 2\sqrt{3} - 2}{4} \\ &= \cfrac{\sqrt{3}}{2} \end{align*}$

The $n = 2$ and $n = 4$ values match my areas for the lower rectangles for those cases... but the $n = 3$ case does not match. I am wondering whether the solution that Life'sHard has is valid.

Life'sHard · Aug 8, 2021

CM_Tutor said:
Ok, I've tested this by calculating the areas of the upper and lower rectangles for a few cases, where I found:

For $n=1$ , the area of the lower rectangle is 0 square units and the area of the upper rectangle is 2 square units.

For $n=2$ , the area of the lower rectangles is 0 square units and the area of the upper rectangle is 2 square units.

For $n=3$ , the area of the lower rectangles is $\cfrac{4\sqrt{2}}{9}$ square units and the area of the upper rectangle is $\cfrac{6 + 4\sqrt{2}}{9}$ square units.

For $n=4$ , the area of the lower rectangles is $\cfrac{\sqrt{3}}{2}$ square units and the area of the upper rectangle is $\cfrac{2 + \sqrt{3}}{2}$ square units.

For $n=5$ , the area of the lower rectangles is $\cfrac{16 + 4\sqrt{6}}{25}$ square units and the area of the upper rectangle is $\cfrac{26 + 8\sqrt{6}}{25}$ square units.

Please, try calculating these for yourselves and see if I have made any errors (certainly possible).

Now, let's take the formula that is shown in @Life'sHard's post, which is:

$4\sum_{k=1}^{n-1} \cfrac{\sqrt{k(n-k)}}{n^2} - \cfrac{2}{n}$

Trying this with sequential values of $n$ , I get:

$\begin{align*} \text{Put $n = 2$} \qquad 4\sum_{k=1}^{n-1} \cfrac{\sqrt{k(n-k)}}{n^2} - \cfrac{2}{n} &= 4\sum_{k=1}^{2-1} \cfrac{\sqrt{k(2-k)}}{2^2} - \cfrac{2}{2} \\ &= 4\sum_{k=1}^{1} \cfrac{\sqrt{k(2-k)}}{2^2} - 1 \\ &= 4 \cfrac{\sqrt{1(2-1)}}{2^2} - 1 \\ &= \cfrac{4 \times 1}{4} - 1 \\ &= 0 \end{align*}$

$\begin{align*} \text{Put $n = 3$} \qquad 4\sum_{k=1}^{n-1} \cfrac{\sqrt{k(n-k)}}{n^2} - \cfrac{2}{n} &= 4\sum_{k=1}^{2} \cfrac{\sqrt{k(3-k)}}{3^2} - \cfrac{2}{3} \\ &= 4\left(\cfrac{\sqrt{1(3-1)}}{3^2} + \cfrac{\sqrt{2(3-2)}}{3^2}\right) - \cfrac{2}{3} \\ &= 4 \times \cfrac{\sqrt{2} + \sqrt{2}}{3^2} - \cfrac{2}{3} \\ &= \cfrac{4 \times 2\sqrt{2}}{9} - \cfrac{2}{3} \\ &= \cfrac{8\sqrt{2} - 6}{9} \end{align*}$

$\begin{align*} \text{Put $n = 4$} \qquad 4\sum_{k=1}^{n-1} \cfrac{\sqrt{k(n-k)}}{n^2} - \cfrac{2}{n} &= 4\sum_{k=1}^{3} \cfrac{\sqrt{k(4-k)}}{4^2} - \cfrac{2}{4} \\ &= 4\left(\cfrac{\sqrt{1(4-1)}}{4^2} + \cfrac{\sqrt{2(4-2)}}{4^2} + \cfrac{\sqrt{3(4-3)}}{4^2}\right) - \cfrac{1}{2} \\ &= 4 \times \cfrac{\sqrt{3} + 2 + \sqrt{3}}{4^2} - \cfrac{1}{2} \\ &= \cfrac{2 + 2\sqrt{3}}{4} - \cfrac{1}{2} \\ &= \cfrac{2 + 2\sqrt{3} - 2}{4} \\ &= \cfrac{\sqrt{3}}{2} \end{align*}$

The $n = 2$ and $n = 4$ values match my areas for the lower rectangles for those cases... but the $n = 3$ case does not match. I am wondering whether the solution that Life'sHard has is valid.

Hmmm very interesting. The solution I have however works with the final part of the question. Your working does make sense. Can you decipher the issue @idkkdi

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CM_Tutor · Aug 8, 2021

Life'sHard said:
Hmmm very interesting. The solution I have however works with the final part of the question. Your working does make sense. Can you decipher the issue @idkkdi
View attachment 31480

I suspect that there are different formulae for $n$ even and $n$ odd, both with the same limit, similar to a situation like

$t_n = 1 + \cfrac{(-1)^n}{2^n}$

which gives

$t_1 = 1 + \cfrac{(-1)^1}{2^1} = 1 - \cfrac{1}{2} = \cfrac{1}{2}$

$t_2 = 1 + \cfrac{(-1)^2}{2^2} = 1 + \cfrac{1}{4} = \cfrac{5}{4}$

$t_3 = 1 + \cfrac{(-1)^3}{2^3} = 1 - \cfrac{1}{8} = \cfrac{7}{8}$

$t_4 = 1 + \cfrac{(-1)^4}{2^4} = 1 + \cfrac{1}{16} = \cfrac{17}{16}$

$t_5 = 1 + \cfrac{(-1)^5}{2^5} = 1 - \cfrac{1}{32} = \cfrac{31}{32}$

$t_6 = 1 + \cfrac{(-1)^6}{2^6} = 1 + \cfrac{1}{64} = \cfrac{65}{64}$

and so we have an increasing sequence $t_1,\ t_3,\ t_5,\ ...$ that is approaching 1 from below and a decreasing sequence $t_2,\ t_4,\ t_6,\ ...$ that is approaching 1 from above.

If you are familiar with the Fibonacci sequence $\{F_n\}$ , the sequence of the ratios of consecutive terms $\cfrac{F_2}{F_1},\ \cfrac{F_3}{F_2},\ \cfrac{F_4}{F_3},\ , ...$ has exactly this type of property, with this sequence approaching its limiting value of $\phi = \cfrac{1 + \sqrt{5}}{2}$ by successive terms alternating above and below the value of $\phi$ .

In this case, if the formula is valid for $n$ even then the limit will still work.

Life'sHard · Aug 8, 2021

CM_Tutor said:
I suspect that there are different formulae for $n$ even and $n$ odd, both with the same limit, similar to a situation like

$t_n = 1 + \cfrac{(-1)^n}{2^n}$

which gives

$t_1 = 1 + \cfrac{(-1)^1}{2^1} = 1 - \cfrac{1}{2} = \cfrac{1}{2}$

$t_2 = 1 + \cfrac{(-1)^2}{2^2} = 1 + \cfrac{1}{4} = \cfrac{5}{4}$

$t_3 = 1 + \cfrac{(-1)^3}{2^3} = 1 - \cfrac{1}{8} = \cfrac{7}{8}$

$t_4 = 1 + \cfrac{(-1)^4}{2^4} = 1 + \cfrac{1}{16} = \cfrac{17}{16}$

$t_5 = 1 + \cfrac{(-1)^5}{2^5} = 1 - \cfrac{1}{32} = \cfrac{31}{32}$

$t_6 = 1 + \cfrac{(-1)^6}{2^6} = 1 + \cfrac{1}{64} = \cfrac{65}{64}$

and so we have an increasing sequence $t_1,\ t_3,\ t_5,\ ...$ that is approaching 1 from below and a decreasing sequence $t_2,\ t_4,\ t_6,\ ...$ that is approaching 1 from above.

If you are familiar with the Fibonacci sequence $\{F_n\}$ , the sequence of the ratios of consecutive terms $\cfrac{F_2}{F_1},\ \cfrac{F_3}{F_2},\ \cfrac{F_4}{F_3},\ , ...$ has exactly this type of property, with this sequence approaching its limiting value of $\phi = \cfrac{1 + \sqrt{5}}{2}$ by successive terms alternating above and below the value of $\phi$ .

In this case, if the formula is valid for $n$ even then the limit will still work.

Yep makes total sense. Limiting sum approaching pi/2. As for the n=odd/even I'm not too sure.

Math Ext 2 - Got Some Questions (1 Viewer)

Life'sHard

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