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chrisplusian

  • one year ago

confirm the integral test can be applied to the series.

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  1. chrisplusian
    • one year ago
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    \[\sum_{1}^{\infty}\frac{ n }{n^{4} +1 }\]

  2. MarcLeclair
    • one year ago
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    You know how to evaluate improper integrals right?

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

  4. MarcLeclair
    • one year ago
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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.

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

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

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

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

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

  10. MarcLeclair
    • one year ago
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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

  11. chrisplusian
    • one year ago
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    ok let me explain why I am confused

  12. chrisplusian
    • one year ago
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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.

  13. MarcLeclair
    • one year ago
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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

  14. chrisplusian
    • one year ago
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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.

  15. chrisplusian
    • one year ago
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    @zepdrix can you help?

  16. chrisplusian
    • one year ago
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    @experimentX ?

  17. experimentX
    • one year ago
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    yes it does

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

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

  20. experimentX
    • one year ago
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    this is too long ... i'll upload a snapshot

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

  22. chrisplusian
    • one year ago
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    Thanks

  23. experimentX
    • one year ago
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    1 Attachment
  24. chrisplusian
    • one year ago
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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.

  25. experimentX
    • one year ago
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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

  26. chrisplusian
    • one year ago
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    ok that statement seems logical.

  27. chrisplusian
    • one year ago
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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

  28. chrisplusian
    • one year ago
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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

  29. experimentX
    • one year ago
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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.

  30. experimentX
    • one year ago
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    also differentiability implies continuity ...

  31. experimentX
    • one year ago
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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.

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

  33. chrisplusian
    • one year ago
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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.

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

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

  36. experimentX
    • one year ago
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    sorry ... i was out

  37. chrisplusian
    • one year ago
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    no problem

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

  39. chrisplusian
    • one year ago
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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.

  40. experimentX
    • one year ago
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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.

  41. chrisplusian
    • one year ago
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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?

  42. experimentX
    • one year ago
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    physics graduate student

  43. experimentX
    • one year ago
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    with a bit experience on analysis

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

  45. chrisplusian
    • one year ago
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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......

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

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

  48. experimentX
    • one year ago
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    this does not converge

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

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

  51. experimentX
    • one year ago
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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.

  52. chrisplusian
    • one year ago
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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

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

  54. experimentX
    • one year ago
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    yes ... that it is

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

  56. experimentX
    • one year ago
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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

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

  58. chrisplusian
    • one year ago
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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?

  59. chrisplusian
    • one year ago
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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.

  60. experimentX
    • one year ago
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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.

  61. experimentX
    • one year ago
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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.

  62. chrisplusian
    • one year ago
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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

  63. experimentX
    • one year ago
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    is calc II differential equatons or multivariable calc?

  64. chrisplusian
    • one year ago
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    Honestly I always wondered what the difference is. lol

  65. experimentX
    • one year ago
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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

  66. chrisplusian
    • one year ago
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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

  67. experimentX
    • one year ago
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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.

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

  69. chrisplusian
    • one year ago
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    what are spivak's calculus?

  70. chrisplusian
    • one year ago
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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

  71. chrisplusian
    • one year ago
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    Then we go on to a class called differential equations.

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

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

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

  75. chrisplusian
    • one year ago
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    So This is the hard part? The analysis?

  76. experimentX
    • one year ago
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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

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

  78. experimentX
    • one year ago
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    no no it's okay ...

  79. experimentX
    • one year ago
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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.

  80. experimentX
    • one year ago
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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|

  81. experimentX
    • one year ago
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    all right good luck.

  82. chrisplusian
    • one year ago
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    Thanks again!!!!!

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