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 $\frac{1}{1 - x}$ can be expressed as a polynomial.

Therefore

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

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

So $1 = a$.

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

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

$1 = b$

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

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

$2 = 2c$

$1 = c$.

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

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

$6 = 6d$

$1 = d$.

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

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

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

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

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

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

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