Question

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

Answer 1

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:

• 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`.

`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 + . . . )`

Answer 2

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`.