2007 Abbotsleigh Q8 (1 Viewer)

asianese · Aug 27, 2012

Need a hint for ii)

Thankyou!

Carrotsticks · Aug 27, 2012

Multiply both sides of (i) by 2n.

I suspect this because the expression in (ii) is the EXACT same thing as (i), except in sigma notation (doesn't change anything) and there's a 2n in front.

seanieg89 · Aug 27, 2012

From previous part, LHS<4n^2/pi=n^2*(4/pi) < n^2*(pi/2).

seanieg89 · Aug 27, 2012

Carrotsticks said:
Multiply both sides of (i) by 2n.

I suspect this because the expression in (ii) is the EXACT same thing as (i), except in sigma notation (doesn't change anything) and there's a 2n in front.

It isn't quite the same. But 2n*(RHS of i) is strictly smaller than RHS of ii.

Carrotsticks · Aug 27, 2012

seanieg89 said:
It isn't quite the same. But 2n*(RHS of i) is strictly smaller than RHS of ii.

Carrotsticks said:
Multiply both sides of (i) by 2n.

I suspect this because the expression in (ii) is the EXACT same thing as (i), except in sigma notation (doesn't change anything) and there's a 2n in front.

.

seanieg89 · Aug 27, 2012

Carrotsticks said:
.

I read what you wrote, the second RHS isn't the first RHS multiplied by 2n, there is an inequality involved.

asianese · Aug 27, 2012

Even if you multiply by 2n it doesn't fall out.

Thanks sean and carrot.

Carrotsticks · Aug 27, 2012

Yep yep, OP wanted a hint so I just gave him a start and hoped he would figure out the rest.

asianese · Aug 27, 2012

Thanks!

Sindivyn · Aug 27, 2012

For part i - did you just integrate the area using the equation of the graph, then state that the rectangles would be less than that?

RealiseNothing · Aug 27, 2012

Sindivyn said:
For part i - did you just integrate the area using the equation of the graph, then state that the rectangles would be less than that?

Integrate the curve to find the area:

$\int_{0}^{n} nsin{\frac{\pi x}{2n}} dx$

$\frac{-2n^2}{\pi}[cos{\frac{\pi x}{2n}}]_0^n$

$\frac{-2n^2}{\pi}[cos{\frac{\pi}{2}} - cos{0}]$

$\frac{-2n^2}{\pi}[-1]$

$\frac{2n^2}{\pi}$

This is the area under the curve.

Now add up the rectangles. They all have area width times length, and the width is always one, so the area of each rectangle is just equal to the height, which is the 'y' value, so just sub in 'x' values into the equation of the curve. The sum of the areas of the rectangles is:

$nsin{\frac{\pi}{2n}} + nsin{\frac{2\pi}{2n}} + nsin{\frac{3\pi}{2n}} + ... + nsin{\frac{\pi(n-1)}{2n}}$

This is however, less than the area under the curve, so:

$nsin{\frac{\pi}{2n}} + nsin{\frac{2\pi}{2n}} + nsin{\frac{3\pi}{2n}} + ... + nsin{\frac{\pi(n-1)}{2n}} < \frac{2n^2}{\pi}$

Divide both sides by 'n':

$sin{\frac{\pi}{2n}} + sin{\frac{2\pi}{2n}} + sin{\frac{3\pi}{2n}} + ... + sin{\frac{\pi(n-1)}{2n}} < \frac{2n}{\pi}$

As required.

bleakarcher · Aug 27, 2012

RealiseNothing said:
Integrate the curve to find the area:

$\int_{0}^{n} nsin{\frac{\pi x}{2n}} dx$

$\frac{-2n^2}{\pi}[cos{\frac{\pi x}{2n}}]_0^n$

$\frac{-2n^2}{\pi}[cos{\frac{\pi}{2}} - cos{0}]$

$\frac{-2n^2}{\pi}[-1]$

$\frac{2n^2}{\pi}$

This is the area under the curve.

Now add up the rectangles. They all have area width times length, and the width is always one, so the area of each rectangle is just equal to the height, which is the 'y' value, so just sub in 'x' values into the equation of the curve. The sum of the areas of the rectangles is:

$nsin{\frac{\pi}{2n}} + nsin{\frac{2\pi}{2n}} + nsin{\frac{3\pi}{2n}} + ... + nsin{\frac{\pi(n-1)}{2n}}$

This is however, less than the area under the curve, so:

$nsin{\frac{\pi}{2n}} + nsin{\frac{2\pi}{2n}} + nsin{\frac{3\pi}{2n}} + ... + nsin{\frac{\pi(n-1)}{2n}} < \frac{2n^2}{\pi}$

Divide both sides by 'n':

$sin{\frac{\pi}{2n}} + sin{\frac{2\pi}{2n}} + sin{\frac{3\pi}{2n}} + ... + sin{\frac{\pi(n-1)}{2n}} < \frac{2n}{\pi}$

As required.

Don't you integrate from x=1 to n?

RealiseNothing · Aug 27, 2012

bleakarcher said:
Don't you integrate from x=1 to n?

No, that would make it much harder.

bleakarcher · Aug 27, 2012

RealiseNothing said:
No, that would make it much harder.

the lower rectangles start at x=1 though?

RealiseNothing · Aug 27, 2012

bleakarcher said:
the lower rectangles start at x=1 though?

$nsin{\frac{\pi}{2n}} + nsin{\frac{2\pi}{2n}} + nsin{\frac{3\pi}{2n}} + ... + nsin{\frac{\pi(n-1)}{2n} < \int_{1}^{n} nsin{\frac{\pi x}{2n}}dx$

$\int_{1}^{n} nsin{\frac{\pi x}{2n}}dx < \int_{0}^{n} nsin{\frac{\pi x}{2n}}dx$

It doesn't matter whether we integrate from 0 to n or from 1 to n, they are both greater than the sum of the rectangles.

bleakarcher · Aug 27, 2012

RealiseNothing said:
$nsin{\frac{\pi}{2n}} + nsin{\frac{2\pi}{2n}} + nsin{\frac{3\pi}{2n}} + ... + nsin{\frac{\pi(n-1)}{2n} < \int_{1}^{n} nsin{\frac{\pi x}{2n}}dx$

$\int_{1}^{n} nsin{\frac{\pi x}{2n}}dx < \int_{0}^{n} nsin{\frac{\pi x}{2n}}dx$

It doesn't matter whether we integrate from 0 to n or from 1 to n, they are both greater than the sum of the rectangles.

oh yeah true lol..

RealiseNothing · Aug 27, 2012

bleakarcher said:
oh yeah true lol..

Integrating from 0 to n just works out nicer.

bleakarcher · Aug 27, 2012

RealiseNothing said:
Integrating from 0 to n just works out nicer.

yeah

2007 Abbotsleigh Q8 (1 Viewer)

asianese

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seanieg89

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