Post by: Chavi on November 28, 2010, 07:21:16 pm
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:
5.
6. For the equation:
Find the value of m for which the equation has infinite solutions.
Post by: taiga on November 28, 2010, 07:23:57 pm
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)?
1. 0
2. 52, 63, 94
It's not very mathsy :P
3. doing...
I subbed in values, and cos(sinx) is bigger becos
Post by: Chavi on November 28, 2010, 07:25:06 pm
2. three terms. . .
Post by: funkyducky on November 28, 2010, 07:28:00 pm
Post by: Chavi on November 28, 2010, 07:28:58 pm
2. Typo? 61 meant to be 16?nope
Post by: stonecold on November 28, 2010, 07:29:25 pm
2. Typo? 61 meant to be 16?
That's what I initially thought haha :P
Post by: taiga on November 28, 2010, 07:31:31 pm
1st one is zero (x-x) ends up in there xD
Post by: Chavi on November 28, 2010, 07:38:14 pm
LOL! niceyou're on a roll mate. :)
1st one is zero (x-x) ends up in there xD
Post by: Chavi on November 28, 2010, 07:39:26 pm
LOL! nice"Sorry, you can't repeat a karma action without waiting 12 hours. " - damn.
1st one is zero (x-x) ends up in there xD
Well +1 here for you again
Post by: iNerd on November 28, 2010, 07:39:35 pm
LOL! niceOMG that is smart. What a sexy question, I'm going to use this on my MHS nerd-friends :P
1st one is zero (x-x) ends up in there xD
Post by: funkyducky on November 28, 2010, 07:48:08 pm
Post by: stonecold on November 28, 2010, 07:49:44 pm
What's the rule for the sequence? It's neither arithmetic or geometric.
i also am intrigued!
Post by: Chavi on November 28, 2010, 07:52:39 pm
What's the rule for the sequence? It's neither arithmetic or geometric.The more maths you know, the harder it is to complete
Post by: funkyducky on November 28, 2010, 07:57:57 pm
Post by: Chavi on November 28, 2010, 08:00:12 pm
x= (13^(1/2) +1)/2haha yep +1.
Hint: for those considering the question: This requires gof(x) (nested functions).
Post by: pi on November 28, 2010, 08:12:58 pm
x= (13^(1/2) +1)/2haha 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...
Post by: funkyducky on November 28, 2010, 08:14:52 pm
Post by: taiga on November 28, 2010, 08:15:23 pm
Post by: funkyducky on November 28, 2010, 08:16:15 pm
Post by: pi on November 28, 2010, 08:17:31 pm
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)
Post by: taiga on November 28, 2010, 08:21:32 pm
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)
Post by: Chavi on November 28, 2010, 08:44:55 pm
3. doing...need a proof - (no trial + error)
I subbed in values, and cos(sinx) is bigger becos
Post by: pi on November 28, 2010, 08:55:46 pm
3. doing...need a proof - (no trial + error)
I subbed in values, and cos(sinx) is bigger becos
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)
(http://www2.wolframalpha.com/Calculate/MSP/MSP302919d82cccigfci7f100005i363d3769gc857e?MSPStoreType=image/gif&s=24&w=299&h=148)
sin(cos x):
(http://www2.wolframalpha.com/Calculate/MSP/MSP151519d82gi024abi176000019fifb3aa9cf8gd9?MSPStoreType=image/gif&s=23&w=299&h=126)
links died, sorry
Post by: Chavi on November 28, 2010, 08:58:51 pm
use trig identities
Post by: TrueTears on November 28, 2010, 09:04:27 pm
Karma for correct answers3 is a simple mathematical exercise.
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:
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?
Post by: pi on November 28, 2010, 09:25:37 pm
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?)
Post by: TrueTears on November 28, 2010, 09:29:21 pm
After simplification a few mathematical arguments yields the proof.
Post by: pi on November 28, 2010, 09:31:05 pm
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)
Post by: TrueTears on November 28, 2010, 09:33:35 pm
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
The function generates the original square numbers, then Chavi's sequence can be generated with some notational matters.
Post by: pi on November 28, 2010, 09:47:29 pm
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 functiongenerates 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!)
Post by: TrueTears on November 28, 2010, 09:54:24 pm
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
Post by: pi on November 28, 2010, 09:59:12 pm
Correct with the expansion, however WHY does the coefficients appear to be all squares? :P 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...
Post by: TrueTears on November 28, 2010, 10:00:43 pm
Post by: pi on November 29, 2010, 01:05:38 pm
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)! :D :D :D :D
(NB. only if x is a real number!)
Done, Chavi!
Post by: pi on November 29, 2010, 02:58:21 pm
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
:D (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!)
Post by: TrueTears on November 29, 2010, 03:00:13 pm
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 :)
Post by: pi on November 29, 2010, 03:03:55 pm
^ 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...)
Post by: TrueTears on November 29, 2010, 03:12:07 pm
Post by: Chavi on November 29, 2010, 04:26:20 pm
Hint: trial and error can be used when algebra gets ugly
Post by: pi on November 30, 2010, 04:36:17 pm
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!
Post by: Chavi on November 30, 2010, 05:06:45 pm
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.
Post by: kamil9876 on December 01, 2010, 11:53:52 am
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
It can not only be proven, but derived with Calculus :P (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 :P
Post by: Chavi on December 01, 2010, 11:58:44 am
Post by: kamil9876 on December 01, 2010, 12:16:54 pm
Post by: pi on January 04, 2011, 08:33:17 pm
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?)
Post by: brightsky on January 04, 2011, 09:34:03 pm
So sub a = 2x and b = 2y in:
4(x+y)^2 = 50x...[1]
36x^4 + 2y^4 = 41xy^2...[2]
Look at equation [1]:
Since x, y are both positive integers, then 50x must be a multiple of 4, which means x = 2q, where q is some integer.
4(2q + y)^2 = 100q
2q + y = 5sqrt(x)
Again, since q and y are both positive integers, we need x = p^2, where p is some integer.
So we have 2q = p^2
The only solution here (besides 0) is q = 2 and p = 2, which means x = 4, which in turn means that y = 6.
Post by: pi on January 04, 2011, 10:14:07 pm
Thanks brightsky (+1 karma well deserved there)
Post by: pi on January 05, 2011, 12:41:22 pm
Using
As
As
As
As
Still have no idea on question 6...