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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?

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  1. anonymous
    • 5 years ago
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    \[\sum_{1}^{\infty}(4^nx^{2n})/n\]

  2. dumbcow
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    Since |x| < 1/2, the series is absolutely convergent for -1/2 < x < 1/2

  6. anonymous
    • 5 years ago
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    Every absolutely convergent series is unconditionally convergent.

  7. anonymous
    • 5 years ago
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    How about at end points? x=-1/2 and 1/2

  8. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Good morning sir~

  10. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    haha morning

  12. anonymous
    • 5 years ago
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    can you check the endpoints for me?

  13. anonymous
    • 5 years ago
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    ok

  14. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    isn't it should be the same because x^(2n)

  16. anonymous
    • 5 years ago
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    shouldn't it be *

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

  18. anonymous
    • 5 years ago
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    \[\frac{4^n \left( \frac{1}{2} \right)^{2n}}{n}=\frac{(2^2)^n \frac{1}{2^{2n}}}{n}=\frac{2^{2n}\frac{1}{2^{2n}}}{n}=\frac{1}{n}\]

  19. anonymous
    • 5 years ago
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    yea i got this how about x=-1/2

  20. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    never mind i know where ive made a mistkae :)

  22. anonymous
    • 5 years ago
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    This is the 'alternating harmonic series' which converges to ln(2). You can test convergence with the alternating series test.

  23. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    LOL alright (Lucas with -.-) i got to go to class now. see you soon haha

  26. anonymous
    • 5 years ago
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    ok

  27. anonymous
    • 5 years ago
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    Have u had another look?

  28. anonymous
    • 5 years ago
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    No...sorry, I'm only back on because someone sent an e-mail. Is this something due now?

  29. anonymous
    • 5 years ago
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    no i am just curious haha just get back to me whenever you can :)

  30. anonymous
    • 5 years ago
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    If you have stuff to do, it is fine :]

  31. anonymous
    • 5 years ago
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    one sec

  32. anonymous
    • 5 years ago
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    It shouldn't be converging for either +1/2 or -1/2

  33. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    *eating

  35. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Book's wrong :)

  37. anonymous
    • 5 years ago
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    Wolfram Alpha agrees

  38. anonymous
    • 5 years ago
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    LOL my professor just told us the book is never wrong =-=

  39. anonymous
    • 5 years ago
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    cuz he wrote part of the textbook we are using

  40. anonymous
    • 5 years ago
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    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).

  41. anonymous
    • 5 years ago
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    how embarrassing :P

  42. anonymous
    • 5 years ago
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    we all make mistakes :)

  43. anonymous
    • 5 years ago
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    It says inconclusive for both tests ..what does that mean?

  44. anonymous
    • 5 years ago
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    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.

  45. anonymous
    • 5 years ago
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    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)

  46. anonymous
    • 5 years ago
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    Alright, so the book is WRONG..

  47. anonymous
    • 5 years ago
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    I'm thinking...and if it's not, I'd like to know why.

  48. anonymous
    • 5 years ago
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    Try a Cauchy Condensation test or something. Or alternating series test.

  49. anonymous
    • 5 years ago
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    it is not an alternating series even if x is negative.. and i haven't learned cauchy condensation test :

  50. anonymous
    • 5 years ago
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    Oh yea do you have time to explain the rearranging theorem ?

  51. anonymous
    • 5 years ago
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    Yeah, exactly, that's my point. I don't think you can throw anything at this to prove convergence.

  52. anonymous
    • 5 years ago
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    hmmm, probably not :(

  53. anonymous
    • 5 years ago
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    I can later.

  54. anonymous
    • 5 years ago
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    okiedokie

  55. anonymous
    • 5 years ago
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    What's your question? At least I can think about it.

  56. anonymous
    • 5 years ago
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    umm for a conditionally convergent series, you can rearrange the terms to get any sums you like?

  57. anonymous
    • 5 years ago
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    the order of the terms

  58. anonymous
    • 5 years ago
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    Sorry, I had to go do that. Some people on here are really starting to pellet me.

  59. anonymous
    • 5 years ago
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    LOL i saw it ..that dude is retarded

  60. anonymous
    • 5 years ago
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    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.

  61. anonymous
    • 5 years ago
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    Okay, so your question...I will scrounge a proof for you.

  62. anonymous
    • 5 years ago
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    COOL buddy, don't waste your time and energy on aXX haha

  63. anonymous
    • 5 years ago
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    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).

  64. anonymous
    • 5 years ago
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    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.

  65. anonymous
    • 5 years ago
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    Because when you get your 'nth' partial sum and send n to infinity, you end up with different limits.

  66. anonymous
    • 5 years ago
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    Is this what you're looking for?

  67. anonymous
    • 5 years ago
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    sort of

  68. anonymous
    • 5 years ago
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    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.

  69. anonymous
    • 5 years ago
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    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?

  70. anonymous
    • 5 years ago
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    Exactly

  71. anonymous
    • 5 years ago
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    oh i feel you ^^ thank you ~~~~~

  72. anonymous
    • 5 years ago
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    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.

  73. anonymous
    • 5 years ago
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    wait how about asolutely convergent series, if we rearrange the oders of terms, the pattern of the series also changes right?

  74. anonymous
    • 5 years ago
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    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

  75. anonymous
    • 5 years ago
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  76. anonymous
    • 5 years ago
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    This is crazy lol

  77. anonymous
    • 5 years ago
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    Has your professor shown you why an infinite series of positive terms doesn't change it's convergence (sum) upon rearrangement?

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

  79. anonymous
    • 5 years ago
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    oh...

  80. anonymous
    • 5 years ago
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    The proof is on that sheet too.

  81. anonymous
    • 5 years ago
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    yea i am looking at it

  82. anonymous
    • 5 years ago
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    ok...i'll leave it with you. I have to go do some work. Have fun with it.

  83. anonymous
    • 5 years ago
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    thanks !!! have fun with your work too

  84. anonymous
    • 5 years ago
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    bye :)

  85. anonymous
    • 5 years ago
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    c ya:)

  86. anonymous
    • 5 years ago
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    LUCAS~~~ I have a question regarding PHYSICS~~

  87. anonymous
    • 5 years ago
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    k

  88. anonymous
    • 5 years ago
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    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

  89. anonymous
    • 5 years ago
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    one sec

  90. anonymous
    • 5 years ago
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    okie okie

  91. anonymous
    • 5 years ago
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    inertia

  92. anonymous
    • 5 years ago
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    sorry, ordered chinese food and guy was at door

  93. anonymous
    • 5 years ago
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    first law of motion

  94. anonymous
    • 5 years ago
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    oh yea chinese food haha~

  95. anonymous
    • 5 years ago
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    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

  96. anonymous
    • 5 years ago
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    what's the question?

  97. anonymous
    • 5 years ago
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    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

  98. anonymous
    • 5 years ago
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    yeah, friction should oppose motion

  99. anonymous
    • 5 years ago
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    the box slides the opposite way of the acceleration becasue of the first law.. right?

  100. anonymous
    • 5 years ago
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    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.

  101. anonymous
    • 5 years ago
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    I can solve this for you, but I need to go offline. Gotta go out. Is this due tomorrow or something?

  102. anonymous
    • 5 years ago
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    nope just get back to me whenever u want :) 3rd law is also involved?

  103. anonymous
    • 5 years ago
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    In terms of the normal force.

  104. anonymous
    • 5 years ago
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    I'll get back to you.

  105. anonymous
    • 5 years ago
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    kk thanks :) see you~

  106. anonymous
    • 5 years ago
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    bye :)

  107. anonymous
    • 5 years ago
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    LOKISAN!!!

  108. anonymous
    • 5 years ago
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    hello

  109. anonymous
    • 5 years ago
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    lol i like how your name sound so japanese

  110. anonymous
    • 5 years ago
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    hehe, me too.

  111. anonymous
    • 5 years ago
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    What did you get for your truck/case question in the end?

  112. anonymous
    • 5 years ago
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    static friction = coeffient*normal force, and the direction is north

  113. anonymous
    • 5 years ago
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    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.

  114. anonymous
    • 5 years ago
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    it is good enough man :)

  115. anonymous
    • 5 years ago
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    the case is moving because the fraction "sticks" it to the floor of the truck. otherwise, it will stay there, by the first law?

  116. anonymous
    • 5 years ago
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    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.

  117. anonymous
    • 5 years ago
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    alright more questions..u mind?

  118. anonymous
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    \[\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n}e^n}{n!} \] \[\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n}n!}{n^n} \] \[\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n}n!}{e^n} \] \[\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n}n^n}{n!} \] 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

  119. anonymous
    • 5 years ago
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    lol, are these due Saturday or something?

  120. anonymous
    • 5 years ago
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    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

  121. anonymous
    • 5 years ago
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    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.

  122. anonymous
    • 5 years ago
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    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 ...

  123. anonymous
    • 5 years ago
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    yea tahts what i need help on

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

  125. anonymous
    • 5 years ago
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    Alrighty~! thank you!

  126. anonymous
    • 5 years ago
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    np

  127. anonymous
    • 5 years ago
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    Did you figure this out?

  128. anonymous
    • 5 years ago
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    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\[\lim_{n \rightarrow \infty}\frac{e^n}{n!}=\lim_{n \rightarrow \infty}\left( \frac{e}{n}.\frac{e}{n-1}.\frac{e}{n-2}....\frac{e}{3}.\frac{e}{2}.\frac{e}{1} \right)\]

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

  130. anonymous
    • 5 years ago
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    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,\[0<\frac{e}{n+1}\frac{e^n}{n!}<\frac{e^n}{n!} \iff 0<\frac{e^{n+1}}{(n+1)!}<\frac{e^n}{n!} \iff a_{n+1}<a_n\]

  131. anonymous
    • 5 years ago
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    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.

  132. anonymous
    • 5 years ago
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    got it.. factorial is the thing that scares me...

  133. anonymous
    • 5 years ago
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    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.

  134. anonymous
    • 5 years ago
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    got it .thanks :)

  135. anonymous
    • 5 years ago
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    ok

  136. anonymous
    • 5 years ago
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    I think you should be right now :)

  137. anonymous
    • 5 years ago
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    gotta go...good luck with it!

  138. anonymous
    • 5 years ago
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    LOL yea that was a very good explaination haha thanks~!!!!

  139. anonymous
    • 5 years ago
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    Hey loki do u have time to help me out with one more problem?

  140. anonymous
    • 5 years ago
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    You can put it down but I have to go...have an insane amount of work to do for tomorrow.

  141. anonymous
    • 5 years ago
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    oh alright !!

  142. anonymous
    • 5 years ago
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    just punch it in...

  143. anonymous
    • 5 years ago
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    \[1 + 4! x + \frac{8! x^2}{(2!)^4} + \frac{12! x^3}{(3!)^4} + \frac{16! x^4}{(4!)^4} + \frac{20! x^5}{(5!)^4}+\cdots\] 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

  144. anonymous
    • 5 years ago
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    The a_n is\[a_n=\frac{(4n)!x^n}{(n!)^4}\]

  145. anonymous
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    n = 0 to infinity.

  146. anonymous
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    Then you can apply the ratio test.

  147. anonymous
    • 5 years ago
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    oh the constant is the 0th term ?

  148. anonymous
    • 5 years ago
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    do you mind doing it so i can comapre my answer with you? i only have one attempt left :(

  149. anonymous
    • 5 years ago
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    Yeah.

  150. anonymous
    • 5 years ago
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    did you get 1/256?

  151. anonymous
    • 5 years ago
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  152. anonymous
    • 5 years ago
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    I haven't finished it off.

  153. anonymous
    • 5 years ago
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    Sorry for taking your precious time away !!!!

  154. anonymous
    • 5 years ago
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    I've rushed through it. Assuming the bottom line is correct, you should be able to finish it off.

  155. anonymous
    • 5 years ago
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    i got that too, as n->ininfty, all i care is n^4 terms right?

  156. anonymous
    • 5 years ago
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    ?

  157. anonymous
    • 5 years ago
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    You have a quartic polynomial above and below.

  158. anonymous
    • 5 years ago
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    so i multiply everything out and take the limit?

  159. anonymous
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    I would multiply the numerator out.

  160. anonymous
    • 5 years ago
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    256 n^4+640 n^3+560 n^2+200 n+24

  161. anonymous
    • 5 years ago
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    or can i just mulitply (4n)^4 ?

  162. anonymous
    • 5 years ago
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    from wolfram

  163. anonymous
    • 5 years ago
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    denominator: n^4+4 n^3+6 n^2+4 n+1

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

  165. anonymous
    • 5 years ago
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    It will be 256.

  166. anonymous
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    |256x|<1 R= 1/256

  167. anonymous
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    \[256|x|<1\]is required for convergence.

  168. anonymous
    • 5 years ago
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    Yes

  169. anonymous
    • 5 years ago
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    THANK YOU !!! And SORRY!!!

  170. anonymous
    • 5 years ago
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    You need to check the end points.

  171. anonymous
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    oh i am just looking for the radius

  172. anonymous
    • 5 years ago
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    OK

  173. anonymous
    • 5 years ago
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    Gotta go.

  174. anonymous
    • 5 years ago
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    i really appreciate your help!!!! WISH you the best!!!

  175. anonymous
    • 5 years ago
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    bye :]

  176. anonymous
    • 5 years ago
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    bye :)

  177. anonymous
    • 5 years ago
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    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=?

  178. anonymous
    • 5 years ago
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    Nevermind loki haha i got it :)

  179. anonymous
    • 5 years ago
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    there are youtube clips that show you what to do

  180. anonymous
    • 5 years ago
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    i used long division ot find the answers

  181. anonymous
    • 5 years ago
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  182. anonymous
    • 5 years ago
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    Can you help me out with this problem?

  183. anonymous
    • 5 years ago
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    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.

  184. anonymous
    • 5 years ago
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    sure it is not due soon :) and i will keep trying~~ haha

  185. anonymous
    • 5 years ago
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    okay...yes, keep trying...

  186. anonymous
    • 5 years ago
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    i;m out.

  187. anonymous
    • 5 years ago
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    good luck~!

  188. anonymous
    • 5 years ago
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    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.

  189. anonymous
    • 5 years ago
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    Actually, I think this version's better:\[f_n=\frac{\phi^n-(1-\phi)^n}{\sqrt{5}}\]

  190. anonymous
    • 5 years ago
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    \[\frac{f_{n+1}}{f_n}=\frac{\frac{\phi^{n+1}-(1-\phi)^{n+1}}{\sqrt{5}}}{\frac{\phi^n-(1-\phi)^n}{\sqrt{5}}}=\frac{\phi^{n+1}-(1-\phi)^{n+1}}{\phi^n-(1-\phi)^n}\]\[=\frac{\phi^{n+1}}{\phi^n}.\frac{1-\left( \frac{1-\phi}{\phi} \right)^{n+1}}{1-\left( \frac{1-\phi}{\phi} \right)^{n}}=\phi \frac{1-\zeta^{n+1}}{1-\zeta^{n}}\]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\[\lim_{n \rightarrow \infty}\frac{f_{n+1}}{f_n}= \phi\]

  191. anonymous
    • 5 years ago
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    The radius of convergence is then,\[\phi |x|<1 \rightarrow |x|<\frac{1}{\phi}\iff -\frac{1}{\phi}<x<\frac{1}{\phi}\]

  192. anonymous
    • 5 years ago
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    Check through it - it's after midnight and I've had about five hours sleep in two days!

  193. anonymous
    • 5 years ago
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    Hope you a good night !!

  194. anonymous
    • 5 years ago
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    good night of sleep*

  195. anonymous
    • 5 years ago
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    wow i would never figure this out... Since i haven;t learned binet's formula

  196. anonymous
    • 5 years ago
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    Thank!! you!! like ______________________________________________________________________________________________________________________this much

  197. anonymous
    • 5 years ago
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    Sorry but i ahve to bother you more!

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  198. anonymous
    • 5 years ago
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    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)

  199. anonymous
    • 5 years ago
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    dichalao, when's this due?

  200. anonymous
    • 5 years ago
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    three days later~~

  201. anonymous
    • 5 years ago
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    have you been waiting all this time?

  202. anonymous
    • 5 years ago
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    nope,, i been doing physics homework, and checked back from time to time

  203. anonymous
    • 5 years ago
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    k

  204. anonymous
    • 5 years ago
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    This thing isn't conceptually hard. What's going to stuff you are mistakes with the algebra.

  205. anonymous
    • 5 years ago
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    Have you tried plugging into wolfram alpha?

  206. anonymous
    • 5 years ago
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    i did... but i got a nonsense answer...as i said

  207. anonymous
    • 5 years ago
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    (2n+2)!! = (2n)!!(2n+2) is this right?

  208. anonymous
    • 5 years ago
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    Yes.

  209. anonymous
    • 5 years ago
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    (4n+7)!!=(4n+7)(4n+5)(4n+3)!!

  210. anonymous
    • 5 years ago
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    Yes

  211. anonymous
    • 5 years ago
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    limit a_(n+1)/(a_n) = 43570(8x+2)/48 |43570(8x+2)/48|<1 |8x|=-1.998898

  212. anonymous
    • 5 years ago
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    I'm going to do it now.

  213. anonymous
    • 5 years ago
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    I'll need coffee...

  214. anonymous
    • 5 years ago
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    POOR loki :(

  215. anonymous
    • 5 years ago
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    are you at home right now?

  216. anonymous
    • 5 years ago
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    yes

  217. anonymous
    • 5 years ago
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    day off

  218. anonymous
    • 5 years ago
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    no school today?

  219. anonymous
    • 5 years ago
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    oh oh haha you should have slept in

  220. anonymous
    • 5 years ago
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    but i should be doing work so that i don't have to spend another week staying up all night :(

  221. anonymous
    • 5 years ago
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    I can't sleep in - never works. Too sunny.

  222. anonymous
    • 5 years ago
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    LOL where is your certain?

  223. anonymous
    • 5 years ago
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    I think my building is on the surface of the sun; nothing stops the light or the heat.

  224. anonymous
    • 5 years ago
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    how about eyepatch~~

  225. anonymous
    • 5 years ago
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    Yeah, but eye masks dry the skin under my eyes.

  226. anonymous
    • 5 years ago
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    Just once a week ,, just apply lotion to it when you get up :P

  227. anonymous
    • 5 years ago
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    lol

  228. anonymous
    • 5 years ago
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    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!!!!!!!!!!!!!!!!!

  229. anonymous
    • 5 years ago
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    k

  230. anonymous
    • 5 years ago
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    you still there?

  231. anonymous
    • 5 years ago
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    yeah

  232. anonymous
    • 5 years ago
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    i found something, but i want to re-check.

  233. anonymous
    • 5 years ago
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    what did you get?

  234. anonymous
    • 5 years ago
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    \[|8x+2|<\frac{24}{21875}\]

  235. anonymous
    • 5 years ago
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    yea...that is what i got i think

  236. anonymous
    • 5 years ago
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    you will have negative number at the right side ...

  237. anonymous
    • 5 years ago
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    ?

  238. anonymous
    • 5 years ago
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    if u subtract 2 from both side..u wil have negative number

  239. anonymous
    • 5 years ago
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    Yes

  240. anonymous
    • 5 years ago
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    divide 8 , then i got -0.25

  241. anonymous
    • 5 years ago
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    i tried that answer before..the system indicated it is incorrect..

  242. anonymous
    • 5 years ago
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    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

  243. anonymous
    • 5 years ago
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    LOL do you mind explain a little?

  244. anonymous
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  245. anonymous
    • 5 years ago
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    There is a tiny radius of convergence.

  246. anonymous
    • 5 years ago
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    oh i see it

  247. anonymous
    • 5 years ago
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    so it centered at x=-0.25 ? taylor series?

  248. anonymous
    • 5 years ago
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    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

  249. anonymous
    • 5 years ago
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    Everything below 24/whatever is what you want (i.e. those x values that generate the function values below the 24/whatever line).

  250. anonymous
    • 5 years ago
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    That's why I said I wanted to check something - I meant, I wanted to use the definition of absolute value.

  251. anonymous
    • 5 years ago
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    so the raidus will just be half of the distance between the two points where they meet?

  252. anonymous
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    yeah

  253. anonymous
    • 5 years ago
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    got it !! Thanks

  254. anonymous
    • 5 years ago
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    is it possible to do this thing without graphing it?

  255. anonymous
    • 5 years ago
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    yes, that's what i wanted to do.

  256. anonymous
    • 5 years ago
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    ok thanks alot i will figure it out ~~!!!

  257. anonymous
    • 5 years ago
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    don't use all your attempts till i check it out.

  258. anonymous
    • 5 years ago
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    I gotta go - need lunch and take care of 'business' :D

  259. anonymous
    • 5 years ago
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    I'll go through the problem again later.

  260. anonymous
    • 5 years ago
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    arlgiht enjoy your day!!!

  261. anonymous
    • 5 years ago
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    sorry for late response i was doing my physics hw

  262. anonymous
    • 5 years ago
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    I tried ...i still couldn't get the right answer

  263. anonymous
    • 5 years ago
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    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...

  264. anonymous
    • 5 years ago
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    it is due before midnight, 21th

  265. anonymous
    • 5 years ago
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    Hey Loki, i got the answer already :) Thank you for helping!!!!!!

  266. anonymous
    • 5 years ago
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    I'm glad you got the answer 'cause I've had no time! :) What was wrong with it in the end?

  267. anonymous
    • 5 years ago
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    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

  268. anonymous
    • 5 years ago
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    Yeah, that's what I got too :)

  269. anonymous
    • 5 years ago
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    cool!! Let me give u a medal haha

  270. anonymous
    • 5 years ago
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    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.

  271. anonymous
    • 5 years ago
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    Loki u got it?

  272. anonymous
    • 5 years ago
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    Hang on

  273. anonymous
    • 5 years ago
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    okie okie you got a new pro pic that doesnt look like you haha

  274. anonymous
    • 5 years ago
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    I move with the times.

  275. anonymous
    • 5 years ago
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    is that your emotion

  276. anonymous
    • 5 years ago
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    why are you sad

  277. anonymous
    • 5 years ago
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    hehe...I'm upset with the new site.

  278. anonymous
    • 5 years ago
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    OH yea...i am so not used to it

  279. anonymous
    • 5 years ago
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    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.

  280. anonymous
    • 5 years ago
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    I don't like all this medal and shield crap.

  281. anonymous
    • 5 years ago
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    I have a headache, so am a little slow right now.

  282. anonymous
    • 5 years ago
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    OH is this s bad time to ask you questions...

  283. anonymous
    • 5 years ago
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    I should be alright.

  284. anonymous
    • 5 years ago
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    ok what you mean by looking at how each term contributes to a place

  285. anonymous
    • 5 years ago
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    \[1-\frac{1}{2!}+\frac{1}{4!}-\frac{1}{6!}+\frac{1}{8!}-\frac{1}{10!}+...\]

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

  287. anonymous
    • 5 years ago
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    oh i see what you mean cuz it is a convergent series

  288. anonymous
    • 5 years ago
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    Yes.

  289. anonymous
    • 5 years ago
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    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?

  290. anonymous
    • 5 years ago
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    can i*

  291. anonymous
    • 5 years ago
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    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).

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

  293. anonymous
    • 5 years ago
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    You have to apply the process to find the first five coefficients.

  294. anonymous
    • 5 years ago
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    Use wolfram to work them out. They're a pain.

  295. anonymous
    • 5 years ago
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    yep..they are..

  296. anonymous
    • 5 years ago
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    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?

  297. anonymous
    • 5 years ago
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    One second. I'm looking at something else.

  298. anonymous
    • 5 years ago
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    Okie

  299. anonymous
    • 5 years ago
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    i should compute the P4 and subtract ln(sec(-0.2) from it?

  300. anonymous
    • 5 years ago
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    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.

  301. anonymous
    • 5 years ago
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    ummm i dont really know..this is an online homework , the problems are given by a third party

  302. anonymous
    • 5 years ago
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    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.

  303. anonymous
    • 5 years ago
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    lol ok.....

  304. anonymous
    • 5 years ago
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    first derivative is tan(x)

  305. anonymous
    • 5 years ago
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    second is sec^2(x)

  306. anonymous
    • 5 years ago
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    third is 2tan(x)sec^2(x)

  307. anonymous
    • 5 years ago
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    oh i computed it on wolfram already thanks ~~

  308. anonymous
    • 5 years ago
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    fourth is -2(cos(2x)-2)sec^4(x)

  309. anonymous
    • 5 years ago
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    oh

  310. anonymous
    • 5 years ago
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    do you mind briefly tell me about the troublesome option?

  311. anonymous
    • 5 years ago
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    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.

  312. anonymous
    • 5 years ago
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    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

  313. anonymous
    • 5 years ago
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    Yeah. If you're studying to become an engineer, numerical methods are what you'll be using.

  314. anonymous
    • 5 years ago
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    numerical methods = ...mesure and calculator~~?

  315. anonymous
    • 5 years ago
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    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.

  316. anonymous
    • 5 years ago
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    Oh..pain in the butt...

  317. anonymous
    • 5 years ago
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    Actually, it's less of a pain to solve things numerically than algebraically. You'll get used to it.

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

  319. anonymous
    • 5 years ago
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    Sorry i got distracted a bit..

  320. anonymous
    • 5 years ago
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    i haven't taken any engineering classes so far

  321. anonymous
    • 5 years ago
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    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.

  322. anonymous
    • 5 years ago
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    but the =-= most troubleosome part is taking derivatives of such function.. omg wolfram haha

  323. anonymous
    • 5 years ago
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    Yeah, there are no easier ways of taking derivatives.

  324. anonymous
    • 5 years ago
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    is that possible to flip a power series,as in 1/ power series

  325. anonymous
    • 5 years ago
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    Yes.

  326. anonymous
    • 5 years ago
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    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

  327. anonymous
    • 5 years ago
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    Hang on. I need to make a phone call.

  328. anonymous
    • 5 years ago
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    brb going to take a quick shower first..

  329. anonymous
    • 5 years ago
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    kkk

  330. anonymous
    • 5 years ago
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    I think it would be more difficult.

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

  332. anonymous
    • 5 years ago
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    slip it?

  333. anonymous
    • 5 years ago
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    flip*..sorry

  334. anonymous
    • 5 years ago
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    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\[f(x)=g(x) \sum c_nx^n= (\sum b_n x^n)(\sum c_n x^n)\]

  335. anonymous
    • 5 years ago
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    oh=-= nvm...

  336. anonymous
    • 5 years ago
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    Yeah, thought so.

  337. anonymous
    • 5 years ago
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    Your only other alternative is to use polynomial division on a few terms ______________________ a_0 + a_1x + a_2x^2+... |1

  338. anonymous
    • 5 years ago
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    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'.

  339. anonymous
    • 5 years ago
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    lol not there yet

  340. anonymous
    • 5 years ago
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    Yeah :)

  341. anonymous
    • 5 years ago
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    will it be easier if i integrate it and find the power series , and then take the derivative of the power series

  342. anonymous
    • 5 years ago
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    are you off today?

  343. anonymous
    • 5 years ago
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    it's easter

  344. anonymous
    • 5 years ago
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    you could try the integration option

  345. anonymous
    • 5 years ago
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    i don't know how much that will help, but it may.

  346. anonymous
    • 5 years ago
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    nvm..it doesnt help much

  347. anonymous
    • 5 years ago
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    i couldnt figure out a pattern for the coefficents

  348. anonymous
    • 5 years ago
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    I'll have a look later. I'm pretty tired.

  349. anonymous
    • 5 years ago
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    okie dokey i am still trying~~

  350. anonymous
    • 5 years ago
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    Which one is it for - the (x..)^(1/3) one or the one after it?

  351. anonymous
    • 5 years ago
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    Someone's spying.

  352. anonymous
    • 5 years ago
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    To find the Maclaurin series for f(x) = x/(8 + 4 x^2)^(1/3) Need to find the radius of convergence~~

  353. anonymous
    • 5 years ago
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    k

  354. anonymous
    • 5 years ago
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    he must be a spy from other famous people here.

  355. anonymous
    • 5 years ago
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    famous?

  356. anonymous
    • 5 years ago
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    lol those that have hella fans like u

  357. anonymous
    • 5 years ago
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    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.

  358. anonymous
    • 5 years ago
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    is he retired or something...

  359. anonymous
    • 5 years ago
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    but he never answers my questions ...

  360. anonymous
    • 5 years ago
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    I'm not sure.

  361. anonymous
    • 5 years ago
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    lol, do you post questions outside of this link?

  362. anonymous
    • 5 years ago
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    yea.....when i have an idea of solving a problem and wants someone to confirm it..

  363. anonymous
    • 5 years ago
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    want*

  364. anonymous
    • 5 years ago
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    So you didn't go home for easter?

  365. anonymous
    • 5 years ago
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    no..too much work...to take care of.. if i go home..i wont get any work done

  366. anonymous
    • 5 years ago
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    I gave you a medal for morale ;p

  367. anonymous
    • 5 years ago
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    LOL yay my first medal in two days~~thanks haha

  368. anonymous
    • 5 years ago
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    how u got your gold shields

  369. anonymous
    • 5 years ago
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    Pfft, randomly. You can do things as random as enter something into chat and you get a medal for talking.

  370. anonymous
    • 5 years ago
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    *shield

  371. anonymous
    • 5 years ago
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    LOL ..how lame

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

  373. anonymous
    • 5 years ago
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    yea

  374. anonymous
    • 5 years ago
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    i just did lol

  375. anonymous
    • 5 years ago
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    cool. maybe i can get medals for everything i've done... -.-

  376. anonymous
    • 5 years ago
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    LOL or u can just ask questions and answer to it,,and i will give u medals

  377. anonymous
    • 5 years ago
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    pfft, this is so easy to rig.

  378. anonymous
    • 5 years ago
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    i like the old way better,, where u can become a fan ~~~

  379. anonymous
    • 5 years ago
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    you can still fan...you just have to move your pointer over a person's icon and fan them there.

  380. anonymous
    • 5 years ago
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    Or do it in the group menu.

  381. anonymous
    • 5 years ago
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    oh=-= lol changes are not always good..

  382. anonymous
    • 5 years ago
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    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..

  383. anonymous
    • 5 years ago
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    lol i just noticed that your icon changed again

  384. anonymous
    • 5 years ago
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    see you

  385. anonymous
    • 5 years ago
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    Loki u figured this out yet.?

  386. anonymous
    • 5 years ago
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    I haven't looked at it yet - sorry :( When's it due? I need a shower.

  387. anonymous
    • 5 years ago
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    no problem, buddy, it is not due anytime soon ,i just want to get it over with haha,

  388. anonymous
    • 5 years ago
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    did u just wake up~~

  389. anonymous
    • 5 years ago
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    ah okay...plenty of time for shower...

  390. anonymous
    • 5 years ago
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    LOL plenty of time for shower?

  391. anonymous
    • 5 years ago
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    I need to go get some food. I haven't eaten anything yet.

  392. anonymous
    • 5 years ago
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    alright~!

  393. anonymous
    • 5 years ago
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    Wow, you're still here.

  394. anonymous
    • 5 years ago
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    yea haha~

  395. anonymous
    • 5 years ago
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    Isn't it after midnight there?

  396. anonymous
    • 5 years ago
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    You should get an icon. The site creators say people with one get more questions answered.

  397. anonymous
    • 5 years ago
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    LOl yea i am going to bed very soon..

  398. anonymous
    • 5 years ago
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    Yeah...I'll look at your question either tonight or tomorrow. I have some things due myself :)

  399. anonymous
    • 5 years ago
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    oh just do it whenever u can and want :) are you off today?

  400. anonymous
    • 5 years ago
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    Yes/no...technically off, but have to finish things.

  401. anonymous
    • 5 years ago
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    oh

  402. anonymous
    • 5 years ago
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    u feel better today?

  403. anonymous
    • 5 years ago
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    yep

  404. anonymous
    • 5 years ago
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    just tired.

  405. anonymous
    • 5 years ago
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    lol same here.i been sleping more than enough ,,but still feel tired

  406. anonymous
    • 5 years ago
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    you work hard, though

  407. anonymous
    • 5 years ago
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    mind work... only lol i dont really exercise much on weekdays

  408. anonymous
    • 5 years ago
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    meh, who does

  409. anonymous
    • 5 years ago
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    how are you going to spend the rest of your day

  410. anonymous
    • 5 years ago
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    some of my friends run and weightlift everyday...

  411. anonymous
    • 5 years ago
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    well, it's nearly 6pm, so probably do some work

  412. anonymous
    • 5 years ago
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    yeah, i've got friends like that. it's boring. i wish i could get into it.

  413. anonymous
    • 5 years ago
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    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

  414. anonymous
    • 5 years ago
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    Do classes start back this week?

  415. anonymous
    • 5 years ago
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    what you mean?

  416. anonymous
    • 5 years ago
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    don't you get time off for easter?

  417. anonymous
    • 5 years ago
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    nope :(

  418. anonymous
    • 5 years ago
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    boo

  419. anonymous
    • 5 years ago
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    i would have gone home if i did

  420. anonymous
    • 5 years ago
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    sad. we get a week.

  421. anonymous
    • 5 years ago
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    yea... thumbs down~~

  422. anonymous
    • 5 years ago
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    oh lol u had fun?

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

  424. anonymous
    • 5 years ago
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    Anyhow...I might log off soon.

  425. anonymous
    • 5 years ago
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    oh haha you should hang out with your friends ~~,,dont get nerdy at home haha

  426. anonymous
    • 5 years ago
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    I went to brush my teeth and get ready to sleep... i am going to bed now..nightzz and have fun :)

  427. anonymous
    • 5 years ago
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    night :)

  428. anonymous
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    I have an answer if you still need it.

  429. anonymous
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    I only had a chance to look at it now.

  430. anonymous
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  431. anonymous
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  432. anonymous
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  433. anonymous
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    ^ 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.

  434. anonymous
    • 5 years ago
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    Let me know if that stuff doesn't come through properly :)

  435. anonymous
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    yea i got the same thing:) thanks anyways!!!

  436. anonymous
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    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.

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  437. anonymous
    • 5 years ago
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    Hey, I don't get that. I get\[\cos \theta = \frac{1}{\sqrt{2626}} \rightarrow \theta \approx 88.88^o\]

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

  440. anonymous
    • 5 years ago
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    Thanks man. How was your break?

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spraguer (Moderator)
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