mate,, u really shouldn't mislead others when u know very well there is no such integration of one to infinity!! Teach the correct procedure,,,, use limits and a constant 't' so it can approach infinity. otherwise, well done.
you have an improper integral in ex. 2, and your just using infinity to plug in? I never learned any method for doing that...I always use the lim as t approaches infinity of the integral from 1 to t.... maybe you know something I don't?
@tadm123 you have to find the critical points. most likely problems will have a critical point that ends up being less than one (but not always.) then, take any point thats larger than that critical point and find what the derivative is at that point. if it's less than 1 you know that its decreasing on the entire interval (cp, infinity) and b/c you have no other critical points.
@tadm123 When the derivative is negative, the original function decreases.
For the first problem, since the denominator was squared, it's always positive, so the derivative is negative wherever -2n is negative. Since n, in this case, is in the positive integers, -2n is always negative, so it's always decreasing in the domain.
mate,, u really shouldn't mislead others when u know very well there is no such integration of one to infinity!! Teach the correct procedure,,,, use limits and a constant 't' so it can approach infinity. otherwise, well done.
lfcforever 4 months ago 2
thanx dude this helped a lot!
randookie3 4 months ago
Thank you, this has been very useful and easy to understand.
LiteHeeto 9 months ago
TO MUCH CAMERA SHAKE! THIS ISNT THE BOURNE IDENTITY! YOU AINT MATT DAMON!
corzocl 11 months ago
@corzocl YES HE IS GODDAMIT! YES HE IS! LEAVE HIM ALONE.
DazzTyler 10 months ago
@DazzTyler THEN WHO AM I BECAUSE I THOUGHT I WAS MATT DAMON?!
MrFarklefreak 10 months ago
@DazzTyler ?
ac7491 10 months ago
you have an improper integral in ex. 2, and your just using infinity to plug in? I never learned any method for doing that...I always use the lim as t approaches infinity of the integral from 1 to t.... maybe you know something I don't?
mpatt79 1 year ago
just b/c we find the first derivative, that means it's decreasing?
mpatt79 1 year ago
@mpatt79 First derivative is -ve => gradient is negative => decreasing.
wowtrax 1 year ago
@mpatt79 Yep
Mihre2 9 months ago
Wow. Great vid
TheBrownScourge 1 year ago
how do you know if its decreasing? i didn't get that part :o. you find that derivative and you plug something in?
tadm123 1 year ago
@tadm123 you have to find the critical points. most likely problems will have a critical point that ends up being less than one (but not always.) then, take any point thats larger than that critical point and find what the derivative is at that point. if it's less than 1 you know that its decreasing on the entire interval (cp, infinity) and b/c you have no other critical points.
dkong13able 1 year ago
@tadm123 When the derivative is negative, the original function decreases.
For the first problem, since the denominator was squared, it's always positive, so the derivative is negative wherever -2n is negative. Since n, in this case, is in the positive integers, -2n is always negative, so it's always decreasing in the domain.
MyOverflow 11 months ago
GOOD VID! this really helped me out a lot since im trying to learn series to Laplace transforms in a about a week.
t0mb03 1 year ago