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March 29, 2024, 12:58:47 pm

Author Topic: Chavi's Mathematical adventure (+Karma for solutions)  (Read 12140 times)  Share 

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pi

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Re: Chavi's Mathematical adventure (+Karma for solutions)
« Reply #15 on: November 28, 2010, 08:12:58 pm »
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x= (13^(1/2) +1)/2
haha yep +1.

Hint: for those considering the question: This requires gof(x) (nested functions).

if this is question 4, also can be done with quadratics...

funkyducky

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Re: Chavi's Mathematical adventure (+Karma for solutions)
« Reply #16 on: November 28, 2010, 08:14:52 pm »
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Yep I used quadratics
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taiga

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Re: Chavi's Mathematical adventure (+Karma for solutions)
« Reply #17 on: November 28, 2010, 08:15:23 pm »
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how to do with quadratics? :X
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Re: Chavi's Mathematical adventure (+Karma for solutions)
« Reply #18 on: November 28, 2010, 08:16:15 pm »
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Square both sides.
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Re: Chavi's Mathematical adventure (+Karma for solutions)
« Reply #19 on: November 28, 2010, 08:17:31 pm »
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Square both sides.

yep, and then minus x from x^2 to get an equation

solve with quadratic formula or complete the square and the answer is made clear


EDIT: A similar question is in Maths Quest 9 (2nd edition) -the ones in that book are a bit easier though (ie. whole solutions)

taiga

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Re: Chavi's Mathematical adventure (+Karma for solutions)
« Reply #20 on: November 28, 2010, 08:21:32 pm »
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Square both sides.

yep, and then minus x from x^2 to get an equation

solve with quadratic formula or complete the square and the answer is made clear


EDIT: A similar question is in Maths Quest 9 (2nd edition) -the ones in that book are a bit easier though (ie. whole solutions)

trueee, for some reason i looked at it and thought it was one of those recurring root ones (which i hate)
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Chavi

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Re: Chavi's Mathematical adventure (+Karma for solutions)
« Reply #21 on: November 28, 2010, 08:44:55 pm »
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3. doing...
I subbed in values, and cos(sinx) is bigger becos
need a proof - (no trial + error)
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pi

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Re: Chavi's Mathematical adventure (+Karma for solutions)
« Reply #22 on: November 28, 2010, 08:55:46 pm »
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3. doing...
I subbed in values, and cos(sinx) is bigger becos
need a proof - (no trial + error)

I'd like to know too, I can't see why (only yr 11 though...)

Very interesting problem



Although when each is graphed, it is clear that cos(sin x) is larger... (I suppose this is a proof though)

cos(sin x):


sin(cos x):


links died, sorry
« Last Edit: November 28, 2010, 10:02:28 pm by Rohitpi »

Chavi

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Re: Chavi's Mathematical adventure (+Karma for solutions)
« Reply #23 on: November 28, 2010, 08:58:51 pm »
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hint for 3.
use trig identities
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TrueTears

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Re: Chavi's Mathematical adventure (+Karma for solutions)
« Reply #24 on: November 28, 2010, 09:04:27 pm »
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Karma for correct answers

1.
2. Complete the next three terms in the sequence: 1, 4, 9, 61,
3, Let x be an element of R. Find which term has a larger value, sin(cos x) or cos(sin x)?
4. Find a value of x such that:
3 is a simple mathematical exercise.

I will give you guys a hint, suppose cos(sin x) > sin(cos x), then think more about cos(sin x) − sin(cos x)

4 is also very simple, hint: square both sides.

my hint for 2 is that it is a permutative sequence, in fact a nice extension to this problem would be this:

can anyone find a generating function which matches that sequence?
« Last Edit: November 28, 2010, 09:18:50 pm by TrueTears »
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Re: Chavi's Mathematical adventure (+Karma for solutions)
« Reply #25 on: November 28, 2010, 09:25:37 pm »
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This is what I got from TT's advice, and using my puny GMA knowledge...  (sorry, but I don't know LaTeX yet...)

cos(sin x) - sin(cos x)
= cos(sin x) - sin(pi/2 -x)
= -2 sin((sin x + cos x +pi/2)/2) sin((sin x + cos x - pi/2)/2)                       <--- using half-angle formula
= −2 sin((−sqrt(2)cos(x + pi/4))/2 + pi/4) sin((sqrt(2)cos(x − pi/4))/2 − pi/4)    <--- using identity: A cos x + B sin x = R cos(x − C), where R = (A2 + B2)^(1/2), and C = arctan(B/A)

What do I do next? (and am I right so far?)
« Last Edit: November 28, 2010, 10:02:53 pm by Rohitpi »

TrueTears

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Re: Chavi's Mathematical adventure (+Karma for solutions)
« Reply #26 on: November 28, 2010, 09:29:21 pm »
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I suggest use radians, also I suggest convert sin(cosx) into cos(pi/2-cosx) then compound angle formula works elegantly.

After simplification a few mathematical arguments yields the proof.

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Re: Chavi's Mathematical adventure (+Karma for solutions)
« Reply #27 on: November 28, 2010, 09:31:05 pm »
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my hint for 2 is that it is a permutative sequence, in fact a nice extension to this problem would be this:

can anyone find a generating function which matches that sequence?

I thought question 2 was all the square numbers backwards: 1, 4, 9, 61 (from 16), 52 (from 25), 63 (from 36), etc....

Maybe it is more complicated than that though...

(as for the function, I have no idea)

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Re: Chavi's Mathematical adventure (+Karma for solutions)
« Reply #28 on: November 28, 2010, 09:33:35 pm »
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correct

as for the function, maybe finding the actual function is out of reach for most VCE students, as an exercise can you prove the generated function generates all square numbers? (and yes you would need to think combinatorially!)

The function generates the original square numbers, then Chavi's sequence can be generated with some notational matters.
« Last Edit: November 28, 2010, 09:41:59 pm by TrueTears »
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Re: Chavi's Mathematical adventure (+Karma for solutions)
« Reply #29 on: November 28, 2010, 09:47:29 pm »
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correct

as for the function, maybe finding the actual function is out of reach for most VCE students, as an exercise can you prove the generated function generates all square numbers? (and yes you would need to think combinatorially!)

The function generates the original square numbers, then Chavi's sequence can be generated with some notational matters.

I don't know any proof, but I have seen it when my GMA teacher when on a tangent about 'generating functions' (not in the course, and I still don't know what they mean).

Nut the one you gave means 1x^1 + 4x^2 + 9x^3 + 16x^4 + 25x^5 ... right? That's all I know about it (luckily I wrote a few of those functions down!)