Finding the Infinite Series Expansion of e^(-x)cos(x)

[jf623]

Homework Statement

Establish an infinite series expansion for the function y=e^(-x)cos(x) from the known series expansions, include terms up to the sixth power.

Homework Equations

The series expansions of e^x and cos(x).

The Attempt at a Solution

I am unsure of how many terms to multiply out to make the value accurate. I have tried the first four terms of e^(-x) multiplied by the first three of cos(x), then I dropped the x^7 term. This feels wrong. I have confidence that wolframalpha has the correct answer. How do I know what terms to keep/drop? I have: 1 -x +x^3/3 -x^4/12 -x^5/12 +x^6/12.

 

[Pengwuino]

You drop x^7 terms. However, for example, $$e^{-x}$$ has a 6th power that must be multiplied by the very first term in the cos(x) expansion which gives a $$x^6$$ term. This would stick around.

 

[Dick]

You should keep all terms up to x^6 in e^(-x) as well, right? Both series start with a '1'. Everything up to x^6 makes some contribution.

 

[jf623]

I have taken 7 terms of e^(-x) and 4 terms of cos(x) (which are both up to x^6) and tactically multiplied them to end up with terms only to x^6, but x^6 itself disappears and I have 1 +x -x^3/3 -x^4/6 -x^5/30 +0x^6. Is this correct? It doesn't agree with what the computer expands it as.

 

[jf623]

Ah it's actually 1 -x +x^3/3 -x^4/6 +x^5/30, I wrote the expansion of e^(-1) incorrectly. Can any of you confirm this to be right?

 

[Dick]

Ah it's actually 1 -x +x^3/3 -x^4/6 +x^5/30, I wrote the expansion of e^(-1) incorrectly. Can any of you confirm this to be right?

That's right. You don't believe Wolfram Alpha?

 

[jf623]

Haha, yes Dick. And thank you, this is the second time you have helped me.