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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?
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{1}^{\infty}(4^nx^{2n})/n\]

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0well 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...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0By 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,\[4x^2 \lt 1 \rightarrow \frac{1}{2} \lt x \lt \frac{1}{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The radius of convergence is, by definition, all x that satisfy \[x < R\] where R is a nonnegative, real number. \[4x^2=4x^2=4x^2<1 \rightarrow x<\frac{1}{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Since x < 1/2, the series is absolutely convergent for 1/2 < x < 1/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Every absolutely convergent series is unconditionally convergent.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How about at end points? x=1/2 and 1/2

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah...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).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you check the endpoints for me?

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0isn't it should be the same because x^(2n)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, when x=1/2, you get an alternating harmonic which I know to be equal to log(2) (natural log).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{4^n \left( \frac{1}{2} \right)^{2n}}{n}=\frac{(2^2)^n \frac{1}{2^{2n}}}{n}=\frac{2^{2n}\frac{1}{2^{2n}}}{n}=\frac{1}{n}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea i got this how about x=1/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, when you sub. x=1/2, you get the same result as above, except for a (1)^n outside the 1/n.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0never mind i know where ive made a mistkae :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is the 'alternating harmonic series' which converges to ln(2). You can test convergence with the alternating series test.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so it conditionally converges at x=1/2.. what i did i squared x , and then simplify the expresion so i always got 1/n :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Actually, (1/2)^(2n) is always positive (I'm eating while doing this so not completely concentrating :( )...so I need to have another look.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL alright (Lucas with .) i got to go to class now. see you soon haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Have u had another look?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No...sorry, I'm only back on because someone sent an email. Is this something due now?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no i am just curious haha just get back to me whenever you can :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you have stuff to do, it is fine :]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It shouldn't be converging for either +1/2 or 1/2

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh, yea i tried..10 times yesterday,,,i still got the same answer,, but the book says it conditionally diverges at x=1/2..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL my professor just told us the book is never wrong ==

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cuz he wrote part of the textbook we are using

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha, well, my professors wrote some books too and I have entire errata sheets handed out from them (i.e. corrections to their printed stuffups).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we all make mistakes :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It says inconclusive for both tests ..what does that mean?

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If 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)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Alright, so the book is WRONG..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm thinking...and if it's not, I'd like to know why.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Try a Cauchy Condensation test or something. Or alternating series test.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is not an alternating series even if x is negative.. and i haven't learned cauchy condensation test :

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh yea do you have time to explain the rearranging theorem ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, exactly, that's my point. I don't think you can throw anything at this to prove convergence.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmmm, probably not :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What's your question? At least I can think about it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0umm for a conditionally convergent series, you can rearrange the terms to get any sums you like?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the order of the terms

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry, I had to go do that. Some people on here are really starting to pellet me.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL i saw it ..that dude is retarded

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, so your question...I will scrounge a proof for you.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0COOL buddy, don't waste your time and energy on aXX haha

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

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Because when you get your 'nth' partial sum and send n to infinity, you end up with different limits.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is this what you're looking for?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The alternating series is a conditionallyconvergent 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you mean the pattern of the series changes if we change the order of the terms, so as n>infinity, the limit is different?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh i feel you ^^ thank you ~~~~~

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'll send something I have. It shows how you can arrange the alternating harmonic to get 1/2 log 2 instead of log 2.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait how about asolutely convergent series, if we rearrange the oders of terms, the pattern of the series also changes right?

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Has your professor shown you why an infinite series of positive terms doesn't change it's convergence (sum) upon rearrangement?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nope he didn't get into any details, he just stated the theorem and said this is not in the scope of this course..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The proof is on that sheet too.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea i am looking at it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok...i'll leave it with you. I have to go do some work. Have fun with it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks !!! have fun with your work too

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LUCAS~~~ I have a question regarding PHYSICS~~

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0a wooden block is at rest in a vehicle that is also at rest, if the vehicle accelerates north, why will the block slide southward

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry, ordered chinese food and guy was at door

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh yea chinese food haha~

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

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, friction should oppose motion

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the box slides the opposite way of the acceleration becasue of the first law.. right?

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I can solve this for you, but I need to go offline. Gotta go out. Is this due tomorrow or something?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nope just get back to me whenever u want :) 3rd law is also involved?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In terms of the normal force.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'll get back to you.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0kk thanks :) see you~

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol i like how your name sound so japanese

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What did you get for your truck/case question in the end?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0static friction = coeffient*normal force, and the direction is north

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is good enough man :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the case is moving because the fraction "sticks" it to the floor of the truck. otherwise, it will stay there, by the first law?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes. It will stay fixed in space (assuming your coordinate system was moving at the original velocity of the case) because of the first law.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0alright more questions..u mind?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\displaystyle \sum_{n=1}^\infty \frac{(1)^{n}e^n}{n!} \] \[\displaystyle \sum_{n=1}^\infty \frac{(1)^{n}n!}{n^n} \] \[\displaystyle \sum_{n=1}^\infty \frac{(1)^{n}n!}{e^n} \] \[\displaystyle \sum_{n=1}^\infty \frac{(1)^{n}n^n}{n!} \] 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, are these due Saturday or something?

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

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

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea tahts what i need help on

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah...kind of have to prove it's decreasing...I'll look at them in my break.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Alrighty~! thank you!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Did you figure this out?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You should look at your sequences like this:\[\frac{e^n}{n!}=\frac{e}{n}.\frac{e}{n1}.\frac{e}{n2}....\frac{e}{3}.\frac{e}{2}.\frac{e}{1}\]There are n copies of e and n terms in the n factorial. Taking the limit on both sides gives\[\lim_{n \rightarrow \infty}\frac{e^n}{n!}=\lim_{n \rightarrow \infty}\left( \frac{e}{n}.\frac{e}{n1}.\frac{e}{n2}....\frac{e}{3}.\frac{e}{2}.\frac{e}{1} \right)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The righthand side can be expanded in limits by the multiplicative limit law:\[\lim _{n \rightarrow \infty} \left( \frac{e}{n}.\frac{e}{n1}.\frac{e}{n2}....\frac{e}{3}.\frac{e}{2}.\frac{e}{1} \right)\]\[\lim_{n \rightarrow \infty}\frac{e}{n}.\lim_{n \rightarrow \infty}\frac{e}{n1}.\lim_{n \rightarrow \infty}\frac{e}{n2}....\lim_{n \rightarrow \infty}\frac{e}{3}.\lim_{n \rightarrow \infty}\frac{e}{2}.\lim_{n \rightarrow \infty}\frac{e}{1} \]\[=0.0.0.....\frac{e}{3}.\frac{e}{2}.\frac{e}{1}\]\[=0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now, 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,\[0<\frac{e}{n+1}\frac{e^n}{n!}<\frac{e^n}{n!} \iff 0<\frac{e^{n+1}}{(n+1)!}<\frac{e^n}{n!} \iff a_{n+1}<a_n\]

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0got it.. factorial is the thing that scares me...

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think you should be right now :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0gotta go...good luck with it!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL yea that was a very good explaination haha thanks~!!!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hey loki do u have time to help me out with one more problem?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can put it down but I have to go...have an insane amount of work to do for tomorrow.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[1 + 4! x + \frac{8! x^2}{(2!)^4} + \frac{12! x^3}{(3!)^4} + \frac{16! x^4}{(4!)^4} + \frac{20! x^5}{(5!)^4}+\cdots\] Use the ratio test to find the radius of convergence of the power series attempt: ((4n4)!x^(n1))/((n1)!)^4 this is the a_n i found for the series

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The a_n is\[a_n=\frac{(4n)!x^n}{(n!)^4}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then you can apply the ratio test.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh the constant is the 0th term ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you mind doing it so i can comapre my answer with you? i only have one attempt left :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I haven't finished it off.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry for taking your precious time away !!!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I've rushed through it. Assuming the bottom line is correct, you should be able to finish it off.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i got that too, as n>ininfty, all i care is n^4 terms right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You have a quartic polynomial above and below.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so i multiply everything out and take the limit?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I would multiply the numerator out.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0256 n^4+640 n^3+560 n^2+200 n+24

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or can i just mulitply (4n)^4 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0denominator: n^4+4 n^3+6 n^2+4 n+1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0factor out n^4 on top and bottom and take the limit.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[256x<1\]is required for convergence.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0THANK YOU !!! And SORRY!!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You need to check the end points.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh i am just looking for the radius

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i really appreciate your help!!!! WISH you the best!!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOki~~ 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=?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Nevermind loki haha i got it :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there are youtube clips that show you what to do

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i used long division ot find the answers

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Can you help me out with this problem?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ooh, 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sure it is not due soon :) and i will keep trying~~ haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay...yes, keep trying...

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Actually, I think this version's better:\[f_n=\frac{\phi^n(1\phi)^n}{\sqrt{5}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{f_{n+1}}{f_n}=\frac{\frac{\phi^{n+1}(1\phi)^{n+1}}{\sqrt{5}}}{\frac{\phi^n(1\phi)^n}{\sqrt{5}}}=\frac{\phi^{n+1}(1\phi)^{n+1}}{\phi^n(1\phi)^n}\]\[=\frac{\phi^{n+1}}{\phi^n}.\frac{1\left( \frac{1\phi}{\phi} \right)^{n+1}}{1\left( \frac{1\phi}{\phi} \right)^{n}}=\phi \frac{1\zeta^{n+1}}{1\zeta^{n}}\]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\[\lim_{n \rightarrow \infty}\frac{f_{n+1}}{f_n}= \phi\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The radius of convergence is then,\[\phi x<1 \rightarrow x<\frac{1}{\phi}\iff \frac{1}{\phi}<x<\frac{1}{\phi}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Check through it  it's after midnight and I've had about five hours sleep in two days!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hope you a good night !!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wow i would never figure this out... Since i haven;t learned binet's formula

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thank!! you!! like ______________________________________________________________________________________________________________________this much

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry but i ahve to bother you more!

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dichalao, when's this due?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0have you been waiting all this time?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nope,, i been doing physics homework, and checked back from time to time

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This thing isn't conceptually hard. What's going to stuff you are mistakes with the algebra.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Have you tried plugging into wolfram alpha?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i did... but i got a nonsense answer...as i said

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(2n+2)!! = (2n)!!(2n+2) is this right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(4n+7)!!=(4n+7)(4n+5)(4n+3)!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0limit a_(n+1)/(a_n) = 43570(8x+2)/48 43570(8x+2)/48<1 8x=1.998898

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm going to do it now.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are you at home right now?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh oh haha you should have slept in

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but i should be doing work so that i don't have to spend another week staying up all night :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I can't sleep in  never works. Too sunny.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL where is your certain?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think my building is on the surface of the sun; nothing stops the light or the heat.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, but eye masks dry the skin under my eyes.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just once a week ,, just apply lotion to it when you get up :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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!!!!!!!!!!!!!!!!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i found something, but i want to recheck.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[8x+2<\frac{24}{21875}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea...that is what i got i think

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you will have negative number at the right side ...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if u subtract 2 from both side..u wil have negative number

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0divide 8 , then i got 0.25

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i tried that answer before..the system indicated it is incorrect..

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL do you mind explain a little?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There is a tiny radius of convergence.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so it centered at x=0.25 ? taylor series?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Everything below 24/whatever is what you want (i.e. those x values that generate the function values below the 24/whatever line).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That's why I said I wanted to check something  I meant, I wanted to use the definition of absolute value.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the raidus will just be half of the distance between the two points where they meet?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is it possible to do this thing without graphing it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, that's what i wanted to do.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok thanks alot i will figure it out ~~!!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0don't use all your attempts till i check it out.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I gotta go  need lunch and take care of 'business' :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'll go through the problem again later.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0arlgiht enjoy your day!!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry for late response i was doing my physics hw

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I tried ...i still couldn't get the right answer

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is due before midnight, 21th

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hey Loki, i got the answer already :) Thank you for helping!!!!!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm glad you got the answer 'cause I've had no time! :) What was wrong with it in the end?

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, that's what I got too :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cool!! Let me give u a medal haha

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okie okie you got a new pro pic that doesnt look like you haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I move with the times.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hehe...I'm upset with the new site.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OH yea...i am so not used to it

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't like all this medal and shield crap.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have a headache, so am a little slow right now.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OH is this s bad time to ask you questions...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok what you mean by looking at how each term contributes to a place

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[1\frac{1}{2!}+\frac{1}{4!}\frac{1}{6!}+\frac{1}{8!}\frac{1}{10!}+...\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0At some point, one of the terms and those that follow stop contributing to the fifth decimal place (else the sum wouldn't converge).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh i see what you mean cuz it is a convergent series

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

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can if your intention is to expand in terms of sec(x)...but that's not what you're asked to do.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You have to apply the process to find the first five coefficients.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Use wolfram to work them out. They're a pain.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Find 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0One second. I'm looking at something else.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i should compute the P4 and subtract ln(sec(0.2) from it?

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ummm i dont really know..this is an online homework , the problems are given by a third party

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0first derivative is tan(x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0third is 2tan(x)sec^2(x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh i computed it on wolfram already thanks ~~

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0fourth is 2(cos(2x)2)sec^4(x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you mind briefly tell me about the troublesome option?

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

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah. If you're studying to become an engineer, numerical methods are what you'll be using.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0numerical methods = ...mesure and calculator~~?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Numerical methods is measurement and numbercrunching. 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh..pain in the butt...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Actually, it's less of a pain to solve things numerically than algebraically. You'll get used to it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0To find the Maclaurin series for f(x) = x/(8 + 4 x^2)^(1/3) ...are there easier ways ...to do it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry i got distracted a bit..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i haven't taken any engineering classes so far

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but the == most troubleosome part is taking derivatives of such function.. omg wolfram haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, there are no easier ways of taking derivatives.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is that possible to flip a power series,as in 1/ power series

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hang on. I need to make a phone call.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0brb going to take a quick shower first..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think it would be more difficult.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if the power seiresi is sum n=0 > infin, 4n(x)^n/n^2 ...if i slip it..what would it be?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It'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\[f(x)=g(x) \sum c_nx^n= (\sum b_n x^n)(\sum c_n x^n)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Your only other alternative is to use polynomial division on a few terms ______________________ a_0 + a_1x + a_2x^2+... 1

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0will it be easier if i integrate it and find the power series , and then take the derivative of the power series

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you could try the integration option

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i don't know how much that will help, but it may.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nvm..it doesnt help much

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i couldnt figure out a pattern for the coefficents

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'll have a look later. I'm pretty tired.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okie dokey i am still trying~~

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Which one is it for  the (x..)^(1/3) one or the one after it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0To find the Maclaurin series for f(x) = x/(8 + 4 x^2)^(1/3) Need to find the radius of convergence~~

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0he must be a spy from other famous people here.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol those that have hella fans like u

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is he retired or something...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but he never answers my questions ...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, do you post questions outside of this link?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea.....when i have an idea of solving a problem and wants someone to confirm it..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So you didn't go home for easter?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no..too much work...to take care of.. if i go home..i wont get any work done

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I gave you a medal for morale ;p

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL yay my first medal in two days~~thanks haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how u got your gold shields

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Pfft, randomly. You can do things as random as enter something into chat and you get a medal for talking.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, yes. see if you can go to this link and give me a medal: http://openstudy.com/groups/mathematics#/updates/4db387a276e08b0bf9c58d28

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cool. maybe i can get medals for everything i've done... .

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL or u can just ask questions and answer to it,,and i will give u medals

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0pfft, this is so easy to rig.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i like the old way better,, where u can become a fan ~~~

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can still fan...you just have to move your pointer over a person's icon and fan them there.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Or do it in the group menu.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh== lol changes are not always good..

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol i just noticed that your icon changed again

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Loki u figured this out yet.?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I haven't looked at it yet  sorry :( When's it due? I need a shower.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no problem, buddy, it is not due anytime soon ,i just want to get it over with haha,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ah okay...plenty of time for shower...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL plenty of time for shower?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I need to go get some food. I haven't eaten anything yet.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Wow, you're still here.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Isn't it after midnight there?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You should get an icon. The site creators say people with one get more questions answered.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOl yea i am going to bed very soon..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah...I'll look at your question either tonight or tomorrow. I have some things due myself :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh just do it whenever u can and want :) are you off today?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes/no...technically off, but have to finish things.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol same here.i been sleping more than enough ,,but still feel tired

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you work hard, though

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0mind work... only lol i dont really exercise much on weekdays

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how are you going to spend the rest of your day

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0some of my friends run and weightlift everyday...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well, it's nearly 6pm, so probably do some work

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, i've got friends like that. it's boring. i wish i could get into it.

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do classes start back this week?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0don't you get time off for easter?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i would have gone home if i did

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0don't know yet  it's just started...well, not really, just means I get to sleep and do work at home instead :( Meh.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Anyhow...I might log off soon.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh haha you should hang out with your friends ~~,,dont get nerdy at home haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I went to brush my teeth and get ready to sleep... i am going to bed now..nightzz and have fun :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have an answer if you still need it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I only had a chance to look at it now.

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let me know if that stuff doesn't come through properly :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea i got the same thing:) thanks anyways!!!

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hey, I don't get that. I get\[\cos \theta = \frac{1}{\sqrt{2626}} \rightarrow \theta \approx 88.88^o\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh yea you are right. 20% grade means rise over run = 1/5 right, that was what confused me

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks man. How was your break?
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