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anonymous
 3 years ago
confirm the integral test can be applied to the series.
anonymous
 3 years ago
confirm the integral test can be applied to the series.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{1}^{\infty}\frac{ n }{n^{4} +1 }\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You know how to evaluate improper integrals right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So I know that you have to show that it is positive, continuos, and decreasing for \[x \ge 1\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and that I have to show that \[A_{n}=f(n) \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Its derivative would be (x^4 + 1)  ( 4x^4 ) which would give you a decreasing function in the denominator.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That is where I run into the problem..... taking the first derivative gives me: \[\frac{ 3x^{4} +1 }{ (x^{4} +1)^2 }\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@MarcLeclair did you use the quotient rule to get the derivative?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah so it would decrease for all x when x is greater than 1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes i did I just wrote the top part :) because the denominator wont dictate if it is negative or not because it is squared

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok let me explain why I am confused

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@zepdrix can you help?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1for a series decreasing to zero, \[ \int_a^b f(x)dx  f(a) \le \sum_{n=a}^b f(n) \le \int_a^b f(x)dx + f(a)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I don't understand that can you explain it?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1this is too long ... i'll upload a snapshot

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0does that follow the epsilon delta proofs? We were told not to do the epsilon delta parts of these

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1integral 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok that statement seems logical.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The 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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1f(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.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1also differentiability implies continuity ...

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1to 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what do you mean "f(x) satisfies all condition, the poles are all complex at z = 1"?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I showed that it was positive the same way you did. assuming that f(x)>0 and proving it that way.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0also why do you say that the function is decreasing to zero?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I am sorry to ask so many questions, just having a difficult time with it

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1sorry ... i was out

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1you just need to show f(n+1)<f(n) ... this shows that function is decreasing

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh 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.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1there 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I have never heard that before. I wish I knew math the way you do. Are you a graduate student or professor?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1physics graduate student

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1with a bit experience on analysis

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thats awesome, I am working on my B.S. majoring in engineering minor in physics.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Do you have to prove the above mentioned statements for ALL x ge 1 or can it be for a different bound ? ex......

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1yes ... those are required condition for integral test. for eg, integral test don't work on alternating series.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{1}^{\infty}\frac{ \ln(n) }{ \sqrt{n} }\]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1this does not converge

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0In this case it is increasing for \[0 \le x \le e^2\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Does the fact that it is increasing automatically show that you cannot use the integral test?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1you 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Some of the questions we have are strictly to decide if the integral test can be used to show convergence or divergence

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So as long as I use the lower bound of integration as 8 then I can use the integral test?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ok can I ask you an unrelated question? It is about physics with calc one and two?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1lol ... 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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1my courses assumes that I have enough experience with calc I, II and III

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1so 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.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1so 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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 corequisite of physics with calculus two

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1is calc II differential equatons or multivariable calc?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Honestly I always wondered what the difference is. lol

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1what did you study then? for first two years at uni http://tutorial.math.lamar.edu/ < this is more than enough

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0here 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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1lol ... 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.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1although i've heard that if you study spivak's calculus, then things are messy

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what are spivak's calculus?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Cacl three topics here are: polar equations vectors and geometry of space vectorvalued functions functions of several variables multiple integration and vector analysis

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Then we go on to a class called differential equations.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1that's calculus book http://www.amazon.com/Calculus4thMichaelSpivak/dp/0914098918

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so what you call analysis is the classes below calculus?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1it's hard part ... more mathematical than usual mathematics you do in physics or engineering. close to pure math.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So This is the hard part? The analysis?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1look at the topics http://www.amazon.com/PrinciplesMathematicalAnalysisThirdWalter/dp/007054235X you will find that it is similar to calculus but not calculus exactly

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks you have been a great help. WIsh I could give you a few more medals lol.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1no no it's okay ...

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1the 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.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1one 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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1all right good luck.
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