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.
How do I know which terms are the "first 3 non-zero terms" of this series?
Hi Jeff! :smile:
(You meant e-x2 *Cos[x] :wink:)
The "first three terms" would be 1 + 0.x - x2/2 .
"non-zero" simply means that you skip over "0.x" and "0.x3" :smile:
Thanks - the answer is attached, I just don't understand how the polynomial, after multiplied out is consolidated at the end.
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 :wink:).
JeffNYC
Aug16-08, 06:46 PM
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
tiny-tim
Aug16-08, 06:59 PM
What is "0.x" and since the 2 original series are infinite, isn't the product of the series infinite as well?
I meant 0 times x.
Yes, it is infinite, but you're only using a few terms at the beginning.
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?
Nooo … you must use all the terms of the three lowest powers (x0 x2 and x4).
Those are the "first 3 non-zero terms". :smile:
JeffNYC
Aug16-08, 07:12 PM
Great - that clarifies it perfectly. Thanks for your help, Tim.