CLUELESS about a Maclaurin series

In summary: the derivative of a constant is simply the constant times the derivative of the variable; in this case, it would be 0+\frac{f^{\prime}(0)}{1!}x+\frac{f^{\prime \prime}(0)}{2!}x^2+\frac{f^{\prime \prime \prime}(0)}{3!}x^3.
  • #1
DivGradCurl
372
0
CLUELESS about a Maclaurin series!

I'm supposed to obtain a Maclaurin series for the function defined by

[tex] f(x) = \left\{ \begin{array}{lc} e^{-1/x^2} & \mbox{ if } x \neq 0 \\
0 & \mbox{ if } x = 0 \end{array} \right. [/tex]

I get immediately stuck as I find:

[tex] f(0) = 0 [/tex]
[tex] f^{\prime}(0) = \mbox{ undefined } [/tex]
[tex] f^{\prime \prime}(0) = \mbox{ undefined } [/tex]
[tex] f^{\prime \prime \prime}(0) = \mbox{ undefined } [/tex]
[tex] f^{(4)}(0) = \mbox{ undefined } [/tex]

So,

[tex] f^{(n)}(0) = \mbox{ undefined } \qquad n > 0 [/tex]

Thus, we may write

[tex] f(x) = f(0) + \frac{f^{\prime}(0)}{1!}x + \frac{f^{\prime \prime}(0)}{2!}x^2 + \frac{f^{\prime \prime \prime}(0)}{3!}x^3 + \cdots [/tex]

Question

How do I handle the undefined results? (I'm supposed to show that the Maclaurin series is not equal to the given function)

Thank you very much. :smile:
 
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  • #2
Not sure if this is right or not. We know that [tex]f'(x) = \left\{ \begin{array}{lc} e^{1/x^3} & \mbox{ if } x \neq 0 \\
0 & \mbox{ if } x = 0 \end{array} \right.[/tex] So [itex]f'(0)=0[/itex], which is well defined.
 
  • #3
You messed up the chain rule, Corneo.


Thiago: are you sure f'(0) is undefined?
 
  • #4
0 is a constant.

What's the derivative of any constant?

--J
 
  • #5
Thanks for your input, guys.

Justin, I don't think I should take the derivative of f'(0) = 0 (and then take the derivative of this constant), but instead find each derivative and later evaluate at x = 0. This is how I got my results.

Hurkyl, here you go:

[tex] f^{\prime} (x) = \frac{d}{dx} \left( e^{-1/x^2} \right) = \frac{2e^{-1/x^2}}{x^3} \Longrightarrow f^{\prime} (0) = \mbox{undefined} [/tex]

I've just evaluated that over again in TI, and this time I also calculated

[tex] f(x) = 0 + \mbox{undefined} +\mbox{undefined} + \mbox{undefined} + \cdots = \mbox{undefined} \neq \left\{ \begin{array}{lc} e^{-1/x^2} & \mbox{ if } x \neq 0 \\
0 & \mbox{ if } x = 0 \end{array} \right. [/tex]

which sounds reasonable, since I need to show they're not equal.
 
  • #6
I'm supposed to show that the Maclaurin series is not equal to the given function

Looks to me like you've done that.
 
  • #7
Ah, but near zero, the function is not defined as [itex]f(x) = e^{-1/x^2}[/itex]!
 
  • #8
that is, at zero, you mean. but yeah, that fact makes the function differentiable at zero (some tinkering indicates the lim of f(x)-f(0)/(x-0) as x goes to zero exists and is zero now that f(0) is defined; should doublecheck, though), so you can write a Maclaurin expansion that'll fail.

thiago_j - i think you're a bit confused about this, the derivative of 0 is not, in general, undefined (though it can be when a function is not differentiable at 0).
 

What is a Maclaurin series?

A Maclaurin series is a type of infinite series that represents a function as a sum of terms with increasing powers of the independent variable, evaluated at a specific point. It is named after Scottish mathematician Colin Maclaurin.

Why is it important to understand Maclaurin series?

Maclaurin series are important in mathematics because they can be used to approximate complicated functions with simpler ones. This allows for easier calculation and analysis of these functions. They are also used in many real-world applications, such as in physics and engineering.

What is the difference between a Maclaurin series and a Taylor series?

A Maclaurin series is a special case of a Taylor series, where the series is centered at the point x=0. In other words, a Maclaurin series is a Taylor series evaluated at a specific point.

How do you find the coefficients of a Maclaurin series?

The coefficients of a Maclaurin series can be found by using the formula:

cn = f(n)(0)/n!

This formula uses the derivatives of the function evaluated at x=0 to determine the coefficients.

What are some common applications of Maclaurin series?

Maclaurin series are commonly used in calculus to approximate functions, calculate integrals and derivatives, and solve differential equations. They also have applications in physics, engineering, and other fields where it is useful to approximate complex functions with simpler ones.

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