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

##### 2 Answers

The **Taylor polynomial** is just another name for the full Taylor series truncated at a finite

Some common errors are:

- Letting
#x = a# within the#(x - a)^n# term. - Taking the derivatives at the same time as writing out the series, which is not necessarily wrong, but in my opinion, it allows more room for silly mistakes.
- Not plugging
#a# in for the#n# th derivative, or plugging in#0# .

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^((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

#=> (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 + . . . )#

We can use the basic geometric series centered at

#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