Taylor Polynomial Error - Page 2

Freemymind · #16 $Old$ May 10th, 2009, 01:50 PM

thank you, I understand know. this is what we are doing in class.

TheEmptySet · #17 $Old$ May 10th, 2009, 01:58 PM

Okay they are expanding the function

$f(x)=(1-x)^{\frac{1}{3}}$

This function is diff at 0

if you expand this using the forumla above you will get

$T(x)=1-\frac{1}{3}x-\frac{1}{9}x^2-\frac{5}{81}x^3-\frac{10}{243}x^4-...$

This is exactly the same series you get if you compose $(1-x)$ with the taylor series I have above

$T(x)=1+\frac{1}{3}(x-1)-\frac{1}{9}(x-1)^2+\frac{5}{81}(x-1)^3-\frac{10}{243}(x-1)^4 +...$

$T(1-x)=1+\frac{1}{3}([1-x]-1)-\frac{1}{9}([1-x]-1)^2+\frac{5}{81}([1-x]-1)^3-\frac{10}{243}([1-x]-1)^4+...$

$T(x)=1-\frac{1}{3}x-\frac{1}{9}x^2-\frac{5}{81}x^3-\frac{10}{243}x^4-...$

The bad thing about this new series is that it does not alternate

So now we have to bound it using taylors remainder theorem (It is alot more work)

Since we are bounding the error on a degree 3 we need to find the maximum of the absolute value of the forth derivative. $f^{(4)}=-\frac{80}{81(1-x)^{-11/3}}$

we are interested in the value of $x=-.3$

Since the fourth derivative is decreasing and negative on the interval $(-\infty,0)$ its max must occur at the right end point of at $x=0$

and $|f^{4}(x)| \le \frac{80}{81}$

So taylors bound on the remainder is

$R=\frac{\frac{80}{81}(0.3)^4}{4!} =0.000\bar 3$

This is identical to the error given by AST estimate in my first post.

Rember is the series is alternating using the AST estimation is alot easier.

I hope this clears up the whole situation