1. ## Analysis

For this problem I need to apply Taylor's Formula
the function is $\displaystyle f(x)= lnx$
where$\displaystyle x0=1$

So the Formula would be:
$\displaystyle lnx= (Sigma) k=0 to n (ak (x-1)^k + M(x-1)^((n+1)))$

first how do I get the general formula for ak and M.
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2.
Let $\displaystyle f(x)= x^2$ and $\displaystyle epsilon=.01$ . Try a value $\displaystyle delta$ for which
$\displaystyle |x-y| < .delta$ implies $\displaystyle |x^2 - y^2| < .01$

for all $\displaystyle x,y in [0,117]$.

For this problem, lets say I choose delta to be 1
then what do we do??
Thank You

2. Originally Posted by Sally_Math
For this problem I need to apply Taylor's Formula
the function is $\displaystyle f(x)= lnx$
where$\displaystyle x0=1$

So the Formula would be:
$\displaystyle lnx= (Sigma) k=0 to n (ak (x-1)^k + M(x-1)^((n+1)))$

first how do I get the general formula for ak and M.

Thank You
First lets calculate some derivatives.

$\displaystyle f(x)=\ln(x)$
$\displaystyle f'(x)=\frac{1}{x}$
$\displaystyle f''(x)=\frac{(-1)}{x^2}$
$\displaystyle f'''(x)=\frac{(-1)(-2)}{x^3}$
$\displaystyle f^{(4)}(x)=\frac{(-1)(-2)(-3)}{x^4}$
Notice the pattern so we get the nth derivative is
$\displaystyle f^{(n)}(x)=\frac{(-1)(-2)\cdots(-n+1)}{x^{n}}=\frac{(-1)^{n-1}(n-1)!}{x^n}$

Since $\displaystyle f(1)=\ln(1)=0$

Now by the defintion of Taylor series we get

$\displaystyle \ln(x)=\sum_{n=1}^{\infty}\frac{f^{(n)}}{n!}(x-1)^n=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}(n-1)!}{n!}(x-1)^n=$

$\displaystyle \ln(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}(x-1)^n$

Notice that the series is alternating and I hope this helps.

3. Originally Posted by TheEmptySet
First lets calculate some derivatives.

$\displaystyle f(x)=\ln(x)$
$\displaystyle f'(x)=\frac{1}{x}$
$\displaystyle f''(x)=\frac{(-1)}{x^2}$
$\displaystyle f'''(x)=\frac{(-1)(-2)}{x^3}$
$\displaystyle f^{(4)}(x)=\frac{(-1)(-2)(-3)}{x^4}$
Notice the pattern so we get the nth derivative is
$\displaystyle f^{(n)}(x)=\frac{(-1)(-2)\cdots(-n+1)}{x^{n}}=\frac{(-1)^{n-1}(n-1)!}{x^n}$

Since $\displaystyle f(1)=\ln(1)=0$

Now by the defintion of Taylor series we get

$\displaystyle \ln(x)=\sum_{n=1}^{\infty}\frac{f^{(n)}}{n!}(x-1)^n=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}(n-1)!}{n!}(x-1)^n=$

$\displaystyle \ln(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}(x-1)^n$

Notice that the series is alternating and I hope this helps.
so can we say that the general formula for ak is
$\displaystyle \ln(x)=\sum_{n=1}^{\infty}\frac{f^{(n)}}{n!}(x-1)^n$

4. Originally Posted by Sally_Math
so can we say that the general formula for ak is
$\displaystyle \ln(x)=\sum_{n=1}^{\infty}\frac{f^{(n)}}{n!}(x-1)^n$
Using the defintion of the Taylor series expansion at the point a

$\displaystyle f(x) =\sum_{n=0}^{\infty}\frac{f^n(a)}{n!}(x-a)^n = \sum_{n=0}^{\infty}a_n(x-a)^n$

So in your case $\displaystyle n \ge 1$

$\displaystyle a_n=\frac{f^{n}(1)}{n!}=\frac{(-1)^{n-1}}{n}$

5. Thanks
I don't know why I am having a hard time with this
so
lets say if x=.75 and n=3, how do I find c:____

and for M is it ((-1)^n (n-1)!)/n!

I'm kind of confused, :S