1. ## taylors series

write the first 4 terms of the taylor's series for f(x)=3^x centered at c=0

2. Originally Posted by jeph
write the first 4 terms of the taylor's series for f(x)=3^x centered at c=0
recall that the Taylor series is of the form:

$\sum_{n=0}^{ \infty} \frac {f^{(n)}(x_0)}{n!} (x - x_0)^n$ where the series is centered at $x_0$

the first four terms of the Taylor series is centered at 0 is given by:
$\sum_{n=0}^{3} \frac {f^{(n)}(0)}{n!} x^n$

$f(x) = 3^x \Rightarrow f(0) = 1$
$f'(x) = ln(3) \cdot 3^x \Rightarrow f'(0) = ln(3)$
$f''(x) = (ln(3))^2 \cdot 3^x \Rightarrow f''(0) = (ln(3))^2$
$f'''(x) = (ln(3))^3 \cdot 3^x \Rightarrow f'''(0) = (ln(3))^3$

So the first 4 terms of the Taylor series is:
$1 + ln(3) x + \frac {(ln(3))^2}{2!}x^2 + \frac {(ln(3))^3}{3!}x^3$

3. when i differentiate 3^x i get 3^x(ln3)

but when i differentiate that again i get

[3^x]' ln3 + 3^x [ln3]'
=3^x(ln3)(ln3) + 3^x 1/3
3^x(ln3)^2 + 3^x 1/3

4. Originally Posted by jeph
when i differentiate 3^x i get 3^x(ln3)

but when i differentiate that again i get

[3^x]' ln3 + 3^x [ln3]'
=3^x(ln3)(ln3) + 3^x 1/3
3^x(ln3)^2 + 3^x 1/3

ln3 is a constant! product rule is not needed here. even if you did the product rule, the derivative of ln3 would be zero, so the second part would dissappear anyway

5. Originally Posted by jeph
write the first 4 terms of the taylor's series for f(x)=3^x centered at c=0
$
f(x) = 3^x = e^{\ln(3)x} = 1 + \ln(3)x + \frac{(\ln(3)x)^2}{2!} + \frac{(\ln(3)x)^3}{3!} + ...
$

from the Taylor series for $\exp(x)$ about $0$

RonL