HSC 2013 MX2 Marathon (archive) (6 Viewers)

[RealiseNothing]

Re: HSC 2013 4U Marathon

Find the other expression for it using part i (just divide by theta then take n to infinity)

As n approaches infinity,



Do you mean like that?

 

[bleakarcher]

Re: HSC 2013 4U Marathon

As n approaches infinity,



Do you mean like that?

What about the 2^n?

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

What about the 2^n?

I would have thought the 0 would dominate, not sure though.

 

[bleakarcher]

Re: HSC 2013 4U Marathon

I would have thought the 0 would dominate, not sure though.

Not sure actually either..

 

[bleakarcher]

Re: HSC 2013 4U Marathon

Wait no it's 1 lol.

 

[bleakarcher]

Re: HSC 2013 4U Marathon

lim[n->infinity] sin(x/2^n)/(x/2^n)=1 which essentially comes from the lim[x->0] sin(x)/x =1 with the substitution u=x/2^n. The cosines approach 1.

Actually, I'm still not sure of this. Some of the cosines are independent of n.

 

[Sy123]

Re: HSC 2013 4U Marathon

lim[n->infinity] sin(x/2^n)/(x/2^n)=1 which essentially comes from the lim[x->0] sin(x)/x =1 with the substitution u=x/2^n. The cosines approach 1.

Actually, I'm still not sure of this. Some of the cosines are independent of n.

Exactly, the expression then becomes all cosines and the sine at the end is eliminated

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

I put it into wolframalpha and it came up "indeterminate".

So unless wolframalpha doesn't do manipulations or something, idk.

 

[bleakarcher]

Re: HSC 2013 4U Marathon

Exactly, the expression then becomes all cosines and the sine at the end is eliminated

What about cos(x/2), where does that go? It doesn't vary with n, that's why I'm still confused.

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

Oh I see what you mean now, so you don't want an actual value, just an expression.

 

[bleakarcher]

Re: HSC 2013 4U Marathon

This question is kind of weird because as goldy said lim[n->infinity] [sin(x)]/x = sin(x)/x.

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

What about cos(x/2), where does that go? It doesn't vary with n, that's why I'm still confused.

I think he wants it to just become:

 

[Sy123]

Re: HSC 2013 4U Marathon

Oh I see what you mean now, so you don't want an actual value, just an expression.

yeah, sorry for the lack of clarity.

 

[Sy123]

Re: HSC 2013 4U Marathon

Oh I see what you mean now, so you don't want an actual value, just an expression.

yeah, sorry for the lack of clarity.

 

[bleakarcher]

Re: HSC 2013 4U Marathon

I think he wants it to just become:

Oh okay.

 

[seanieg89]

Re: HSC 2013 4U Marathon

Yep, Sy you would want to phrase that question as asking the student to evaluate the infinite product of cosines, what you have written has exactly the problem Goldy mentioned, theta is independent of n.

 

[Sy123]

Re: HSC 2013 4U Marathon

Yep, Sy you would want to phrase that question as asking the student to evaluate the infinite product of cosines, what you have written has exactly the problem Goldy mentioned, theta is independent of n.

Yeah, I rephrased it:

MkIII

 

[bleakarcher]

Re: HSC 2013 4U Marathon

Yeah, I rephrased it:

lol, are we like the only 4U maths students in the state?

For part iii), use the fact that lim[n->infinity] sin(x)/x is equal to the limit of the product of the cosine functions as n->infinite. Let x=pi/2, rearrange for pi and use the identity cos(2x)=2cos^2(x)-1 to evaluate the cos(pi/8), cos(pi/16) etc since we know cos(pi/4)=1/sqrt(2).

 

[Sy123]

Re: HSC 2013 4U Marathon

lol, are we like the only 4U maths students in the state?

For part iii), use the fact that lim[n->infinity] sin(x)/x is equal to the limit of the product of the cosine functions as n->infinite. Let x=pi/2, rearrange for pi and use the identity cos(2x)=2cos^2(x)-1 to evaluate the cos(pi/8), cos(pi/16) etc since we know cos(pi/4)=1/sqrt(2).

Correct, good job - and what do you mean by your first sentence lol?

 

[bleakarcher]

Re: HSC 2013 4U Marathon

Correct, good job - and what do you mean by your first sentence lol?

It's as if this thread is invisible to everyone besides the usual bosers. Usually you'd have some other randoms participate in marathon threads.