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Limit of Bounding Products (Ext 2 Problem 16 style) (1 Viewer)

vuhung

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I've composed a problem so student can practise with problem 16s.
Let me know what do you think.

bounding product.png
 

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WeiWeiMan

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I've composed a problem so student can practise with problem 16s.
Let me know what do you think.

View attachment 51038
(i)
Let f(x) = ln(1+x) - x + x^2/2
f(0) =ln(1) - 0 + 0 =0
f’(x) = 1/(x+1) -1 + x
= x+1 + 1/(x+1) -2
> 2 -2 {by AM/GM for x>0, no equality since x =/= 0 => x+1 =/= 1/(x+1)
= 0
f’(x) > 0 for x > 0 and f(0) = 0
Thus, f(x) > 0 for x > 0
=> 0 < ln(1+x) -x + x^2/2
=> x - x^2/2 < ln(1+x) QED

(ii)
Let x = k/n^2 for k = 1,2,3,…, n

Via x-x^2/2 < ln(1+x) < x: [given]
1/n^2 - 1/2n^4 < ln(1+1/n^2) < 1/n^2
2/n^2 - 2^2/2n^4 < ln(1+2/n^2) < 2/n^2

n/n^2 - n^2/2n^2 < ln(1+n/n^2) < n/n^2

Adding these:
(1/n^2 + 2/n^2 + … + n/n^2) - (1/2n^4 + 2^2/2n^4 + … + n^2/2n^2) < ln[(1+1/n^2)(1+2/n^2)…(1+n/n^2)] < 1/n^2 + 2/n^2 + … + n/n^2

Factorising and simplifying a bit:

1/n^2 (1 + 2 + … + n) -1/2n^4 (1 + 2^2 + … + n^2) < ln(Pn) < 1/n^2 (1+2+… + n)

I am far too lazy to prove 1+2^2 + … + n^2 = n(n+1)(2n+1)/6 so I’ll just accept it as is

Using this alongside 1+2+…+n = n(n+1)/2 (Via AP):

n(n+1)/2n^2 - n(n+1)(2n+1)/12n^2 < ln(Pn) < n(n+1)/2n^2

Exponentiating all 3:
e^[n(n+1)/2n^2 - n(n+1)(2n+1)/12n^2] < Pn < e^[n(n+1)/2n^2]

I’ll do (iii) and (iv) later since I’m doing these on phone and it’s a genuine pain to type it out

My rough ideas are:

Sandwich/Squeeze, looks like sqrt(e)
Question literally tells you what to do
 
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vuhung

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Nice work Wei Wei.

You'll find some interesting findings working with part (iv), linking to part (i), Taylor expansion and the squeeze theorem.
 

WeiWeiMan

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Nice work Wei Wei.

You'll find some interesting findings working with part (iv), linking to part (i), Taylor expansion and the squeeze theorem.
Yeah I’ll get it done later when I’m bothered
 

WeiWeiMan

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is this even in the syllbus 💀
Almost certainly
(i), (ii) and (iii) can be very reasonably asked in 4U, however, (ii) is a bit annoying to do without the realistic ability to quote without proof that (1+2^2+3^2+...+n^2) = n(n+1)(2n+1)/6.

(iv) is nice to think about, however, not super HSC style I'm pretty sure. I think it's still a reasonable question since it tells you how you should arrive at the proof though.
 

ivanradoszyce

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Is this problem from a text book? If so, can you supply the title and author(s).

Thanks
 

vuhung

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I am the author of the booklet “The Last Resorts” which the problem is taken from.
You can download the booklet from my profile page.
 

ivanradoszyce

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Excellent material.

Thank-you so much for the time and effort in creating this resource.
 

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