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

Dec 7, 2016

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 ${\left(x - a\right)}^{n}$ 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 $f$.

${f}^{\left(0\right)} \left(x\right) = f \left(x\right) = \frac{x}{1 + x}$

$f ' \left(x\right) = x \cdot \left(- {\left(1 + x\right)}^{- 2}\right) + \frac{1}{1 + x}$

$= - \frac{x}{1 + x} ^ 2 + \frac{1}{1 + x}$

$= - \frac{x}{1 + x} ^ 2 + \frac{1 + x}{1 + x} ^ 2$

$= \frac{1}{1 + x} ^ 2$

$f ' ' \left(x\right) = - 2 {\left(1 + x\right)}^{3} = - \frac{2}{1 + x} ^ 3$

$f ' ' ' \left(x\right) = \frac{6}{1 + x} ^ 4$

$f ' ' ' ' \left(x\right) = - \frac{24}{1 + x} ^ 5$

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 + . . .

$= \frac{a}{1 + a} + \frac{1}{1 + a} ^ 2 \left(x - a\right) + \frac{- \frac{2}{1 + a} ^ 3}{2} {\left(x - a\right)}^{2} + \frac{\frac{6}{1 + a} ^ 4}{6} {\left(x - a\right)}^{3} + \frac{- \frac{24}{1 + a} ^ 5}{24} {\left(x - a\right)}^{4} + . . .$

$= \textcolor{b l u e}{\frac{a}{1 + a} + \frac{1}{1 + a} ^ 2 \left(x - a\right) - \frac{1}{1 + a} ^ 3 {\left(x - a\right)}^{2} + \frac{1}{1 + a} ^ 4 {\left(x - a\right)}^{3} - \frac{1}{1 + a} ^ 5 {\left(x - a\right)}^{4} + . . .}$

Mar 11, 2017

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

$\frac{1}{1 - \left(x - a\right)} = {\sum}_{n = 0}^{\infty} {\left(x - a\right)}^{n}$

Or:

$\frac{1}{1 + \left(a - x\right)} = {\sum}_{n = 0}^{\infty} {\left(x - a\right)}^{n}$

Then:

$\frac{a - x}{1 + \left(a - x\right)} = \left(a - x\right) {\sum}_{n = 0}^{\infty} {\left(x - a\right)}^{n}$

Flipping the sign and moving into the series:

$\frac{a - x}{1 + \left(a - x\right)} = - {\sum}_{n = 0}^{\infty} {\left(x - a\right)}^{n + 1}$

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