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Any Math Geniuses Here? (pg. 4)
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| GelatinPufF |
| quote: | Originally posted by Noisician
1) if 2 < x < 26/11, then 11π/2 < 11πx/4 < 13π/2 and thus cos(11πx/4) > 0; meaning that on this particular interval
8/x-x²-cos(11πx/4)<8/x-x²<0 (since x>2)
2) if 26/11 ≤ x < 3, then 8/x-x²≤11·8/26-(26/11)²<44/13-5<-1 (since f(x)= 8/x-x² is monotonically decreasing when x>0). hence,
8/x-x²-cos(11πx/4)<-1-cos(11πx/4)≤0
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:conf::confused:
:nervous: |
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| Resnick |
| quote: | Originally posted by politicsofdancin
Problem 1:
Bear, Squirrel, Rabbit, Deer, and Owl are seating themselves around a campfire to roast marshmallows. However, they have to be careful about who sits where!
Bear's favorite snack is rabbit, therefore they cannot sit together.
Every time Owl hoots Deer gets nervous so they do not sit next to each other.
Squirrel's arms are really short so Bear is next to her to help with the marshmallows.
Owl is allergic to chestnuts so he cannot be next to Squirrel because she smells like them.
Question: In what order are they sitting around the campfire?
Extra: After the campfire everyone shakes "hands" (paws, hooves, talons) before going home. How many shakes are exchanged?
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Problem 2:
A sequence {an} is bounded above if there is some K such that anK for all n. We say that K is an upper bound for the sequence.
There is a similar definition of a sequence being bounded below.
The number K may not be the best or smallest upper bound. All we know from the definition is that it will be an upper bound.
The sequence an = 2 + (- 1)n is bounded above.
ok solve this
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um going clockwise (or anti, doesnt matter ) -> Bear,Owl,Squirrel,Deer,Rabbit? cant remeber last ones name
and 10 handshakes
and ur #2 isnt asking a question..at least i didnt notice it as that |
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| Ste |
| quote: | Originally posted by Noisician
well, here's one way to solve it:
first, by plugging in 0 in place of x, we can find all values of a that give the initial equation the solution x=0:
if x=0, we get (a-1)√8=0 ⇔ a=1
thus, x=0 will be (one of) the initial equation's solution(s) if, and only if, a=1. we can also determine whether there are an odd number of solutions on the given interval by plugging in 1 instead of a. we'll get the equation:
(1-x²-cos(11πx/4)√(8-x)=0
since √(8-x) is strictly positive on the interval x∈[-2,3], on this interval the above equation can be rewritten as
x²+cos(11πx/4)=1
since f(x)=x²+cos(11πx/4) is an even function, the latter equation will have an odd number of solutions on the interval x∈[-2,2] (because even if x1 is its solution, then (-x1) will be one just as well. besides, x=0 is also its solution, which was already determined earlier. so the total number will always be ODD). in addition, that equation obviously has no solutions on the interval x∈[2,3], since on this specific interval
x²+cos(11πx/4)≥4-1=3>1
thus, if a=1, the initial equation has an odd number of solutions on the interval x∈[-2,3]. that means a=1 is one of the answers to this problem.
next, let's see what happens when a equals 0. if a=0, the initial equation can be rewritten as
x²+cos(11πx/4)=0
as has been noted earlier, the function on the left side is even, but then it follows that, on the interval x∈[-2,2], the above equation will have either an EVEN number of solutions, since x=0 is no longer among the solutions (remember, in this case, a=0, not 1!), or none at all. in addition, the above equation has no *additional* solutions on the interval x∈[2,3] because
x²+cos(11πx/4)≥4-1=3>0
thus, if a=0, the initial equation has an even number of solutions on the given interval. so a=0 should be excluded from the final answer.
next, for each and every value of parameter a such that a≠0 and a≠1, the initial equation has the solution x=8/a. in order for this solution to be within the given interval (otherwise it would be invalid), the double inequality
-2≤8/a≤3
must be true for all such values of x. when u solve it, u'll get two possible intervals for parameter a: a≤-4 and a≥8/3.
thus, all we need to do is study 3 intervals involving a. after having done that, the initial problem can safely be considered done. the 3 intervals are:
i. a≤-4
ii. -4< a<8/3, a≠0, a≠1
iii. a≥8/3
i. if a≤-4, the initial equation has the solution x=8/a on the given interval. we now need to determine if it has any more of them. it will have additional solutions if, and only if, the equation
x²+cos(11πx/4)=a
has solutions when a≤-4. but since x²+cos(11πx/4)≥cos(11πx/4)≥-1>-4, it appears that the above equation is unsolvable on the interval a≤-4. thus, when a≤-4, the initial equation has only one solution, x=8/a. that means any a≤-4 is among the right answers to the initial problem.
ii. if -4 < a < 8/3, a ≠ 0, a ≠1, then √(8-ax)≠0 on the interval x∈[-2,3] (see above). that means on this particular interval for a, the initial equation is equivalent to the equation
x²+cos(11πx/4)=a
if x∈[-2,2], it will have either an even number of solutions or none at all (see way above). if x∈(2,3], it will have no solutions whatsoever because, on this interval,
x²+cos(11πx/4)≥x²-1≥4-1=3>8/3
thus, when -4< a<8/3, a≠0, a≠1, the initial equation either has no solutions at all or has an EVEN number of solutions. that means, all values within the above interval should be excluded from the final answer.
iii. if a≥8/3, the initial equation will have AT LEAST one solution (x=8/a) on the interval x∈[-2,3], which was determined earlier. but, unlike with case i, on this interval for a, all eligible values must satisfy not one but two conditions:
1). since we already know that, on this interval, the initial equation has x=8/a as its solution (see above), to make the total number of solutions ODD, the equation
x²+cos(11πx/4)=a
must have an EVEN number of solutions on this interval or not have any at all. however, all such solutions must also meet a second condition:
2). for all such values of a, (8-ax) must be non-negative.
so in this particular case, the initial problem can be restated as:
on the interval a∈[8/3,∞), find all values of a, such that the equation
x²+cos(11πx/4)=a
has an EVEN number of solutions on the interval x∈[-2,8/a).
iii-a. let's see what happens if a=4 (i picked 4 because it makes for a whole number in the above interval). in this case, the above equation can be rewritten as
x²+cos(11πx/4)=4, and the interval itself is x∈[-2,8/4=2)
on the interval x∈(-2,2), that equation does, in fact, have an even number of solutions. however, x=-2 is also its solution (u can check it by plugging it in in the above equation). that means, when a=4, the above equation actually has an odd number of solutions, making the grand total even. thus, a=4 must be excluded from the final answer.
iii-b. if a>4, then 8/a<2. and since the equation "x²+cos(11πx/4)=a" has an even number of solutions on x∈[-2,2], the valid answers to the initial problem can only be the values of parameter a (a>4) such that the latter equation, on the interval x∈[8/a,2], has either an even number of solutions or no solutions at all. in other words, if we can prove that it is unsolvable on x∈[8/a,2], we will prove that all a>4 belong among the right answers to the initial problem. now, since x=2 is not a solution in this case (a≠4) and since a-x²-cos(11πx/4)≥8/x-x²-cos(11πx/4), all we need to do is prove that "8/x-x²-cos(11πx/4)" is positive on the interval x∈(0,2).
1). if 18/11 < x < 2, then 9π/2 < 11πx/4 < 11π/2 and therefore cos(11πx/4)<0; meaning that on this particular interval
8/x-x²-cos(11πx/4)>8/x-x²>0 (since x<2)
2). if 0 < x ≤ 18/11, then 8/x-x²≥11·8/18-(18/11)²>44/9-3>1 (since f(x)= 8/x-x² is monotonically decreasing when x>0). hence,
8/x-x²-cos(11πx/4)>1-cos(11πx/4)≥0
iii-c. if 8/3 ≤ a < 4, then 8/a > 2. this case is almost identical to iii-b. we need to find such values of a, for which the equation "x²+cos(11πx/4)=a" will have either an even number of solutions or none on the interval x∈(2,8/a). and just like with iii-b, all we need to do is prove that it is unsolvable on x∈(2,8/a) when 8/3 ≤ a < 4. only in this case (because, on this interval, a<8/x and, therefore, a-x²-cos(11πx/4)<8/x-x²-cos(11πx/4)) we only need to prove that "8/x-x²-cos(11πx/4)" is negative on x∈(2,3).
1) if 2 < x < 26/11, then 11π/2 < 11πx/4 < 13π/2 and thus cos(11πx/4) > 0; meaning that on this particular interval
8/x-x²-cos(11πx/4)<8/x-x²<0 (since x>2)
2) if 26/11 ≤ x < 3, then 8/x-x²≤11·8/26-(26/11)²<44/13-5<-1 (since f(x)= 8/x-x² is monotonically decreasing when x>0). hence,
8/x-x²-cos(11πx/4)<-1-cos(11πx/4)≤0
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so if i haven't made any mistakes, the final answer should be: a∈(-∞,-4]∪{1}∪[8/3,4)∪(4,∞) |
a post worthy of the word "pwn3d" |
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| smokeape |
I don't need a real genius here. Just give me the BPM!
[[[smoke]]]
Electrique Boutique ft Taz - Heal |
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| Noisician |
| quote: | Originally posted by GelatinPufF
:conf::confused:
:nervous: |
what part of that seems confusing to u? i shall explain. |
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| Flyboy217 |
| quote: | Originally posted by Noisician
however, all such solutions must also meet a second condition:
2). for all such values of a, (8-ax) must be non-negative.
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Sorry, it's been a long while since I've been doing math competitions, but can you clear this up for me? Why must constraint (2) hold? If (8-ax) is negative, then although the square root will be complex, the product of the functions will still be zero, and we will still have a valid root.
I agree a<=-4 is a solution, as is a=1. Now, since f(x) = a-x^2-cos(11pi*x/4) is monotonically decreasing on x > 11/16 and x < -11/16 (since |2x| > 11/8 = the wave's maximum derivative), f(x) can have (at most) one zero on those same intervals. And because we're only concerned with 2
Solving for a, we see that
a = x^2+cos(11pi*x/4) so that
4 + cos(11pi/2) < a <= 9 + cos(33pi/4) or
4 < a <= 9+sqrt(2)/2
are the values for which f(x) will yield one root on (2,3]. However, since all the values in this interval are greater than 8/3, the total number of roots will be even. Thus, we need to exclude the interval
(4, 9+sqrt(2)/2]
We also see that, for -2<=x<=2, there are double roots to worry about, wherever
a = 8/x = x^2+cos(11pi*x/4)
a=4 is one such example which we need to exclude. It appears there should be no others, but I'm sure you've verified that.
Therefore, it looks like a<=-4, a=1, 8/3<=a<4, 9+sqrt(2)/2
Apologies if I've duplicated effort and/or been unclear. Lemme know what you think. |
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| Vigilante |
| quote: | Originally posted by mentalbarter
its ing maths damnit |
Word!
It pisses me off when North Americans call it "math".
The full name is "mathematics", so the abbreviated word should be maths, which denotes the plural. |
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| Flyboy217 |
| quote: | Originally posted by Vigilante
Word!
It pisses me off when North Americans call it "math".
The full name is "mathematics", so the abbreviated word should be maths, which denotes the plural. |
Tee hee... you just shot yourself in the foot buddy. "Mathematics" is a singular noun ("Mathematics is a subject" instead of "Mathematics are subjects"). Of course, "maths" is also used in the singular (at least, by you funny-speaking folk). But I'll give it to you on the grounds that "maths" sounds all foreign and exotic and thus more fun ;) |
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| AnonymousRacer |
...just wondering, where,when, and how can this stuff be applied to normal, everyday life?
It can't, huh? :D
All I need to know is all I need to get by. |
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| starglider |
| quote: | Originally posted by AnonymousRacer
...just wondering, where,when, and how can this stuff be applied to normal, everyday life?
It can't, huh? :D
All I need to know is all I need to get by. |
Fair enough, you rarely need a working knowledge of math(s) to flip patties at Burger King. |
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| DJ Mikey Mike |
| quote: | Originally posted by Wicked Neo
OMG :eyes: :eyes: :eyes: :eyes:
this thread is going to the archives to be preserved in TA history for the most 'INTELLIGENT' thread ever in the history of the Chill Out Forum !
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We do this in Pure 4 mate, it's not rocket science :p Nicely explained tho Acid Junkie, and I dunno how you could be arsed with the symbols :p I remember doing my Mechanics 4 coursework back in the day, and putting in symbols like that was a ing nightmare. This topic is making me excited about the possibility of doing Maths at uni this year :) |
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