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- Aug 22nd 2007, 03:34 AMAmanda-UKMaclaurin series help
- Aug 22nd 2007, 04:43 AMtopsquark
A Maclauren series for a function is just the Taylor series at an argument of 0.

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

$\displaystyle f(x) = ln(1 + x) \implies f(0) = 0$

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

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

etc.

So

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

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

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

-Dan - Aug 22nd 2007, 06:43 AMAmanda-UK
- Aug 22nd 2007, 07:39 AMThePerfectHacker
Now the second one show that $\displaystyle f(1)$ and $\displaystyle f(2)$ have different signs. That means, by intermediate value theorem there is a number $\displaystyle x_0\in (1,2)$ so that $\displaystyle f(x_0)=0$.

- Aug 22nd 2007, 04:16 PMtopsquark