# Multiplication of Maclaurin Series

1. Aug 16, 2008

### JeffNYC

I have the following problem:

find the first 3 non-zero terms in the Maclaurin series for the function:

e-x2 + Cos[x]

I know in this case, the series behave like polynomials and I have done the following. The left expression is the first 3 terms of the e portion of the problem, and the second expression is the first 3 terms of Cosx.

(1 - x2+ x4/2)(1 - x2/2 + x4/24)

this =

1 - x2/2 + x4/24 - x2 - x4/2 - x6/24 + x4 - x6/4 + x8/48

How do I know which terms are the "first 3 non-zero terms" of this series?

Thanks - the answer is attached, I just don't understand how the polynomial, after multiplied out is consolidated at the end.

Jeff

2. Aug 16, 2008

### tiny-tim

Hi Jeff!

(You meant e-x2 *Cos[x] )

The "first three terms" would be 1 + 0.x - x2/2 .

"non-zero" simply means that you skip over "0.x" and "0.x3"
You can always change the order of the terms of a series (except if you're using an infinite number of terms, in which case there are rules to follow ).

3. Aug 16, 2008

### JeffNYC

Tiny-Tim,

Thanks, and yes - not sure how "+" found its way in there :-)

2 Questions -

What is "0.x" and since the 2 original series are infinite, isn't the product of the series infinite as well?

So, just to confirm: If I take the first 3 terms of each Taylor polynomial and multiply through (line 4 in the image), I can use any 3 non zero terms of that product? What convention compelled them to use:

1 - 1.5x2 +25/24x4 as the answer to the question?

Many Thanks,

Jeff

4. Aug 16, 2008

### tiny-tim

I meant 0 times x.

Yes, it is infinite, but you're only using a few terms at the beginning.
Nooo … you must use all the terms of the three lowest powers (x0 x2 and x4).

Those are the "first 3 non-zero terms".

5. Aug 16, 2008

### JeffNYC

Great - that clarifies it perfectly. Thanks for your help, Tim.

Jeff