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 2 years ago
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?
 2 years ago
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?

This Question is Closed

telliott99
 2 years ago
Best ResponseYou've already chosen the best response.1Well, could you start with the infinite series for cosine x?

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

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.0\[ \int \frac{\cos x}{x}dx  \int \frac{1}{x}dx \]

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

agentx5
 2 years ago
Best ResponseYou've already chosen the best response.0That's the MacLaurin equivalent for cosine

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

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.0Is it? Did I look at the wrong one? (:

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.0oh yes that's the hyperbolic one! SOrry!

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

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.0lets both check again then hehe.

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.0yes I believe I have the right one

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

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

agentx5
 2 years ago
Best ResponseYou've already chosen the best response.0Antiderivative power rule?

Spacelimbus
 2 years ago
Best 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 \[ \frac{\cos x}{x} = \sum_{n=0}^\infty \frac{(1)^nx^{2n1}}{(2n)!} \]

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.0yes that's it @agentx5

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.0where @mahmit2012 ?

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.3all !but it is time to pray i wll come back and show you

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

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1lol ... where did i make mistake??

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.0Did we do the same @experimentX ? besides that you completed it already.

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.0yes but obviously "ALL" is wrong :D

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.3first 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.

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.3dw:1343670857318:dw

agentx5
 2 years ago
Best ResponseYou've already chosen the best response.0*tilting head sideways to read* :D

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.3dw:1343670990609:dw

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.3dw:1343671203407:dw

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

agentx5
 2 years ago
Best ResponseYou've already chosen the best response.0Well the final answer is probably going to be lnx + something, I think...

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

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1can't we cancel 1 and start from n=1 instead?

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.3yes but if you start with 1cosx in num so you wont confront the mistake.

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.3take a look my solution and figure out the mistake.

Spacelimbus
 2 years ago
Best 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.

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.0Better: Substitute it directly and derive the formula in form of a series representation?

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1yeah ... my case (1)^n was mistake ..

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.3and for the last this problem is originally has defined to 0 to inf.

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.3dw:1343671601188:dw

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.0I understand, thanks @mahmit2012

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1perhaps this is related to http://en.wikipedia.org/wiki/Dirichlet_integral

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.3Not exactly but a little bit !

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

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

agentx5
 2 years ago
Best ResponseYou've already chosen the best response.0actually that needs to be n \(\pm\) 1on the alternating part

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

agentx5
 2 years ago
Best ResponseYou've already chosen the best response.0@experimentX you mean shifting the index eh?

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1since the one gets cancelled ... the after that x is divided.

agentx5
 2 years ago
Best ResponseYou've already chosen the best response.0Odd 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

agentx5
 2 years ago
Best ResponseYou've already chosen the best response.0Then 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)

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

agentx5
 2 years ago
Best ResponseYou've already chosen the best response.0The 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

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1i 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.

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1this gives the value of dw:1343673440350:dw

agentx5
 2 years ago
Best ResponseYou've already chosen the best response.0I've never even heard of Laplace Transform until now :3

telliott99
 2 years ago
Best ResponseYou've already chosen the best response.1Can 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?

agentx5
 2 years ago
Best ResponseYou've already chosen the best response.0I 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)

telliott99
 2 years ago
Best ResponseYou've already chosen the best response.1OK. 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.

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