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

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- May 22nd 2007, 10:36 PMjephtaylors series
write the first 4 terms of the taylor's series for f(x)=3^x centered at c=0

- May 22nd 2007, 10:56 PMJhevon
recall that the Taylor series is of the form:

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

the first four terms of the Taylor series is centered at 0 is given by:

$\displaystyle \sum_{n=0}^{3} \frac {f^{(n)}(0)}{n!} x^n$

$\displaystyle f(x) = 3^x \Rightarrow f(0) = 1$

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

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

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

So the first 4 terms of the Taylor series is:

$\displaystyle 1 + ln(3) x + \frac {(ln(3))^2}{2!}x^2 + \frac {(ln(3))^3}{3!}x^3 $ - May 23rd 2007, 07:48 AMjeph
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 - May 23rd 2007, 11:56 AMJhevon
- May 23rd 2007, 01:05 PMCaptainBlack