# Thread: Taylor Polynomials

1. ## Taylor Polynomials

I missed the class when my professor went over this. Can anyone show how this is to be done?

Produce a general formula for the degree n Taylor polynomials using a = 0 as the point of approximation.

a) 1 / (1-x)

b) Square Root of (1+x)

Thanks for the help!

2. Originally Posted by jzellt
I missed the class when my professor went over this. Can anyone show how this is to be done?

Produce a general formula for the degree n Taylor polynomials using a = 0 as the point of approximation.

a) 1 / (1-x)

b) Square Root of (1+x)

Thanks for the help!
Assume that $\displaystyle \frac{1}{1 - x}$ can be expressed as a polynomial.

Therefore

$\displaystyle \frac{1}{1 - x} = a + bx + cx^2 + dx^3 + ex^4 + \dots$.

To find $\displaystyle a$, let $\displaystyle x = 0$.

So $\displaystyle 1 = a$.

To find $\displaystyle b$, take the derivative of both sides, then set $\displaystyle x = 0$.

So $\displaystyle \frac{1}{(1 - x)^2} = b + 2cx + 3dx^2 + 4ex^3 + 5fx^4 + \dots$

$\displaystyle 1 = b$

To find $\displaystyle c$, take the derivative of both sides, then set $\displaystyle x = 0$.

So $\displaystyle \frac{2}{(1 - x)^3} = 2c + 6dx + 12ex^2 + 20fx^3 + 30gx^4 + \dots$

$\displaystyle 2 = 2c$

$\displaystyle 1 = c$.

To find $\displaystyle d$, take the derivative of both sides, then set $\displaystyle x = 0$

So $\displaystyle \frac{6}{(1 - x)^4} = 6d + 24ex + 60fx^2 + 120gx^3 + \dots$

$\displaystyle 6 = 6d$

$\displaystyle 1 = d$.

I think it's relatively easy to see that all the coefficients will be equal to 1.

So $\displaystyle \frac{1}{1 - x} = 1 + x + x^2 + x^3 + x^4 + \dots$.

We can also realise this using the fact that for a convergent infinite geometric series

$\displaystyle S_{\infty} = \frac{a}{1 - r}$ provided $\displaystyle |r| < 1$.

If we take $\displaystyle a = 1$ and $\displaystyle r = x$, we find

$\displaystyle \frac{1}{1 - x} = 1 + x + x^2 + x^3 + x^4 + \dots$, provided that $\displaystyle |x| < 1$.

3. Originally Posted by jzellt
I missed the class when my professor went over this. Can anyone show how this is to be done?

Produce a general formula for the degree n Taylor polynomials using a = 0 as the point of approximation.

a) 1 / (1-x)

b) Square Root of (1+x)

Thanks for the help!
See the Wikipedia article on Taylor series, then degree n Taylor polynomial is the series truncated at the degree n term.

CB