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\[\int\limits e^{(sinx)}sinx dx\]
HEY LOKI
Evening :)

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Make a substitution,\[u=\sin x\]
du=cos x
Yeah, hang on...
haha i tried..by parts..and substitution..
If you're looking for a solution in terms of standard integrals, I'm pretty sure it doesn't exist. You either have to write your answer in terms of the Bessel and Struve functions, or take a series solution, where you take the series expansion for sin(x), the expansion for e^(sin(x)) and then take the Cauchy product of the two series, and integrate term-by-term (which you can do since the series is uniformly convergent). Where did this problem come from?
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 :)
A geometric series is one where there's a constant ratio between each successive terms. A power series is different in that it is not always the case that the ratio between two successive terms is constant. In a geometric series, you'd have to successive terms,\[\frac{a_{n+1}(x-x_0)^{n+1}}{a_n(x-x_0)^n}=\frac{a_{n+1}}{a_n}(x-x_0)\]which is only constant if a_(n+1) and a_n give some number independent of n. In general, this isn't the case.
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~
Hey loki, i have a question, what test should i use to determine the convergence/divergence of the series sin(x/9)
\[\sum_{1}^{\infty}\sin(9/x)\]
I don't think this converges. Let me check.
ok
I'm distracted because some moron in my building has been playing the pellettest music I've ever heard, all day, all loud and I'm about to go break down doors ><
Dude you are the boss, just order your followers to do it for you
lol, i wish :(
I'm thinking you can set the problem up as proving it divergent using something like a limit comparison test.
Sorry, had to go talk to the neighbor.
no problem :) i went through every single tests...but none of them worked
To use the limit comparison, we need a series that is divergent (since we're suspicious this thing won't converge). When I think 'non-convergence', the first series I look to to test against is the harmonic series.
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
Form the ratio\[\frac{1/n}{\sin \frac{9}{n}}\]and note that, in the limit, you get an indeterminate form, \[\frac{0}{0}\]
This is great, because you now have access to L'Hopital's rule.
So\[\lim_{n \rightarrow \infty}\frac{1/n}{\sin 9/n}=\frac{\lim_{n \rightarrow \infty}1/n}{\lim_{n \rightarrow \infty}\sin 9/n}=\frac{\lim_{n \rightarrow \infty}-\frac{1}{n^2}}{\lim_{n \rightarrow \infty}\frac{-9\cos 9/n}{n^2}}=\frac{1}{9}\]
since cos(9/n) goes to 1 as n goes to infinity.
Now, since your ratio was that for a known divergent series with your own series, and because this ratio converged, you know that your series is divergent also.
So at the beginning, you have to guess whether this series diverges or converges?
Pretty much...it's okay, since you then turn towards something that can either prove or disprove your case.
A lot of mathematics is experimental.
But you yourself said you had a feeling that it was divergent...
On that integral before, you'd have to find more than just a couple of series expansions. I over-simplified it. If you ever see something like that in practice (i.e. engineering) you would use an approximation rule, like Simpson's or something.
I know it behaves like p-series as n->infinity, so if i had compared it to 1/n^2, it wouldn't follow any of the outcomes for the test?
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:
If \[\lim_{n \rightarrow \infty}a_n \ne0 \] or if this limit does not exist as n tends to infinity, then\[\sum_{n}^{\infty}a_n \]does not converge.
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
Now we come to 'instinct', which is a function of experience. It means you just need practice. You're looking at good questions. Trust your teachers - they'll show you all sorts of things you need to consider.
LOL what my math professors have taught me the most is HUMOR
Well, that *can* help..? :~S
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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
Ah, there's an order to it. You can see these orders if you make plots of the corresponding functions. Exponential functions will dominate polynomials at some point. You'll see, I'll write out how I worked out your question.
can you please do number 4 also :P
\[\frac{2^n+n^5+2}{19^{2n}+6^n+3} \iff \frac{2^n}{(19^2)^n}\]since in the numerator, the exponential dominates the quintic and 2, and in the denominator, (19^2)^n dominates 6^n+3 (19^2 > 6).
The equivalence sign should be replaced with ~. For some reason it doesn't show up in the equation editor.
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.
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You asked for tricks in math - I love asymptotic equivalence when it comes to limits - along with L'Hopital's rule. It saves your butt.
\[\frac{2^n/(19^2)^n}{r^n}=\left( \frac{2}{361r} \right)^n\]
Then choose r = 2/361 to ensure a convergent limit that's non-zero.
You can show that the original expression is asymptotically equivalent to the one we use in the end by taking the ratio between the two and showing the ratio approaches 1 as n approaches infinity.
Does exponential function always dominate quadratic, quintic ?
\[a^n>n^m\] for all a >1 and positive exponent. You have positive exponent because of the problem, and here you have the numbers 2, 6, 19. It is the case,\[19^n>6^n>2^n\]
Remember, when you're dealing with limits, the definition of limits makes use of "for *large* n...so you have to be thinking about the general properties of functions once they go through the motions and their domains start heading to infinity.
:P can you prove a^n >n^m
yes, but I'll show you a plot...
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They're all monotonic increasing functions, so when it gets to a point where the rate of one outstrips the other, the one with the slower rate of increase is dead in the water.
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.
http://www.youtube.com/watch?v=-C3v3aThfFA just when you have time loki, check out the electric car my friends from the university i wanna go built, it won xprize 2nd place
LOL when is your presentation ? i forgot~~
@ oktal Nice car ~ haha
I'll check it out - I saved the link. And presentations on Wed and Fri >< And to both of you - there's a site you should check out if you haven't: wolframalpha.com. You can check out math questions, like integrals and such.
yeah, when I found out they were builing it, i could not believe it, and when it won google x-prize 2nd place, I was stunned
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
Hi Loki, can you help me out with this problem
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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 \[a=\frac{v^2}{r}\]
11.7 is what i got from that equation..but it is wrong..probably becasue this is not uniform circular motion
This is a vector problem, and you're going to have to subtract an initial vector from a final vector...no, this is uniform circular motion...
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
and we know, from definition of angular speed and uniform circular motion that,\[\omega = \frac{\theta }{t} \rightarrow \theta= \omega t\]
So \[v=(-r \omega \sin (\omega t), r \omega \cos (\omega t))\]
There's a minus sign from when we take the derivative of the x-component of displacement with respect to time. That component is a cosine.
Since\[v = r \omega \rightarrow \omega = \frac{v}{r}\]so\[V=(-v \sin \frac{vt}{r}, v \cos \frac{vt}{r})=v(\sin \frac{vt}{r}, \cos \frac{vt}{r})\]
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.
It goes like this: let the arc of a circle of radius r have length (distance) l. Then that distance is\[l=r \theta\]when theta s in radians. Take the derivative with respect to time on both sides to get\[\frac{dl}{dt}=\frac{d(r \theta)}{dt}=r \frac{d \theta }{dt}+ \theta \frac{dr}{dt}=r \frac{d \theta }{dt}:=r \omega\]
dr/dt is zero by construction of the problem -> the radius is fixed and therefore unchanging in time.
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 -.-
i am going to ask my professor... see if there is another way to do it..since we haven't covered angular velocity....
Okay. At least it worked. I hate online submission. I used to hate it as an undergrad.; you can never argue a point if it's wrong.
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 ;)

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