How this comes?


by waqarrashid33
Tags: None
waqarrashid33
waqarrashid33 is offline
#1
May5-12, 03:13 AM
P: 77
Let me know how 1/2 comes from it.see attachemet
Attached Thumbnails
prob.JPG  
Phys.Org News Partner Mathematics news on Phys.org
Researchers help Boston Marathon organizers plan for 2014 race
'Math detective' analyzes odds for suspicious lottery wins
Pseudo-mathematics and financial charlatanism
Mentallic
Mentallic is offline
#2
May5-12, 03:21 AM
HW Helper
P: 3,436
Have you tried anything for yourself?

Solve the integral and see if you can simplify the limit to transform the expression into a form that you know the answer to.
waqarrashid33
waqarrashid33 is offline
#3
May5-12, 03:33 AM
P: 77
i tried a lot but answer goes wrong..
i didn't touched with calculus since long time...May be it is because of this...
For the integral of 1+cos(2t) gives 2T+sin(2T).

Mentallic
Mentallic is offline
#4
May5-12, 03:46 AM
HW Helper
P: 3,436

How this comes?


Quote Quote by waqarrashid33 View Post
i tried a lot but answer goes wrong..
i didn't touched with calculus since long time...May be it is because of this...
For the integral of 1+cos(2t) gives 2T+sin(2T).
Yes that would probably be the problem then.
[tex]\int1+\cos(2t)dt=t+\frac{1}{2}\sin(2t)[/tex]

Start from there.
chiro
chiro is offline
#5
May5-12, 03:46 AM
P: 4,570
Quote Quote by waqarrashid33 View Post
i tried a lot but answer goes wrong..
i didn't touched with calculus since long time...May be it is because of this...
For the integral of 1+cos(2t) gives 2T+sin(2T).
Try solving your improper integral (for 1 + cos(2t) you will get anti-derivative t + 1/2sin(2t)) and then expand into F(T) - F(-T), collect your constant outside of the integral (1/2T) bring it together and you will get a limit expression in terms of T where T goes to infinity.

There are limit theorems you can use to solve this also.
waqarrashid33
waqarrashid33 is offline
#6
May5-12, 05:33 AM
P: 77
Thanks....
DonAntonio
DonAntonio is offline
#7
May5-12, 08:25 AM
P: 606
Quote Quote by waqarrashid33 View Post
Thanks....


Interesting: you don't even need [itex]\,\,T\to\infty\,\,[/itex]. It is 1/2 for any [itex]\,\,T\neq 0\,[/itex].

DonAntonio
Mentallic
Mentallic is offline
#8
May5-12, 08:40 AM
HW Helper
P: 3,436
Quote Quote by DonAntonio View Post
Interesting: you don't even need [itex]\,\,T\to\infty\,\,[/itex]. It is 1/2 for any [itex]\,\,T\neq 0\,[/itex].

DonAntonio
Not quite, the final steps of the solution are to simplify [tex]\frac{1}{2}\left(1+\lim_{T\to a}\frac{\sin(2T)}{2T}\right)[/tex]

and that expression is only equal to 1/2 if [tex]\lim_{T\to a}\frac{\sin(2T)}{2T}=0[/tex] which only happens for [itex]a=\infty[/itex]
DonAntonio
DonAntonio is offline
#9
May5-12, 10:42 AM
P: 606
Quote Quote by Mentallic View Post
Not quite, the final steps of the solution are to simplify [tex]\frac{1}{2}\left(1+\lim_{T\to a}\frac{\sin(2T)}{2T}\right)[/tex]

and that expression is only equal to 1/2 if [tex]\lim_{T\to a}\frac{\sin(2T)}{2T}=0[/tex] which only happens for [itex]a=\infty[/itex]


I don't know how you got that. I get
[tex]\int_{-T}^T \cos^2(t)dt=\left[\frac{t+\cos t\sin t}{2}\right]_{-T}^T=\frac{1}{2}\left[T+\cos T\sin T-\left(-T-\cos(-T)\sin(-T)\right)\right]=\frac{2T}{2}=T[/tex]
as [itex]\,\,\cos(-T)\sin(-T)=-\cos T\sin T\,\,[/itex] , and then
[tex]\frac{1}{2T}\int^T_{-T}\cos^2 t\,dt=\frac{1}{2}[/tex]
like that, without limit...

DonAntonio
Mentallic
Mentallic is offline
#10
May5-12, 11:28 AM
HW Helper
P: 3,436
Quote Quote by DonAntonio View Post
I don't know how you got that. I get
[tex]\int_{-T}^T \cos^2(t)dt=\left[\frac{t+\cos t\sin t}{2}\right]_{-T}^T[/tex]
How did you get that?

[tex]\cos^2t=\frac{1}{2}\left(1+\cos(2t)\right)[/tex]

Oh ok I see what you have, after integrating you converted sin(2t) to 2sin(t)cos(t)

Quote Quote by DonAntonio View Post
[tex]\left(-T-\cos(-T)\sin(-T)\right)[/tex]
This should be

[tex]\left(-T+\cos(-T)\sin(-T)\right)[/tex]
DonAntonio
DonAntonio is offline
#11
May5-12, 01:17 PM
P: 606
Quote Quote by Mentallic View Post
How did you get that?

[tex]\cos^2t=\frac{1}{2}\left(1+\cos(2t)\right)[/tex]

Oh ok I see what you have, after integrating you converted sin(2t) to 2sin(t)cos(t)



This should be

[tex]\left(-T+\cos(-T)\sin(-T)\right)[/tex]

Yes indeed. So much worrying about the change of sign in the sine of -T that I forgot I had that minus sign out of the parentheses. Thanx.

DonAntonio


Register to reply