Register to reply

How this comes?

by waqarrashid33
Tags: None
Share this thread:
waqarrashid33
#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
Heat distributions help researchers to understand curved space
Professor quantifies how 'one thing leads to another'
Team announces construction of a formal computer-verified proof of the Kepler conjecture
Mentallic
#2
May5-12, 03:21 AM
HW Helper
P: 3,540
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
#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
#4
May5-12, 03:46 AM
HW Helper
P: 3,540
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
#5
May5-12, 03:46 AM
P: 4,573
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
#6
May5-12, 05:33 AM
P: 77
Thanks....
DonAntonio
#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
#8
May5-12, 08:40 AM
HW Helper
P: 3,540
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
#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
#10
May5-12, 11:28 AM
HW Helper
P: 3,540
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
#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