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chrisplusian

confirm the integral test can be applied to the series.

  • 10 months ago
  • 10 months ago

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  1. chrisplusian
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    \[\sum_{1}^{\infty}\frac{ n }{n^{4} +1 }\]

    • 10 months ago
  2. MarcLeclair
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    You know how to evaluate improper integrals right?

    • 10 months ago
  3. chrisplusian
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    So I know that you have to show that it is positive, continuos, and decreasing for \[x \ge 1\]

    • 10 months ago
  4. MarcLeclair
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    Well first you have to check the the series is decreasing and approaching 0. So if you aren't sure it is decreasing, find its derivative to show it is decreasing over x.

    • 10 months ago
  5. chrisplusian
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    and that I have to show that \[A_{n}=f(n) \]

    • 10 months ago
  6. MarcLeclair
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    Its derivative would be (x^4 + 1) - ( 4x^4 ) which would give you a decreasing function in the denominator.

    • 10 months ago
  7. chrisplusian
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    That is where I run into the problem..... taking the first derivative gives me: \[\frac{ -3x^{4} +1 }{ (x^{4} +1)^2 }\]

    • 10 months ago
  8. chrisplusian
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    @MarcLeclair did you use the quotient rule to get the derivative?

    • 10 months ago
  9. MarcLeclair
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    Yeah so it would decrease for all x when x is greater than 1

    • 10 months ago
  10. MarcLeclair
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    Yes i did I just wrote the top part :) because the denominator wont dictate if it is negative or not because it is squared

    • 10 months ago
  11. chrisplusian
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    ok let me explain why I am confused

    • 10 months ago
  12. chrisplusian
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    the only way I know to show that the function is decreasing is by taking the first derivative and then finding the critical values. Then I use a sign chart to determine if it is increasing or decreasing. The problem with this one is finding the critical values. Maybe my algebra is a little rusty.

    • 10 months ago
  13. MarcLeclair
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    Well this is weird, when I was doing it ( I just finished cal 2 here, took my final few weeks ago) I did it in my book and it only found the first derivative. It didn't go all the way showing its critical values, it just showed the numerator was decreasing over x when x was greater than a certain value. If you're doing it by finding those I guess I'm the wrong person to help :P

    • 10 months ago
  14. chrisplusian
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    well the critical values of the first derivative are the places where the graph can change from increasing to decreasing so we are required in my class to show that there are either no critical values that are greater or equal to one, or that if there are they are continuosly decreasing.

    • 10 months ago
  15. chrisplusian
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    @zepdrix can you help?

    • 10 months ago
  16. chrisplusian
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    @experimentX ?

    • 10 months ago
  17. experimentX
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    yes it does

    • 10 months ago
  18. experimentX
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    for a series decreasing to zero, \[ \int_a^b f(x)dx - f(a) \le \sum_{n=a}^b f(n) \le \int_a^b f(x)dx + f(a)\]

    • 10 months ago
  19. chrisplusian
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    I don't understand that can you explain it?

    • 10 months ago
  20. experimentX
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    this is too long ... i'll upload a snapshot

    • 10 months ago
  21. chrisplusian
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    does that follow the epsilon delta proofs? We were told not to do the epsilon delta parts of these

    • 10 months ago
  22. chrisplusian
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    Thanks

    • 10 months ago
  23. experimentX
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    • 10 months ago
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  24. chrisplusian
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    I will have to go print that at office max.... I have a macbook air and it is to big to show on this lillte screen. Hopefully that will help thank you.

    • 10 months ago
  25. experimentX
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    integral test is stronger than ratio test or root test in general. for you case, \[ \int_1^\infty \frac{x}{x^4 + 1 }dx < \int_1^\infty \frac{x}{x^4}dx\] is less than inifinity

    • 10 months ago
  26. chrisplusian
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    ok that statement seems logical.

    • 10 months ago
  27. chrisplusian
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    The problem I am running into is that my professor wants me to show four steps before I can use the integral test and there will be problems on our test that say specifically to use the integral test, which in turn means to show these four things: 1)f(x) is positive for x ge 1. 2) f(x) is continuous for x ge 1. 3) f is decreasing for x ge 1. 4)\[a_{n}=f(n) \] for n= positive integer

    • 10 months ago
  28. chrisplusian
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    The problem I ma running into over and over is that the functions I am given are very hard to find the critical values of the f prime function and thereby make it very difficult to show that the original function is decreasing for all x ge 1

    • 10 months ago
  29. experimentX
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    f(x) satisfies all condition, the poles are all complex at |z| = 1, the function is decreasing to zero ... to show that take derivative, put x>0 ... lim x->infty f(x) = 0 is not hard to show.

    • 10 months ago
  30. experimentX
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    also differentiability implies continuity ...

    • 10 months ago
  31. experimentX
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    to show positive, assume, f(x)>0, it immediately gives you that for all x>0, the f(x) is positive since x^4 + 1 is never negative.

    • 10 months ago
  32. chrisplusian
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    what do you mean "f(x) satisfies all condition, the poles are all complex at |z| = 1"?

    • 10 months ago
  33. chrisplusian
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    I showed that it was positive the same way you did. assuming that f(x)>0 and proving it that way.

    • 10 months ago
  34. chrisplusian
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    also why do you say that the function is decreasing to zero?

    • 10 months ago
  35. chrisplusian
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    I am sorry to ask so many questions, just having a difficult time with it

    • 10 months ago
  36. experimentX
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    sorry ... i was out

    • 10 months ago
  37. chrisplusian
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    no problem

    • 10 months ago
  38. experimentX
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    you just need to show f(n+1)<f(n) ... this shows that function is decreasing

    • 10 months ago
  39. chrisplusian
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    oh that makes sense..... but fairly long in this problem, yet that makes sense. its alost like you are showing it is monotonic for that interval.

    • 10 months ago
  40. experimentX
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    there is a thing called singularity ... when the function becomes infinite, for that x^4 +1 = 0 => all roots are complex. there is singularity on real line.

    • 10 months ago
  41. chrisplusian
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    I have never heard that before. I wish I knew math the way you do. Are you a graduate student or professor?

    • 10 months ago
  42. experimentX
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    physics graduate student

    • 10 months ago
  43. experimentX
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    with a bit experience on analysis

    • 10 months ago
  44. chrisplusian
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    thats awesome, I am working on my B.S. majoring in engineering minor in physics.

    • 10 months ago
  45. chrisplusian
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    Do you have to prove the above mentioned statements for ALL x ge 1 or can it be for a different bound ? ex......

    • 10 months ago
  46. experimentX
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    yes ... those are required condition for integral test. for eg, integral test don't work on alternating series.

    • 10 months ago
  47. chrisplusian
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    \[\sum_{1}^{\infty}\frac{ \ln(n) }{ \sqrt{n} }\]

    • 10 months ago
  48. experimentX
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    this does not converge

    • 10 months ago
  49. chrisplusian
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    In this case it is increasing for \[0 \le x \le e^2\]

    • 10 months ago
  50. chrisplusian
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    Does the fact that it is increasing automatically show that you cannot use the integral test?

    • 10 months ago
  51. experimentX
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    you know that you can always partition the series ... consider two separate cases. 1) n < e^2 and n>e^2 for finite sum always converges given that there is no singularity. for infinite value, use integral test.

    • 10 months ago
  52. chrisplusian
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    Some of the questions we have are strictly to decide if the integral test can be used to show convergence or divergence

    • 10 months ago
  53. chrisplusian
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    So as long as I use the lower bound of integration as 8 then I can use the integral test?

    • 10 months ago
  54. experimentX
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    yes ... that it is

    • 10 months ago
  55. chrisplusian
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    Ok can I ask you an unrelated question? It is about physics with calc one and two?

    • 10 months ago
  56. experimentX
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    lol ... calc is heavy sub on physics. calc I and II are for undergrad juniors. I had mathematical physics which focuses on many areas of math mostly non linear DE, PDE, Complex analysis, Tensors, Linear algebra, Statistics, Numerical analysis

    • 10 months ago
  57. experimentX
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    my courses assumes that I have enough experience with calc I, II and III

    • 10 months ago
  58. chrisplusian
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    I am in a small school and have completed physics with calc one and physics with calc two. I was grouped in with the regular physics students because of the class size (only three for physics with calc one, and two students for physics with calc two). Our professor never taught a single thing in regards to calculus based physics. My textbook did not correspond to what he was teaching because of the calculus element. I have to go on and take thermal dynamics, statics, etc. Plus I am going to have to take 18 hours in physics to get the minor in physics. I am contemplating filing a grievance and asking to be taught the material with the calculus element because the school did not meet the requirements of the statewide numbering system (Florida). But my professor is saying not to worry about it because it is only important that I understand the concepts, that the calc is not that important. What do you think? Is it a mistake to proceed without knowing the calculus side?

    • 10 months ago
  59. chrisplusian
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    I am transferring to University of Florida in five months, and I saw some of their phyisics with calculus questions from their study review....... I had no clue.

    • 10 months ago
  60. experimentX
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    so far i know calc I is elementay calculus, calc II is differential equations and calc III is multivariable calc ... these are basic courses commonly taught for all physics/engg student. you should have mastery of it. I recommend it. it seems that you are currently doing analysis. this is a bit hard part.

    • 10 months ago
  61. experimentX
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    so far i have seen, this is probably with linear algebra this is usually taught to freshmen or sophomore. of course math gets messy on III and IV years, concept is important and so it problem solving.

    • 10 months ago
  62. chrisplusian
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    I have taken calculus one, this is the end of calculus two, and I have already completed physics with calculus one and physics with calculus two at my college physics with calculus two is just a co-requisite of physics with calculus two

    • 10 months ago
  63. experimentX
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    is calc II differential equatons or multivariable calc?

    • 10 months ago
  64. chrisplusian
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    Honestly I always wondered what the difference is. lol

    • 10 months ago
  65. experimentX
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    what did you study then? for first two years at uni http://tutorial.math.lamar.edu/ <-- this is more than enough

    • 10 months ago
  66. chrisplusian
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    here they use a numbering system and to complete a class you have to complete certain objectives. In this class CALC TWO you have to go through all the way to the ratio and root test. First part of the semester we learned the

    • 10 months ago
  67. experimentX
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    lol ... this is not calc, it's a part of analysis. http://en.wikipedia.org/wiki/Real_analysis when calc get's worse, usually analysis comes to rescue.

    • 10 months ago
  68. experimentX
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    although i've heard that if you study spivak's calculus, then things are messy

    • 10 months ago
  69. chrisplusian
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    what are spivak's calculus?

    • 10 months ago
  70. chrisplusian
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    Cacl three topics here are: polar equations vectors and geometry of space vector-valued functions functions of several variables multiple integration and vector analysis

    • 10 months ago
  71. chrisplusian
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    Then we go on to a class called differential equations.

    • 10 months ago
  72. experimentX
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    that's calculus book http://www.amazon.com/Calculus-4th-Michael-Spivak/dp/0914098918

    • 10 months ago
  73. chrisplusian
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    so what you call analysis is the classes below calculus?

    • 10 months ago
  74. experimentX
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    it's hard part ... more mathematical than usual mathematics you do in physics or engineering. close to pure math.

    • 10 months ago
  75. chrisplusian
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    So This is the hard part? The analysis?

    • 10 months ago
  76. experimentX
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    look at the topics http://www.amazon.com/Principles-Mathematical-Analysis-Third-Walter/dp/007054235X you will find that it is similar to calculus but not calculus exactly

    • 10 months ago
  77. chrisplusian
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    Thanks you have been a great help. WIsh I could give you a few more medals lol.

    • 10 months ago
  78. experimentX
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    no no it's okay ...

    • 10 months ago
  79. experimentX
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    the thing is calculus is too smooth like integration, derivatives, etc ... with analysis you will understand the underlying theory behind it. take a look ... it's more rigorous than calc but don't spend much time in it.

    • 10 months ago
  80. experimentX
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    one eg: if you have infinite nested roots like this one, it's easy to calculate it's value, it's easy part ... the hard part comes when you have to show that this value exists using some simple theorems. basically analysis deals with these stuff. |dw:1369672079150:dw|

    • 10 months ago
  81. experimentX
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    all right good luck.

    • 10 months ago
  82. chrisplusian
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    Thanks again!!!!!

    • 10 months ago
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