Simple taylor expansion? (Complex)

**scorpion007** · May 14th 2010, 01:42 AM

Using

$\frac{1}{1+z} = \frac{1}{3+(z-2)}$

expand $\frac{1}{1+z}$ is a Taylor series about z=2.

---
Do I just find all the derivatives or something? Perhaps I need to use a geometric series here? Partial fractions? How do I start?

EDIT: It also mentions I need the first four non-zero terms of the series.

**tonio** · May 14th 2010, 02:13 AM

Originally Posted by scorpion007

Using

$\frac{1}{1+z} = \frac{1}{3+(z-2)}$

expand $\frac{1}{1+z}$ is a Taylor series about z=2.

---
Do I just find all the derivatives or something? Perhaps I need to use a geometric series here? Partial fractions? How do I start?

$\frac{1}{1+z}=1-z+z^2-z^3+\ldots=\sum^\infty_{n=0}(-1)^{n+1}z^n\,,\,|z|<1\Longrightarrow$ $\frac{1}{1+z}=\frac{1}{3}\,\frac{1}{1+\frac{z-2}{3}}=$ $\frac{1}{3}\left(1-\frac{z-2}{3}+\frac{(z-2)^2}{3^2}-\ldots\right)=$ $\sum^\infty_{n=0}\frac{(z-2)^n}{3^{n+1}}$ , $\left|\frac{z-2}{3}\right|<1\iff |z-2|<3$ .

Tonio

**ejgmath** · May 14th 2010, 02:19 AM

Taylor series - Wikipedia, the free encyclopedia

**Prove It** · May 14th 2010, 02:19 AM

Originally Posted by scorpion007

Using

$\frac{1}{1+z} = \frac{1}{3+(z-2)}$

expand $\frac{1}{1+z}$ is a Taylor series about z=2.

---
Do I just find all the derivatives or something? Perhaps I need to use a geometric series here? Partial fractions? How do I start?

Assume you can write $\frac{1}{1 + z}$ as a Taylor polynomial.

So $\frac{1}{3 + (z - 2)} = c_0 + c_1(z - 2) + c_2(z - 2)^2 + c_3(z - 2)^3 + \dots$ .

By letting $z = 2$ we can see that $c_0 = \frac{1}{3}$ .

Differentiate both sides

$-\frac{1}{[3 + (z - 2)]^2} = c_1 + 2c_2(z - 2) + 3c_3(z - 2)^2 + 4c_4(z - 2)^3 + \dots$ .

By letting $z = 2$ we can see that $c_1 = -\frac{1}{3^2}$ .

Differentiate both sides

$\frac{2}{[3 + (z - 2)]^3} = 2c_2 + 3\cdot 2c_3(z - 2) + 4\cdot 3c_4(z - 2)^2 + 5\cdot 4c_5(z - 2)^3 + \dots$ .

By letting $z = 2$ we can see that $c_2 = \frac{1}{3^3}$ .

Differentiate both sides

$-\frac{3\cdot 2}{[3 + (z - 2)]^4} = 3\cdot 2c_3 + 4\cdot 3 \cdot 2c_4(z - 2) + 5\cdot 4 \cdot 3 c_5(z - 2)^2 + 6\cdot 5 \cdot 4 c_6(z - 2)^3 + \dots$ .

By letting $z = 2$ we can see that $c_3 = -\frac{1}{3^4}$ .

Differentiate both sides

$\frac{4\cdot 3 \cdot 2}{[3 + (z - 2)]^5} = 4 \cdot 3 \cdot 2c_4 + 5 \cdot 4 \cdot 3 \cdot 2c_5(z - 2) + 6\cdot 5 \cdot 4 \cdot 3 c_6 (z - 2)^2 + 7\cdot 6 \cdot 5 \cdot 4 c_7(z - 2)^3 + \dots$ .

By letting $z = 2$ we can see that $c_4 = \frac{1}{3^5}$ .

I think you can see that the series is taking the form

$\frac{1}{1 + z} = \frac{1}{3} - \frac{1}{3^2}(z - 2) + \frac{1}{3^3}(z - 2)^2 - \frac{1}{3^4}(z - 2)^3 + \frac{1}{3^5}(z - 2)^4 + \dots - \dots$

$\frac{1}{1 + z} = \sum_{k = 0}^{\infty}\frac{(-1)^k}{3^{k + 1}}(z - 2)^k$ .

Now you need to check the values for which this series will converge.

Since this is a geometric series with common ratio $r = -\frac{1}{3}(z - 2)$ , it only converges for

$\left|-\frac{1}{3}(z - 2)\right| < 1$

$-1 < -\frac{1}{3}(z - 2) < 1$

$3 > z - 2 > -3$

$-3 < z - 2 < 3$

$-1 < z < 5$ .

**scorpion007** · May 14th 2010, 02:25 AM

Thanks a lot guys! Tonio, did you mean
$\sum^\infty_{n=0}(-1)^{n}z^n$ ?

and
$<br /> \sum^\infty_{n=0}\frac{(-1)^n(z-2)^n}{3^{n+1}}<br />$

**tonio** · May 14th 2010, 03:44 AM

Originally Posted by scorpion007

Thanks a lot guys! Tonio, did you mean
$\sum^\infty_{n=0}(-1)^{n}z^n$ ?

and
$<br /> \sum^\infty_{n=0}\frac{(-1)^n(z-2)^n}{3^{n+1}}<br />$

Yes. I just forgot the sum begin at zero and not at 1...

Tonio

**dave0147** · November 14th 2012, 12:33 PM

multiply top and bottom by 1/3 and re-express the function as 1/3*(-1/(2-z)/3) now substitute (2-z)/3 into the geometric expansion and the rest is algebra! God I wish I could express myself better symbolically. Much less messy than ProveIt's approach!

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