Calculating Maclaurin Series for f(x) = (10x^2) e^(-2x)

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Homework Help Overview

The discussion revolves around calculating the Maclaurin series for the function f(x) = (10x^2) e^(-2x). Participants are exploring the derivatives of the function to identify non-zero terms in the series expansion.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the first and second derivatives of the function, questioning why both yield zero at x=0. There is also a mention of the Maclaurin series for e^(-2x) and how it relates to the function in question.

Discussion Status

The conversation is ongoing, with participants providing insights into the differentiation process and the nature of the Maclaurin series. Some express confusion regarding the evaluation of terms and the expectations of the homework program, while others clarify the relationship between the series and its evaluation point.

Contextual Notes

There is mention of an online homework program that may have specific requirements for inputting terms, particularly regarding zero values in the series expansion.

chrono210
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For the function f(x) = (10x^2) e^(-2x), I calculated the first term of the Maclaurin series to be 0.

However, for the second term, I also calculated it to be 0, but apparently this is wrong. Shouldn't the second term be f'(x) = 10x^2 * -2e^(-2x) + e^(-2x) * 20x? Or what am I doing incorrectly here?
 
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No, that is not wrong, apparently or not! If [itex]f(x)= 10x^2e^{-2x}[/itex] then, yes, f(0)= 0. [itex]f'(x)= 20xe^{-2x}- 20x^2e^{-x}= 20(x-x^2)e^{-x}[/itex] which is also 0 at x= 0. You don't get a non-zero term until the second derivative: [itex]f"(x)= 20(1- 2x)e^{-x}- 40(x-x^2)e^{-x}[/itex] which is 20 when x= 0. The first non-zero term of the series is (20/2)x2.

Actually, you don't need to do any differentiations for this. Presumably you know that the MacLaurin series for ex is [itex]\Sum x^n/n![/itex] so the MacLaurin series for e-2x is [itex]\Sum (-2x)^n/n!= \Sum (-2)^nx^n/n![/itex] and so the MacLaurin series for [itex]10x^2e^{-2x}[/itex] is [itex]\Sum 10(-2)^n x^{n+2}/n![/itex].
 
Is x the point at which the series is evaluated? Or is it the argument of the series, e.g. the x in Taylor series f(a) + f'(a)(x - a) + ...?
 
HallsofIvy said:
No, that is not wrong, apparently or not! If [itex]f(x)= 10x^2e^{-2x}[/itex] then, yes, f(0)= 0. [itex]f'(x)= 20xe^{-2x}- 20x^2e^{-x}= 20(x-x^2)e^{-x}[/itex] which is also 0 at x= 0. You don't get a non-zero term until the second derivative: [itex]f"(x)= 20(1- 2x)e^{-x}- 40(x-x^2)e^{-x}[/itex] which is 20 when x= 0. The first non-zero term of the series is (20/2)x2.

Actually, you don't need to do any differentiations for this. Presumably you know that the MacLaurin series for ex is [itex]\Sum x^n/n![/itex] so the MacLaurin series for e-2x is [itex]\Sum (-2x)^n/n!= \Sum (-2)^nx^n/n![/itex] and so the MacLaurin series for [itex]10x^2e^{-2x}[/itex] is [itex]\Sum 10(-2)^n x^{n+2}/n![/itex].
Interesting. Granted, this is an online-based homework program, so I wouldn't be surprised if it had an error in it (although it's been fine so far this semester). When I put in 0 for the first term, it indicated "correct", but when I put in 0 for the second term, it indicated "incorrect". Thanks for the help!
EnumaElish said:
Is x the point at which the series is evaluated? Or is it the argument of the series, e.g. the x in Taylor series f(a) + f'(a)(x - a) + ...?
Not sure exactly what you mean by this, but the question just asks to find the first five terms of the Maclaurin series for the function.

EDIT: I figured out the problem. Apparently if a term is 0, it wants you to just skip to the next one...but it still wants the first term at 0? *wishes this thing would make sense*
 
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