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anonymous

  • 5 years ago

For each of the following, determine whether the series is convergent or divergent. 1: sum x=1 to x=infinity 6/(n(n+3)) 2: sum x=1 to x=infinity ln(n)/(2n) 3: sum x=1 to x=infinity 1/(3+n) 4: sum x=1 to infinity 6/(n^2-65) 5: sum x=1 to x=infinity (1+6^n)/(3+6^n)

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  1. anonymous
    • 5 years ago
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    It would be nice if you can also tell me which test is used to determine the convergence/divergence of the series

  2. anonymous
    • 5 years ago
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    You can use the integral test on the first. It suffices to show that the integral of \[\lim_{c-> \infty}\int\limits_{1}^{c}\frac{6}{x(x+3)}dx \]is convergent. When you integrate this out, you get\[\lim_{c-> \infty}[2\log (x) - 2 \log (3 + x)|_1^c=\lim_{c-> \infty}[2\log \frac{x}{3+x}|_1^c\]\[=\lim_{c-> \infty}[2\log \frac{1}{3/x+1}|_1^c=2 \log 1 - 2 \log \frac{1}{4}=4 \log 2<\infty\]

  3. anonymous
    • 5 years ago
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    You familiar with this test?

  4. anonymous
    • 5 years ago
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    yes

  5. anonymous
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    It works if the terms of your series are monotonic decreasing.

  6. anonymous
    • 5 years ago
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    I have a feeling you could use this test for the others. The second one is divergent using the same test. Don't wast time - set it up and just let wolframalpha do the integral so you can check it out.

  7. anonymous
    • 5 years ago
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    The only catch is that the terms of your sequence have to be monotone decreasing.

  8. anonymous
    • 5 years ago
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    How about the fifth series

  9. anonymous
    • 5 years ago
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    I'll look at it.

  10. anonymous
    • 5 years ago
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    Also for the 4th one, you can't use the integral test right?

  11. anonymous
    • 5 years ago
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    Yeah...you don't need to use the integral test for all of them. For the fifth, you can use the limit test. The test says that if the limit as n approaches infinity of a_n does not equal zero, the series will diverge. All you have to do is show the limit of (1+6^n)/(3+6^n) does not equal zero.

  12. anonymous
    • 5 years ago
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    is that the nth term test?

  13. anonymous
    • 5 years ago
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    might be - different names

  14. anonymous
    • 5 years ago
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    Thanks :) How about the 4th one. Which test should i use? I just have difficult to decide what test should be used

  15. anonymous
    • 5 years ago
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    I'll have a look.

  16. anonymous
    • 5 years ago
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    sorry for bothering you :[

  17. anonymous
    • 5 years ago
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    You're not bothering me :)

  18. anonymous
    • 5 years ago
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    After a while, 6/(n^2-65) will just behave like 6/n^2 which diverges. Is that reasonable?

  19. anonymous
    • 5 years ago
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    I'm just going through them now - had to do something.

  20. anonymous
    • 5 years ago
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    :) NP

  21. anonymous
    • 5 years ago
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    hi guys. i have midterm from same chapter. what is your advices about which method must be use for each equation. how can i decide it ? sorry for my poor english i hope you understand it.

  22. anonymous
    • 5 years ago
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    I have some advice - just give me a minute.

  23. anonymous
    • 5 years ago
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    ok thnx (:

  24. anonymous
    • 5 years ago
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    I know 4 diverges, and because of its form (polynomial fraction) I'm looking into the comparison test, or the limit comparison test. I'm trying to find an appropriate divergent sequence. I have to do this because the ratio test, etc. are inconclusive, and the integral (from the integral test) does not exist over all the domain we're interested in...so hang on...a little longer...

  25. anonymous
    • 5 years ago
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    Thanks for your effort :)

  26. anonymous
    • 5 years ago
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    Okay, sorry, got distracted again...but, I think I have your answer. Look at what happens when n=5 -> the denominator does not exist. I was looking at this series and thinking, "Why is it not convergent if it's asymptotically equivalent to 1/n^2 (which is a p-series with p>1)...well, yeah, you're diving be zero at some point, which is undefined. If you exclude n=5, you'll get a convergent sum.

  27. anonymous
    • 5 years ago
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    Well, the denominator exists, but the quotient doesn't.

  28. anonymous
    • 5 years ago
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    You comfortable with that, dichalao?

  29. anonymous
    • 5 years ago
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    What you mean n=5 ->denominator does not exist?

  30. anonymous
    • 5 years ago
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    You can't have a denominator equal to 0. When n is 5, the denominator is 5^2-25 = 25 - 25 = 0. I corrected myself above: the denominator exists, it's the quotient that doesn't (at this point).

  31. anonymous
    • 5 years ago
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    (n^2-65) it is n^2-65

  32. anonymous
    • 5 years ago
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    \[\frac{1}{0}= \infty\]You can remember it loosely as that.

  33. anonymous
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    and n can only be integers so ...it should be continious

  34. anonymous
    • 5 years ago
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    Lhunardaien, here is a good run-down/strategy sheet. http://tutorial.math.lamar.edu/Classes/CalcII/SeriesStrategy.aspx If I find any more, I'll post them here.

  35. anonymous
    • 5 years ago
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    This is what's happening with your series (summing the first seven terms):

  36. anonymous
    • 5 years ago
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    \[-\frac{1}{24}-\frac{1}{21}-\frac{1}{16}-\frac{1}{9}+\frac{1}{0}+\frac{1}{11}+\frac{1}{24}+...\]

  37. anonymous
    • 5 years ago
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    \[\lim_{c->0}\frac{1}{c}=\infty\]

  38. anonymous
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    Assuming the rest of the series is convergent, say to limit s, then you'd have, basically,\[s+\infty=\infty\]

  39. anonymous
    • 5 years ago
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    It's therefore not convergent.

  40. anonymous
    • 5 years ago
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    For what number of n you got how u got 1/0

  41. anonymous
    • 5 years ago
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    you got 1/0

  42. anonymous
    • 5 years ago
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    \[\frac{6}{n^2-25}\rightarrow \frac{6}{5^2-25}=\frac{6}{0}=6\times \frac{1}{0}\]

  43. anonymous
    • 5 years ago
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    Sorry, was the 1 bit doing you in?

  44. anonymous
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    Since multiplicative constants don't do much, I just leave them out whenever it's appropriate. Sorry if that added to some confusion.

  45. anonymous
    • 5 years ago
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    wait the original problem is (6/(n^2-65)) not 25

  46. anonymous
    • 5 years ago
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    Oh God...I misread it...

  47. anonymous
    • 5 years ago
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    It's convergent then

  48. anonymous
    • 5 years ago
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    LOL time to get pair of glasses :P

  49. anonymous
    • 5 years ago
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    At least you learnt something else!

  50. anonymous
    • 5 years ago
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    Let me make sure properly, though.

  51. anonymous
    • 5 years ago
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    Basically, it's asymptotically equivalent to 1/n^2 which is convergent (it's a p-series with p>1).

  52. anonymous
    • 5 years ago
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    yea haha so 1,4 converges and the rest diverges?

  53. anonymous
    • 5 years ago
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    so if i want to show how it is convergent. Which test should i use?

  54. anonymous
    • 5 years ago
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    Number 4?

  55. anonymous
    • 5 years ago
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    p-series test

  56. anonymous
    • 5 years ago
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    after acknowledging that, in the limit,\[\frac{6}{n^2-65} \]is asymptotically equivalent to \[\frac{6}{n^2}=6\frac{1}{n^2}\]

  57. anonymous
    • 5 years ago
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    Got It haha:) i might have one more problem would you mind help me out too

  58. anonymous
    • 5 years ago
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    ok

  59. anonymous
    • 5 years ago
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    you need to check the other problems i didn't explicitly cover here. i didn't look through all of them.

  60. anonymous
    • 5 years ago
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    yea i did check the ones that you didn't cover :)

  61. anonymous
    • 5 years ago
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    We might think that a ball that is dropped from a height of 10 feet and rebounds to a height 7/8 of its previous height at each bounce keeps bouncing forever since it takes infinitely many bounces. This is not true! We examine this idea in this problem. A. Show that a ball dropped from a height h feet reaches the floor in {1\over4}\sqrt{h} seconds. Then use this result to find the time, in seconds, the ball has been bouncing when it hits the floor for the first, second, third and fourth times: time at first bounce = time at second bounce = time at third bounce = time at fourth bounce = B. How long, in seconds, has the ball been bouncing when it hits the floor for the nth time (find a closed form expression)? time at nth bounce = C. What is the total time that the ball bounces for? total time =

  62. anonymous
    • 5 years ago
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    {1\over4}\sqrt{h} = sqrt(h)/4

  63. anonymous
    • 5 years ago
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    I got part A..but i couldn't figure out part B

  64. anonymous
    • 5 years ago
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    Can you type out the expressions in the equation editor?

  65. anonymous
    • 5 years ago
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    We might think that a ball that is dropped from a height of 10 feet and rebounds to a height 7/8 of its previous height at each bounce keeps bouncing forever since it takes infinitely many bounces. A. Show that a ball dropped from a height h feet reaches the floor in \[{1\over4}\sqrt{h}\] seconds. Then use this result to find the time, in seconds, the ball has been bouncing when it hits the floor for the first, second, third and fourth times: time at first bounce = time at second bounce = time at third bounce = time at fourth bounce = B. How long, in seconds, has the ball been bouncing when it hits the floor for the nth time (find a closed form expression)? time at nth bounce = C. What is the total time that the ball bounces for? total time =

  66. anonymous
    • 5 years ago
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    Okay, I was thrown for a minute then remembered you're using imperial units.

  67. anonymous
    • 5 years ago
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    Haha Take your time :)

  68. anonymous
    • 5 years ago
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    Okay...I have an answer.

  69. anonymous
    • 5 years ago
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    COOL!!

  70. anonymous
    • 5 years ago
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    When setting up your times, you have to keep in mind, for the first time, the ball falls only to the ground (from where it's dropped) BUT AFTER THAT, the time you calculate needs to be double at each step because it spends time going back up, and then back down to mark the next turn.

  71. anonymous
    • 5 years ago
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    Are you expected to use mathematical induction to prove you have the correct series, or can it just be 'obvious'?

  72. anonymous
    • 5 years ago
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    I got part A,so i tried to come up with an expression ,,,i got I did that.. i got sqrt10/4+(sqrt10)(sqrt(7/8))^(n-1)/2 for part B...

  73. anonymous
    • 5 years ago
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    No just the answer is fine

  74. anonymous
    • 5 years ago
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    I wrote the accumulated times up to 4 for you, then skipped to s_n. I'll scan it.

  75. anonymous
    • 5 years ago
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    OH MY GOD you are the NICEST person i have ever met online :OOOOO

  76. anonymous
    • 5 years ago
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    1 Attachment
  77. anonymous
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    After the line for s_n, I jumped to the limit of an infinite sum. The formula on the second page is the accumulated time for the nth step.

  78. anonymous
    • 5 years ago
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    Well, according to my fried head. I'm *very* tired, so please double-check everything.

  79. anonymous
    • 5 years ago
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    HAHA it is all good. Thanks for all your help. This really helps me out alot

  80. anonymous
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    Okay. Just check the accumulated sum. I usually derive the sums of geometric series on a case-by-case basis, but this time, just pulled the formula. I don't think I'm wrong in my analysis, though, like I said, I'm fried, so the mathematics needs to be checked. Good luck.

  81. anonymous
    • 5 years ago
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    Lhunardaien, did you get the link?

  82. anonymous
    • 5 years ago
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    I am looking through it :)

  83. anonymous
    • 5 years ago
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    i just looked now and thanx for link im looking now.

  84. anonymous
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    Lok,,can you please explain how to derive equation for the accumulated time for n step~

  85. anonymous
    • 5 years ago
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    Yeah, look at the powers of r in relation to the subscript on s.

  86. anonymous
    • 5 years ago
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    When n=2, the highest power is 1/2 n=3, the highest power is 2/2 n=4, the highest power is 3/2 ... n=k, the highest power is (k-1)/2

  87. anonymous
    • 5 years ago
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    brb

  88. anonymous
    • 5 years ago
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    \[1+r ^{1/2}+r+^{3/2}+.....) ->1(1-r^{n/2})/(1-r) this is the most confusing part

  89. anonymous
    • 5 years ago
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    Okay, this is just applying the formula for the sum of a geometric series. For r<1, the sum of the first n terms of the series a+ar+ar^2+...+ar^n is \[\frac{a(1-r^{n+1})}{1-r}\]

  90. anonymous
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    Here, a=1, and r is the r we're using anyway.

  91. anonymous
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    7/8, I think...

  92. anonymous
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    I input that equation , the system indicated that the equation is incorrect

  93. anonymous
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    hmm

  94. anonymous
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    is this a test?

  95. anonymous
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    This is a homework problem. (Webwork)

  96. anonymous
    • 5 years ago
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    I think your analysis makes very much sense.

  97. anonymous
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    The accumulated sum is correct tho

  98. anonymous
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    I think it came down to a mismatch between the time-step wanted in the question and the one I used.

  99. anonymous
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    Ah...that confirms it.

  100. anonymous
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    The actual mathematics isn't wrong, it's just your instructor named the time-steps differently.

  101. anonymous
    • 5 years ago
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    Siht

  102. anonymous
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    Sorry

  103. anonymous
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    Do you get another crack at it?

  104. anonymous
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    Did we start with n=1 right for the geometric series?

  105. anonymous
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    n=1 for *

  106. anonymous
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    Yes

  107. anonymous
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    If the accumulated sum (the limit) is correct, the analysis of the physics of the situation is correct, and there's been a difference in labeling.

  108. anonymous
    • 5 years ago
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    http://mathworld.wolfram.com/GeometricSeries.html check out equation 8

  109. anonymous
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    Yeah, that's the case because the first term is r.

  110. anonymous
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    does that mean we use r^n instead of r^(n+1) for the partial sum

  111. anonymous
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    This is why I derive these things from first principles...because of this kind of crap that can happen. Bear with me.

  112. anonymous
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    So that would be (1-r^((n-2)/2)) /(1-r) instead of (1-r^(n/2)in this case?

  113. anonymous
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    Oh no...I know what happened - I should have derived it...

  114. anonymous
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    In the last formula, instead of \[\frac{1-r^{n/2}}{1-r}\]it should have been\[\frac{1-r^{\frac{n-1}{2}}}{1-r^{1/2}}\]

  115. anonymous
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    \[s_n=\frac{1}{4}\sqrt{h_0}\left[ 1+2r^{1/2}\frac{1-r^{\frac{n-1}{2}}}{1-r^{1/2}} \right]\]

  116. anonymous
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    the nth term is \[r^{n-2/2} \] With the partial sum formula, how you got r^[(n-1)/2]

  117. anonymous
    • 5 years ago
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    did you get 23.69s for the limiting time?

  118. anonymous
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    yea

  119. anonymous
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    the rate is r^(1/2) in this case.. so i think it suppose to be (1-r^(n/2)/(1-r^(1/2))

  120. anonymous
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    Okay...I'm going to write out what I have\[s_n=\frac{1}{4}\sqrt{h_0}+\frac{1}{2}\sqrt{h_0}r^{1/2}(1+r^{1/2}+r^{2/2}+...+r^{(n-2)/2})\]\[=\frac{1}{4}\sqrt{h_0}+\frac{1}{2}\sqrt{h_0}r^{1/2}.\frac{1-r^{(n-1)/2}}{1-r^{1/2}}\]

  121. anonymous
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    how you got to\[ 1-r^{(n-1)/2}\]

  122. anonymous
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    okay

  123. anonymous
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    let the sum in brackets be equal to p_n,\[p_n=1+r^{1/2}+r^{2/2}+...+r^{(n-2)/2}\]then multiply through by r^{1/2},\[r^{1/2}p_n=r^{1/2}+r^{2/2}+r^{3/2}+...+r^{(n-1)/2}\]

  124. anonymous
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    subtract the second from the first,\[(1-r^{1/2})p_n=1-r^{(n-1)/2} \rightarrow p_n=\frac{1-r^{(n-1)/2}}{1-r^{1/2}}\]

  125. anonymous
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    OH nice proof~!!!

  126. anonymous
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    You should get the same answer for your limiting time.

  127. anonymous
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    as n-> infinity , it should give the same result~

  128. anonymous
    • 5 years ago
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    Yeah, p_n should approach \[\frac{1}{1-r^{1/2}}\]

  129. anonymous
    • 5 years ago
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    \[\frac{1}{4}\sqrt{h_0}+\frac{1}{2}\sqrt{h_0}r^{1/2}.\frac{1-r^{(n-1)/2}}{1-r^{1/2}}\]

  130. anonymous
    • 5 years ago
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    if i plug in n=1,2,3, it should give answers to PART A as well?

  131. anonymous
    • 5 years ago
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    try that

  132. anonymous
    • 5 years ago
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    that would have been a useful check

  133. anonymous
    • 5 years ago
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    YEP it works you are hero ~~!!!

  134. anonymous
    • 5 years ago
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    oh good...

  135. anonymous
    • 5 years ago
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    thank God

  136. anonymous
    • 5 years ago
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    so does that mean you'll get it right for your homework?

  137. anonymous
    • 5 years ago
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    i am typing in right now. Last attempt

  138. anonymous
    • 5 years ago
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    sure you don't want to check some more...

  139. anonymous
    • 5 years ago
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    ?

  140. anonymous
    • 5 years ago
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    I'm reeaaally tired and making mistakes at this hour.

  141. anonymous
    • 5 years ago
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    It seems to work fine i tried it with n=1-4 ,and they all worked out fine

  142. anonymous
    • 5 years ago
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    ok, if you're happy.

  143. anonymous
    • 5 years ago
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    lol should i submit it..

  144. anonymous
    • 5 years ago
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    check the one you put in last time that was wrong - check to see you DON'T get the right answers to part A.

  145. anonymous
    • 5 years ago
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    it ..shouldn't work since (n-2)/2 will be negative when n =1

  146. anonymous
    • 5 years ago
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    If you're happy with it. Is it worth much to your final grade?

  147. anonymous
    • 5 years ago
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    i guess not

  148. anonymous
    • 5 years ago
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    HIGH FIVE..IT IS CORRECT! What would i have done if this was on my midterm

  149. anonymous
    • 5 years ago
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    AWESOME!!!!!!!!!!!!!!

  150. anonymous
    • 5 years ago
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    i was nervous

  151. anonymous
    • 5 years ago
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    LOL sorry for making you nervous~~haha

  152. anonymous
    • 5 years ago
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    My advice - learn to derive the results for geometric sums instead of using a formula. See what happened when I just used a formula?

  153. anonymous
    • 5 years ago
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    yea.. we should always try to derive equations

  154. anonymous
    • 5 years ago
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    phew

  155. anonymous
    • 5 years ago
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    are you good at physics as well?

  156. anonymous
    • 5 years ago
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    i can sleep well

  157. anonymous
    • 5 years ago
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    LOL i really should go to bed now..it is 5 in the morning here :)

  158. anonymous
    • 5 years ago
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    omg

  159. anonymous
    • 5 years ago
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    haha nice meeting you man YOU ARE GREAT !! Have a good one

  160. anonymous
    • 5 years ago
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    no probs. good luck!

  161. anonymous
    • 5 years ago
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    nice meeting you too

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spraguer (Moderator)
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