Maclaurin Series: Find e^(3x) + e(-3x)

[physstudent1]

Homework Statement

find the maclaurin series of e^(3x) + e(-3x)

Homework Equations

The Attempt at a Solution

I'm not sure about finding taylor and maclaurin series, I understand perfectly how to find the terms of the series...But how do I put it into a general term do I just have to recognize the pattern by looking at it? anyway...

to attempt this problem I was thinking you use the known maclaurin series for e^x which has a general term of x^n / n! it will be (3x)^n / n! for the first term and (-3x)^n / n! for the 2nd term I'm unsure of combining these The answer is 2*3^2n*x^2n/(2n)! What I do not understand is how they got (2n)! I thought it should be 2(n!) and also I thought there should be a negative in the numerator...can anyone explain to me these parts I do not understand...my textbook has one small paragraph on these series and it does not explain much...

 

[Pere Callahan]

Hi,

what happens to the terms in your two sums with n even, what to the terms with n odd?

 

[olgranpappy]

the terms with odd 'n' cancel each other, so the sum is only over terms n=0,2,4,6,... etc

then you can rewrite that as a sum over m=0,1,2,3,... etc
then rename m->n

[edit: whoops sorry about stealing your thunder, Pere. You posted too quick for me.]

 

[physstudent1]

ahh I see thank you, I'm wondering though this series looks a little strange if there was one like this that I didn't have a known series for I don't think I would have ever found the general term by just looking at it...and also why does the denominator become (2n)! i thought it would be 2(n!)

 

[Pere Callahan]

I don't see how you could get a 2(n!) in the denominator. How did you get it?

As olgranpappy said, all the odd-n terms cancel. So the n=1 term of the first sum cancels with the n=1 term of the second sum and so on. What happens to the terms with n even..? Show some more work.

 

[physstudent1]

well actually I don't think its 2(n!) anymore either I got the general term by adding the two modified E^x general terms together which would still have just a single n! in the denominator..the even terms get multiplied by 2 but that is why I thought it would be 2*(n!) instead of (2n)!

 

[Pere Callahan]

True, you get the n.th general term by adding the n.th general terms of the modified E^x sums, as you put it.

The result of this "adding" is however zero for odd n.. do you agree?

So in your final sum, all terms for n odd are zero ...so practically you're only summing over even integers n ...so you can just as well replace n by 2m and some over all integers m ... think about it, write down the first 20 terms or so, it's not hard

 

[physstudent1]

I agree that the odd terms are zero because they cancel; I think I am thinking of this (2n)! in the wrong way...so if it was say the 2nd term it would be (2!) correct? why wouldn't it be (4!) because (2*2)! = 4! ? that's where I'm confused...I understand that only the even terms have value so we're basically summing the even terms

EDIT: I think I am finally getting it because the odd terms drop out then what was once the 3rd term is basically now the 2nd term so then the bottom is (2*1)! which is why it works like this thanks a lot

 

[physstudent1]

I think I finally get it because the odd terms drop out what was once the 3rd term is now basically the 2nd term since so now the 2nd term will have (2*1)! and the third term will have (2*2)! because the third term will be the term where n=4

 

[djeitnstine]

I don't know if this clarifies anything for you but (2*n)! simple means the number is even and likewise (2n+1)! means its odd.