ATAR Notes: Forum

VCE Stuff => VCE University of Melbourne Extension Program => VCE University Enhancement Programs => VCE Subjects + Help => VCE Mathematics => Topic started by: Ahmad on October 20, 2007, 12:24:53 pm

Title: Recreational Problems
Post by: Ahmad on October 20, 2007, 12:24:53 pm

This is the UMEP equivalent of the spesh thread. Questions here tend to require UMEP knowledge.

I'll start with a couple of integrals from the exercise book, then we'll move out from there.

Title: Recreational Problems
Post by: bilgia on October 20, 2007, 09:50:46 pm

lol u sure have alot of spare time to play with maths

Title: Recreational Problems
Post by: Ahmad on October 20, 2007, 09:54:45 pm

Mainly because I choose to ignore school work.  Not recommended! 8)

Title: Recreational Problems
Post by: Collin Li on October 21, 2007, 02:19:30 am

Question 1
f(t) = t^4 * (1 - t^4) / (1 + t^2)
= t^4 (1 - t^2)(1 + t^2) / (1 + t^2)
= t^4 (1 - t^2)
= t^4 - t^6

Integration (with respect to t) yields:
F(t) = 1/5*t^5 - 1/6*t^6 + C

Therefore, integration from 0 to 1:
F(1) - F(0) = (1/5 - 1/6) - 0
= 1/30

Title: Recreational Problems
Post by: Ahmad on October 21, 2007, 07:04:29 am

Of course, you're right. Except I accidently mistyped the problem! I have corrected it now.  :oops:

Title: Recreational Problems
Post by: Collin Li on October 21, 2007, 09:16:00 am

Quote from: "Ahmad"

Of course, you're right. Except I accidently mistyped the problem! I have corrected it now.  :oops:

Wow, I'm surprised that the accident actually yielded a possible problem.

Title: Recreational Problems
Post by: Defiler on October 30, 2007, 12:55:49 pm

Right.

1. int (0 to 1) t^4 * (1 - t)^4 / (1 + t^2) dt (Corrected version)

Expanding the top:

(t^8 - 4t^2 + 6t^6 - 4t^5 + t^4) / (1+t^2)

Taking a factor of (1+t^2) from the numerator:

((1+t^2)(-4/(t^2+1)+t^6-4t^5+5t^4-4t^2+4)) / (1+t^2)

Hence int (0 to 1) t^4 * (1 - t)^4 / (1 + t^2) dt = int (0 to 1) ( -4/(t^2+1)+t^6-4t^5+5t^4-4t^2+4) )

Which can be evaluated.

= [ -4tan-1(t) + t^7/7 - 2t^6/3 + t^5 - 4t^3/3 + 4t ] (0 to 1)

= [ -4*(pi/4) + 1/7 - 2/3 + 1 - 4/3 +4 ] - (zero)

= -pi + 22/7

= 22/7 - pi


(I know there are probably more efficient ways to deduce the answer, but hey... whatever works?)

Title: Recreational Problems
Post by: Ahmad on October 30, 2007, 01:08:37 pm

Correct. I'll post more when I have time/can be bothered. :)

Title: Recreational Problems
Post by: enwiabe on October 30, 2007, 03:27:09 pm

Quote from: "Defiler"

Right.

1. int (0 to 1) t^4 * (1 - t)^4 / (1 + t^2) dt (Corrected version)

Expanding the top:

(t^8 - 4t^2 + 6t^6 - 4t^5 + t^4) / (1+t^2)

Taking a factor of (1+t^2) from the numerator:

((1+t^2)(-4/(t^2+1)+t^6-4t^5+5t^4-4t^2+4)) / (1+t^2)

Hence int (0 to 1) t^4 * (1 - t)^4 / (1 + t^2) dt = int (0 to 1) ( -4/(t^2+1)+t^6-4t^5+5t^4-4t^2+4) )

Which can be evaluated.

= [ -4tan-1(t) + t^7/7 - 2t^6/3 + t^5 - 4t^3/3 + 4t ] (0 to 1)

= [ -4*(pi/4) + 1/7 - 2/3 + 1 - 4/3 +4 ] - (zero)

= -pi + 22/7

= 22/7 - pi


(I know there are probably more efficient ways to deduce the answer, but hey... whatever works?)

What's interesting is that 22/7 is the fraction approximation for pi that we learn in like year 8. :P so 22/7 - pi according to year 8 maths is 0. Ah, the lies we're told. :(

Title: Recreational Problems
Post by: Ahmad on October 30, 2007, 03:31:06 pm

Yup, furthermore you can yield a bound on pi by noting int (0 to 1) t^4 * (1 - t)^4 / (1 + t^2) > int (0 to 1) t^4*(1-t)^4/2


(since (1+t^2) <= 2 over the interval).

It's not the best bound though.. but it works :D

Title: Recreational Problems
Post by: Freitag on November 01, 2007, 08:16:26 pm

Quote from: "enwiabe"

Quote from: "Defiler"

Right.

1. int (0 to 1) t^4 * (1 - t)^4 / (1 + t^2) dt (Corrected version)

Expanding the top:

(t^8 - 4t^2 + 6t^6 - 4t^5 + t^4) / (1+t^2)

Taking a factor of (1+t^2) from the numerator:

((1+t^2)(-4/(t^2+1)+t^6-4t^5+5t^4-4t^2+4)) / (1+t^2)

Hence int (0 to 1) t^4 * (1 - t)^4 / (1 + t^2) dt = int (0 to 1) ( -4/(t^2+1)+t^6-4t^5+5t^4-4t^2+4) )

Which can be evaluated.

= [ -4tan-1(t) + t^7/7 - 2t^6/3 + t^5 - 4t^3/3 + 4t ] (0 to 1)

= [ -4*(pi/4) + 1/7 - 2/3 + 1 - 4/3 +4 ] - (zero)

= -pi + 22/7

= 22/7 - pi


(I know there are probably more efficient ways to deduce the answer, but hey... whatever works?)

What's interesting is that 22/7 is the fraction approximation for pi that we learn in like year 8. :P so 22/7 - pi according to year 8 maths is 0. Ah, the lies we're told. :(

Up until this year I still thought that 22/7 was an extremely close approximation of pi.

It was until i got bored at work one day and started dividing 22 by 7 mentally that i realised how quickly they differ in decimal places. :-/

Title: Recreational Problems
Post by: Ahmad on November 01, 2007, 09:28:22 pm

Well, if I remember correctly it's part of the continued fraction for pi, so in a sense it's the "best" approximation. Try 355/113, which is also part of the continued fractions..

Title: Recreational Problems
Post by: Ahmad on November 12, 2007, 10:10:28 pm

No ones did the second problem, and it's been a while. It's in the exercise book as well. Come on you guys!! Don't back down the challenge! Want a hint?

Title: Recreational Problems
Post by: Despondent on November 23, 2007, 01:49:02 am

(2x)/(1+(cos(x))^2) is odd so we only need to consider the integral int(-pi to pi) (2x*sin(x))/(1+(cos(x))^2). Integrate by parts - diff the 2x and integrate the rest (I don't do the whole set u = whatever, v = something else because that's n00bish).

Suggestion: Please implement LaTeX if possible. :D

I know the subject's pretty much over for you guys, apart from exams I think but I just thought I'd chip in.;)

Title: Recreational Problems
Post by: Ahmad on November 23, 2007, 02:36:10 pm

That's an idea, but you didn't finish your solution. I haven't tried it, but it's not obvious to me how to proceed with this method. What do you do after integrating by parts? Please provide a complete solution.  :)

Title: Recreational Problems
Post by: Despondent on November 23, 2007, 03:55:24 pm

It was more of a hint to others as to how to proceed rather than a personal attempt to get the answer out. One thing I observed whilst actually writing this one out was that over use of properties of the integrand can lead to dead ends. :)

The integral is
 
I = int(-pi,pi) (2x+2x*sin(x))/(1+(cos(x))^2) dx

= int(-pi,pi) ((2x)/(1+(cos(x))^2))dx + int(-pi,pi) (2x*sin(x))/(1+(cos(x))^2)dx

The first integral is zero because the integrand is odd and so

I = int(-pi,pi) (2x*sin(x))/(1+(cos(x))^2) dx

Rewriting the integral as 2 times an integral from 0 to pi here will actually end up making things more complicated.

= int(-pi,pi) (2x)*((sin(x))/(1+(cos(x))^2)dx

int ((sin(x))/(1+(cos(x))^2) dx = -arctan(cos(x)) + c

=> I = [-2x*arctan(cos(x))]|(-pi,pi) + 2*int(-pi,pi)(arctan(cos(x))) dx

= -2[x*arctan(cos(x))] |(-pi,pi) + 0 since the second integrand is odd

= -2(-(((pi)^2)/4)-(((pi)^2)/4))

So I = (pi)^2.

Title: Recreational Problems
Post by: Ahmad on November 23, 2007, 04:12:23 pm

Can you justify why:
2*int(-pi,pi)(arctan(cos(x))) dx = 0, you say the integrand is odd, but it isn't. :)

I'll get more problems up when I have time.  8)

Title: Recreational Problems
Post by: Despondent on November 23, 2007, 04:19:34 pm

If you've still got some exams to do it's probably best that you concentrate on those for now. Extra questions can come at any time.

Anyway, arctan(x) is an odd function. As long as its argument is continuous, any integral over arctan from (-b,b) (with b a real number) will always be zero.

Edit: Let me check.

Title: Recreational Problems
Post by: Ahmad on November 23, 2007, 04:21:36 pm

Yes, but for a function to be odd:

f(-x) = -f(x)

Here you have f(x) = arctan[cos(x)]

But, arctan[cos(-x)] isn't equal to -arctan[cos(x)].

You have to use a special argument!  :wink:

Title: Recreational Problems
Post by: Despondent on November 23, 2007, 04:43:18 pm

int(-pi,pi) arctan(cos(x)) dx

= 2*int(0,pi) arctan(cos(x)) dx

= 2*int(0,pi/2) arctan(cos(x)) dx + 2*int(pi/2,pi) arctan(cos(x)) dx

Let u = pi - x in the first integral then

arctan(cos(x)) = arctan(cos(pi-u)) = arctan(-cos(u))=-arctan(cos(u)) and so

I = 2*int(pi,pi/2) arctan(cos(u)) du + 2*int(pi/2,pi) arctan(cos(x)) dx

= -2*int*(pi/2,pi) arctan(cos(u)) du + 2*int(pi/2,pi) arctan(cos(x)) dx

= 0

Title: Recreational Problems
Post by: Ahmad on November 23, 2007, 04:48:28 pm

If you need a hint, do ask.  It's really great how far you've got already, really well done! :)

Title: Recreational Problems
Post by: Despondent on November 23, 2007, 05:08:28 pm

I should hardly need a hint considering that I did this sort of thing two years ago. Periodicity, it took a while to kick in. I'm a bit rusty.;)

Title: Recreational Problems
Post by: Ahmad on November 23, 2007, 05:11:50 pm

That's correct. Well done. My solution is a little different, but it doesn't matter, they're both good.

Two years ago? You did this in year 10? Good stuff!  :wink:

Title: Recreational Problems
Post by: Despondent on November 23, 2007, 05:17:32 pm

I'm in 3rd year uni. :D  My original intention was to just suggest a way to start this problem since there weren't any attempts. I didn't plan on doing the whole thing.

Title: Recreational Problems
Post by: Ahmad on November 23, 2007, 05:21:35 pm

Cool. What are you taking in university?

Title: Recreational Problems
Post by: Despondent on November 23, 2007, 05:24:48 pm

I'm taking BE/BSc.

Which course do you plan on taking?

Title: Recreational Problems
Post by: Ahmad on November 23, 2007, 05:27:37 pm

Actuarial studies, so that's BComm I think. Although maths is my hobby/passion!  8)

Title: Recreational Problems
Post by: Despondent on November 23, 2007, 05:33:29 pm

Well I hope you have fun taking it. Not sure if you know this already, but after first year a lot of the 'maths' that you would be doing will be prob/stats.

I've always disliked stats but I guess some people would enjoy it.