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\[\sum_{1}^{\infty}(4^nx^{2n})/n\]

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~

haha morning

can you check the endpoints for me?

ok

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).

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 :)

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

*eating

Book's wrong :)

Wolfram Alpha agrees

LOL my professor just told us the book is never wrong =-=

cuz he wrote part of the textbook we are using

how embarrassing :P

we all make mistakes :)

It says inconclusive for both tests ..what does that mean?

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.

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

Okay, so your question...I will scrounge a proof for you.

COOL buddy, don't waste your time and energy on aXX haha

Is this what you're looking for?

sort of

Exactly

oh i feel you ^^ thank you ~~~~~

This is crazy lol

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~~

one sec

okie okie

inertia

sorry, ordered chinese food and guy was at door

first law of motion

oh yea chinese food haha~

what's the question?

yeah, friction should oppose motion

the box slides the opposite way of the acceleration becasue of the first law.. right?

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

it is good enough man :)

alright more questions..u mind?

lol, are these due Saturday or something?

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?

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}

got it.. factorial is the thing that scares me...

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...

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?

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 :)

Nevermind loki haha i got it :)

there are youtube clips that show you what to do

i used long division ot find the answers

Can you help me out with this problem?

sure it is not due soon :) and i will keep trying~~ haha

okay...yes, keep trying...

i;m out.

good luck~!

Actually, I think this version's better:\[f_n=\frac{\phi^n-(1-\phi)^n}{\sqrt{5}}\]

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

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

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

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..

LOL do you mind explain a little?

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

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

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?

Yeah, that's what I got too :)

cool!! Let me give u a medal haha

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

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

\[1-\frac{1}{2!}+\frac{1}{4!}-\frac{1}{6!}+\frac{1}{8!}-\frac{1}{10!}+...\]

oh i see what you mean cuz it is a convergent series

Yes.

can i*

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..

One second. I'm looking at something else.

Okie

i should compute the P4 and subtract ln(sec(-0.2) from it?

ummm i dont really know..this is an online homework , the problems are given by a third party

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?

Yeah. If you're studying to become an engineer, numerical methods are what you'll be using.

numerical methods = ...mesure and calculator~~?

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

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.

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

oh=-= nvm...

Yeah, thought so.

lol not there yet

Yeah :)

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~~

he must be a spy from other famous people here.

famous?

lol those that have hella fans like u

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

*shield

LOL ..how lame

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..

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.

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?

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.

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.

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?