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

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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.
  • #1
jf623
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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.
 
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  • #2
You drop x^7 terms. However, for example, [tex]e^{-x}[/tex] has a 6th power that must be multiplied by the very first term in the cos(x) expansion which gives a [tex]x^6[/tex] term. This would stick around.
 
  • #3
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.
 
  • #4
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.
 
  • #5
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?
 
  • #6
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?
 
  • #7
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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