# Taylor series...

• May 2nd 2009, 11:08 PM
Raidan
Taylor series...
use the Taylor series formula to find the first five terms of the taylor series for the function f(x)=1/(1+x/3) about x=0. Find the general formula for the n th term or the series and use ratio test to find the intervals on which the series converges absolutely.

How do we find the general formula for the nth term??(Wondering)
• May 2nd 2009, 11:41 PM
Prove It
Quote:

Originally Posted by Raidan
use the Taylor series formula to find the first five terms of the taylor series for the function f(x)=1/(1+x/3) about x=0. Find the general formula for the n th term or the series and use ratio test to find the intervals on which the series converges absolutely.

How do we find the general formula for the nth term??(Wondering)

Do you know how to set up a Taylor Series?

Assume that you can express $\displaystyle \frac{1}{1 + \frac{1}{3}x}$ as a polynomial. Also note that this can be expressed as $\displaystyle \left(1 + \frac{1}{3}x\right)^{-1}$

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

Since we are finding the Taylor Series about $\displaystyle x = 0$, let $\displaystyle x = 0$. This gives $\displaystyle a = 1$.

Now take the derivative of both sides.

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

Let $\displaystyle x = 0$ to find $\displaystyle b = -\frac{1}{3}$.

Take the derivative again.

$\displaystyle \frac{2}{9}\left(1 + \frac{1}{3}x\right)^{-3} = 2c + 6dx + 12ex^2 + 20fx^3 + 30gx^4 + \dots$.

Let $\displaystyle x = 0$ to find $\displaystyle c = \frac{1}{9}$.

Take the derivative again.

$\displaystyle -\frac{2}{9}\left(1 + \frac{1}{3}x\right)^{-4} = 6d + 24ex + 60fx^2 + 120gx^3 + \dots$.

Let $\displaystyle x = 0$ to find $\displaystyle d = -\frac{1}{27}$.

Let's have a look at the Polynomial.

$\displaystyle \left(1 + \frac{1}{3}x\right)^{-1} = 1 - \frac{1}{3}x + \frac{1}{9}x^2 - \frac{1}{27}x^3 + \dots$

$\displaystyle = \left(-\frac{1}{3}\right)^0x^0 + \left(-\frac{1}{3}\right)^1x^1 + \left(-\frac{1}{3}\right)^2x^2 + \left(-\frac{1}{3}\right)^3x^3 + \dots$

$\displaystyle = \sum_{i = 0}^{\infty}{\left(-\frac{1}{3}\right)^ix^i}$.

Can you see what the $\displaystyle n^{\textrm{th}}$ term must be?
• May 2nd 2009, 11:51 PM
diroga
• May 3rd 2009, 12:11 AM
Raidan
Thanx prove it...I jst have some doubt about your values for c and d..how do you get 1/9 and -1/27? obviously we substitue the x=0 but I had different values>>c=2/9 and d=-2/9

Did i go wrong somewhere>?
• May 3rd 2009, 12:33 AM
Raidan
ahh I see now,thats your final values for the taylor series..Thanks guys..(Clapping)
• May 3rd 2009, 03:59 PM
Prove It
Quote:

Originally Posted by Raidan
Thanx prove it...I jst have some doubt about your values for c and d..how do you get 1/9 and -1/27? obviously we substitue the x=0 but I had different values>>c=2/9 and d=-2/9

Did i go wrong somewhere>?

If you substitute $\displaystyle x = 0$ you get

$\displaystyle \frac{2}{9}\left[1 + \frac{1}{3}(0)\right] = 2c$

$\displaystyle \frac{2}{9} = 2c$

$\displaystyle \frac{1}{9} = c$.

Similarly you should find

$\displaystyle -\frac{2}{9} = 6d$

$\displaystyle -\frac{1}{27} = d$.