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agentx5
Group Title
Topic: \(\textbf{Calculus 2}\)
Evaluate the indefinite integral as an infinite series.
\[\int \frac{\cos x − 1}{x} \ \ \ dx \]
What the heck? I mean putting these in Taylor series forms are hard enough, but this? Not even sure where to start here >_< It worries me because the final is coming up and this is a hole in my knowledge... So, um, help?
 one year ago
 one year ago
agentx5 Group Title
Topic: \(\textbf{Calculus 2}\) Evaluate the indefinite integral as an infinite series. \[\int \frac{\cos x − 1}{x} \ \ \ dx \] What the heck? I mean putting these in Taylor series forms are hard enough, but this? Not even sure where to start here >_< It worries me because the final is coming up and this is a hole in my knowledge... So, um, help?
 one year ago
 one year ago

This Question is Closed

telliott99 Group TitleBest ResponseYou've already chosen the best response.1
Well, could you start with the infinite series for cosine x?
 one year ago

telliott99 Group TitleBest ResponseYou've already chosen the best response.1
You'd have (1/x) (x^2/2! + x^4/4! .....)
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
\[ \int \frac{\cos x}{x}dx  \int \frac{1}{x}dx \]
 one year ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
Let's see... \[\sum_{n=0}^{\infty} (1)^n \frac{x^{2n+1}}{(2n+1)!}, \text{for all x = } \mathbb{R}\]
 one year ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
That's the MacLaurin equivalent for cosine
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
\[ \frac{ \cos x}{x} = \frac{1}{x} \sum_{n=0}^\infty (1)^n \frac{x^{2n}}{(2n)!} \]
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
Is it? Did I look at the wrong one? (:
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
oh yes that's the hyperbolic one! SOrry!
 one year ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
Oh no wait I have the wrong one, I wrote sine by mistake
 one year ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
Arg soo confusing lol
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
lets both check again then hehe.
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
yes I believe I have the right one
 one year ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
I thought you couldn't just do 1/x , you'd have to convert that to a series as well, yes?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
\[ \cos x = \sum_{n = 0}^\infty {(1)^nx^{2n}\over (2n)!}\] which makes \[ \int \sum_{n = 1}^\infty {(1)^nx^{2n1}\over (2n)!} dx = {(1)^nx^{2n}\over 2n(2n)!}\]
 one year ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
Antiderivative power rule?
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
\[ \frac{ \cos x}{x} = \frac{1}{x} \sum_{n=0}^\infty (1)^n \frac{x^{2n}}{(2n)!} \] Distribute it out \[ \frac{\cos x}{x} = \sum_{n=0}^\infty \frac{(1)^nx^{2n1}}{(2n)!} \]
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
yes that's it @agentx5
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.3
all has mistake!
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
where @mahmit2012 ?
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.3
all !but it is time to pray i wll come back and show you
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
\[ \int \sum_{n = 1}^\infty {(1)^nx^{2n1}\over (2n)!} dx = \sum_{n=1}^\infty {(1)^nx^{2n}\over 2n(2n)!} \]
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.3
after pray !
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
lol ... where did i make mistake??
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
Did we do the same @experimentX ? besides that you completed it already.
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
I think we did.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
n=1, n=0
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
yes but obviously "ALL" is wrong :D
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.3
first of all the original question has interval [0,inf) and it has many different solution to get the answer. and if you are looking for undefined integral so just a matching series is the answer ! but with a log function.
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.3
dw:1343670857318:dw
 one year ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
*tilting head sideways to read* :D
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.3
dw:1343670990609:dw
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.3
dw:1343671203407:dw
 one year ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
In that first step you just used the cosine equivalent from MacLaurin right? Then you're dividing that polynomial by x, then... Pattern?
 one year ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
Well the final answer is probably going to be lnx + something, I think...
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.3
yes but you didn't allow to separate 1 and cos because of in cosx/x you have a log function.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
can't we cancel 1 and start from n=1 instead?
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.3
yes but if you start with 1cosx in num so you wont confront the mistake.
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.3
take a look my solution and figure out the mistake.
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
@mahmit2012, I think I understand the main problem, it was the 'splitting' up of terms right? Separating the 1 or cos(x) from the remaining term.
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
Better: Substitute it directly and derive the formula in form of a series representation?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
yeah ... my case (1)^n was mistake ..
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.3
and for the last this problem is originally has defined to 0 to inf.
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.3
dw:1343671601188:dw
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
I understand, thanks @mahmit2012
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
perhaps this is related to http://en.wikipedia.org/wiki/Dirichlet_integral
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.3
Not exactly but a little bit !
 one year ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
Um, how did you do this step? \[\frac{x^2}{2\cdot 2!}\frac{x^4}{4\cdot 4!}+\frac{x^4}{6\cdot 6!}... = \sum_{n=1}^{\infty} (1)^n \cdot \frac{1}{2n(2n!)} \cdot x^{2n}\]
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
in the series of cosine ... there is 1 ... cancel out 1 ... it's simply ... shifting n=0 to n=1
 one year ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
actually that needs to be n \(\pm\) 1on the alternating part
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
and note that you are making same mistake as I previously did, (1)^n+1 or n1 ... note that this also shifts.
 one year ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
@experimentX you mean shifting the index eh?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
since the one gets cancelled ... the after that x is divided.
 one year ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
Odd though, because it's showing @mahmit2012's answer, put into the index (by increasing all n's by n+2) they want, as incorrect. They have the sigma upper and lower bounds fixed at n=2 to n=\(\infty\) ________ + C
 one year ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
Then of course I'll need to use a calculator to find a certain number of significant figures blah blah but it's this part that's costing me 15 points on this homework (everything else is correct but this one problem, which of course would have to be the most points)
 one year ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
I read that link you posted @experimentX , I didn't quite understand it. New concepts I've not encountered before so a little overwhelming lol
 one year ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
The most I got out of it was that some things like that (sin(x))/x function are equal to 90 degrees, or \(\pi\)/2 radians
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
i don't understand it too well ... it's the second time i'm seeing the the definite integral with Laplace transform giving it's value.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
this gives the value of dw:1343673440350:dw
 one year ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
I've never even heard of Laplace Transform until now :3
 one year ago

telliott99 Group TitleBest ResponseYou've already chosen the best response.1
Can one of you guys summarize this for me, because I'm not getting it? I understand where the series went wrong I don't undertand Laplace Transformis that where I need to go? I didn't find @mahmit2012 answer, just some demos of incorrect proofs. Are you, in the end, converting integral ( (cos x 1)/x) dx into integral (sin x / x) dx How can that be?
 one year ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
I agree with @telliott99 , I'm confused and don't understand what's going on with that, or how it answers my original question (or if it even does)
 one year ago

telliott99 Group TitleBest ResponseYou've already chosen the best response.1
OK. I think the thing to do is to learn the Laplace Transform. I have these notes: http://tutorial.math.lamar.edu/Classes/DE/LaplaceDefinition.aspx But it's going to take a while. I know calculus pretty well and some LA but no DEs.
 one year ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
I haven't learned either of those yet, the section from a chapter of the text on infinite series.
 one year ago
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