VCE Stuff > VCE Mathematics

Recreational Problems

<< < (3/6) > >>

Freitag:

--- 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?)
--- End quote ---


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. :(
--- End quote ---



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. :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.;)

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.  :)

Navigation

[0] Message Index

[#] Next page

[*] Previous page