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HSC Mathematics Marathon (3 Viewers)

apollo1

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you mean |x| < 1, negative values of x and zero still work

to make sure the GP converges, do it like this:



So if we let
which converges as n->infinity if |z| < 1, and diverges if |z|>1. If |z|=1, i.e. x = 1 or -1, it neither converges nor diverges: it just sorta oscillates around the place (although, since 1 is not a rational multiple of pi, its not actually periodic but rather slightly chaotic). Clearly |z| = |x|, so it converges iff |x|<1, goes weird if |x| = 1 and diverges if |x| > 1.

Cool question!

EDIT: oh wait I kinda messed it up because I showed the cis sum diverges, not Re(cissum). Its pretty easy to do that though I think:



Now note that as n goes to infinity, the demoninator remains the same while the numerator converges if |x|<1. For |x|>=1, its a lot more complicated but it certainly doesnt converge; it doesnt necessarily "increase without bound though", it sorta oscillates whilst "diverging" at the same time (I dont actually know the technical definition of diverging so yeah, in this case a function like xsinx diverges even though it has zeroes for arbitrarily large x)
wouldnt it be z^n+1 becuase there are n+1 terms?
 

largarithmic

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can someone help me with this question:
Okay um do it like this: Let .

We're given that this sequence has a limit, also clearly as , L_n is always less than half, and it is also clearly always positive.

Now rearrange that equation there to be in terms of n:




Now send n to infinity: if L_n goes to limit L where L is NOT equal to a half (and clearly 0 ≤ L ≤ 1/2 anyway), then the RHS has limit which is a finite limit as L=/=2: by contrast the LHS obviously does not have a finite limit. You then require the limit to be half.

You can get rid of the question's assumption that there is a limit by proving that the Ln sequence is monotonic increasing (not too hard, start with L_n+1 > L_n, which would be true if an only if blahhh, which would be true if an only if blahhh, and each time you get rid of the squares so that you get down to a fourth degree polynomial inequation which after expanding out is true. You then know Ln is monotonic and its bounded as well from what I showed earlier, so it has a limit
 

largarithmic

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Exactly, and your mention of the chaotic nature of the partial sums for certain x brings up an interesting point...consider the same problem if you replace the series by:



(Considerably harder than the geometric series case)
That questions pretty similar to finding the integral of (sinx)/x isnt it?
 

Trebla

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I think you guys are scaring some students a little bit with hardcore questions which are highly unlikely to appear in the HSC...lol keep it relatively simple so ordinary students can participate
 

I like maths

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I remember seeing this question earlier this year (perhaps in a past SGS trial?) it is dealing with Chebyshev polynomials which makes it an intrinsically excellent question but perhaps slightly divorced from the approximation theory/differential equations that make them important. Anyway you have made a copyist error, cos(kx)=2cos(x)cos((k-1)x)-cos((k-2)x)
 

AAEldar

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Did this one the other day. Pretty sure there were 2 substitutions that I got to work, and .
 

hup

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Did this one the other day. Pretty sure there were 2 substitutions that I got to work,
this is good
if you have a surd on bottom letting u = surd makes it so that when you differentiate it it easily goes away
 

slyhunter

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You can either use a trigonometric or algebraic substitution. I personally found letting the easiest.
 

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