Find Maclaurin Series for f(x) = (e^x - cos(x))/x

[physstudent1]

Homework Statement

find the Maclaurin Series:

f(x)=(e^x - cos(x))/x

Homework Equations

The Attempt at a Solution

I'm not really sure what to do I do know the Maclaurin series of cos(x) I was thinking that using this known series then doing e^x - the known series then dividinig the entire thing by x. I'm just not sure if this is allowed. Do you have to do the known series of e^x - the cos(x) series ? Someone please let me know.

 

[ice109]

you know i didn't expect that to work but it does. [series(exp(x))-series(cos(x))]/x is the right answer

 

[Gib Z]

Either your question wanted you to derive the series for Taylors theorem straight, or more likely not, show that each of those functions are analytic, as you can only do that trick for analytic functions. Most functions studied in analysis, however, are analytic =]

 

[HallsofIvy]

Homework Statement

find the Maclaurin Series:

f(x)=(e^x - cos(x))/x

Homework Equations

The Attempt at a Solution

I'm not really sure what to do I do know the Maclaurin series of cos(x) I was thinking that using this known series then doing e^x - the known series then dividinig the entire thing by x. I'm just not sure if this is allowed. Do you have to do the known series of e^x - the cos(x) series ? Someone please let me know.

Are you saying you don't know the Maclaurin series for e^x? Surely not! The crucial point here is that the first (constant) term of the Maclaurin series for both e^x and cos(x) is 1- subtracting them cancels the constant term so that you can then divide by x and still have a Maclaurin series left!

 

[physstudent1]

I do know the maclaurin series for e^x and for cos(x) :) oh yes you are right; but I do not know if it is acceptable to leave my answer in a form with just like the terms of the series or if it is standard to put the answer into a summation. I can't really figure out the pattern:( because the x^4 terms cancel out and then I am left with 0+ 1 + x + x^2/3! + x^5/^6! so what I did and I do not know if this is correct is subtracted the general terms and then divided by x to get: (x^(n-1))/n!)-((-1)^n*(x^2n-1))/(2n)!

 

[ice109]

Either your question wanted you to derive the series for Taylors theorem straight, or more likely not, show that each of those functions are analytic, as you can only do that trick for analytic functions. Most functions studied in analysis, however, are analytic =]

1/x isn't analytic

 

[physstudent1]

well there's an example in the book with (1-cos(x^2))/x and they did it the same way. I don't really know what an "analytic function is" but I am pretty sure you can do it by subtracting the two series I just don't know if I subtract the general terms or should subtract like the first 5 terms from each series? Because when I subtracted the first 5 terms from each I was left with something kind of odd that I couldn't see a pattern into put into a summation form.

the terms I end up getting are:

1 + x + (x^2 / 3!) + (x^5 / 6!) - (x^7/ 8!)

 

[HallsofIvy]

1/x isn't analytic

No, it isn't. However, since all of the functions in this problem are I don't see how that is relevant!

 

[HallsofIvy]

You know that the general term of the Maclaurin series for e^x is 1/n! xⁿ and that the general term of the Maclaurin series for cos(x) is (-1)ⁿ/(2n)!x²ⁿ or: "0 if n is odd (-1)^n/2/n! xⁿ if n is even". The general term of the Maclaurin series for e^x- cos(x) is "(1/n!)xⁿ if n is odd, [1/n!- (-1)^n/2/(2n+1)!]xⁿ if n is even". Notice that when n is 0 (even), that is 1- 1= 0. Dividing each term by x gives "(1/n!)x^n-1 if n is odd, [1/n!- (-1)^n/2/(2n+1)!]x^n-1 if n is even"

 

[physstudent1]

ah yes I understand what you are saying here would it be acceptable to write it like that though ? I do not understnad how you got (2n +1) however how did it go from (2n)! to (2n+1)! Since the cos part will be 0 for odd terms can you just say the series is the series you wrote for the even terms since when its odd the cos part will drop out anyway and will give you basically the odd series?

 

[ice109]

No, it isn't. However, since all of the functions in this problem are I don't see how that is relevant!

how is the function in question analytic if it is (expx - cosx)*1/x? are you saying simply because it is (expx - cosx ) divided by x f(x) is analytic?

 

[Hootenanny]

how is the function in question analytic if it is (expx - cosx)*1/x? are you saying simply because it is (expx - cosx ) divided by x f(x) is analytic?

What I believe Halls is saying is that if each of the individual functions are analytic, then one can find a series representation of the entire function, whether the entire function is analytic or not.

 

[physstudent1]

well this is definitely the way to do it we went over it today in class I asked for help as I was still a little confused but thanks guys I sort of get it now? lol i know that you need two summations one for the even and one for the odd getting to them is a little weird

 

[ice109]

What I believe Halls is saying is that if each of the individual functions are analytic, then one can find a series representation of the entire function, whether the entire function is analytic or not.

no i understand that but my question still stands.

 

[Dick]

no i understand that but my question still stands.

I'm not sure what the question is, but (exp(x)-cos(x))/x only has a removable singularity at x=0. Once you plug that whole the resulting function is analytic.