wildareal's questions thread

[kamil9876]

Actually, just realised there is a really simple way. Can use a trick I used to use all the time when I did more calculus/analysis. Can't believe I missed it 

[Mao]

Actually, just realised there is a really simple way. Can use a trick I used to use all the time when I did more calculus/analysis. Can't believe I missed it 

Now I want to know what it is. Post please.

[kamil9876]

Using the previous notation we have hence

is a decreasing function, hence is a increasing. Now suppose that then since f(f(a)) is increasing we have and hence . Similairly we get that if    then . So we have shown that the sequences a_1,a_3... and a_2,a_4... are monotonic (don't go up and down) hence they both have a limit (they are bounded). Now it only suffices to find both of these limits by solving the equation f(f(a))=a (you get this equation by "taking the limit of both sides" in the reccurence and of course using the fact that f(a) is continous) and showing that the two limits are indeed the same, hence the whole sequence converges.

too bad no latex, may be an annoying read

[Mao]

Using the previous notation we have hence

is a decreasing function, hence is a increasing. Now suppose that then since f(f(a)) is increasing we have and hence . Similairly we get that if    then . So we have shown that the sequences a_1,a_3... and a_2,a_4... are monotonic (don't go up and down) hence they both have a limit (they are bounded). Now it only suffices to find both of these limits by solving the equation f(f(a))=a (you get this equation by "taking the limit of both sides" in the reccurence and of course using the fact that f(a) is continous) and showing that the two limits are indeed the same, hence the whole sequence converges.

too bad no latex, may be an annoying read

Nice!

I got to is an increasing function. I just did not make the connection that this would mean

Kicking myself at the moment. Haha

[wildareal]

Does anyone know anything about Bessel equations?

[wildareal]

...and differentiation using eigenvectors?

[wildareal]

A train that is moving with uniform acceleration is observed to take 20 and 30 seconds to travel successive 500m intervals. How much farther will it travel before coming to rest if the acceleration remains constant?

[b^3]

So for the first 500m
u=u
a=a
s=500
t=20
using s=ut+1/2*at^2
500=20u+200a
u=(500-200a)/20
u=25-10a
also v=u+at
v=u+20a
For the second 500m
u₂=v₁=u+20a
a=a
s=500
t=30
so 500=u+20a+(1/2)*a*900
500=u+20a+450a
500=u+470a
sub in u=25-10a
500=25-10a+470a
475=460a
a=95/92m/s^2
Now from here until it stops.
First find u (for the first 500m)
u=25-10a
=25-10*95/92
=14.68 (use fraction not approx)
so u for the second 500m
u=u+20*95/92
=14.68+20.65
=35.33
Find the final velocity after the second 500m and that will become u below.

a=95/92
s=?
v=0
u=...
Subtitute in and solve for s.

I think there is a simpler way and I have a feeling that this may be wrong, but I don't have a calculator on me right now.