Find the sereis's radius and interval of convergence. For waht values of x does the series converge absolutely or conditionally?

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Find the sereis's radius and interval of convergence. For waht values of x does the series converge absolutely or conditionally?

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

\[\sum_{1}^{\infty}(4^nx^{2n})/n\]
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...
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,\[4x^2 \lt 1 \rightarrow -\frac{1}{2} \lt x \lt \frac{1}{2}\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question