How do you find the Maclaurin series for #f(x)= e^(-1/(x^2)) # centered at 0?

Question

Samantha Adkison · Accepted Answer

The Maclaurin series (i.e. Taylor series centred at #0#) for #f(x)# is #0#, which only matches #f(x)# at #x=0#.

In other words, there is no useful Maclaurin series for #f(x)# First, take note of this: #e^(-1/x^2)# is undefined when #x = 0#, but: #lim_(x->0) e^(-1/x^2) = 0# So we could patch up the definition of #f(x)# as: #f(x) = { (0, " if " x = 0), (e^(-1/x^2), " otherwise") :}# In general, if #r(x)# is any rational function in #x# then: #d/(dx) (r(x) e^(-1/x^2)) = r'(x) e^(-1/x^2) + r(x) 2/x^3 e^(-1/x^2)# #color(white)(d/(dx) (r(x) e^(-1/x^2))) = (r'(x) + (2r(x))/x^3) e^(-1/x^2)# and #r'(x) + (2r(x))/x^3# is also a rational function in #x# So #f(x)# and all of its derivatives are of the form #r(x) e^(-1/x^2)# for some rational function #r(x)# and all satisfy: #lim_(x->0) f^((n))(x) = 0# Hence all of the terms in the Maclaurin series are #0#, so the series only matches the value of #f(x)# at #x = 0#. Here's what the graph of #f(x)# looks like: graph{-2.5, 2.5, -1.25, 1.25]} graph{e^(-1/x^2) What then makes Taylor's theorem invalid? If we examine #e^(-1/x^2)# for Complex values of #x#, then we find that it has an essential singularity at #x=0#: In any open neighbourhood of #0#, no matter how small, you can find a value of #x# such that #f(x)# is arbitrarily close to any chosen Complex value. So as a Complex function #f(x) = e^(-1/x^2)# is not even close to being differentiable at #x = 0#.

Christopher Agresta · Answer

undefined To find the Maclaurin series for $ f(x) = e^{-1/(x^2)} $ centered at 0, follow these steps:

1. Calculate the derivatives of $ f(x) $ up to the desired order.
2. Evaluate each derivative at $ x = 0 $.
3. Use the formula for the Maclaurin series expansion:

$$ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n $$

where $ f^{(n)}(0) $ denotes the nth derivative of $ f(x) $ evaluated at $ x = 0 $.

For $ f(x) = e^{-1/(x^2)} $, the derivatives up to the third order are:

$$ f(x) = e^{-1/(x^2)} $$
$$ f'(x) = \frac{2e^{-1/(x^2)}}{x^3} $$
$$ f''(x) = \frac{4e^{-1/(x^2)}}{x^6} - \frac{6e^{-1/(x^2)}}{x^4} $$
$$ f'''(x) = \frac{8e^{-1/(x^2)}}{x^9} - \frac{72e^{-1/(x^2)}}{x^7} + \frac{120e^{-1/(x^2)}}{x^5} $$

Evaluating each derivative at $ x = 0 $ yields:

$$ f(0) = 1 $$
$$ f'(0) = 0 $$
$$ f''(0) = 0 $$
$$ f'''(0) = 0 $$

Now, substitute these values into the Maclaurin series formula:

$$ f(x) = 1 + 0x + 0x^2 + 0x^3 + \cdots $$

So, the Maclaurin series for $ f(x) = e^{-1/(x^2)} $ centered at 0 is simply $ 1 $.