anonymous 5 years ago Find the sereis's radius and interval of convergence. For waht values of x does the series converge absolutely or conditionally?

1. anonymous

$\sum_{1}^{\infty}(4^nx^{2n})/n$

2. anonymous

well the lim of x^2n as n->inf = 0 when x < 1causing the sum to converge at some point since it will start adding 0+0+0...

3. anonymous

By the ratio test, if the result of the following is less than 1, them series is guaranteed to converge:$\lim_{n \rightarrow \infty}\left| \frac{4^{n+1}x^{2(n+1)}/(n+1)}{4^{n}x^{2n}/n} \right|=\lim_{n \rightarrow \infty}\left| \frac{4^{n+1}x^{2n+2}}{4^nx^{2n}}\frac{n}{n+1} \right|$$=\lim_{n \rightarrow \infty}\left| \frac{4.4^{n}x^{2n}x^{2}}{4^nx^{2n}}\frac{n}{n+1} \right|=\lim_{n \rightarrow \infty}\left| 4x^2\frac{1}{1+1/n} \right|$$=\left| 4x^2 \right|=4x^2$So for convergence, we require,$4x^2 \lt 1 \rightarrow -\frac{1}{2} \lt x \lt \frac{1}{2}$

4. anonymous

The radius of convergence is, by definition, all x that satisfy $|x| < R$ where R is a non-negative, real number. $|4x^2|=4|x^2|=4|x|^2<1 \rightarrow |x|<\frac{1}{2}$

5. anonymous

Since |x| < 1/2, the series is absolutely convergent for -1/2 < x < 1/2

6. anonymous

Every absolutely convergent series is unconditionally convergent.

7. anonymous

How about at end points? x=-1/2 and 1/2

8. anonymous

I checked the end points, they both diverges, but the textbooks says it absolutely convergent for -1/2<x<1/2, and conditionally converges at x=-1/2

9. anonymous

Good morning sir~

10. anonymous

Yeah...it was late when I answered and hoped you wouldn't be considering the case where the magnitude of the ratio is equal to 1 (since anything can happen there).

11. anonymous

haha morning

12. anonymous

can you check the endpoints for me?

13. anonymous

ok

14. anonymous

When x= 1/2 (i.e. sub in 1/2) and simplify, you end up with the harmonic series, which is divergent. I'll do x=-1/2 now.

15. anonymous

isn't it should be the same because x^(2n)

16. anonymous

shouldn't it be *

17. anonymous

Yeah, when x=-1/2, you get an alternating harmonic which I know to be equal to log(2) (natural log).

18. anonymous

$\frac{4^n \left( \frac{1}{2} \right)^{2n}}{n}=\frac{(2^2)^n \frac{1}{2^{2n}}}{n}=\frac{2^{2n}\frac{1}{2^{2n}}}{n}=\frac{1}{n}$

19. anonymous

yea i got this how about x=-1/2

20. anonymous

Well, when you sub. x=-1/2, you get the same result as above, except for a (-1)^n outside the 1/n.

21. anonymous

never mind i know where ive made a mistkae :)

22. anonymous

This is the 'alternating harmonic series' which converges to ln(2). You can test convergence with the alternating series test.

23. anonymous

so it conditionally converges at x=-1/2.. what i did i squared x , and then simplify the expresion so i always got 1/n :(

24. anonymous

Actually, (-1/2)^(2n) is always positive (I'm eating while doing this so not completely concentrating :( )...so I need to have another look.

25. anonymous

LOL alright (Lucas with -.-) i got to go to class now. see you soon haha

26. anonymous

ok

27. anonymous

28. anonymous

No...sorry, I'm only back on because someone sent an e-mail. Is this something due now?

29. anonymous

no i am just curious haha just get back to me whenever you can :)

30. anonymous

If you have stuff to do, it is fine :]

31. anonymous

one sec

32. anonymous

It shouldn't be converging for either +1/2 or -1/2

33. anonymous

Maybe the person who plugged in the answer for the book was easting and not paying attention like me before and ignored the '2' in 2n, and just focused on the 'n'.

34. anonymous

*eating

35. anonymous

oh, yea i tried..10 times yesterday,,,i still got the same answer,, but the book says it conditionally diverges at x=-1/2..

36. anonymous

Book's wrong :)

37. anonymous

Wolfram Alpha agrees

38. anonymous
39. anonymous

LOL my professor just told us the book is never wrong =-=

40. anonymous

cuz he wrote part of the textbook we are using

41. anonymous

haha, well, my professors wrote some books too and I have entire errata sheets handed out from them (i.e. corrections to their printed stuff-ups).

42. anonymous

how embarrassing :P

43. anonymous

we all make mistakes :)

44. anonymous

It says inconclusive for both tests ..what does that mean?

45. anonymous

It just says it used those two tests and found no answers...just like when we use tests and don't find answers and have to figure out another way.

46. anonymous

If this thing converges, it will be ignoring the fact that (-1/2)^(2n) equals (-1)^(2n)(1/2)^(2n) = ((-1)^2)^n times (1/2)^(2n) = (1/2)^(2n)

47. anonymous

Alright, so the book is WRONG..

48. anonymous

I'm thinking...and if it's not, I'd like to know why.

49. anonymous

Try a Cauchy Condensation test or something. Or alternating series test.

50. anonymous

it is not an alternating series even if x is negative.. and i haven't learned cauchy condensation test :

51. anonymous

Oh yea do you have time to explain the rearranging theorem ?

52. anonymous

Yeah, exactly, that's my point. I don't think you can throw anything at this to prove convergence.

53. anonymous

hmmm, probably not :(

54. anonymous

I can later.

55. anonymous

okiedokie

56. anonymous

57. anonymous

umm for a conditionally convergent series, you can rearrange the terms to get any sums you like?

58. anonymous

the order of the terms

59. anonymous

Sorry, I had to go do that. Some people on here are really starting to pellet me.

60. anonymous

LOL i saw it ..that dude is retarded

61. anonymous

So...anyways...yeah, I do it because I think sites like this are important and there are people on here (like you and a few others) who actually want to learn this stuff. The ones that piss me off are the ones crying for help on huge questions and then bugger off without even saying thank you.

62. anonymous

Okay, so your question...I will scrounge a proof for you.

63. anonymous

COOL buddy, don't waste your time and energy on aXX haha

64. anonymous

The order of terms doesn't matter for a sum whose terms are positive: all rearrangements of a positive series converge to the same sum. It can make a difference if the series is alternating. For example, you can rearrange the terms of the alternating harmonic series to get 1/2 log 2 vs what it 'should' be (i.e. log 2).

65. anonymous

It comes about when you start forming the partial sums: you can get different arrangements or patterns forming. This is okay if the sum is finite, but when you introduce infinity, it's a different game.

66. anonymous

Because when you get your 'nth' partial sum and send n to infinity, you end up with different limits.

67. anonymous

Is this what you're looking for?

68. anonymous

sort of

69. anonymous

The alternating series is a conditionally-convergent series, and it's also a series where you can get different sums depending on the rearrangements...so, yes, you can get different sums in a conditionally convergent series.

70. anonymous

do you mean the pattern of the series changes if we change the order of the terms, so as n->infinity, the limit is different?

71. anonymous

Exactly

72. anonymous

oh i feel you ^^ thank you ~~~~~

73. anonymous

I'll send something I have. It shows how you can arrange the alternating harmonic to get 1/2 log 2 instead of log 2.

74. anonymous

wait how about asolutely convergent series, if we rearrange the oders of terms, the pattern of the series also changes right?

75. anonymous

All the terms in an abs. conv. series are positive, so as I said before, they will sum to the same limit no matter the arrangement. It's the introduction of two things that stuffs infinite sums up: 1) minus sign 2) infinity

76. anonymous

77. anonymous

This is crazy lol

78. anonymous

Has your professor shown you why an infinite series of positive terms doesn't change it's convergence (sum) upon rearrangement?

79. anonymous

nope he didn't get into any details, he just stated the theorem and said this is not in the scope of this course..

80. anonymous

oh...

81. anonymous

The proof is on that sheet too.

82. anonymous

yea i am looking at it

83. anonymous

ok...i'll leave it with you. I have to go do some work. Have fun with it.

84. anonymous

thanks !!! have fun with your work too

85. anonymous

bye :)

86. anonymous

c ya:)

87. anonymous

LUCAS~~~ I have a question regarding PHYSICS~~

88. anonymous

k

89. anonymous

a wooden block is at rest in a vehicle that is also at rest, if the vehicle accelerates north, why will the block slide southward

90. anonymous

one sec

91. anonymous

okie okie

92. anonymous

inertia

93. anonymous

sorry, ordered chinese food and guy was at door

94. anonymous

first law of motion

95. anonymous

oh yea chinese food haha~

96. anonymous

A 34.0kg packing case is initially at rest on the floor of a 1380kg pickup truck. The coefficient of static friction between the case and the truck floor is 0.30, and the coefficient of kinetic friction is 0.20. Before each acceleration given below, the truck is traveling due north at constant speed. a1=2 m/s^2

97. anonymous

what's the question?

98. anonymous

umm find the magnitude and direction of the frictional force i found both them..but i found it surprise that the frictional force is in the same direction as the acceceltraion

99. anonymous

yeah, friction should oppose motion

100. anonymous

the box slides the opposite way of the acceleration becasue of the first law.. right?

101. anonymous

Well, technically, the box wants to stay where it is in space, and moves only because of the force applied to it by the floor of the truck.

102. anonymous

I can solve this for you, but I need to go offline. Gotta go out. Is this due tomorrow or something?

103. anonymous

nope just get back to me whenever u want :) 3rd law is also involved?

104. anonymous

In terms of the normal force.

105. anonymous

I'll get back to you.

106. anonymous

kk thanks :) see you~

107. anonymous

bye :)

108. anonymous

LOKISAN!!!

109. anonymous

hello

110. anonymous

lol i like how your name sound so japanese

111. anonymous

hehe, me too.

112. anonymous

What did you get for your truck/case question in the end?

113. anonymous

static friction = coeffient*normal force, and the direction is north

114. anonymous

Yeah. The force exerted on the case isn't enough to overcome the maximum static friction, so its direction will be that of the truck. When I said friction opposes motion, it's easy to see when you're dealing with kinetic friction, but if you were to hop in the coordinate frame that was travelling at the initial velocity of the truck, and there was no friction between the truck floor and case, as the truck accelerated, you'd see the case not move (it would stay fixed in space). So its motion would be opposite that of the truck...the friction, when 'activated', acts in the opposite direction to the motion of the object - here, it acts in the direction of the truck. I think I made a meal of that explanation - got distracted + tired.

115. anonymous

it is good enough man :)

116. anonymous

the case is moving because the fraction "sticks" it to the floor of the truck. otherwise, it will stay there, by the first law?

117. anonymous

Yes. It will stay fixed in space (assuming your coordinate system was moving at the original velocity of the case) because of the first law.

118. anonymous

alright more questions..u mind?

119. anonymous

$\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n}e^n}{n!}$ $\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n}n!}{n^n}$ $\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n}n!}{e^n}$ $\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n}n^n}{n!}$ I need to determine whether i can apply Alternating series test on this series, how do i show u_n->0 and it is a decreasing

120. anonymous

lol, are these due Saturday or something?

121. anonymous

oh nope. i am just trying to finish some hw tonite, cuz i will be out all day tmr so yea..it is fine just get back to me later if u can

122. anonymous

So you need help on showing why a_n->0 and is decreasing? The application of the test is just a checklist thing based on the assumptions of the theorem.

123. anonymous

i found the answers already.. in an informal way.. I just plug in numbers for n to check if it is decreasing, and i just guess it goes to 0 ...

124. anonymous

yea tahts what i need help on

125. anonymous

Yeah...kind of have to prove it's decreasing...I'll look at them in my break.

126. anonymous

Alrighty~! thank you!

127. anonymous

np

128. anonymous

Did you figure this out?

129. anonymous

You should look at your sequences like this:$\frac{e^n}{n!}=\frac{e}{n}.\frac{e}{n-1}.\frac{e}{n-2}....\frac{e}{3}.\frac{e}{2}.\frac{e}{1}$There are n copies of e and n terms in the n factorial. Taking the limit on both sides gives$\lim_{n \rightarrow \infty}\frac{e^n}{n!}=\lim_{n \rightarrow \infty}\left( \frac{e}{n}.\frac{e}{n-1}.\frac{e}{n-2}....\frac{e}{3}.\frac{e}{2}.\frac{e}{1} \right)$

130. anonymous

The right-hand side can be expanded in limits by the multiplicative limit law:$\lim _{n \rightarrow \infty} \left( \frac{e}{n}.\frac{e}{n-1}.\frac{e}{n-2}....\frac{e}{3}.\frac{e}{2}.\frac{e}{1} \right)$$\lim_{n \rightarrow \infty}\frac{e}{n}.\lim_{n \rightarrow \infty}\frac{e}{n-1}.\lim_{n \rightarrow \infty}\frac{e}{n-2}....\lim_{n \rightarrow \infty}\frac{e}{3}.\lim_{n \rightarrow \infty}\frac{e}{2}.\lim_{n \rightarrow \infty}\frac{e}{1}$$=0.0.0.....\frac{e}{3}.\frac{e}{2}.\frac{e}{1}$$=0$

131. anonymous

Now, to find for monotonic decreasing, you need to show $a_{n+1}<a_n$for all n. For n such that n+1>e, you have$0<\frac{e}{n+1}<1$and so,$0<\frac{e}{n+1}\frac{e^n}{n!}<\frac{e^n}{n!} \iff 0<\frac{e^{n+1}}{(n+1)!}<\frac{e^n}{n!} \iff a_{n+1}<a_n$

132. anonymous

The condition on this proof was that n+1 needs to be greater than e, but this is the case for all n greater than or equal to 2. You only need to check 'manually' now for terms n=0 and n=1 to see if a_0 > a_1 > a_2 and after that, the argument above takes over.

133. anonymous

got it.. factorial is the thing that scares me...

134. anonymous

The other one is very similar and you should try it. For the last two, you'll notice that the terms of the summand are reciprocals of summands above$\frac{e^n}{n!}$is the reciprocal of $\frac{n!}{e^n}$so if one converges to 0, the other will blow out to infinity.

135. anonymous

got it .thanks :)

136. anonymous

ok

137. anonymous

I think you should be right now :)

138. anonymous

gotta go...good luck with it!

139. anonymous

LOL yea that was a very good explaination haha thanks~!!!!

140. anonymous

Hey loki do u have time to help me out with one more problem?

141. anonymous

You can put it down but I have to go...have an insane amount of work to do for tomorrow.

142. anonymous

oh alright !!

143. anonymous

just punch it in...

144. anonymous

$1 + 4! x + \frac{8! x^2}{(2!)^4} + \frac{12! x^3}{(3!)^4} + \frac{16! x^4}{(4!)^4} + \frac{20! x^5}{(5!)^4}+\cdots$ Use the ratio test to find the radius of convergence of the power series attempt: ((4n-4)!x^(n-1))/((n-1)!)^4 this is the a_n i found for the series

145. anonymous

The a_n is$a_n=\frac{(4n)!x^n}{(n!)^4}$

146. anonymous

n = 0 to infinity.

147. anonymous

Then you can apply the ratio test.

148. anonymous

oh the constant is the 0th term ?

149. anonymous

do you mind doing it so i can comapre my answer with you? i only have one attempt left :(

150. anonymous

Yeah.

151. anonymous

did you get 1/256?

152. anonymous

153. anonymous

I haven't finished it off.

154. anonymous

Sorry for taking your precious time away !!!!

155. anonymous

I've rushed through it. Assuming the bottom line is correct, you should be able to finish it off.

156. anonymous

i got that too, as n->ininfty, all i care is n^4 terms right?

157. anonymous

?

158. anonymous

You have a quartic polynomial above and below.

159. anonymous

so i multiply everything out and take the limit?

160. anonymous

I would multiply the numerator out.

161. anonymous

256 n^4+640 n^3+560 n^2+200 n+24

162. anonymous

or can i just mulitply (4n)^4 ?

163. anonymous

from wolfram

164. anonymous

denominator: n^4+4 n^3+6 n^2+4 n+1

165. anonymous

factor out n^4 on top and bottom and take the limit.

166. anonymous

It will be 256.

167. anonymous

|256x|<1 R= 1/256

168. anonymous

$256|x|<1$is required for convergence.

169. anonymous

Yes

170. anonymous

THANK YOU !!! And SORRY!!!

171. anonymous

You need to check the end points.

172. anonymous

oh i am just looking for the radius

173. anonymous

OK

174. anonymous

Gotta go.

175. anonymous

i really appreciate your help!!!! WISH you the best!!!

176. anonymous

bye :]

177. anonymous

bye :)

178. anonymous

LOki~~ Please help~ Suppose that f(x) and g(x) are given by the power series $g(x) = 21 + 33 x + 59 x^2 + 59 x^3 + \cdots .$ By dividing power series, find the first few terms of the series for the quotient $h(x)=\displaystyle \frac{g(x)}{f(x)} = c_0+c_1 x +c_2 x^2 + c_3 x^3 + \cdots.$ c_0, c_1,c_2,c_3=?

179. anonymous

Nevermind loki haha i got it :)

180. anonymous

there are youtube clips that show you what to do

181. anonymous

i used long division ot find the answers

182. anonymous

183. anonymous

Can you help me out with this problem?

184. anonymous

Ooh, probably not at the moment. I have something that needs to be finished before I leave for the university. I can look at it tonight, though.

185. anonymous

sure it is not due soon :) and i will keep trying~~ haha

186. anonymous

okay...yes, keep trying...

187. anonymous

i;m out.

188. anonymous

good luck~!

189. anonymous

When you test out your radius of convergence using the ratio test, you should end up with seeking x such that$\lim_{n \rightarrow\infty}\frac{f_{n+1}}{f_n}|x|<1$The Fibonacci sequence is a linear, iterative recurrence which has a closed form solution, given by Binet's formula,$f_n=\frac{\phi^n -(-\frac{1}{\phi})^n}{\sqrt{5}}$where$\phi = \frac{1+\sqrt{5}}{2}$(the golden ratio). You can use this formula for f_(n+1) and f_n, manipulate the expression and send it to the limit.

190. anonymous

Actually, I think this version's better:$f_n=\frac{\phi^n-(1-\phi)^n}{\sqrt{5}}$

191. anonymous

$\frac{f_{n+1}}{f_n}=\frac{\frac{\phi^{n+1}-(1-\phi)^{n+1}}{\sqrt{5}}}{\frac{\phi^n-(1-\phi)^n}{\sqrt{5}}}=\frac{\phi^{n+1}-(1-\phi)^{n+1}}{\phi^n-(1-\phi)^n}$$=\frac{\phi^{n+1}}{\phi^n}.\frac{1-\left( \frac{1-\phi}{\phi} \right)^{n+1}}{1-\left( \frac{1-\phi}{\phi} \right)^{n}}=\phi \frac{1-\zeta^{n+1}}{1-\zeta^{n}}$where$\zeta = \frac{1-\phi}{\phi}$Given the definition of phi, th golden ratio, it should be seen that$|\zeta|<1$and as a consequence, as n is taken to infinity$\lim_{n \rightarrow \infty}\frac{f_{n+1}}{f_n}= \phi$

192. anonymous

The radius of convergence is then,$\phi |x|<1 \rightarrow |x|<\frac{1}{\phi}\iff -\frac{1}{\phi}<x<\frac{1}{\phi}$

193. anonymous

Check through it - it's after midnight and I've had about five hours sleep in two days!

194. anonymous

Hope you a good night !!

195. anonymous

good night of sleep*

196. anonymous

wow i would never figure this out... Since i haven;t learned binet's formula

197. anonymous

Thank!! you!! like ______________________________________________________________________________________________________________________this much

198. anonymous

Sorry but i ahve to bother you more!

199. anonymous

I tried with the ratio test, and somehow i got negative number radius which is nonsense. and i am not sure the way i factor double factorials is correct or not . for instance, (2n+2)!! = (2n)!!*(2n+2)

200. anonymous

dichalao, when's this due?

201. anonymous

three days later~~

202. anonymous

have you been waiting all this time?

203. anonymous

nope,, i been doing physics homework, and checked back from time to time

204. anonymous

k

205. anonymous

This thing isn't conceptually hard. What's going to stuff you are mistakes with the algebra.

206. anonymous

Have you tried plugging into wolfram alpha?

207. anonymous

i did... but i got a nonsense answer...as i said

208. anonymous

(2n+2)!! = (2n)!!(2n+2) is this right?

209. anonymous

Yes.

210. anonymous

(4n+7)!!=(4n+7)(4n+5)(4n+3)!!

211. anonymous

Yes

212. anonymous

limit a_(n+1)/(a_n) = 43570(8x+2)/48 |43570(8x+2)/48|<1 |8x|=-1.998898

213. anonymous

I'm going to do it now.

214. anonymous

I'll need coffee...

215. anonymous

POOR loki :(

216. anonymous

are you at home right now?

217. anonymous

yes

218. anonymous

day off

219. anonymous

no school today?

220. anonymous

oh oh haha you should have slept in

221. anonymous

but i should be doing work so that i don't have to spend another week staying up all night :(

222. anonymous

I can't sleep in - never works. Too sunny.

223. anonymous

224. anonymous

I think my building is on the surface of the sun; nothing stops the light or the heat.

225. anonymous

226. anonymous

Yeah, but eye masks dry the skin under my eyes.

227. anonymous

Just once a week ,, just apply lotion to it when you get up :P

228. anonymous

lol

229. anonymous

i am gonna go grab some food before dining hall is closed.. Sorry for laboring you !!! I will be back in 30 mins or so !~~ Thanks!!!!!!!!!!!!!!!!!

230. anonymous

k

231. anonymous

you still there?

232. anonymous

yeah

233. anonymous

i found something, but i want to re-check.

234. anonymous

what did you get?

235. anonymous

$|8x+2|<\frac{24}{21875}$

236. anonymous

yea...that is what i got i think

237. anonymous

you will have negative number at the right side ...

238. anonymous

?

239. anonymous

if u subtract 2 from both side..u wil have negative number

240. anonymous

Yes

241. anonymous

divide 8 , then i got -0.25

242. anonymous

i tried that answer before..the system indicated it is incorrect..

243. anonymous

When you plot y=|8x+2| and y=21/21875, you'll see a section of the absolute value function under the 21/21875 line, which means there is a radius of convergence. I think I know what's going on, but I was too lazy to do it properly :S

244. anonymous

LOL do you mind explain a little?

245. anonymous

246. anonymous

There is a tiny radius of convergence.

247. anonymous

oh i see it

248. anonymous

so it centered at x=-0.25 ? taylor series?

249. anonymous

No no...go back to high school when you were asked to find the region for x such that, y<x, say. Here we're looking for all x such that |8x+2|<24/whatever

250. anonymous

Everything below 24/whatever is what you want (i.e. those x values that generate the function values below the 24/whatever line).

251. anonymous

That's why I said I wanted to check something - I meant, I wanted to use the definition of absolute value.

252. anonymous

so the raidus will just be half of the distance between the two points where they meet?

253. anonymous

yeah

254. anonymous

got it !! Thanks

255. anonymous

is it possible to do this thing without graphing it?

256. anonymous

yes, that's what i wanted to do.

257. anonymous

ok thanks alot i will figure it out ~~!!!

258. anonymous

don't use all your attempts till i check it out.

259. anonymous

I gotta go - need lunch and take care of 'business' :D

260. anonymous

I'll go through the problem again later.

261. anonymous

262. anonymous

sorry for late response i was doing my physics hw

263. anonymous

I tried ...i still couldn't get the right answer

264. anonymous

OK. I'll take another look at it. Just let me know when this is due (i.e. date and time (pacific time)). I may have some time tomorrow...

265. anonymous

it is due before midnight, 21th

266. anonymous

Hey Loki, i got the answer already :) Thank you for helping!!!!!!

267. anonymous

I'm glad you got the answer 'cause I've had no time! :) What was wrong with it in the end?

268. anonymous

Sorry for the late response i been busy with midterms~~ (8x+2)<24/21875, and divided both side by 8 (x+1/4)<24/(8)21875 the right side is the R, becasue -1/4 is the center of the power series

269. anonymous

Yeah, that's what I got too :)

270. anonymous

cool!! Let me give u a medal haha

271. anonymous

What is the minimal degree Taylor polynomial about x= 0 that you need to calculate cos(1) to 5 decimal places? First, i use the calculator to solve for cos (1), so i can see what the remainder/error is. And then, since the power series of cosx is an alternating series, so i applied the alternating series estimation theorem, but i could not solve for n without trial and error i also tried taylor's remainder , i coudnt solve for n without trial and error as well. I just wonder if i misunderstand some concepts about taylor series, whereby ended up solving the question with trial and error.

272. anonymous

Loki u got it?

273. anonymous

Hang on

274. anonymous

okie okie you got a new pro pic that doesnt look like you haha

275. anonymous

I move with the times.

276. anonymous

277. anonymous

278. anonymous

hehe...I'm upset with the new site.

279. anonymous

OH yea...i am so not used to it

280. anonymous

Taylor's remainder will only give you the maximum number of terms needed to guarantee a certain level of accuracy. You could just look at the expansion and look at how each term contributes to a place in the decimal system.

281. anonymous

I don't like all this medal and shield crap.

282. anonymous

I have a headache, so am a little slow right now.

283. anonymous

284. anonymous

I should be alright.

285. anonymous

ok what you mean by looking at how each term contributes to a place

286. anonymous

$1-\frac{1}{2!}+\frac{1}{4!}-\frac{1}{6!}+\frac{1}{8!}-\frac{1}{10!}+...$

287. anonymous

At some point, one of the terms and those that follow stop contributing to the fifth decimal place (else the sum wouldn't converge).

288. anonymous

oh i see what you mean cuz it is a convergent series

289. anonymous

Yes.

290. anonymous

The Taylor series for f(x) = ln(sec(x)) at a = 0 is \sum n=0 to n=infty c_n( x )^n. To find the first 5 coefficients, can you just use the taylor series for ln(x) and then replace x with sec x?

291. anonymous

can i*

292. anonymous

The fifth decimal places is the 100,000ths column in the decimal representation. 1/8! = 1/40320 (hundred thousandths place) and then 1/10! = 1/362880 (10 millionth place).

293. anonymous

You can if your intention is to expand in terms of sec(x)...but that's not what you're asked to do.

294. anonymous

You have to apply the process to find the first five coefficients.

295. anonymous

Use wolfram to work them out. They're a pain.

296. anonymous

yep..they are..

297. anonymous

Find the exact error in approximating ln(sec(-0.2)) by its fourth degree Taylor polynomial at a = 0 . I can't use the remainder formula ..right cuz the remainder will only give the error bound?

298. anonymous

One second. I'm looking at something else.

299. anonymous

Okie

300. anonymous

i should compute the P4 and subtract ln(sec(-0.2) from it?

301. anonymous

The problem is, I don't know what the exact intentions of your lecturer are. Is implying 'exact' ln(sec(-0.2)) is what you get on your calculator, or something algebraic. That's what I'm looking at.

302. anonymous

ummm i dont really know..this is an online homework , the problems are given by a third party

303. anonymous

Oh, well, if it's all online and it's asking for a numerical answer, I would take the easy option and calculate the terms in your expansion and compare with wolfram.

304. anonymous

lol ok.....

305. anonymous

first derivative is tan(x)

306. anonymous

second is sec^2(x)

307. anonymous

third is 2tan(x)sec^2(x)

308. anonymous

oh i computed it on wolfram already thanks ~~

309. anonymous

fourth is -2(cos(2x)-2)sec^4(x)

310. anonymous

oh

311. anonymous

do you mind briefly tell me about the troublesome option?

312. anonymous

Well, there isn't actually something tangible, it was something I was about to look at, maybe with the Lagrange remainder. The problem with that is you have to be a bit tricky, which, due to my headache, I didn't pursue.

313. anonymous

oh it is cool.. and i haven't learned about the Lagrange remainder...so i think it is not what they are trying to test us on

314. anonymous

Yeah. If you're studying to become an engineer, numerical methods are what you'll be using.

315. anonymous

numerical methods = ...mesure and calculator~~?

316. anonymous

Numerical methods is measurement and number-crunching. But I'm talking about numerical analysis, which is a branch of mathematics that allows us to extract numerical answers from equations that either have no algebraic solution, or are too much of a pain to bother with.

317. anonymous

Oh..pain in the butt...

318. anonymous

Actually, it's less of a pain to solve things numerically than algebraically. You'll get used to it.

319. anonymous

To find the Maclaurin series for f(x) = x/(8 + 4 x^2)^(1/3) ...are there easier ways ...to do it

320. anonymous

Sorry i got distracted a bit..

321. anonymous

i haven't taken any engineering classes so far

322. anonymous

Maclaurin series is expanded about zero, so you might want to take the first few derivatives and see what you get when you sub. in 0...there might be a pattern. If it looks like it, you can hypothesize the form of the nth derivative and use induction to prove it.

323. anonymous

but the =-= most troubleosome part is taking derivatives of such function.. omg wolfram haha

324. anonymous

Yeah, there are no easier ways of taking derivatives.

325. anonymous

is that possible to flip a power series,as in 1/ power series

326. anonymous

Yes.

327. anonymous

can i do it here... i find the power series for (8 + 4 x^2)^(1/3) first and then divide x by it? because i need to find the raidus of convergence..so i have to find a patern

328. anonymous

Hang on. I need to make a phone call.

329. anonymous

brb going to take a quick shower first..

330. anonymous

kkk

331. anonymous

I think it would be more difficult.

332. anonymous

if the power seiresi is sum n=0 -> infin, 4n(x)^n/n^2 ...if i slip it..what would it be?

333. anonymous

slip it?

334. anonymous

flip*..sorry

335. anonymous

It's not a trivial task. If you have two series$f(x)=\sum a_n x^n$$g(x)=\sum b_nx^n$and you want$\frac{f(x)}{g(x)}=\sum c_n x^n$then you have to find those c_n by forming the Cauchy product of$f(x)=g(x) \sum c_nx^n= (\sum b_n x^n)(\sum c_n x^n)$

336. anonymous

oh=-= nvm...

337. anonymous

Yeah, thought so.

338. anonymous

Your only other alternative is to use polynomial division on a few terms ______________________ a_0 + a_1x + a_2x^2+... |1

339. anonymous

But that's not going to lead you to anything general. That's only useful in Laurent series (an extended version of a Taylor series) for calculating things called, 'residues' and 'poles'.

340. anonymous

lol not there yet

341. anonymous

Yeah :)

342. anonymous

will it be easier if i integrate it and find the power series , and then take the derivative of the power series

343. anonymous

are you off today?

344. anonymous

it's easter

345. anonymous

you could try the integration option

346. anonymous

i don't know how much that will help, but it may.

347. anonymous

nvm..it doesnt help much

348. anonymous

i couldnt figure out a pattern for the coefficents

349. anonymous

I'll have a look later. I'm pretty tired.

350. anonymous

okie dokey i am still trying~~

351. anonymous

Which one is it for - the (x..)^(1/3) one or the one after it?

352. anonymous

Someone's spying.

353. anonymous

To find the Maclaurin series for f(x) = x/(8 + 4 x^2)^(1/3) Need to find the radius of convergence~~

354. anonymous

k

355. anonymous

he must be a spy from other famous people here.

356. anonymous

famous?

357. anonymous

lol those that have hella fans like u

358. anonymous

I'm hardly on anymore, so the fan accumulation rate has gone down. Amistre64 is on here ALL the time...I don't know where people find the time.

359. anonymous

is he retired or something...

360. anonymous

but he never answers my questions ...

361. anonymous

I'm not sure.

362. anonymous

lol, do you post questions outside of this link?

363. anonymous

yea.....when i have an idea of solving a problem and wants someone to confirm it..

364. anonymous

want*

365. anonymous

So you didn't go home for easter?

366. anonymous

no..too much work...to take care of.. if i go home..i wont get any work done

367. anonymous

I gave you a medal for morale ;p

368. anonymous

LOL yay my first medal in two days~~thanks haha

369. anonymous

how u got your gold shields

370. anonymous

Pfft, randomly. You can do things as random as enter something into chat and you get a medal for talking.

371. anonymous

*shield

372. anonymous

LOL ..how lame

373. anonymous

lol, yes. see if you can go to this link and give me a medal: http://openstudy.com/groups/mathematics#/updates/4db387a276e08b0bf9c58d28

374. anonymous

yea

375. anonymous

i just did lol

376. anonymous

cool. maybe i can get medals for everything i've done... -.-

377. anonymous

LOL or u can just ask questions and answer to it,,and i will give u medals

378. anonymous

pfft, this is so easy to rig.

379. anonymous

i like the old way better,, where u can become a fan ~~~

380. anonymous

you can still fan...you just have to move your pointer over a person's icon and fan them there.

381. anonymous

Or do it in the group menu.

382. anonymous

oh=-= lol changes are not always good..

383. anonymous

alright it is 3 in the morning here...i am heading to bed right now...Thanks for the help :) Feel better and have a great day..

384. anonymous

lol i just noticed that your icon changed again

385. anonymous

see you

386. anonymous

Loki u figured this out yet.?

387. anonymous

I haven't looked at it yet - sorry :( When's it due? I need a shower.

388. anonymous

no problem, buddy, it is not due anytime soon ,i just want to get it over with haha,

389. anonymous

did u just wake up~~

390. anonymous

ah okay...plenty of time for shower...

391. anonymous

LOL plenty of time for shower?

392. anonymous

I need to go get some food. I haven't eaten anything yet.

393. anonymous

alright~!

394. anonymous

Wow, you're still here.

395. anonymous

yea haha~

396. anonymous

Isn't it after midnight there?

397. anonymous

You should get an icon. The site creators say people with one get more questions answered.

398. anonymous

LOl yea i am going to bed very soon..

399. anonymous

Yeah...I'll look at your question either tonight or tomorrow. I have some things due myself :)

400. anonymous

oh just do it whenever u can and want :) are you off today?

401. anonymous

Yes/no...technically off, but have to finish things.

402. anonymous

oh

403. anonymous

u feel better today?

404. anonymous

yep

405. anonymous

just tired.

406. anonymous

lol same here.i been sleping more than enough ,,but still feel tired

407. anonymous

you work hard, though

408. anonymous

mind work... only lol i dont really exercise much on weekdays

409. anonymous

meh, who does

410. anonymous

how are you going to spend the rest of your day

411. anonymous

some of my friends run and weightlift everyday...

412. anonymous

well, it's nearly 6pm, so probably do some work

413. anonymous

yeah, i've got friends like that. it's boring. i wish i could get into it.

414. anonymous

yea.. i really hope i have that kind of determination and time.... once i sit down, i dont really want to get up unless i have to

415. anonymous

Do classes start back this week?

416. anonymous

what you mean?

417. anonymous

don't you get time off for easter?

418. anonymous

nope :(

419. anonymous

boo

420. anonymous

i would have gone home if i did

421. anonymous

422. anonymous

yea... thumbs down~~

423. anonymous

424. anonymous

don't know yet - it's just started...well, not really, just means I get to sleep and do work at home instead :( Meh.

425. anonymous

Anyhow...I might log off soon.

426. anonymous

oh haha you should hang out with your friends ~~,,dont get nerdy at home haha

427. anonymous

I went to brush my teeth and get ready to sleep... i am going to bed now..nightzz and have fun :)

428. anonymous

night :)

429. anonymous

I have an answer if you still need it.

430. anonymous

I only had a chance to look at it now.

431. anonymous

432. anonymous

433. anonymous

434. anonymous

^ is an OpenOffice spreadsheet so you can see in numbers. I didn't check endpoints - you said radius of convergence. I didn't load an Excel spreadsheet since the file was too large.

435. anonymous

Let me know if that stuff doesn't come through properly :)

436. anonymous

yea i got the same thing:) thanks anyways!!!

437. anonymous

Loki got question! A water main is to be constructed with a 20% grade in thenorth direction and a 10% grade in the east direction. Determine the angle θ required in the water main for the turnfrom north to east using vectors. I got 91.2 degree, but i am not sure about it.

438. anonymous

Hey, I don't get that. I get$\cos \theta = \frac{1}{\sqrt{2626}} \rightarrow \theta \approx 88.88^o$

439. anonymous

440. anonymous

oh yea you are right. 20% grade means rise over run = 1/5 right, that was what confused me

441. anonymous

Thanks man. How was your break?