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

The Maclaurin series (i.e. Taylor series centred at

In other words, there is no useful Maclaurin series for

First, take note of this:

graph{-2.5, 2.5, -1.25, 1.25]} graph{e^(-1/x^2)

What then makes Taylor's theorem invalid?

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

- Calculate the derivatives of ( f(x) ) up to the desired order.
- Evaluate each derivative at ( x = 0 ).
- 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 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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