HSC Mathematics Marathon (2 Viewers)

math man

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Yeah I know what the MVT is, but I don't really get how your proof of seanieg89's question actually works (like how do you assume f'(c) is constant o_O, presumably as you range x and y)
if f(x) is my function then f'(x) is my derivative, so subbing c into the derivative, i.e f'(c), will give me a constant
 

hup

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you cannot assume that
 
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hup

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now is a solution like this accepted?
I also do not have another question...someone else can put one up.
idk if you can use induction like that it's like taking the easy way out
 
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Wight

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yeah neither do I...
also I skip too many steps -.-
 
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Wight

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start with

as for then .
So as any irrational (< 1)^n is even smaller, then as . Apply that to my solution as relevant and it should be good (too lazy to do it again).

Yeah the above will need alot more work to suit a proper solution...just a general idea.

I'm still curious as to whether you can use induction...it's obviously the longer process, but the easier way out when you don't think of the obvious...can anyone clarify?
 
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apollo1

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question similar to this in 2005 hsc where induction was used.
 

Wight

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Hmm okay. Personally I don't see why it can't be used...but I guess that's the marker's decision. Not mine.
 

hup

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you use the series

S = 1+xcisx+(xcisx)^2+....+(xcisx)^n

then sum it using a gp and equate the real parts

for it to conerge |r| if r is the common ratio must be <1

but since r is xcisx then |r| = x

so x<1 for it to converge?
 

largarithmic

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you use the series

S = 1+xcisx+(xcisx)^2+....+(xcisx)^n

then sum it using a gp and equate the real parts

for it to conerge |r| if r is the common ratio must be <1

but since r is xcisx then |r| = x

so x<1 for it to converge?
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)
 
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seanieg89

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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!
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)
 

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