MHB 242.13.3 Taylor remander formula

[karush]

$\tiny{242.13.3}$
$\textsf{1. Using the known series expansion of
$\displaystyle e^x = \sum_{n=0}^{\infty}$, find the series representation of}\\$
$\textsf{a. $e^{-3x}$}\\$
$\textsf{b. $e^{x^3}$}$

 

[I like Serena]

$\tiny{242.13.3}$
$\textsf{1. Using the known series expansion of
$\displaystyle e^x = \sum_{n=0}^{\infty}$, find the series representation of}\\$
$\textsf{a. $e^{-3x}$}\\$

Hi karush!

I think you've already found that:
$$e^x=1+x+\frac 12 x^2 + \frac 1{3!}x^3 + ...$$
So for $e^{-3x}$, we get:
$$e^{-3x}=1+(-3x)+\frac 12 (-3x)^2 + \frac 1{3!}(-3x)^3 + ...$$

 

[topsquark]

$\textsf{b. $e^{x^3}$}$

This is the way I'd suggest as it gives you more practice.

[math]f(x) = e^{x^3}[/math]

[math]f'(x) = e^{x^3} \cdot 3x^2[/math]

[math]f''(x) = \left [ \left ( e^{x^3} \right ) \cdot \left ( 3x^2 \right ) \right ] '[/math]

[math]f''(x) = \left ( e^{x^3} \cdot 3x^2 \right ) \cdot 3x^2 + \left ( e^{x^3} \right ) \cdot 6x = \left ( 9x^4 + 6x \right ) \cdot e^{x^3}[/math]

Rinse and repeat.

Or you could do it I Like Serena's way:
[math]e^x = 1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3 + \text{ ...}[/math]

Thus
[math]e^{x^3} = 1 + \left ( x^3 \right ) + \frac{1}{2} \left ( x^3 \right ) ^2 + \text{ ...}[/math]

-Dan

 

[karush]

much mahalo...

I was struggling looking for examples but wasn't to clear

the help here has really helped me a lot

I'm only able to attend the class once a week so I am kinda on the fringe.