anonymous
  • anonymous
Find the sereis's radius and interval of convergence. For waht values of x does the series converge absolutely or conditionally?
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\sum_{1}^{\infty}(4^nx^{2n})/n\]
dumbcow
  • dumbcow
well 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
  • anonymous
By 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,\[4x^2 \lt 1 \rightarrow -\frac{1}{2} \lt x \lt \frac{1}{2}\]

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anonymous
  • anonymous
The radius of convergence is, by definition, all x that satisfy \[|x| < R\] where R is a non-negative, real number. \[|4x^2|=4|x^2|=4|x|^2<1 \rightarrow |x|<\frac{1}{2}\]
anonymous
  • anonymous
Since |x| < 1/2, the series is absolutely convergent for -1/2 < x < 1/2
anonymous
  • anonymous
Every absolutely convergent series is unconditionally convergent.
anonymous
  • anonymous
How about at end points? x=-1/2 and 1/2
anonymous
  • anonymous
I checked the end points, they both diverges, but the textbooks says it absolutely convergent for -1/2
anonymous
  • anonymous
Good morning sir~
anonymous
  • anonymous
Yeah...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
  • anonymous
haha morning
anonymous
  • anonymous
can you check the endpoints for me?
anonymous
  • anonymous
ok
anonymous
  • anonymous
When 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
  • anonymous
isn't it should be the same because x^(2n)
anonymous
  • anonymous
shouldn't it be *
anonymous
  • anonymous
Yeah, when x=-1/2, you get an alternating harmonic which I know to be equal to log(2) (natural log).
anonymous
  • anonymous
\[\frac{4^n \left( \frac{1}{2} \right)^{2n}}{n}=\frac{(2^2)^n \frac{1}{2^{2n}}}{n}=\frac{2^{2n}\frac{1}{2^{2n}}}{n}=\frac{1}{n}\]
anonymous
  • anonymous
yea i got this how about x=-1/2
anonymous
  • anonymous
Well, when you sub. x=-1/2, you get the same result as above, except for a (-1)^n outside the 1/n.
anonymous
  • anonymous
never mind i know where ive made a mistkae :)
anonymous
  • anonymous
This is the 'alternating harmonic series' which converges to ln(2). You can test convergence with the alternating series test.
anonymous
  • anonymous
so 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
  • anonymous
Actually, (-1/2)^(2n) is always positive (I'm eating while doing this so not completely concentrating :( )...so I need to have another look.
anonymous
  • anonymous
LOL alright (Lucas with -.-) i got to go to class now. see you soon haha
anonymous
  • anonymous
ok
anonymous
  • anonymous
Have u had another look?
anonymous
  • anonymous
No...sorry, I'm only back on because someone sent an e-mail. Is this something due now?
anonymous
  • anonymous
no i am just curious haha just get back to me whenever you can :)
anonymous
  • anonymous
If you have stuff to do, it is fine :]
anonymous
  • anonymous
one sec
anonymous
  • anonymous
It shouldn't be converging for either +1/2 or -1/2
anonymous
  • anonymous
Maybe 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
  • anonymous
*eating
anonymous
  • anonymous
oh, yea i tried..10 times yesterday,,,i still got the same answer,, but the book says it conditionally diverges at x=-1/2..
anonymous
  • anonymous
Book's wrong :)
anonymous
  • anonymous
Wolfram Alpha agrees
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=sum+from+n%3D1+to+infinity+of+%284^n*%28-1%2F2%29^%282n%29%29%2F%28n%29
anonymous
  • anonymous
LOL my professor just told us the book is never wrong =-=
anonymous
  • anonymous
cuz he wrote part of the textbook we are using
anonymous
  • anonymous
haha, well, my professors wrote some books too and I have entire errata sheets handed out from them (i.e. corrections to their printed stuff-ups).
anonymous
  • anonymous
how embarrassing :P
anonymous
  • anonymous
we all make mistakes :)
anonymous
  • anonymous
It says inconclusive for both tests ..what does that mean?
anonymous
  • anonymous
It 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
  • anonymous
If 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
  • anonymous
Alright, so the book is WRONG..
anonymous
  • anonymous
I'm thinking...and if it's not, I'd like to know why.
anonymous
  • anonymous
Try a Cauchy Condensation test or something. Or alternating series test.
anonymous
  • anonymous
it is not an alternating series even if x is negative.. and i haven't learned cauchy condensation test :
anonymous
  • anonymous
Oh yea do you have time to explain the rearranging theorem ?
anonymous
  • anonymous
Yeah, exactly, that's my point. I don't think you can throw anything at this to prove convergence.
anonymous
  • anonymous
hmmm, probably not :(
anonymous
  • anonymous
I can later.
anonymous
  • anonymous
okiedokie
anonymous
  • anonymous
What's your question? At least I can think about it.
anonymous
  • anonymous
umm for a conditionally convergent series, you can rearrange the terms to get any sums you like?
anonymous
  • anonymous
the order of the terms
anonymous
  • anonymous
Sorry, I had to go do that. Some people on here are really starting to pellet me.
anonymous
  • anonymous
LOL i saw it ..that dude is retarded
anonymous
  • anonymous
So...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
  • anonymous
Okay, so your question...I will scrounge a proof for you.
anonymous
  • anonymous
COOL buddy, don't waste your time and energy on aXX haha
anonymous
  • anonymous
The 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
  • anonymous
It 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
  • anonymous
Because when you get your 'nth' partial sum and send n to infinity, you end up with different limits.
anonymous
  • anonymous
Is this what you're looking for?
anonymous
  • anonymous
sort of
anonymous
  • anonymous
The alternating series is a conditionally-convergent 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
  • anonymous
do 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
  • anonymous
Exactly
anonymous
  • anonymous
oh i feel you ^^ thank you ~~~~~
anonymous
  • anonymous
I'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
  • anonymous
wait how about asolutely convergent series, if we rearrange the oders of terms, the pattern of the series also changes right?
anonymous
  • anonymous
All 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
  • anonymous
anonymous
  • anonymous
This is crazy lol
anonymous
  • anonymous
Has your professor shown you why an infinite series of positive terms doesn't change it's convergence (sum) upon rearrangement?
anonymous
  • anonymous
nope he didn't get into any details, he just stated the theorem and said this is not in the scope of this course..
anonymous
  • anonymous
oh...
anonymous
  • anonymous
The proof is on that sheet too.
anonymous
  • anonymous
yea i am looking at it
anonymous
  • anonymous
ok...i'll leave it with you. I have to go do some work. Have fun with it.
anonymous
  • anonymous
thanks !!! have fun with your work too
anonymous
  • anonymous
bye :)
anonymous
  • anonymous
c ya:)
anonymous
  • anonymous
LUCAS~~~ I have a question regarding PHYSICS~~
anonymous
  • anonymous
k
anonymous
  • anonymous
a 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
  • anonymous
one sec
anonymous
  • anonymous
okie okie
anonymous
  • anonymous
inertia
anonymous
  • anonymous
sorry, ordered chinese food and guy was at door
anonymous
  • anonymous
first law of motion
anonymous
  • anonymous
oh yea chinese food haha~
anonymous
  • anonymous
A 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
  • anonymous
what's the question?
anonymous
  • anonymous
umm 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
  • anonymous
yeah, friction should oppose motion
anonymous
  • anonymous
the box slides the opposite way of the acceleration becasue of the first law.. right?
anonymous
  • anonymous
Well, 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
  • anonymous
I can solve this for you, but I need to go offline. Gotta go out. Is this due tomorrow or something?
anonymous
  • anonymous
nope just get back to me whenever u want :) 3rd law is also involved?
anonymous
  • anonymous
In terms of the normal force.
anonymous
  • anonymous
I'll get back to you.
anonymous
  • anonymous
kk thanks :) see you~
anonymous
  • anonymous
bye :)
anonymous
  • anonymous
LOKISAN!!!
anonymous
  • anonymous
hello
anonymous
  • anonymous
lol i like how your name sound so japanese
anonymous
  • anonymous
hehe, me too.
anonymous
  • anonymous
What did you get for your truck/case question in the end?
anonymous
  • anonymous
static friction = coeffient*normal force, and the direction is north
anonymous
  • anonymous
Yeah. 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
  • anonymous
it is good enough man :)
anonymous
  • anonymous
the case is moving because the fraction "sticks" it to the floor of the truck. otherwise, it will stay there, by the first law?
anonymous
  • anonymous
Yes. 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
  • anonymous
alright more questions..u mind?
anonymous
  • anonymous
\[\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n}e^n}{n!} \] \[\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n}n!}{n^n} \] \[\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n}n!}{e^n} \] \[\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n}n^n}{n!} \] 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
  • anonymous
lol, are these due Saturday or something?
anonymous
  • anonymous
oh 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
  • anonymous
So 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
  • anonymous
i 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
  • anonymous
yea tahts what i need help on
anonymous
  • anonymous
Yeah...kind of have to prove it's decreasing...I'll look at them in my break.
anonymous
  • anonymous
Alrighty~! thank you!
anonymous
  • anonymous
np
anonymous
  • anonymous
Did you figure this out?
anonymous
  • anonymous
You should look at your sequences like this:\[\frac{e^n}{n!}=\frac{e}{n}.\frac{e}{n-1}.\frac{e}{n-2}....\frac{e}{3}.\frac{e}{2}.\frac{e}{1}\]There are n copies of e and n terms in the n factorial. Taking the limit on both sides gives\[\lim_{n \rightarrow \infty}\frac{e^n}{n!}=\lim_{n \rightarrow \infty}\left( \frac{e}{n}.\frac{e}{n-1}.\frac{e}{n-2}....\frac{e}{3}.\frac{e}{2}.\frac{e}{1} \right)\]
anonymous
  • anonymous
The right-hand side can be expanded in limits by the multiplicative limit law:\[\lim _{n \rightarrow \infty} \left( \frac{e}{n}.\frac{e}{n-1}.\frac{e}{n-2}....\frac{e}{3}.\frac{e}{2}.\frac{e}{1} \right)\]\[\lim_{n \rightarrow \infty}\frac{e}{n}.\lim_{n \rightarrow \infty}\frac{e}{n-1}.\lim_{n \rightarrow \infty}\frac{e}{n-2}....\lim_{n \rightarrow \infty}\frac{e}{3}.\lim_{n \rightarrow \infty}\frac{e}{2}.\lim_{n \rightarrow \infty}\frac{e}{1} \]\[=0.0.0.....\frac{e}{3}.\frac{e}{2}.\frac{e}{1}\]\[=0\]
anonymous
  • anonymous
Now, to find for monotonic decreasing, you need to show \[a_{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}
anonymous
  • anonymous
The 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
  • anonymous
got it.. factorial is the thing that scares me...
anonymous
  • anonymous
The 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
  • anonymous
got it .thanks :)
anonymous
  • anonymous
ok
anonymous
  • anonymous
I think you should be right now :)
anonymous
  • anonymous
gotta go...good luck with it!
anonymous
  • anonymous
LOL yea that was a very good explaination haha thanks~!!!!
anonymous
  • anonymous
Hey loki do u have time to help me out with one more problem?
anonymous
  • anonymous
You can put it down but I have to go...have an insane amount of work to do for tomorrow.
anonymous
  • anonymous
oh alright !!
anonymous
  • anonymous
just punch it in...
anonymous
  • anonymous
\[1 + 4! x + \frac{8! x^2}{(2!)^4} + \frac{12! x^3}{(3!)^4} + \frac{16! x^4}{(4!)^4} + \frac{20! x^5}{(5!)^4}+\cdots\] Use the ratio test to find the radius of convergence of the power series attempt: ((4n-4)!x^(n-1))/((n-1)!)^4 this is the a_n i found for the series
anonymous
  • anonymous
The a_n is\[a_n=\frac{(4n)!x^n}{(n!)^4}\]
anonymous
  • anonymous
n = 0 to infinity.
anonymous
  • anonymous
Then you can apply the ratio test.
anonymous
  • anonymous
oh the constant is the 0th term ?
anonymous
  • anonymous
do you mind doing it so i can comapre my answer with you? i only have one attempt left :(
anonymous
  • anonymous
Yeah.
anonymous
  • anonymous
did you get 1/256?
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
I haven't finished it off.
anonymous
  • anonymous
Sorry for taking your precious time away !!!!
anonymous
  • anonymous
I've rushed through it. Assuming the bottom line is correct, you should be able to finish it off.
anonymous
  • anonymous
i got that too, as n->ininfty, all i care is n^4 terms right?
anonymous
  • anonymous
?
anonymous
  • anonymous
You have a quartic polynomial above and below.
anonymous
  • anonymous
so i multiply everything out and take the limit?
anonymous
  • anonymous
I would multiply the numerator out.
anonymous
  • anonymous
256 n^4+640 n^3+560 n^2+200 n+24
anonymous
  • anonymous
or can i just mulitply (4n)^4 ?
anonymous
  • anonymous
from wolfram
anonymous
  • anonymous
denominator: n^4+4 n^3+6 n^2+4 n+1
anonymous
  • anonymous
factor out n^4 on top and bottom and take the limit.
anonymous
  • anonymous
It will be 256.
anonymous
  • anonymous
|256x|<1 R= 1/256
anonymous
  • anonymous
\[256|x|<1\]is required for convergence.
anonymous
  • anonymous
Yes
anonymous
  • anonymous
THANK YOU !!! And SORRY!!!
anonymous
  • anonymous
You need to check the end points.
anonymous
  • anonymous
oh i am just looking for the radius
anonymous
  • anonymous
OK
anonymous
  • anonymous
Gotta go.
anonymous
  • anonymous
i really appreciate your help!!!! WISH you the best!!!
anonymous
  • anonymous
bye :]
anonymous
  • anonymous
bye :)
anonymous
  • anonymous
LOki~~ 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
  • anonymous
Nevermind loki haha i got it :)
anonymous
  • anonymous
there are youtube clips that show you what to do
anonymous
  • anonymous
i used long division ot find the answers
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
Can you help me out with this problem?
anonymous
  • anonymous
Ooh, 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
  • anonymous
sure it is not due soon :) and i will keep trying~~ haha
anonymous
  • anonymous
okay...yes, keep trying...
anonymous
  • anonymous
i;m out.
anonymous
  • anonymous
good luck~!
anonymous
  • anonymous
When 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
  • anonymous
Actually, I think this version's better:\[f_n=\frac{\phi^n-(1-\phi)^n}{\sqrt{5}}\]
anonymous
  • anonymous
\[\frac{f_{n+1}}{f_n}=\frac{\frac{\phi^{n+1}-(1-\phi)^{n+1}}{\sqrt{5}}}{\frac{\phi^n-(1-\phi)^n}{\sqrt{5}}}=\frac{\phi^{n+1}-(1-\phi)^{n+1}}{\phi^n-(1-\phi)^n}\]\[=\frac{\phi^{n+1}}{\phi^n}.\frac{1-\left( \frac{1-\phi}{\phi} \right)^{n+1}}{1-\left( \frac{1-\phi}{\phi} \right)^{n}}=\phi \frac{1-\zeta^{n+1}}{1-\zeta^{n}}\]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
  • anonymous
The radius of convergence is then,\[\phi |x|<1 \rightarrow |x|<\frac{1}{\phi}\iff -\frac{1}{\phi}
anonymous
  • anonymous
Check through it - it's after midnight and I've had about five hours sleep in two days!
anonymous
  • anonymous
Hope you a good night !!
anonymous
  • anonymous
good night of sleep*
anonymous
  • anonymous
wow i would never figure this out... Since i haven;t learned binet's formula
anonymous
  • anonymous
Thank!! you!! like ______________________________________________________________________________________________________________________this much
anonymous
  • anonymous
Sorry but i ahve to bother you more!
1 Attachment
anonymous
  • anonymous
I 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
  • anonymous
dichalao, when's this due?
anonymous
  • anonymous
three days later~~
anonymous
  • anonymous
have you been waiting all this time?
anonymous
  • anonymous
nope,, i been doing physics homework, and checked back from time to time
anonymous
  • anonymous
k
anonymous
  • anonymous
This thing isn't conceptually hard. What's going to stuff you are mistakes with the algebra.
anonymous
  • anonymous
Have you tried plugging into wolfram alpha?
anonymous
  • anonymous
i did... but i got a nonsense answer...as i said
anonymous
  • anonymous
(2n+2)!! = (2n)!!(2n+2) is this right?
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
(4n+7)!!=(4n+7)(4n+5)(4n+3)!!
anonymous
  • anonymous
Yes
anonymous
  • anonymous
limit a_(n+1)/(a_n) = 43570(8x+2)/48 |43570(8x+2)/48|<1 |8x|=-1.998898
anonymous
  • anonymous
I'm going to do it now.
anonymous
  • anonymous
I'll need coffee...
anonymous
  • anonymous
POOR loki :(
anonymous
  • anonymous
are you at home right now?
anonymous
  • anonymous
yes
anonymous
  • anonymous
day off
anonymous
  • anonymous
no school today?
anonymous
  • anonymous
oh oh haha you should have slept in
anonymous
  • anonymous
but i should be doing work so that i don't have to spend another week staying up all night :(
anonymous
  • anonymous
I can't sleep in - never works. Too sunny.
anonymous
  • anonymous
LOL where is your certain?
anonymous
  • anonymous
I think my building is on the surface of the sun; nothing stops the light or the heat.
anonymous
  • anonymous
how about eyepatch~~
anonymous
  • anonymous
Yeah, but eye masks dry the skin under my eyes.
anonymous
  • anonymous
Just once a week ,, just apply lotion to it when you get up :P
anonymous
  • anonymous
lol
anonymous
  • anonymous
i 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
  • anonymous
k
anonymous
  • anonymous
you still there?
anonymous
  • anonymous
yeah
anonymous
  • anonymous
i found something, but i want to re-check.
anonymous
  • anonymous
what did you get?
anonymous
  • anonymous
\[|8x+2|<\frac{24}{21875}\]
anonymous
  • anonymous
yea...that is what i got i think
anonymous
  • anonymous
you will have negative number at the right side ...
anonymous
  • anonymous
?
anonymous
  • anonymous
if u subtract 2 from both side..u wil have negative number
anonymous
  • anonymous
Yes
anonymous
  • anonymous
divide 8 , then i got -0.25
anonymous
  • anonymous
i tried that answer before..the system indicated it is incorrect..
anonymous
  • anonymous
When 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
  • anonymous
LOL do you mind explain a little?
anonymous
  • anonymous
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anonymous
  • anonymous
There is a tiny radius of convergence.
anonymous
  • anonymous
oh i see it
anonymous
  • anonymous
so it centered at x=-0.25 ? taylor series?
anonymous
  • anonymous
No no...go back to high school when you were asked to find the region for x such that, y
anonymous
  • anonymous
Everything below 24/whatever is what you want (i.e. those x values that generate the function values below the 24/whatever line).
anonymous
  • anonymous
That's why I said I wanted to check something - I meant, I wanted to use the definition of absolute value.
anonymous
  • anonymous
so the raidus will just be half of the distance between the two points where they meet?
anonymous
  • anonymous
yeah
anonymous
  • anonymous
got it !! Thanks
anonymous
  • anonymous
is it possible to do this thing without graphing it?
anonymous
  • anonymous
yes, that's what i wanted to do.
anonymous
  • anonymous
ok thanks alot i will figure it out ~~!!!
anonymous
  • anonymous
don't use all your attempts till i check it out.
anonymous
  • anonymous
I gotta go - need lunch and take care of 'business' :D
anonymous
  • anonymous
I'll go through the problem again later.
anonymous
  • anonymous
arlgiht enjoy your day!!!
anonymous
  • anonymous
sorry for late response i was doing my physics hw
anonymous
  • anonymous
I tried ...i still couldn't get the right answer
anonymous
  • anonymous
OK. 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
  • anonymous
it is due before midnight, 21th
anonymous
  • anonymous
Hey Loki, i got the answer already :) Thank you for helping!!!!!!
anonymous
  • anonymous
I'm glad you got the answer 'cause I've had no time! :) What was wrong with it in the end?
anonymous
  • anonymous
Sorry 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
  • anonymous
Yeah, that's what I got too :)
anonymous
  • anonymous
cool!! Let me give u a medal haha
anonymous
  • anonymous
What 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
  • anonymous
Loki u got it?
anonymous
  • anonymous
Hang on
anonymous
  • anonymous
okie okie you got a new pro pic that doesnt look like you haha
anonymous
  • anonymous
I move with the times.
anonymous
  • anonymous
is that your emotion
anonymous
  • anonymous
why are you sad
anonymous
  • anonymous
hehe...I'm upset with the new site.
anonymous
  • anonymous
OH yea...i am so not used to it
anonymous
  • anonymous
Taylor'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
  • anonymous
I don't like all this medal and shield crap.
anonymous
  • anonymous
I have a headache, so am a little slow right now.
anonymous
  • anonymous
OH is this s bad time to ask you questions...
anonymous
  • anonymous
I should be alright.
anonymous
  • anonymous
ok what you mean by looking at how each term contributes to a place
anonymous
  • anonymous
\[1-\frac{1}{2!}+\frac{1}{4!}-\frac{1}{6!}+\frac{1}{8!}-\frac{1}{10!}+...\]
anonymous
  • anonymous
At some point, one of the terms and those that follow stop contributing to the fifth decimal place (else the sum wouldn't converge).
anonymous
  • anonymous
oh i see what you mean cuz it is a convergent series
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
The 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
  • anonymous
can i*
anonymous
  • anonymous
The 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
  • anonymous
You can if your intention is to expand in terms of sec(x)...but that's not what you're asked to do.
anonymous
  • anonymous
You have to apply the process to find the first five coefficients.
anonymous
  • anonymous
Use wolfram to work them out. They're a pain.
anonymous
  • anonymous
yep..they are..
anonymous
  • anonymous
Find 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
  • anonymous
One second. I'm looking at something else.
anonymous
  • anonymous
Okie
anonymous
  • anonymous
i should compute the P4 and subtract ln(sec(-0.2) from it?
anonymous
  • anonymous
The 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
  • anonymous
ummm i dont really know..this is an online homework , the problems are given by a third party
anonymous
  • anonymous
Oh, 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
  • anonymous
lol ok.....
anonymous
  • anonymous
first derivative is tan(x)
anonymous
  • anonymous
second is sec^2(x)
anonymous
  • anonymous
third is 2tan(x)sec^2(x)
anonymous
  • anonymous
oh i computed it on wolfram already thanks ~~
anonymous
  • anonymous
fourth is -2(cos(2x)-2)sec^4(x)
anonymous
  • anonymous
oh
anonymous
  • anonymous
do you mind briefly tell me about the troublesome option?
anonymous
  • anonymous
Well, 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
  • anonymous
oh 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
  • anonymous
Yeah. If you're studying to become an engineer, numerical methods are what you'll be using.
anonymous
  • anonymous
numerical methods = ...mesure and calculator~~?
anonymous
  • anonymous
Numerical methods is measurement and number-crunching. 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
  • anonymous
Oh..pain in the butt...
anonymous
  • anonymous
Actually, it's less of a pain to solve things numerically than algebraically. You'll get used to it.
anonymous
  • anonymous
To find the Maclaurin series for f(x) = x/(8 + 4 x^2)^(1/3) ...are there easier ways ...to do it
anonymous
  • anonymous
Sorry i got distracted a bit..
anonymous
  • anonymous
i haven't taken any engineering classes so far
anonymous
  • anonymous
Maclaurin 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
  • anonymous
but the =-= most troubleosome part is taking derivatives of such function.. omg wolfram haha
anonymous
  • anonymous
Yeah, there are no easier ways of taking derivatives.
anonymous
  • anonymous
is that possible to flip a power series,as in 1/ power series
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
can 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
  • anonymous
Hang on. I need to make a phone call.
anonymous
  • anonymous
brb going to take a quick shower first..
anonymous
  • anonymous
kkk
anonymous
  • anonymous
I think it would be more difficult.
anonymous
  • anonymous
if the power seiresi is sum n=0 -> infin, 4n(x)^n/n^2 ...if i slip it..what would it be?
anonymous
  • anonymous
slip it?
anonymous
  • anonymous
flip*..sorry
anonymous
  • anonymous
It'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
  • anonymous
oh=-= nvm...
anonymous
  • anonymous
Yeah, thought so.
anonymous
  • anonymous
Your only other alternative is to use polynomial division on a few terms ______________________ a_0 + a_1x + a_2x^2+... |1
anonymous
  • anonymous
But 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
  • anonymous
lol not there yet
anonymous
  • anonymous
Yeah :)
anonymous
  • anonymous
will it be easier if i integrate it and find the power series , and then take the derivative of the power series
anonymous
  • anonymous
are you off today?
anonymous
  • anonymous
it's easter
anonymous
  • anonymous
you could try the integration option
anonymous
  • anonymous
i don't know how much that will help, but it may.
anonymous
  • anonymous
nvm..it doesnt help much
anonymous
  • anonymous
i couldnt figure out a pattern for the coefficents
anonymous
  • anonymous
I'll have a look later. I'm pretty tired.
anonymous
  • anonymous
okie dokey i am still trying~~
anonymous
  • anonymous
Which one is it for - the (x..)^(1/3) one or the one after it?
anonymous
  • anonymous
Someone's spying.
anonymous
  • anonymous
To find the Maclaurin series for f(x) = x/(8 + 4 x^2)^(1/3) Need to find the radius of convergence~~
anonymous
  • anonymous
k
anonymous
  • anonymous
he must be a spy from other famous people here.
anonymous
  • anonymous
famous?
anonymous
  • anonymous
lol those that have hella fans like u
anonymous
  • anonymous
I'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
  • anonymous
is he retired or something...
anonymous
  • anonymous
but he never answers my questions ...
anonymous
  • anonymous
I'm not sure.
anonymous
  • anonymous
lol, do you post questions outside of this link?
anonymous
  • anonymous
yea.....when i have an idea of solving a problem and wants someone to confirm it..
anonymous
  • anonymous
want*
anonymous
  • anonymous
So you didn't go home for easter?
anonymous
  • anonymous
no..too much work...to take care of.. if i go home..i wont get any work done
anonymous
  • anonymous
I gave you a medal for morale ;p
anonymous
  • anonymous
LOL yay my first medal in two days~~thanks haha
anonymous
  • anonymous
how u got your gold shields
anonymous
  • anonymous
Pfft, randomly. You can do things as random as enter something into chat and you get a medal for talking.
anonymous
  • anonymous
*shield
anonymous
  • anonymous
LOL ..how lame
anonymous
  • anonymous
lol, yes. see if you can go to this link and give me a medal: http://openstudy.com/groups/mathematics#/updates/4db387a276e08b0bf9c58d28
anonymous
  • anonymous
yea
anonymous
  • anonymous
i just did lol
anonymous
  • anonymous
cool. maybe i can get medals for everything i've done... -.-
anonymous
  • anonymous
LOL or u can just ask questions and answer to it,,and i will give u medals
anonymous
  • anonymous
pfft, this is so easy to rig.
anonymous
  • anonymous
i like the old way better,, where u can become a fan ~~~
anonymous
  • anonymous
you can still fan...you just have to move your pointer over a person's icon and fan them there.
anonymous
  • anonymous
Or do it in the group menu.
anonymous
  • anonymous
oh=-= lol changes are not always good..
anonymous
  • anonymous
alright 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
  • anonymous
lol i just noticed that your icon changed again
anonymous
  • anonymous
see you
anonymous
  • anonymous
Loki u figured this out yet.?
anonymous
  • anonymous
I haven't looked at it yet - sorry :( When's it due? I need a shower.
anonymous
  • anonymous
no problem, buddy, it is not due anytime soon ,i just want to get it over with haha,
anonymous
  • anonymous
did u just wake up~~
anonymous
  • anonymous
ah okay...plenty of time for shower...
anonymous
  • anonymous
LOL plenty of time for shower?
anonymous
  • anonymous
I need to go get some food. I haven't eaten anything yet.
anonymous
  • anonymous
alright~!
anonymous
  • anonymous
Wow, you're still here.
anonymous
  • anonymous
yea haha~
anonymous
  • anonymous
Isn't it after midnight there?
anonymous
  • anonymous
You should get an icon. The site creators say people with one get more questions answered.
anonymous
  • anonymous
LOl yea i am going to bed very soon..
anonymous
  • anonymous
Yeah...I'll look at your question either tonight or tomorrow. I have some things due myself :)
anonymous
  • anonymous
oh just do it whenever u can and want :) are you off today?
anonymous
  • anonymous
Yes/no...technically off, but have to finish things.
anonymous
  • anonymous
oh
anonymous
  • anonymous
u feel better today?
anonymous
  • anonymous
yep
anonymous
  • anonymous
just tired.
anonymous
  • anonymous
lol same here.i been sleping more than enough ,,but still feel tired
anonymous
  • anonymous
you work hard, though
anonymous
  • anonymous
mind work... only lol i dont really exercise much on weekdays
anonymous
  • anonymous
meh, who does
anonymous
  • anonymous
how are you going to spend the rest of your day
anonymous
  • anonymous
some of my friends run and weightlift everyday...
anonymous
  • anonymous
well, it's nearly 6pm, so probably do some work
anonymous
  • anonymous
yeah, i've got friends like that. it's boring. i wish i could get into it.
anonymous
  • anonymous
yea.. 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
  • anonymous
Do classes start back this week?
anonymous
  • anonymous
what you mean?
anonymous
  • anonymous
don't you get time off for easter?
anonymous
  • anonymous
nope :(
anonymous
  • anonymous
boo
anonymous
  • anonymous
i would have gone home if i did
anonymous
  • anonymous
sad. we get a week.
anonymous
  • anonymous
yea... thumbs down~~
anonymous
  • anonymous
oh lol u had fun?
anonymous
  • anonymous
don't know yet - it's just started...well, not really, just means I get to sleep and do work at home instead :( Meh.
anonymous
  • anonymous
Anyhow...I might log off soon.
anonymous
  • anonymous
oh haha you should hang out with your friends ~~,,dont get nerdy at home haha
anonymous
  • anonymous
I went to brush my teeth and get ready to sleep... i am going to bed now..nightzz and have fun :)
anonymous
  • anonymous
night :)
anonymous
  • anonymous
I have an answer if you still need it.
anonymous
  • anonymous
I only had a chance to look at it now.
anonymous
  • anonymous
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anonymous
  • anonymous
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anonymous
  • anonymous
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anonymous
  • anonymous
^ 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
  • anonymous
Let me know if that stuff doesn't come through properly :)
anonymous
  • anonymous
yea i got the same thing:) thanks anyways!!!
anonymous
  • anonymous
Loki 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.
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anonymous
  • anonymous
Hey, I don't get that. I get\[\cos \theta = \frac{1}{\sqrt{2626}} \rightarrow \theta \approx 88.88^o\]
anonymous
  • anonymous
anonymous
  • anonymous
oh yea you are right. 20% grade means rise over run = 1/5 right, that was what confused me
anonymous
  • anonymous
Thanks man. How was your break?

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