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HSC 2015 MX2 Marathon ADVANCED (archive) (1 Viewer)

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integral95

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Re: HSC 2015 4U Marathon - Advanced Level

Well I dunno if this is too hard for the integration thread, but i'll post it here anyway.

From UNSW integration bee.

 

seanieg89

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Re: HSC 2015 4U Marathon - Advanced Level























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I don't think I made any mistakes, is that the solution you're looking for?
Your logic is only 1-directional when you introduce the arbitrary odd function O.

Indeed any f must be of such a form (apart from the constant solution 1), but a generic odd O will not do.

Eg O(x)=0 does not work.
 

SilentWaters

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Re: HSC 2015 4U Marathon - Advanced Level

Damn, difficult question. I agree with you up to about the final line.

By substituting x = -k, we obtain the constants:



So the final answer should be:



EDIT: Note the fractional exponent.

@InteGrand: 0! = 1
 
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FrankXie

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Re: HSC 2015 4U Marathon - Advanced Level

Let ABCD be a trapezium with AB||CD such that

(i) its vertices A,B,C,D lie in a circle with centre O ,
(ii) its diagonals AC and BD intersect at a point M and ∠AMD=60 ∘ ,
(iii) MO=10 .

Find the difference between the lengths of AB and CD .
 

FrankXie

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Re: HSC 2015 4U Marathon - Advanced Level

it is clear that the trapezium is isosceles and angle ACD=angle BCD=30 degree. So angle BOC=60 degree and triangle BOC is equilateral. Note that the centre O must lie inside the trapezium. Let the radius=R and angle OCD=\alpha.



 
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seanieg89

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Re: HSC 2015 4U Marathon - Advanced Level

Do you insist that the domain contains 0? Because that forces otherwise we blow up at 0.

And for then we tend to 0 from above at both 0 and infinity, which makes injectivity impossible.

For this is of course just a decreasing exponential function that is certainly injective because it is monotone.

If you only require the function to be defined on the positive reals, then words as well. In this case w is the product of two decreasing positive functions and is hence decreasing (hence injective). In fact in this case w is a bijection from the positive reals to themselves because it blows up at 0 and exponentially decays to 0 in the large x limit.
 

InteGrand

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Re: HSC 2015 4U Marathon - Advanced Level

Do you insist that the domain contains 0? Because that forces otherwise we blow up at 0.
I want the domain to contain 0 if it doesn't blow up (), but not to contain 0 if it does blow up ().

But it doesn't seem to matter to the question if 0 is excluded.
 

InteGrand

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Re: HSC 2015 4U Marathon - Advanced Level

Determine whether converges.
 

SilentWaters

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Re: HSC 2015 4U Marathon - Advanced Level

Alternatively, observing the standard limit:



we can say that as x tends to infinity this summation tends to the harmonic series. By looking at the upper bound rectangle sum for y = 1/x from x = 1 onwards, we establish the relationship:



where k is positive integral. Now note the improper integral



So since this latter summation diverges, the former summation of the arc-tangent must do the same.
 
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InteGrand

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Re: HSC 2015 4U Marathon - Advanced Level

Hence if we consider the sum of upper-bound rectangles of unit width from x = 1 onwards, noting that the corresponding exact area beneath the graph is infinite, we can say the summation diverges.
How did you know the area under the arctan graph was infinite (without doing the integration)?

(Or did you just mean by doing an integration, you see the area is infinite? Because without doing an integration, it doesn't seem obvious to me.)
 
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InteGrand

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Re: HSC 2015 4U Marathon - Advanced Level

Isn't the nth partial sum of the sequence S(n) = b1 + b2 + .. + bn?

Therefore, S(1) = b1?
He's saying that your current formula for doesn't agree with .
 
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