# Taylor Series

• October 25th 2012, 10:09 AM
Preston019
Taylor Series
Find the Taylor Polynomial P_n(x) for the function f with the given values of c and n. Then give a bound on the error that is incurred if P_n(x) is used to approximate f(x) on the given interval. f(x)=e^(3x) c=1, n=4, [1, 1.1]

I found P_4(x):

P_4(x)=(e^3) + 3(e^3)(x-1) + (9/2)(e^3)(x-1)^2 + (9/2)(e^3)(x-1)^3 + (27/8)(e^3)(x-1)^4

but the problem is also asking for the absolute value of R_4(x) less than or equal to...

How do I find R_4(x)? The answer they gave is .000549
• October 25th 2012, 10:47 AM
skeeter
Re: Taylor Series
Lagrange form of the remainder ...

$|R_n(x)| \le \, max|f^{(n+1)}(z)| \cdot \frac{|x-c|^{n+1}}{(n+1)!}$

$R_4(x) \le \frac{81e^{3.3}(1.1 - 1)^5}{40} = 0.000549$
• October 25th 2012, 02:49 PM
Preston019
Re: Taylor Series
Quote:

Originally Posted by skeeter
Lagrange form of the remainder ...

$|R_n(x)| \le \, max|f^{(n+1)}(z)| \cdot \frac{|x-c|^{n+1}}{(n+1)!}$

$R_4(x) \le \frac{81e^{3.3}(1.1 - 1)^5}{40} = 0.000549$

Where did the 3.3 come from? And I thought that form was saying to use the 5th derivative? Or are you really just using the 4th derivative?
• October 25th 2012, 03:10 PM
skeeter
Re: Taylor Series
$R_4(x)$ is the remainder of a 4th degree Taylor polynomial.

$|R_4(x)| \le \, max|f^5(z)| \cdot \frac{|x-1|^5}{5!}$

$243e^{3.3}$ is the maximum possible value of $f^5(z)$ , where $z \in [1,1.1]$
• October 25th 2012, 03:54 PM
Preston019
Re: Taylor Series
Quote:

Originally Posted by skeeter
$R_4(x)$ is the remainder of a 4th degree Taylor polynomial.

$|R_4(x)| \le \, max|f^5(z)| \cdot \frac{|x-1|^5}{5!}$

$243e^{3.3}$ is the maximum possible value of $f^5(z)$ , where $z \in [1,1.1]$

Oh! Okay, I see what i was doing wrong: I was plugging in 1 to e^3x instead of 1.1

Thank You!