## anonymous 5 years ago .....

1. anonymous

$\int\limits e^{(sinx)}sinx dx$

2. anonymous

HEY LOKI

3. anonymous

Evening :)

4. anonymous

Make a substitution,$u=\sin x$

5. anonymous

du=cos x

6. anonymous

Yeah, hang on...

7. anonymous

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

8. 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?

9. anonymous

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

10. anonymous

Haha, well, boss says that.

11. anonymous

oh dam be humble :P

12. anonymous

lol, I am ;)

13. anonymous

LOL sure boss :P

14. anonymous

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

15. anonymous

16. anonymous

You should chillax, it's Friday.

17. anonymous

question!!

18. anonymous

What is the difference between power and geometric series!

19. anonymous

I am chillaxing by chit chating with my roomate

20. anonymous

Take your time on the wheel question :)

21. 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.

22. anonymous

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

23. anonymous

This is research :)

24. anonymous

LOL just go do your work!

25. anonymous

Yeah I know...

26. anonymous

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

27. anonymous

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

28. anonymous

Hey loki, i have a question, what test should i use to determine the convergence/divergence of the series sin(x/9)

29. anonymous

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

30. anonymous

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

31. anonymous

ok

32. 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 ><

33. anonymous

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

34. anonymous

lol, i wish :(

35. anonymous

I'm thinking you can set the problem up as proving it divergent using something like a limit comparison test.

36. anonymous

Sorry, had to go talk to the neighbor.

37. anonymous

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

38. 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.

39. anonymous

Should I wait for you to finish typing?

40. anonymous

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

41. anonymous

Lol, yes, you would be :D

42. anonymous

Form the ratio$\frac{1/n}{\sin \frac{9}{n}}$and note that, in the limit, you get an indeterminate form, $\frac{0}{0}$

43. anonymous

44. anonymous

So$\lim_{n \rightarrow \infty}\frac{1/n}{\sin 9/n}=\frac{\lim_{n \rightarrow \infty}1/n}{\lim_{n \rightarrow \infty}\sin 9/n}=\frac{\lim_{n \rightarrow \infty}-\frac{1}{n^2}}{\lim_{n \rightarrow \infty}\frac{-9\cos 9/n}{n^2}}=\frac{1}{9}$

45. anonymous

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

46. 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.

47. anonymous

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

48. anonymous

Pretty much...it's okay, since you then turn towards something that can either prove or disprove your case.

49. anonymous

A lot of mathematics is experimental.

50. anonymous

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

51. 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.

52. 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?

53. anonymous

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

54. anonymous

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

55. anonymous

Ahhhhh....no....

56. anonymous

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

57. 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.

58. anonymous

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

59. anonymous

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

60. anonymous

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

61. anonymous

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

62. anonymous

SUSPICION

63. 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.

64. anonymous

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

65. anonymous

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

66. anonymous

67. anonymous

i am stuck at number 3

68. anonymous

you mean 4?

69. anonymous

70. anonymous

Does this have to be submitted now?

71. anonymous

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

72. anonymous

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

73. anonymous

OH alright :) have fun!

74. anonymous

Bye..

75. anonymous

Did you figure it out?

76. anonymous

Nope haha you told me to chillax

77. anonymous

Slacker... ;p

78. anonymous

I'll look now.

79. anonymous

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

80. anonymous

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

81. anonymous

Have you covered asymptotic equivalence?

82. anonymous

what is that? might have..not sure

83. anonymous

brb

84. anonymous

kay kay

85. anonymous

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

86. anonymous

Yeah, basically. The definition is this:

87. anonymous

But how you determine which term is going to infinity faster

88. 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.

89. anonymous

can you please do number 4 also :P

90. 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).

91. anonymous

The equivalence sign should be replaced with ~. For some reason it doesn't show up in the equation editor.

92. anonymous

brb

93. anonymous

i thought n^5 is going to infinity faster

94. anonymous

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

95. anonymous

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

96. anonymous

97. 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.

98. anonymous

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

99. anonymous

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

100. 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.

101. anonymous

Does exponential function always dominate quadratic, quintic ?

102. 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,$19^n>6^n>2^n$

103. 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.

104. anonymous

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

105. anonymous

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

106. anonymous

107. 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.

108. anonymous

is that a free program haha

109. anonymous

Yes it's free.

110. anonymous

GeoGebra

111. anonymous
112. anonymous

thanks !!

113. anonymous

how do i input multiple graphs

114. anonymous

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

115. anonymous

Got it hohoho

116. anonymous

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

117. anonymous

yea i did it haha

118. anonymous

what you up to right now

119. anonymous

I should be doing my work...

120. anonymous

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

121. 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

122. anonymous

LOL when is your presentation ? i forgot~~

123. anonymous

@ oktal Nice car ~ haha

124. 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.

125. 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

126. anonymous

ok, i will see it, thanks bye

127. anonymous

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

128. anonymous

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

129. anonymous

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

130. anonymous

Nice car, oktal

131. anonymous

not mine :(

132. anonymous

my friend's

133. anonymous

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

134. anonymous

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

135. anonymous

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

136. anonymous

:)

137. anonymous

you would haha

138. anonymous

139. anonymous

they did it few times, drove around town :)

140. anonymous

ok

141. anonymous

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

142. anonymous

nice, will check it

143. anonymous

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

144. anonymous

Laters ~~ i am going to to bed too

145. anonymous

nice meeting you okatal :)

146. anonymous

ok, see you, nice to meet you too dichalao

147. anonymous

Hi Loki, can you help me out with this problem

148. anonymous

What did you enter before?

149. anonymous

11.7

150. anonymous

brb...when's this due?

151. anonymous

umm not soon :)

152. anonymous

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

153. anonymous

it is due on tuesday before lecture

154. anonymous

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

155. anonymous

11.7 is what i got from that equation..but it is wrong..probably becasue this is not uniform circular motion

156. 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...

157. 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.

158. anonymous

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

159. anonymous

why is there a negative sign for V_x

160. anonymous

and we know, from definition of angular speed and uniform circular motion that,$\omega = \frac{\theta }{t} \rightarrow \theta= \omega t$

161. anonymous

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

162. 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.

163. anonymous

Since$v = r \omega \rightarrow \omega = \frac{v}{r}$so$V=(-v \sin \frac{vt}{r}, v \cos \frac{vt}{r})=v(\sin \frac{vt}{r}, \cos \frac{vt}{r})$

164. anonymous

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

165. anonymous

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

166. anonymous

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

167. anonymous

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

168. 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$\frac{dl}{dt}=\frac{d(r \theta)}{dt}=r \frac{d \theta }{dt}+ \theta \frac{dr}{dt}=r \frac{d \theta }{dt}:=r \omega$

169. anonymous

dr/dt is zero by construction of the problem -> the radius is fixed and therefore unchanging in time.

170. anonymous

Angular speed is the rate of change in the angle.

171. anonymous

brb

172. anonymous

kk

173. anonymous

Did you work it out?

174. anonymous

kk

175. anonymous

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

176. anonymous

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

177. anonymous

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

178. anonymous

I would plug 0 and 0.6.

179. anonymous

180. anonymous

Always.

181. anonymous

Let me check something before you waste a turn.

182. anonymous

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

183. anonymous

Not off the top of my head.

184. anonymous

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

185. anonymous

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

186. anonymous

Thanks!!!!

187. anonymous

What, did it work?

188. anonymous

yep~

189. anonymous

Good -.-

190. anonymous

i am going to ask my professor... see if there is another way to do it..since we haven't covered angular velocity....

191. 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.

192. anonymous

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

193. anonymous

LOL you meant physicsing?

194. anonymous

All the same...

195. anonymous

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

196. anonymous

At least they run you into the ground.

197. anonymous

LOL true~ see ya ~buddy

198. anonymous

cu ;)