# Taylor Polynomial

• Jul 27th 2011, 09:46 AM
Arron
Taylor Polynomial
I need to calculate the taylor polynomial T2(x) at 1 for the function

f(x) = x/1+2x

I have the following, with some blanks can anyone help?

For this function

f(x) = x/1+2x, f(1) = 1/3
F'(x) = ? f'(1) = ?
F"(x) = ? f"(1) = ?

Hence the Taylor polynomial of degree 2 at 1 for f is

T2(x) = 1/3 ? (x-1) ? (x-1)^2

Then I need to show that T2(x) approximates f(x) with an error less than 1/30 on the interval [0.5,1.5].

I have with I=[0.5,1.5], a=1,r=1,n=2.

1.First, f^(3) (c) = ?

2. Thus |f^(3) (c)|= ? for c E [0.5,1.5]

so we take M =

3. Using the remainder estimate (M/(n+1)!r^n+1,

we obtain

|f(x)-T2(x)| = |R2(x)|

equalor more than M/(2+1)! r^2+1

=1/3! * ? * 1^3

= ? = ? for x E [0.5,1.5]

Thus T2(x) approximates f(x) with an error less than 1/30 on [0.5,1.5].
• Jul 27th 2011, 09:55 AM
TKHunny
Re: Taylor Polynomial
Again, please try to remember your early lessons. There are reasons why you studied the Order of Operations.

These are NOT the same:

x/1+2x

and

x/(1+2x)

You really should know this, or you really have no business in calculus. Step up your game if you want to do well.

Anyway, try long division. $\displaystyle \frac{1}{2}\cdot\left(1 - \frac{1}{1+2x}\right)\;=\;\frac{1}{2}\cdot\left(1 - \frac{1}{1-(-2x)}\right)$

Now, why would I do such a silly thing?

Unique series don't care how you find them.

If you don't like that, just start in with your derivatives from the first form after long division. That should make your derivatives easier.
• Jul 27th 2011, 12:08 PM
Arron
Re: Taylor Polynomial
Can anyone help me with the derivatives?
• Jul 27th 2011, 12:42 PM
Siron
Re: Taylor Polynomial
We can help if you're more clear (like TkHunny already said):
If you have to calculate the first derivative of:
$\displaystyle f(x)=\frac{x}{1+2x}$
Use the quotientrule:
$\displaystyle D\left[\frac{f(x)}{g(x)}\right]=\frac{D[f(x)]\cdot g(x)-f(x)\cdot D[g(x)]}{[g(x)]^2}$
(To calculate the second derivative try to simplify the first derivative as much as you can because then the calculating of the second derivative will be easier.)
• Jul 27th 2011, 01:24 PM
Arron
Re: Taylor Polynomial
Would this mean 1 * 1+2x - 1 * 2/(1+2x)^2

= 2x / (1+2x)^2 for f' ?
• Jul 27th 2011, 01:29 PM
Siron
Re: Taylor Polynomial
You made a little mistake in the numerator:
$\displaystyle f'(x)=\frac{(1+2x)-2x}{(1+2x)^2}=\frac{1}{(1+2x)^2}=(1+2x)^{-2}$
Now determine $\displaystyle f'(1),f''(x),f''(1)$
• Jul 27th 2011, 01:36 PM
Arron
Re: Taylor Polynomial
Would f" = -2(1+2x)^-3.

f'(1) = 1/9

f"(1) = not sure
• Jul 27th 2011, 01:37 PM
Arron
Re: Taylor Polynomial
f"(1) = -2/27
• Jul 27th 2011, 01:42 PM
Siron
Re: Taylor Polynomial
$\displaystyle f'(1)=\frac{1}{9}$ is correct, but if you want to calculate the second derivative you have to use the chain rule:
$\displaystyle f''(x)=D\left[(1+2x)^-2\right]=-2\cdot (1+2x)^{-3}\cdot D(1+2x)=\frac{-4}{(1+2x)^3}$
So $\displaystyle f''(1)=...$

Now you can determine the taylor polynomial.
• Jul 27th 2011, 01:50 PM
Arron
Re: Taylor Polynomial
Would that be -4/27.

The taylor polynomial I have is;

1/3 + 1/9(x-1) - 4/27(x-1)^2
• Jul 27th 2011, 01:51 PM
Siron
Re: Taylor Polynomial
Correct!
• Jul 27th 2011, 01:58 PM
Arron
Re: Taylor Polynomial
could you help me with f(3) the third derivative.
• Jul 27th 2011, 02:11 PM
Siron
Re: Taylor Polynomial
$\displaystyle f^{3}(x)=-4\cdot D[(1+2x)^{-3}]=-4\cdot \left[-3\cdot (1+2x)^{-4}\cdot D(1+2x)\rigt]=...$
Try to complete.
• Jul 27th 2011, 02:16 PM
Arron
Re: Taylor Polynomial
Would that be 4/(1+2x)^4
• Jul 27th 2011, 03:53 PM
mr fantastic
Re: Taylor Polynomial
Quote:

Originally Posted by Arron
Would that be 4/(1+2x)^4

Look, if you're studying Taylor polynomials you should already know how to differentiate. We cannot hold your hand every step of the way.

You can check your derivatives here (and click on Show steps to see working): differentiate -4&#47;&#40;1 &#43; 2x&#41;&#94;3 - Wolfram|Alpha