Maclaurin series help

• Aug 22nd 2007, 04:34 AM
Amanda-UK
Maclaurin series help
• Aug 22nd 2007, 05:43 AM
topsquark
A Maclauren series for a function is just the Taylor series at an argument of 0.

So for $f(x) = ln(1 + x)$:
$f(x) = ln(1 + x) \implies f(0) = 0$

$f^{\prime}(x) = \frac{1}{1 + x} \implies f^{\prime}(0) = 1$

$f^{\prime \prime}(x) = -\frac{1}{(1 + x)^2} \implies f^{\prime \prime}(0) = -1$

etc.

So
$f(x) \approx f(0) + \frac{1}{1!}f^{\prime}(0)x + \frac{1}{2!}f^{\prime \prime}(0)x^2 + ~ ...$

$f(x) \approx 0 + \frac{1}{1} \cdot 1 \cdot x + \frac{1}{2} \cdot (-1) \cdot x^2 + ~ ...$

$f(x) \approx x - \frac{1}{2}x^2 + ~ ...$

-Dan
• Aug 22nd 2007, 07:43 AM
Amanda-UK
Quote:

Originally Posted by topsquark
A Maclauren series for a function is just the Taylor series at an argument of 0.

So for $f(x) = ln(1 + x)$:
$f(x) = ln(1 + x) \implies f(0) = 0$

$f^{\prime}(x) = \frac{1}{1 + x} \implies f^{\prime}(0) = 1$

$f^{\prime \prime}(x) = -\frac{1}{(1 + x)^2} \implies f^{\prime \prime}(0) = -1$

etc.

So
$f(x) \approx f(0) + \frac{1}{1!}f^{\prime}(0)x + \frac{1}{2!}f^{\prime \prime}(0)x^2 + ~ ...$

$f(x) \approx 0 + \frac{1}{1} \cdot 1 \cdot x + \frac{1}{2} \cdot (-1) \cdot x^2 + ~ ...$

$f(x) \approx x - \frac{1}{2}x^2 + ~ ...$

-Dan

would you be able to go into more detail wit this please?also stuck on B
thank u
Amandax
• Aug 22nd 2007, 08:39 AM
ThePerfectHacker
Now the second one show that $f(1)$ and $f(2)$ have different signs. That means, by intermediate value theorem there is a number $x_0\in (1,2)$ so that $f(x_0)=0$.
• Aug 22nd 2007, 05:16 PM
topsquark
Quote:

Originally Posted by Amanda-UK
would you be able to go into more detail wit this please?
Amandax

:confused: More detail with what? What specifically are you having a problem with?

-Dan