anonymous
  • anonymous
.....
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\int\limits e^{(sinx)}sinx dx\]
anonymous
  • anonymous
HEY LOKI
anonymous
  • anonymous
Evening :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Make a substitution,\[u=\sin x\]
anonymous
  • anonymous
du=cos x
anonymous
  • anonymous
Yeah, hang on...
anonymous
  • anonymous
haha i tried..by parts..and substitution..
anonymous
  • anonymous
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?
anonymous
  • anonymous
LOL umm someone asked this question yesterday, and i couldn't solve it.So i am asking you, the boss
anonymous
  • anonymous
Haha, well, boss says that.
anonymous
  • anonymous
oh dam be humble :P
anonymous
  • anonymous
lol, I am ;)
anonymous
  • anonymous
LOL sure boss :P
anonymous
  • anonymous
Well, I'm signing out...have a *lot* of work to do ><
anonymous
  • anonymous
I'll answer your ferris wheel question later :)
anonymous
  • anonymous
You should chillax, it's Friday.
anonymous
  • anonymous
question!!
anonymous
  • anonymous
What is the difference between power and geometric series!
anonymous
  • anonymous
I am chillaxing by chit chating with my roomate
anonymous
  • anonymous
Take your time on the wheel question :)
anonymous
  • anonymous
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.
anonymous
  • anonymous
Thank you. I will try to do a little bit more research on my ow :)
anonymous
  • anonymous
This is research :)
anonymous
  • anonymous
LOL just go do your work!
anonymous
  • anonymous
Yeah I know...
anonymous
  • anonymous
I have Lammy below trying to figure something out. When he/she's done, I'm out... -.-
anonymous
  • anonymous
Alright i will ask questions later if have any :) Peace~
anonymous
  • anonymous
Hey loki, i have a question, what test should i use to determine the convergence/divergence of the series sin(x/9)
anonymous
  • anonymous
\[\sum_{1}^{\infty}\sin(9/x)\]
anonymous
  • anonymous
I don't think this converges. Let me check.
anonymous
  • anonymous
ok
anonymous
  • anonymous
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 ><
anonymous
  • anonymous
Dude you are the boss, just order your followers to do it for you
anonymous
  • anonymous
lol, i wish :(
anonymous
  • anonymous
I'm thinking you can set the problem up as proving it divergent using something like a limit comparison test.
anonymous
  • anonymous
Sorry, had to go talk to the neighbor.
anonymous
  • anonymous
no problem :) i went through every single tests...but none of them worked
anonymous
  • anonymous
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.
anonymous
  • anonymous
Should I wait for you to finish typing?
anonymous
  • anonymous
lol what if i am suspicious this thing converge... then i will be in trouble
anonymous
  • anonymous
Lol, yes, you would be :D
anonymous
  • anonymous
Form the ratio\[\frac{1/n}{\sin \frac{9}{n}}\]and note that, in the limit, you get an indeterminate form, \[\frac{0}{0}\]
anonymous
  • anonymous
This is great, because you now have access to L'Hopital's rule.
anonymous
  • anonymous
So\[\lim_{n \rightarrow \infty}\frac{1/n}{\sin 9/n}=\frac{\lim_{n \rightarrow \infty}1/n}{\lim_{n \rightarrow \infty}\sin 9/n}=\frac{\lim_{n \rightarrow \infty}-\frac{1}{n^2}}{\lim_{n \rightarrow \infty}\frac{-9\cos 9/n}{n^2}}=\frac{1}{9}\]
anonymous
  • anonymous
since cos(9/n) goes to 1 as n goes to infinity.
anonymous
  • anonymous
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.
anonymous
  • anonymous
So at the beginning, you have to guess whether this series diverges or converges?
anonymous
  • anonymous
Pretty much...it's okay, since you then turn towards something that can either prove or disprove your case.
anonymous
  • anonymous
A lot of mathematics is experimental.
anonymous
  • anonymous
But you yourself said you had a feeling that it was divergent...
anonymous
  • anonymous
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.
anonymous
  • anonymous
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?
anonymous
  • anonymous
1/n^2 is convergent, whereas your suspicion is that sin(9/n) is divergent. It wouldn't have helped.
anonymous
  • anonymous
i used the nth term test, and the limit goes to 0, so i thought it should be convergent.
anonymous
  • anonymous
Ahhhhh....no....
anonymous
  • anonymous
If you read the statement of the nth term test, it says:
anonymous
  • anonymous
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.
anonymous
  • anonymous
The nth term test doesn't say that if your sequence tends to zero, your series converges.
anonymous
  • anonymous
It says if it DOESN'T tend to zero, your series DIVERGES.
anonymous
  • anonymous
Take 1/n for example. The limit is 0 as n goes to infinity, but the series itself does not converge.
anonymous
  • anonymous
yea, so , as a normal college kid, my suspicion would be convergent
anonymous
  • anonymous
SUSPICION
anonymous
  • anonymous
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.
anonymous
  • anonymous
LOL what my math professors have taught me the most is HUMOR
anonymous
  • anonymous
Well, that *can* help..? :~S
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
i am stuck at number 3
anonymous
  • anonymous
you mean 4?
anonymous
  • anonymous
3..that answer is wrong
anonymous
  • anonymous
Does this have to be submitted now?
anonymous
  • anonymous
Nope, but soon~~->>>> tomorrow night
anonymous
  • anonymous
Oh, okay. I need to go out, that's all. I'll take a look later.
anonymous
  • anonymous
OH alright :) have fun!
anonymous
  • anonymous
Bye..
anonymous
  • anonymous
Did you figure it out?
anonymous
  • anonymous
Nope haha you told me to chillax
anonymous
  • anonymous
Slacker... ;p
anonymous
  • anonymous
I'll look now.
anonymous
  • anonymous
take your time. just look at it whenever u feel like it
anonymous
  • anonymous
The first one can be 2/361. You have 1/361.
anonymous
  • anonymous
Have you covered asymptotic equivalence?
anonymous
  • anonymous
what is that? might have..not sure
anonymous
  • anonymous
brb
anonymous
  • anonymous
kay kay
anonymous
  • anonymous
I know that as n-> infin, some terms become negligible
anonymous
  • anonymous
Yeah, basically. The definition is this:
anonymous
  • anonymous
But how you determine which term is going to infinity faster
anonymous
  • anonymous
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.
anonymous
  • anonymous
can you please do number 4 also :P
anonymous
  • anonymous
\[\frac{2^n+n^5+2}{19^{2n}+6^n+3} \iff \frac{2^n}{(19^2)^n}\]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).
anonymous
  • anonymous
The equivalence sign should be replaced with ~. For some reason it doesn't show up in the equation editor.
anonymous
  • anonymous
brb
anonymous
  • anonymous
i thought n^5 is going to infinity faster
anonymous
  • anonymous
No, if you plot n^5 and 2^n, you'll see 2^n overtake pretty quickly.
anonymous
  • anonymous
But listen, you can always check domination by using the definition. I'll scan in something.
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
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.
anonymous
  • anonymous
\[\frac{2^n/(19^2)^n}{r^n}=\left( \frac{2}{361r} \right)^n\]
anonymous
  • anonymous
Then choose r = 2/361 to ensure a convergent limit that's non-zero.
anonymous
  • anonymous
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.
anonymous
  • anonymous
Does exponential function always dominate quadratic, quintic ?
anonymous
  • anonymous
\[a^n>n^m\] 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,\[19^n>6^n>2^n\]
anonymous
  • anonymous
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.
anonymous
  • anonymous
:P can you prove a^n >n^m
anonymous
  • anonymous
yes, but I'll show you a plot...
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
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.
anonymous
  • anonymous
is that a free program haha
anonymous
  • anonymous
Yes it's free.
anonymous
  • anonymous
GeoGebra
anonymous
  • anonymous
http://www.geogebra.org/cms/
anonymous
  • anonymous
thanks !!
anonymous
  • anonymous
how do i input multiple graphs
anonymous
  • anonymous
Just type them in on the one page. y=x^2, y=x^4, etc. They should show up.
anonymous
  • anonymous
Got it hohoho
anonymous
  • anonymous
I don't know if you've done the last one, but I have \[r=\left( \frac{2^{7/10}}{19} \right)^3\]
anonymous
  • anonymous
yea i did it haha
anonymous
  • anonymous
what you up to right now
anonymous
  • anonymous
I should be doing my work...
anonymous
  • anonymous
Seriously going offline...it's getting dark and I've done nothing.
anonymous
  • anonymous
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
anonymous
  • anonymous
LOL when is your presentation ? i forgot~~
anonymous
  • anonymous
@ oktal Nice car ~ haha
anonymous
  • anonymous
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.
anonymous
  • anonymous
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
anonymous
  • anonymous
ok, i will see it, thanks bye
anonymous
  • anonymous
lol you still have few days to prepare for it no worries. you should chill with us
anonymous
  • anonymous
no, let him work, he has a life to manage...
anonymous
  • anonymous
LOL he has troublesome neighbors to manage as matter of fact :P
anonymous
  • anonymous
Nice car, oktal
anonymous
  • anonymous
not mine :(
anonymous
  • anonymous
my friend's
anonymous
  • anonymous
and actually it belongs to the university now, as it funded the building of it
anonymous
  • anonymous
it's pretty cool. they should at least be able to take it out on weekends ;p
anonymous
  • anonymous
"Hey ladies, check out my revolutionary design..." ;)
anonymous
  • anonymous
:)
anonymous
  • anonymous
you would haha
anonymous
  • anonymous
lol. oktal, check out a nerd link above for free plotting software, GeoGebra.
anonymous
  • anonymous
they did it few times, drove around town :)
anonymous
  • anonymous
ok
anonymous
  • anonymous
You could have used it to check out that absolute value problem.
anonymous
  • anonymous
nice, will check it
anonymous
  • anonymous
Okay, I'm out...seriously need to get sorted. Later!
anonymous
  • anonymous
Laters ~~ i am going to to bed too
anonymous
  • anonymous
nice meeting you okatal :)
anonymous
  • anonymous
ok, see you, nice to meet you too dichalao
anonymous
  • anonymous
Hi Loki, can you help me out with this problem
1 Attachment
anonymous
  • anonymous
What did you enter before?
anonymous
  • anonymous
11.7
anonymous
  • anonymous
brb...when's this due?
anonymous
  • anonymous
umm not soon :)
anonymous
  • anonymous
oh, are these your physics questions that aren't due till the end of the week or something?
anonymous
  • anonymous
it is due on tuesday before lecture
anonymous
  • anonymous
Your value is what you get when you work out magnitude, since \[a=\frac{v^2}{r}\]
anonymous
  • anonymous
11.7 is what i got from that equation..but it is wrong..probably becasue this is not uniform circular motion
anonymous
  • anonymous
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...
anonymous
  • anonymous
So,\[a=\frac{\Delta v}{\Delta t}=\frac{v_2-v_1}{t_2-t_1}\]where the a and v's are vectors.
anonymous
  • anonymous
The velocity in uniform circular motion is\[v=(-r \omega \sin \theta, r \omega \cos \theta)\]
anonymous
  • anonymous
why is there a negative sign for V_x
anonymous
  • anonymous
and we know, from definition of angular speed and uniform circular motion that,\[\omega = \frac{\theta }{t} \rightarrow \theta= \omega t\]
anonymous
  • anonymous
So \[v=(-r \omega \sin (\omega t), r \omega \cos (\omega t))\]
anonymous
  • anonymous
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.
anonymous
  • anonymous
Since\[v = r \omega \rightarrow \omega = \frac{v}{r}\]so\[V=(-v \sin \frac{vt}{r}, v \cos \frac{vt}{r})=v(\sin \frac{vt}{r}, \cos \frac{vt}{r})\]
anonymous
  • anonymous
You should be able to work out two vectors at different times and then subtract.
anonymous
  • anonymous
Is that making sense? I hope it is 'cause I just got up and I'm doing the on-the-fly.
anonymous
  • anonymous
in fact, we haven't "learned" angular speed .
anonymous
  • anonymous
There's a minus sign on the last expression outside sine...typo.
anonymous
  • anonymous
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\[\frac{dl}{dt}=\frac{d(r \theta)}{dt}=r \frac{d \theta }{dt}+ \theta \frac{dr}{dt}=r \frac{d \theta }{dt}:=r \omega\]
anonymous
  • anonymous
dr/dt is zero by construction of the problem -> the radius is fixed and therefore unchanging in time.
anonymous
  • anonymous
Angular speed is the rate of change in the angle.
anonymous
  • anonymous
brb
anonymous
  • anonymous
kk
anonymous
  • anonymous
Did you work it out?
anonymous
  • anonymous
kk
anonymous
  • anonymous
is there something wrong with this site..i didn't reply"kk"
anonymous
  • anonymous
do i just plug in t=0,and t=0.6?
anonymous
  • anonymous
it came through...and there's always something wrong with this site.
anonymous
  • anonymous
I would plug 0 and 0.6.
anonymous
  • anonymous
is vt/r in radians?
anonymous
  • anonymous
Always.
anonymous
  • anonymous
Let me check something before you waste a turn.
anonymous
  • anonymous
is there a way solving this problem without using angular speed?
anonymous
  • anonymous
Not off the top of my head.
anonymous
  • anonymous
\[\Delta v = 5.3(-\sin (1.325), \cos (1.325)-1)\]
anonymous
  • anonymous
\[a=\frac{\Delta v}{\Delta t}\approx \frac{1}{0.6}(-5.14,-4.01) \approx (-8.57, -6.68)\]
anonymous
  • anonymous
Thanks!!!!
anonymous
  • anonymous
What, did it work?
anonymous
  • anonymous
yep~
anonymous
  • anonymous
Good -.-
anonymous
  • anonymous
i am going to ask my professor... see if there is another way to do it..since we haven't covered angular velocity....
anonymous
  • anonymous
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.
anonymous
  • anonymous
Anyways, I have to sign out and get some work done. Happy mathing :)
anonymous
  • anonymous
LOL you meant physicsing?
anonymous
  • anonymous
All the same...
anonymous
  • anonymous
alright thanks for helping!! have a great day...still have an essay to do..
anonymous
  • anonymous
At least they run you into the ground.
anonymous
  • anonymous
LOL true~ see ya ~buddy
anonymous
  • anonymous
cu ;)

Looking for something else?

Not the answer you are looking for? Search for more explanations.