taylors series

**jeph** · May 22nd 2007, 10:36 PM

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

**Jhevon** · May 22nd 2007, 10:56 PM

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:
$1 + ln(3) x + \frac {(ln(3))^2}{2!}x^2 + \frac {(ln(3))^3}{3!}x^3$

**jeph** · May 23rd 2007, 07:48 AM

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

**Jhevon** · May 23rd 2007, 11:56 AM

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

**CaptainBlack** · May 23rd 2007, 01:05 PM

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

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