How do you find the Maclaurin series for #f(t) =t^3(e^(-t^2))# centered at 0?

Question

Isabella Ackerman · Accepted Answer

#t^3e^(-t^2) = sum_(n=0)^oo (-1)^n t^(2n+3)/(n!)#

Let us begin with the exponential function's MacLaurin series:

#e^x = sum_(n=0)^oo x^n/(n!)#

Substitute #x =-t^2#

#e^(-t^2) = sum_(n=0)^oo (-t^2)^n/(n!) = sum_(n=0)^oo (-1)^nt^(2n)/(n!) #

Then multiply by #t^3# term by term:

#t^3e^(-t^2) =sum_(n=0)^oo (-1)^n t^3t^(2n)/(n!) = sum_(n=0)^oo (-1)^n t^(2n+3)/(n!)#

I would also note that a MacLaurin series is always centered at #x=0#, otherwise it is called a Taylor series.

William Abdalla · Answer

undefined To find the Maclaurin series for $ f(t) = t^3e^{-t^2} $ centered at 0, you need to compute the derivatives of $ f(t) $ and evaluate them at $ t = 0 $. Then, you can use the Maclaurin series formula:

$$ f(t) = f(0) + f'(0)t + \frac{f''(0)}{2!}t^2 + \frac{f'''(0)}{3!}t^3 + \cdots $$

First, find the derivatives of $ f(t) $:

$$ f'(t) = 3t^2e^{-t^2} - 2t^4e^{-t^2} $$
$$ f''(t) = 6te^{-t^2} - 6t^3e^{-t^2} - 4t^3e^{-t^2} + 4t^5e^{-t^2} $$
$$ f'''(t) = (6 - 12t^2)e^{-t^2} - (18t^2 - 12t^4)e^{-t^2} + (12t^4 - 20t^6)e^{-t^2} $$

Now, evaluate these derivatives at $ t = 0 $ to find the coefficients:

$$ f(0) = 0 $$
$$ f'(0) = 0 $$
$$ f''(0) = 6 $$
$$ f'''(0) = 6 $$

Now, plug these coefficients into the Maclaurin series formula:

$$ f(t) = 0 + 0t + \frac{6}{2!}t^2 + \frac{6}{3!}t^3 + \cdots $$

Simplify the series:

$$ f(t) = 3t^2 + t^3 + \cdots $$

So, the Maclaurin series for $ f(t) = t^3e^{-t^2} $ centered at 0 is $ 3t^2 + t^3 + \cdots $.