|
Any Math Geniuses Here? (pg. 5)
|
View this Thread in Original format
| Ste |
| mecahnics was mint, pure was ing and i hated it. |
|
|
| Flyboy217 |
| 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. |
Agreed-- if your profession is not directly related to math(s), this specific knowledge may do you no good. However, (nearly) regardless of your profession, it will be beneficial for you to use clear, insightful reasoning. I honestly believe that most of you here can solve this problem. The concepts are not difficult; instead, the formulation with symbols looks daunting.
From years of tutoring, I've found that most people are far smarter than they give themselves credit for. It's all in how you think about it... don't let symbols scare you off! Give it some real thought and you'll see the simplicity. I'm with Noisician... if you want clarification, say something! |
|
|
| LoCa |
| quote: | Originally posted by Flyboy217
Agreed-- if your profession is not directly related to math(s), this specific knowledge may do you no good. However, (nearly) regardless of your profession, it will be beneficial for you to use clear, insightful reasoning. I honestly believe that most of you here can solve this problem. The concepts are not difficult; instead, the formulation with symbols looks daunting.
From years of tutoring, I've found that most people are far smarter than they give themselves credit for. It's all in how you think about it... don't let symbols scare you off! Give it some real thought and you'll see the simplicity. I'm with Noisician... if you want clarification, say something! |
i would love to try and understand but i have no idea what all those symbols mean apart from a^2 being a squared? (i think thats it anyways, correct me if i'm wrong)
I have a full semester of mathematics for computing coming up and im scaaaaarrrreeddddd :nervous: :nervous: :nervous: :nervous: |
|
|
| Noisician |
| quote: | Originally posted by Flyboy217
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]
|
hold on a second... that's not exactly what i got there. how is it monotonically decreasing? consider the function f(x) = x² + cos (11πx/4) once again. if u play around with its derivative a little, u should see that it actually has both a local maximum AND a local minimum (in that order) on the interval x∈[2.15,2.5] and those are not the end points. what is important here is that the abscissas of those points are within the interval (2,3), while their ordinates are actually within (4,√2/2+9). imo, the actual fragment should look something like this
note that both a1 and a2 yield exactly 2 roots for the equation x² + cos (11πx/4) = a on the interval x∈[2,3] (u can check that by plugging in the actual values in place of a). also note that both a values are in (4,√2/2+9). so i don't think excluding all values of a within the interval (4, √2/2+9] would be valid. 2 different solutions would turn out missing. |
|
|
| LoCa |
holy !!!
you even made a graph!!
gah i can only hope to one day understand that :wtf: |
|
|
| Fundamental |
| quote: | Originally posted by Noisician
what is important here is that the abscissas... |
abscissa
"The coordinate representing the position of a point along a line perpendicular to the y-axis in a plane Cartesian coordinate system."
Now we know. |
|
|
| Flyboy217 |
| quote: | Originally posted by Noisician
hold on a second... that's not exactly what i got there. how is it monotonically decreasing? |
Yarrrr, matey! I seem to have lost a 2π in there. With that little modification:
d/dx(11πx/4) = 11π/4 < 2x --> x > 11π/8
So it's monotonically decreasing above 11π/8 (of course not a tight bound), which is quite safely outside of [-2,3].
Luckily, all is not lost! Because the function only has two local extrema on the interval [2,3] (as you so nicely demonstrated with the graph:)), those are the only two values in the interval (4, 9+√2/2) that must be included. The rest of the interval must still be excluded. Agree?
You also agree that we needn't ignore x-values (abcissas!;)) for which √(8ax) is complex and where f(x) = 0?
So the solution should be something like this:
a<=-4, a=1, 8/3<=a<4, a=a1, a=a2, a>9+√2/2
So we were both off... but at least I was only off by measure zero ;) |
|
|
| <--ME--> |
| quote: | Originally posted by Resnick
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 |
Think you're wrong about the order. First of all, Squirrel HAS to be next to Bear, and Squirrel CANNOT be next to Owl. The order I got is Squirrel, Bear, Owl, Rabbit and Deer (although who would put an own next to a rabbit??). You are right about the 10 handshakes though. |
|
|
| mongeone |
Does the (*) symbol used in the original question mean times?
as in 3*4=12
Or is it some crazy maths symbol I have never seen before!
I miss maths it is such an interesting subject. I havn`t studied it for 2 years now and i have forgotten almost everything (besides the basics ofcourse). Unless that example above is wrong then dam me to hell. |
|
|
| Flyboy217 |
| quote: | Originally posted by mongeone
Does the (*) symbol used in the original question mean times?
as in 3*4=12
Or is it some crazy maths symbol I have never seen before!
I miss maths it is such an interesting subject. I havn`t studied it for 2 years now and i have forgotten almost everything (besides the basics ofcourse). Unless that example above is wrong then dam me to hell. |
Yes, it is a multiplication symbol. However, I believe you mean "Unless that example above is RIGHT...," as your statement is equivalent to "If that example is right, damn me to hell" :) |
|
|
| ali92 |
| 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
-----------------------------------------------------------------------------------------------
so if i haven't made any mistakes, the final answer should be: a∈(-∞,-4]∪{1}∪[8/3,4)∪(4,∞) |
WOOOOOOW!!!!!!!!! I never had anything past Geometry & Algebra 2 back when I was in secondary school. How'd you type all of those symbols? Did you need some type of 'maths keyboard' for all of that? Or did you take hours to type all of that through a point & click interface like Windows Character Map? Almost all of that looks really alien to me, as I haven't even had Trigonometry yet but, I know it isn't really complex maths. It just looks complex... I can't wait to learn that stuff, ias it looks pretty cool! I love Greek letters and various symbols from different languages... They fascinate me... |
|
|
| Noisician |
| quote: | Originally posted by Flyboy217
So we were both off... but at least I was only off by measure zero ;) |
fair enough. as for the other roots, that's just the way we are taught to perceive "all solutions" in my country. the difference is that, where i studied those things, "find all solutions" practically always implies "find all real and only real solutions" for the simple reason that we learn parametric equations (and parametric analysis in general) *long* before we do complex numbers. unless it specifically says "find all real and imaginary solutions," we are expected (required, even) to only consider the real ones by default. i'm not quite sure how it is in the states. but judging by your reaction, i guess it's different there. so yeah, in that case, complex roots certainly shouldn't be ruled out. |
|
|
|
|
