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\[\int\limits e^{(sinx)}sinx dx\]

HEY LOKI

Evening :)

Make a substitution,\[u=\sin x\]

du=cos x

Yeah, hang on...

haha i tried..by parts..and substitution..

LOL umm someone asked this question yesterday, and i couldn't solve it.So i am asking you, the boss

Haha, well, boss says that.

oh dam be humble :P

lol, I am ;)

LOL sure boss :P

Well, I'm signing out...have a *lot* of work to do ><

I'll answer your ferris wheel question later :)

You should chillax, it's Friday.

question!!

What is the difference between power and geometric series!

I am chillaxing by chit chating with my roomate

Take your time on the wheel question :)

Thank you. I will try to do a little bit more research on my ow :)

This is research :)

LOL just go do your work!

Yeah I know...

I have Lammy below trying to figure something out. When he/she's done, I'm out... -.-

Alright i will ask questions later if have any :) Peace~

\[\sum_{1}^{\infty}\sin(9/x)\]

I don't think this converges. Let me check.

ok

Dude you are the boss, just order your followers to do it for you

lol, i wish :(

Sorry, had to go talk to the neighbor.

no problem :) i went through every single tests...but none of them worked

Should I wait for you to finish typing?

lol what if i am suspicious this thing converge... then i will be in trouble

Lol, yes, you would be :D

This is great, because you now have access to L'Hopital's rule.

since cos(9/n) goes to 1 as n goes to infinity.

So at the beginning, you have to guess whether this series diverges or converges?

A lot of mathematics is experimental.

But you yourself said you had a feeling that it was divergent...

1/n^2 is convergent, whereas your suspicion is that sin(9/n) is divergent. It wouldn't have helped.

i used the nth term test, and the limit goes to 0, so i thought it should be convergent.

Ahhhhh....no....

If you read the statement of the nth term test, it says:

The nth term test doesn't say that if your sequence tends to zero, your series converges.

It says if it DOESN'T tend to zero, your series DIVERGES.

Take 1/n for example. The limit is 0 as n goes to infinity, but the series itself does not converge.

yea, so , as a normal college kid, my suspicion would be convergent

SUSPICION

LOL what my math professors have taught me the most is HUMOR

Well, that *can* help..? :~S

i am stuck at number 3

you mean 4?

3..that answer is wrong

Does this have to be submitted now?

Nope, but soon~~->>>> tomorrow night

Oh, okay. I need to go out, that's all. I'll take a look later.

OH alright :) have fun!

Bye..

Did you figure it out?

Nope haha you told me to chillax

Slacker... ;p

I'll look now.

take your time. just look at it whenever u feel like it

The first one can be 2/361. You have 1/361.

Have you covered asymptotic equivalence?

what is that? might have..not sure

brb

kay kay

I know that as n-> infin, some terms become negligible

Yeah, basically. The definition is this:

But how you determine which term is going to infinity faster

can you please do number 4 also :P

brb

i thought n^5 is going to infinity faster

No, if you plot n^5 and 2^n, you'll see 2^n overtake pretty quickly.

But listen, you can always check domination by using the definition. I'll scan in something.

\[\frac{2^n/(19^2)^n}{r^n}=\left( \frac{2}{361r} \right)^n\]

Then choose r = 2/361 to ensure a convergent limit that's non-zero.

Does exponential function always dominate quadratic, quintic ?

:P can you prove a^n >n^m

yes, but I'll show you a plot...

is that a free program haha

Yes it's free.

GeoGebra

http://www.geogebra.org/cms/

thanks !!

how do i input multiple graphs

Just type them in on the one page. y=x^2, y=x^4, etc. They should show up.

Got it hohoho

I don't know if you've done the last one, but I have \[r=\left( \frac{2^{7/10}}{19} \right)^3\]

yea i did it haha

what you up to right now

I should be doing my work...

Seriously going offline...it's getting dark and I've done nothing.

LOL when is your presentation ? i forgot~~

@ oktal Nice car ~ haha

ok, i will see it, thanks bye

lol you still have few days to prepare for it no worries. you should chill with us

no, let him work, he has a life to manage...

LOL he has troublesome neighbors to manage as matter of fact :P

Nice car, oktal

not mine :(

my friend's

and actually it belongs to the university now, as it funded the building of it

it's pretty cool. they should at least be able to take it out on weekends ;p

"Hey ladies, check out my revolutionary design..." ;)

:)

you would haha

lol. oktal, check out a nerd link above for free plotting software, GeoGebra.

they did it few times, drove around town :)

ok

You could have used it to check out that absolute value problem.

nice, will check it

Okay, I'm out...seriously need to get sorted. Later!

Laters ~~ i am going to to bed too

nice meeting you okatal :)

ok, see you, nice to meet you too dichalao

What did you enter before?

11.7

brb...when's this due?

umm not soon :)

oh, are these your physics questions that aren't due till the end of the week or something?

it is due on tuesday before lecture

Your value is what you get when you work out magnitude, since \[a=\frac{v^2}{r}\]

So,\[a=\frac{\Delta v}{\Delta t}=\frac{v_2-v_1}{t_2-t_1}\]where the a and v's are vectors.

The velocity in uniform circular motion is\[v=(-r \omega \sin \theta, r \omega \cos \theta)\]

why is there a negative sign for V_x

So \[v=(-r \omega \sin (\omega t), r \omega \cos (\omega t))\]

You should be able to work out two vectors at different times and then subtract.

Is that making sense? I hope it is 'cause I just got up and I'm doing the on-the-fly.

in fact, we haven't "learned" angular speed .

There's a minus sign on the last expression outside sine...typo.

Angular speed is the rate of change in the angle.

brb

kk

Did you work it out?

kk

is there something wrong with this site..i didn't reply"kk"

do i just plug in t=0,and t=0.6?

it came through...and there's always something wrong with this site.

I would plug 0 and 0.6.

is vt/r in radians?

Always.

Let me check something before you waste a turn.

is there a way solving this problem without using angular speed?

Not off the top of my head.

\[\Delta v = 5.3(-\sin (1.325), \cos (1.325)-1)\]

\[a=\frac{\Delta v}{\Delta t}\approx \frac{1}{0.6}(-5.14,-4.01) \approx (-8.57, -6.68)\]

Thanks!!!!

What, did it work?

yep~

Good -.-

Anyways, I have to sign out and get some work done. Happy mathing :)

LOL you meant physicsing?

All the same...

alright thanks for helping!! have a great day...still have an essay to do..

At least they run you into the ground.

LOL true~ see ya ~buddy

cu ;)