# Taylor Polynomial question

1. Apr 16, 2013

### izelkay

1. The problem statement, all variables and given/known data
Here's a screenshot of the problem: http://puu.sh/2Bta5 [Broken]

2. Relevant equations

3. The attempt at a solution
As can be seen by the screenshot, the answer's already given, but I'm not sure how to go about getting it. This one has me stumped since e-4x and sin(5x) are both Maclaurin series that I can easily find. But I'm not sure where to go from there, or if I'm even supposed to use both of the Maclaurin series.

Last edited by a moderator: May 6, 2017
2. Apr 16, 2013

### Staff: Mentor

It looks like you are supposed to write a few terms of the Maclaurin series for e-4x in one area and a few terms for the series for sin(5x) in the other area.

Last edited by a moderator: May 6, 2017
3. Apr 16, 2013

### Dick

Using both the Maclaurin series is probably the easiest way to do it. Just multiply them together. Ignore products that will give you powers greater than 3.

Last edited by a moderator: May 6, 2017
4. Apr 16, 2013

### izelkay

Ok, that's what I thought I should do but wasn't sure if that was allowed or not.
Is this correct so far? :

http://puu.sh/2BySD [Broken]

I'll try this too, thanks.

Last edited by a moderator: May 6, 2017
5. Apr 16, 2013

### Dick

No, not correct. You don't multiply series term by term. You multiply them like polynomials. Write out each series up to power x^3 and then foil them. Like polynomials. Drop powers bigger than 3.

Last edited by a moderator: May 6, 2017
6. Apr 16, 2013

### izelkay

Ah, ok. I think I understand now.

7. Apr 16, 2013

### izelkay

Never mind, still not getting it. Looking at these series for e^-4x and sin(5x), wouldn't that mean I need 4 terms (n=0 to n=3) for e^-4x to make that 3rd degree, and 2 terms (n=0 to n=1) for sin(5x) to make that 3rd degree?

So for e^-4x I have:
1 - 4x + 8x - (64/6)x3

For sin(5x):
5x + (125/6)x3

FOILing those out and ignoring powers greater than 3, I get:

5x + (125/6)x3 - 20x2 + 40x3

Is there something else I'm doing wrong?

8. Apr 16, 2013

### Dick

For sin(5x):

5x - (125/6)x3

Not plus.

9. Apr 16, 2013

### izelkay

Ok, never mind, found my error. That + (125/6)x3 should actually be a - (125/6)x3

10. Apr 16, 2013

### izelkay

Just caught it. Thanks for your help, much appreciated.