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

• jf623
In summary, the conversation discusses finding an infinite series expansion for the function y=e^(-x)cos(x) using known series expansions for e^x and cos(x). The participants discuss the number of terms to include and how to determine which terms to keep or drop. Eventually, it is confirmed that the correct expansion is 1 -x +x^3/3 -x^4/6 +x^5/30.
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.

Last edited:
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.

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?

jf623 said:
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.

## What is the expansion of e-xcos(x)?

The expansion of e-xcos(x) is a mathematical expression that represents the sum of all possible terms in a power series. It is written as:
e-xcos(x) = 1 - x + (x2/2!) - (x3/3!) + (x4/4!) - ...

## Why is the expansion of e-xcos(x) important?

The expansion of e-xcos(x) is important because it allows us to approximate the value of the function for a given input. This can be useful in various mathematical and scientific applications, such as solving differential equations or modeling physical systems.

## What is the relationship between the expansion of e-xcos(x) and the Maclaurin series?

The expansion of e-xcos(x) is a specific case of the Maclaurin series, which is a power series representation of a function around x = 0. The Maclaurin series for e-xcos(x) is the same as its power series expansion, as shown in the first question.

## How is the expansion of e-xcos(x) derived?

The expansion of e-xcos(x) can be derived using the Taylor series expansion formula, which allows us to express a function as an infinite sum of its derivatives evaluated at a specific point. By using this formula and evaluating the derivatives of e-xcos(x) at x = 0, we can obtain the power series expansion.

## Can the expansion of e-xcos(x) be used to solve differential equations?

Yes, the expansion of e-xcos(x) can be used to solve certain types of differential equations. By substituting the expansion into the differential equation, we can often find a solution by comparing coefficients of like powers of x. However, this method may not always work for more complex differential equations.

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