taylors series

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May 22nd 2007, 10:36 PM
jeph

taylors 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 PM
Jhevon

Quote:

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:
$1 + ln(3) x + \frac {(ln(3))^2}{2!}x^2 + \frac {(ln(3))^3}{3!}x^3$
May 23rd 2007, 07:48 AM
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
May 23rd 2007, 11:56 AM
Jhevon

Quote:

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
May 23rd 2007, 01:05 PM
CaptainBlack

Quote:

Originally Posted by jeph

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

$<br /> f(x) = 3^x = e^{\ln(3)x} = 1 + \ln(3)x + \frac{(\ln(3)x)^2}{2!} + \frac{(\ln(3)x)^3}{3!} + ...<br />$

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

RonL