Chavi's Mathematical adventure (+Karma for solutions)

[TrueTears]

Correct with the expansion, however WHY does the coefficients appear to be all squares? In fact can you prove the coefficients will continue to be squares (induction!)? If you can prove the latter the exercise is complete. Strong induction or normal induction will suffice.

generating functions is a great area of mathematics, it seems VERY abstract at first but it is so useful.

You GMA teacher must be a gun, must be an awesome dude if he goes on about generating functions in class haha

[pi]

Correct with the expansion, however WHY does the coefficients appear to be all squares? In fact can you prove the coefficients will continue to be squares (induction!)? If you can prove the latter the exercise is complete. Strong induction or normal induction will suffice.

generating is a great area of mathematics, it seems VERY abstract at first but it is so useful.

You GMA teacher must be a gun, must be an awesome dude if he goes on about generating functions in class haha

I have no idea about proofs (yr11 maths doesn't go that far I think), he just told us the expansions of random things. And yes, he is a genius, starts going on about gamma functions to find the factorials of decimal numbers (eg. (0.5)!) and he mentioned the factorial of i (i!) but said he would corrupt our brains if he told us the answer... Didn't understand most of what he said during these rare times...

[TrueTears]

haha what an awesome guy

[pi]

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?)

Okay, I think I have done this now... (continued from the above quote)

As pi/4 > sqrt(2)/2,  0 < −sqrt(2)cos(x + pi/4))/2 + pi/4 < pi/2.
Therefore sin(−sqrt(2)cos(x + pi/4))/2 + pi/4) > 0

pi/2 < sqrt(2)cos(x − pi/4))/2 − pi/4 < 0
Therefore sin(sqrt(2)cos(x − pi/4))/2 − pi/4) < 0

THEREFORE −2 sin((−sqrt(2)cos(x + pi/4))/2 + pi/4) sin((sqrt(2)cos(x − pi/4))/2 − pi/4) > 0    

Therefore cos(sin x) > sin(cos x) as the difference is > 0 for cos(sin x) - sin(cos x)!        

(NB. only if x is a real number!)

Done, Chavi!

[pi]

Full solution for question 3: Let x be an element of R. Find which term has a larger value, sin(cos x) or cos(sin x)

cos(sin x) - sin(cos x)

= cos(sin x) - sin(pi/2 -x)         <--- we should all know that identity

= -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)

As pi/4 > sqrt(2)/2,  0 < −sqrt(2)cos(x + pi/4))/2 + pi/4 < pi/2.           <--- basic arithmetic

Therefore sin(−sqrt(2)cos(x + pi/4))/2 + pi/4) > 0

pi/2 < sqrt(2)cos(x − pi/4))/2 − pi/4 < 0

Therefore sin(sqrt(2)cos(x − pi/4))/2 − pi/4) < 0

THEREFORE: −2 sin((−sqrt(2)cos(x + pi/4))/2 + pi/4) sin((sqrt(2)cos(x − pi/4))/2 − pi/4) > 0    

Therefore cos(sin x) > sin(cos x) as the difference is > 0 for cos(sin x) - sin(cos x), only if x is a real number
 
  (I think we have solved them all -all so far- Chavi!)

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

Proved myself wrong (with the help of TT and Chavi of course!)

[TrueTears]

^ correct, great work.

now is any other elegant methods? i am not sure myself, the thought of finding the difference only occurred to me because that is a good tactic to prove which is larger, but i am also interested in any other cool methods

[pi]

^ correct, great work.

now is any other elegant methods? i am not sure myself, the thought of finding the difference only occurred to me because that is a good tactic to prove which is larger, but i am also interested in any other cool methods

What do you mean by 'elegant'? (I have seen this word in a number of posts by uni maths students as well...)

[TrueTears]

elegant u know... doesn't have to be short or anything but like strikes you as beautiful. so looking at the solution is like seeing this hot babe walk past you or something lol

[Chavi]

new questions up

Hint: trial and error can be used when algebra gets ugly

[pi]

new questions up

Hint: trial and error can be used when algebra gets ugly

the only trial and error I got was in Q5: a=0 and b=0  (but that answer was obvious)

But I know there are more solutions, can you use simultaneous equations? I tried substitution but things got really ugly!

[Chavi]

Hints:
Firstly, you know that the greatest common divisor between a and b is 2.
So let a = 2x and b = 2y. Plug these in to the equation, and you'll find it easier with (x , y) = 1/

Generate a sequence of integers (e.g. -1, 0, 1, 2) for the independent variable (x) and then double check to see if it works with y.
There should be one obvious solution.

[kamil9876]

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!)

It can not only be proven, but derived with Calculus (ie it is better to find something and prove it than it is to prove something given to you)


Now if you differentiate you get:



Now multiply by :



Now differentiate again and viola

[Chavi]

The sequence wasn't intended to be very mathsy

[kamil9876]

It's not me who hijacked the thread

[pi]

Sorry to revive this thread, but I really don't get question 5...

All I can get is a = b = 0, in this case a and b are both integers, but I don't think this satisfies (a, b) = 2.

This is how far I got (cbs LaTeX):
As (a, b) = 2
Let a = 2x, and let b = 2y
Sub into second equation
18(2x)^4 + (2y)^4 = 41(2x)(2y)^2
288x^4 - 328xy^2 + 16y^4 = 0
36x^4 - 41xy^2 + 2y^4 = 0
Let y^2 = q
Using quadratic equation to solve for q1 and q2 (not going to type it up, its ugly)
Square root answers to find y1 and y2
Sub into other equation and then solve for x1 and x2 (had to use calculator here)
Sub these into (separately) into earlier solutions of y1 and y2
Solve for y again (with calc)
Find a (a = 2x)
Find b (b = 2y)

All I can get is a = 0 and b = 0


Solving the original equation on a graphics calc, it is clear that a = 4 and b = 6...

How do you solve this?!? (am I even on the right track?)