anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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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
Well, could you start with the infinite series for cosine x?
anonymous
  • anonymous
You'd have (1/x) (-x^2/2! + x^4/4! .....)
anonymous
  • anonymous
\[ \int \frac{\cos x}{x}dx - \int \frac{1}{x}dx \]

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anonymous
  • anonymous
Let's see... \[\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}, \text{for all x = } \mathbb{R}\]
anonymous
  • anonymous
That's the MacLaurin equivalent for cosine
anonymous
  • anonymous
\[ \frac{ \cos x}{x} = \frac{1}{x} \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!} \]
anonymous
  • anonymous
Is it? Did I look at the wrong one? (-:
anonymous
  • anonymous
oh yes that's the hyperbolic one! SOrry!
anonymous
  • anonymous
Oh no wait I have the wrong one, I wrote sine by mistake
anonymous
  • anonymous
Arg soo confusing lol
anonymous
  • anonymous
lets both check again then hehe.
anonymous
  • anonymous
yes I believe I have the right one
anonymous
  • anonymous
I thought you couldn't just do 1/x , you'd have to convert that to a series as well, yes?
experimentX
  • experimentX
\[ \cos x = \sum_{n = 0}^\infty {(-1)^nx^{2n}\over (2n)!}\] which makes \[ \int \sum_{n = 1}^\infty {(-1)^nx^{2n-1}\over (2n)!} dx = {(-1)^nx^{2n}\over 2n(2n)!}\]
experimentX
  • experimentX
+c
anonymous
  • anonymous
Anti-derivative power rule?
anonymous
  • anonymous
\[ \frac{ \cos x}{x} = \frac{1}{x} \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!} \] Distribute it out \[ \frac{\cos x}{x} = \sum_{n=0}^\infty \frac{(-1)^nx^{2n-1}}{(2n)!} \]
anonymous
  • anonymous
yes that's it @agentx5
anonymous
  • anonymous
all has mistake!
anonymous
  • anonymous
where @mahmit2012 ?
anonymous
  • anonymous
all !but it is time to pray i wll come back and show you
experimentX
  • experimentX
\[ \int \sum_{n = 1}^\infty {(-1)^nx^{2n-1}\over (2n)!} dx = \sum_{n=1}^\infty {(-1)^nx^{2n}\over 2n(2n)!} \]
anonymous
  • anonymous
after pray !
experimentX
  • experimentX
lol ... where did i make mistake??
anonymous
  • anonymous
Did we do the same @experimentX ? besides that you completed it already.
anonymous
  • anonymous
I think we did.
experimentX
  • experimentX
n=1, n=0
anonymous
  • anonymous
yes but obviously "ALL" is wrong :D
anonymous
  • anonymous
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.
anonymous
  • anonymous
|dw:1343670857318:dw|
anonymous
  • anonymous
*tilting head sideways to read* :-D
anonymous
  • anonymous
|dw:1343670990609:dw|
anonymous
  • anonymous
|dw:1343671203407:dw|
anonymous
  • anonymous
In that first step you just used the cosine equivalent from MacLaurin right? Then you're dividing that polynomial by x, then... Pattern?
anonymous
  • anonymous
Well the final answer is probably going to be ln|x| + something, I think...
anonymous
  • anonymous
yes but you didn't allow to separate 1 and cos because of in cosx/x you have a log function.
experimentX
  • experimentX
can't we cancel 1 and start from n=1 instead?
anonymous
  • anonymous
yes but if you start with 1-cosx in num so you wont confront the mistake.
anonymous
  • anonymous
take a look my solution and figure out the mistake.
anonymous
  • anonymous
@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.
anonymous
  • anonymous
Better: Substitute it directly and derive the formula in form of a series representation?
experimentX
  • experimentX
yeah ... my case (-1)^n was mistake ..
anonymous
  • anonymous
and for the last this problem is originally has defined to 0 to inf.
anonymous
  • anonymous
|dw:1343671601188:dw|
anonymous
  • anonymous
I understand, thanks @mahmit2012
experimentX
  • experimentX
perhaps this is related to http://en.wikipedia.org/wiki/Dirichlet_integral
anonymous
  • anonymous
Not exactly but a little bit !
anonymous
  • anonymous
*
anonymous
  • anonymous
Um, how did you do this step? \[\frac{x^2}{2\cdot 2!}-\frac{x^4}{4\cdot 4!}+\frac{x^4}{6\cdot 6!}-... = \sum_{n=1}^{\infty} (-1)^n \cdot \frac{1}{2n(2n!)} \cdot x^{2n}\]
experimentX
  • experimentX
in the series of cosine ... there is 1 ... cancel out 1 ... it's simply ... shifting n=0 to n=1
anonymous
  • anonymous
actually that needs to be n \(\pm\) 1on the alternating part
experimentX
  • experimentX
and note that you are making same mistake as I previously did, (-1)^n+1 or n-1 ... note that this also shifts.
anonymous
  • anonymous
@experimentX you mean shifting the index eh?
experimentX
  • experimentX
yep
experimentX
  • experimentX
since the one gets cancelled ... the after that x is divided.
anonymous
  • anonymous
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
anonymous
  • anonymous
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)
anonymous
  • anonymous
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
anonymous
  • anonymous
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
experimentX
  • experimentX
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.
experimentX
  • experimentX
this gives the value of |dw:1343673440350:dw|
anonymous
  • anonymous
I've never even heard of Laplace Transform until now :3
anonymous
  • anonymous
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 Transform---is 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?
anonymous
  • anonymous
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)
anonymous
  • anonymous
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.
anonymous
  • anonymous
I haven't learned either of those yet, the section from a chapter of the text on infinite series.

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