# Expansion of e^(-x)cos(x)

## 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.

Last edited:

Pengwuino
Gold Member
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
Homework Helper
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.

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.

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
Homework Helper
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?

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