Recreational Problems

[Ahmad]

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.

[bilgia]

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

[Ahmad]

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

[Collin Li]

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

[Ahmad]

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

[Collin Li]

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.

[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?)

[Ahmad]

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

[enwiabe]

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. so 22/7 - pi according to year 8 maths is 0. Ah, the lies we're told.

[Ahmad]

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

[Freitag]

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. 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. :-/

[Ahmad]

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..

[Ahmad]

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?

[Despondent]

(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.

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

[Ahmad]

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.