How do you derive taylor polynomial for #x/(1+x)#?

Question

Charles Arocha · Accepted Answer

We can use the basic geometric series centered at #x=a#:

#1/(1-(x-a))=sum_(n=0)^oo(x-a)^n#

Or:

#1/(1+(a-x))=sum_(n=0)^oo(x-a)^n#

Then:

#(a-x)/(1+(a-x))=(a-x)sum_(n=0)^oo(x-a)^n#

Flipping the sign and moving into the series:

#(a-x)/(1+(a-x))=-sum_(n=0)^oo(x-a)^(n+1)#

If we let #r=a-x# we see this is the form you requested, just at the generalized central point #x=a#.

Emily Ammerman · Answer

The Taylor polynomial is just another name for the full Taylor series truncated at a finite #n#. In other words, it is a partial Taylor series (i.e. one we could write down in a reasonable amount of time).

Some common errors are:

The formula to write out the series was:

#sum_(n=0)^(oo) (f^((n))(a))/(n!)(x-a)^n#

So, we would have to take some derivatives of #f#.

#f^((0))(x) = f(x) = x/(1 + x)#

#f'(x) = x*(-(1 + x)^(-2)) + 1/(1+x)#

#= -x/(1 + x)^2 + 1/(1+x)#

#= -x/(1 + x)^2 + (1 + x)/(1+x)^2#

#= 1/(1+x)^2#

#f''(x) = -2(1 + x)^3 = -2/(1+x)^3#

#f'''(x) = 6/(1 + x)^4#

#f''''(x) = -24/(1 + x)^5#

etc.

So plugging things in gives (truncated at #n = 4#):

#=> (f^((0))(a))/(0!)(x-a)^0 + (f'(a))/(1!)(x-a)^1 + (f''(a))/(2!)(x-a)^2 + (f'''(a))/(3!)(x-a)^3 + (f''''(a))/(4!)(x-a)^4 + . . . #

#= a/(1 + a) + 1/(1+a)^2(x-a) + (-2/(1+a)^3)/(2)(x-a)^2 + (6/(1+a)^4)/(6)(x-a)^3 + (-24/(1 + a)^5)/(24)(x-a)^4 + . . . #

#= color(blue)(a/(1 + a) + 1/(1+a)^2(x-a) - 1/(1+a)^3(x-a)^2 + 1/(1+a)^4(x-a)^3 - 1/(1 + a)^5(x-a)^4 + . . . )#

Scarlett Allison · Answer

undefined To derive the Taylor polynomial for $ \frac{x}{1+x} $, we follow these steps:

1. Find the derivatives of $ \frac{x}{1+x} $ up to the desired order.
2. Evaluate these derivatives at the point where we want to center our Taylor polynomial, usually $ a = 0 $ (the Maclaurin series).
3. Use the Taylor series formula to write the polynomial using these derivatives and their evaluations.

Let's go through each step:

1. Derivatives of $ \frac{x}{1+x} $:
   - First derivative: $ \frac{d}{dx}(\frac{x}{1+x}) = \frac{1}{(1+x)^2} $
   - Second derivative: $ \frac{d^2}{dx^2}(\frac{x}{1+x}) = \frac{-2}{(1+x)^3} $
   - Third derivative: $ \frac{d^3}{dx^3}(\frac{x}{1+x}) = \frac{6}{(1+x)^4} $
   - Fourth derivative: $ \frac{d^4}{dx^4}(\frac{x}{1+x}) = \frac{-24}{(1+x)^5} $

2. Evaluate these derivatives at $ x = 0 $:
   - First derivative: $ \frac{1}{(1+0)^2} = 1 $
   - Second derivative: $ \frac{-2}{(1+0)^3} = -2 $
   - Third derivative: $ \frac{6}{(1+0)^4} = 6 $
   - Fourth derivative: $ \frac{-24}{(1+0)^5} = -24 $

3. Write the Taylor polynomial using the derivatives and their evaluations:
$$ T_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \dotsb $$
$$ T_3(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 $$
$$ T_3(x) = 1 + (-2)x + \frac{6}{2!}x^2 + \frac{-24}{3!}x^3 $$
$$ T_3(x) = 1 - 2x + 3x^2 - 4x^3 $$

So, the Taylor polynomial of degree 3 for $ \frac{x}{1+x} $ centered at $ x = 0 $ (the Maclaurin polynomial) is $ 1 - 2x + 3x^2 - 4x^3 $.